m=20
m=20 kg
Initially a box at rest. A man is applying the Force F to box. Mass of the box is M (kg). Kinetic friction between box and ground 0.3. Determine the power supplied by man when the time t=3 s. M= öğr

Answers

Answer 1

The power supplied by the man when t = 3 s is approximately 4498.93 watts.

Given:

M = 45 kg

F = 500 N

μ = 0.3

t = 3 s

g = 9.8 m/s²

Calculate the net force:

F(friction) = μ × M × g

F(friction) = 0.3 × 45 × 9.8 = = 132.3 N

F(net) = F - F(friction) = 500 - 132.3 = 367.7 N

Calculate the acceleration:

a = F(net) / M

a = 367.7 / 45

a =  8.17 m/s²

Calculate the distance covered:

d = (1/2) × a × t²

d = (1/2) × 8.17 × (3)²

d = 36.75 m

Calculate the work done:

W = F(net) × d

W= 367.7 × 36.75

W = 13,496.78 J

Calculate the power supplied:

P = W / t

P = 13,496.78 / 3

P = 4498.93 W

Therefore, the power supplied by the man when t = 3 s is approximately 4498.93 watts.

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Answer 2

The power supplied by man when the time t=3 s is 134.94 W.

Given:

Mass of the box, m = 20 kg

Time, t = 3 s

Coefficient of kinetic friction between box and ground, μk = 0.3

Acceleration due to gravity, g = 9.8 m/s²

We can calculate the acceleration of the box as follows:

a = (F - μkmg)/m

where F is the force applied by the man.

The power supplied by the man is given as:

P = Fv

Let's calculate the velocity of the box, using the formula:

v = u + at

As the box is at rest initially, the initial velocity, u = 0.

Substituting the given values, we get:

a = (F - μkmg)/m = F/m - μkg

Now, let's solve for F:

F = ma + μkmg

Substituting the given values, we get:

F = (20)((9.8) + (0.3)(9.8)(20))/20 = 67.86 N

Using the formula:

v = u + at

Substituting the values:

a = (F - μkmg)/m = (67.86 - (0.3)(20)(9.8))/(20) = 1.496 m/s²

v = u + at = 0 + (1.496)(3) = 4.488 m/s

Using the formula:

P = ma(at)

Substituting the values:

P = (20)(1.496)(4.488) = 134.94 W

Therefore, the power supplied by the man when the time t = 3 s is 134.94 W.

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Related Questions

Two resistors with values of 29Ω and 48Ω, respectively, are connected in parallel and hooked to a 10 V batter. (a) What would be the current in the circuit? A (b) What would be the power expended in the circuit? W

Answers

According to the question (a) The current in the circuit is approximately 0.552A. (b) The power expended in the circuit is approximately 5.52W.

(a) The current in the circuit can be calculated using Ohm's Law for the total resistance in a parallel circuit:

[tex]\( I = \frac{V}{R_{\text{total}}} \)[/tex]

where V is the voltage and [tex]\( R_{\text{total}} \)[/tex] is the total resistance.

To calculate [tex]\( R_{\text{total}} \)[/tex], we use the formula for resistors connected in parallel:

[tex]\( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \)[/tex]

Substituting the given values:

[tex]\( \frac{1}{R_{\text{total}}} = \frac{1}{29\Omega} + \frac{1}{48\Omega} \)[/tex]

[tex]\( \frac{1}{R_{\text{total}}} \approx 0.0345 + 0.0208 \)[/tex]

[tex]\( \frac{1}{R_{\text{total}}} \approx 0.0553 \)[/tex]

[tex]\( R_{\text{total}} \approx \frac{1}{0.0553} \)[/tex]

[tex]\( R_{\text{total}} \approx 18.09\Omega \)[/tex]

Now we can calculate the current:

[tex]\( I = \frac{V}{R_{\text{total}}} = \frac{10V}{18.09\Omega} \approx 0.552A \)[/tex]

Therefore, the current in the circuit is approximately 0.552A.

(b) The power expended in the circuit can be calculated using the formula:

[tex]\( P = IV \)[/tex]

Substituting the known values:

[tex]\( P = 0.552A \times 10V \)[/tex]

[tex]\( P \approx 5.52W \)[/tex]

Therefore, the power expended in the circuit is approximately 5.52W.

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Light with a wavelength of 655 nm (6.55 x 107 m) is incident upon a double slit with a
separation of 0.9 mm (9 x 10+ m). A screen is location 2.5 m from the double slit. (a) At what distance from the center of the screen will the first bright fringe beyond the center
fringe appear?

Answers

Given:
Wavelength of light = 655 nm
Separation between double slits = 0.9 mm = 9 x 10^-4 m
Distance of screen from double slits = 2.5 m

Find the distance from the center of the screen to the first bright fringe beyond the center fringe.

The distance between the central maximum and the next bright spot is given by:tanθ = y / L Where, y is the distance of the bright fringe from the central maximum, L is the distance from the double slits to the screen and θ is the angle between the central maximum and the bright fringe.

The bright fringes occur when the path difference between the two waves is equal to λ, 2λ, 3λ, ....nλ.The path difference between the two waves of the double-slit experiment is given by

d = Dsinθ Where D is the distance between the two slits, d is the path difference between the two waves and θ is the angle between the path difference and the line perpendicular to the double slit.

Using the relation between path difference and angle

θ = λ/d = λ/(Dsinθ)y = Ltanθ = L(λ/d) = Lλ/Dsinθ

Substituting the given values, we get:

y = 2.5 x 655 x 10^-9 / (9 x 10^-4) = 0.018 m = 1.8 cm.

Therefore, the first bright fringe beyond the center fringe will appear at a distance of 1.8 cm from the center of the screen.

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A 870 kg cylindrical metal block of specific gravity 2.7 is place in a tank in which is poured a
liquid with a specific gravity 13.6. If the cross section of the cylinder is 16 inches, to what depth must the
tank be filled before the normal force on the block goes to zero.

Answers

To determine the depth to which the tank must be filled for the normal force on the block to go to zero, we need to consider the balance of forces acting on the block.

The normal force exerted on the block is equal to its weight, which is the gravitational force acting on it. In this case, the weight of the block is equal to its mass multiplied by the acceleration due to gravity.

Given the specific gravity of the block and the liquid, we can calculate their respective densities. The density of the block is equal to the product of its specific gravity and the density of water. The density of the liquid is equal to the product of its specific gravity and the density of water.

Next, we calculate the weight of the block and the buoyant force acting on it. The buoyant force is equal to the weight of the liquid displaced by the block. The block will experience a net upward force when the buoyant force exceeds its weight.

By equating the weight of the block and the buoyant force, we can solve for the depth of the liquid. The depth is calculated as the ratio of the block's cross-sectional area to the cross-sectional area of the tank multiplied by the height of the tank.

By performing these calculations, we can determine the depth to which the tank must be filled before the normal force on the block goes to zero.

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In a step-up transformer (select all that
apply): • A. The induced EMF in the secondary coil is smaller than the applied EMF in the
primary coil B. The number of turns in the secondary coil must be greater than the number of
turns in the primary coil
C. The induced EMF in the secondary coil is larger than the applied EMF in the
primary coil > D. The number of turns in the primary coil must be greater than the number of
turns in the secondary coil

Answers

In a step-up transformer, the induced EMF in the secondary coil is larger than the applied EMF in the primary coil (Option C), and the number of turns in the secondary coil must be greater than the number of turns in the primary coil (Option B).

A step-up transformer is designed to increase the voltage from the primary coil to the secondary coil. This is achieved by having more turns in the secondary coil compared to the primary coil.

As a result, the induced electromotive force (EMF) in the secondary coil is greater than the applied EMF in the primary coil. This increase in voltage allows for efficient power transmission over long distances and is a fundamental principle of transformers.

Option C is correct because the induced EMF in the secondary coil is larger than the applied EMF in the primary coil. This is due to the ratio of the number of turns in the secondary coil to the number of turns in the primary coil.

Option B is also correct because in order to achieve a step-up transformation, the number of turns in the secondary coil must be greater than the number of turns in the primary coil. This ensures that the voltage is increased in the secondary coil.

Therefore, both options C and B are true for a step-up transformer.

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3. Mike owes James the following obligations: 1. P10,000 due at the end of 4 years II. P1,500 due at the end of 6 years with accumulated interest from today at (0.06, m = 2) Mike will be allowed to replace his total obligation by a payment at P2,000 at the end of 2 years and a second payment at the end of 5 years, with money worth 5%. a) Find the unknown payment. Comparison date: at the end of 5 years. b) Mike wishes to replace the obligations by a first payment at the end of 2 years and twice as much at the end of 6 years with money worth 2 1/2%. Find the unknown payments at a comparison date at the end of 5 years.

Answers

a) Unknown payment: P5,180.47 b) First payment: P4,442.27, Second payment: P8,884.54

a) To find the unknown payment at the end of 5 years, we need to calculate the present value of the existing obligations and equate it to the present value of the proposed payment schedule.

For the first obligation: P10,000 due at the end of 4 years.

Present Value (PV1) = P10,000 / (1 + 0.06/2)^(4*2) = P7,348.36

For the second obligation: P1,500 due at the end of 6 years with accumulated interest.

Present Value (PV2) = P1,500 / (1 + 0.06/2)^(6*2) = P1,104.90

Now, let's calculate the present value of the proposed payment schedule:

First payment: P2,000 at the end of 2 years.

Present Value (PV3) = P2,000 / (1 + 0.05/2)^(2*2) = P1,822.70

Second payment: Unknown payment at the end of 5 years.

Present Value (PV4) = Unknown payment / (1 + 0.05/2)^(5*2) = Unknown payment / (1.025)^10

Since Mike wants to replace his total obligation, we can set up the equation:

PV1 + PV2 = PV3 + PV4

P7,348.36 + P1,104.90 = P1,822.70 + Unknown payment / (1.025)^10

Simplifying the equation, we can solve for the unknown payment:

Unknown payment = (P7,348.36 + P1,104.90 - P1,822.70) * (1.025)^10

Unknown payment = P5,180.47

Therefore, the unknown payment at the end of 5 years is P5,180.47.

b) Similarly, to find the unknown payments at the end of 5 years under the new proposal, we can follow the same approach.

First payment: End of 2 years

Present Value (PV5) = Unknown payment / (1 + 0.025/2)^(2*2)

Second payment: Twice as much at the end of 6 years

Present Value (PV6) = 2 * Unknown payment / (1 + 0.025/2)^(6*2)

Setting up the equation with the present value of existing obligations:

PV1 + PV2 = PV5 + PV6

P7,348.36 + P1,104.90 = PV5 + PV6

Unknown payment = (P7,348.36 + P1,104.90 - PV5 - PV6)

By substituting the present value calculations, we can find the unknown payments at the end of 5 years.

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An EM wave has an electric field given by E = (200 V/m) [sin ((0.5m^−1)x − (5 × 10^6 rad/s)t)]ˆj. Find
a) Find the wavelength of the wave.
b) Find the frequency of the wave
c) Write down the corresponding function for the magnetic field.

Answers

a) The wavelength of the wave is approximately 12.57 meters. This can be calculated using the formula λ = 2π / k, where k is the wave number. In the given electric field expression, the wave number is (0.5 m^−1).

b) The frequency of the wave can be determined using the formula c = λ * f, where c is the speed of light, λ is the wavelength, and f is the frequency. Rearranging the formula, we find f = c / λ. Since the speed of light is approximately 3 × 10^8 meters per second, and the wavelength is approximately 12.57 meters, the frequency of the wave is approximately 2.39 × 10^7 hertz or 23.9 megahertz.

c) The corresponding function for the magnetic field can be obtained by applying the relationship between the electric and magnetic fields in an electromagnetic wave. The magnetic field (B) is related to the electric field (E) by the equation B = (1 / c) * E, where c is the speed of light. In this case, the magnetic field function would be B = (1 / (3 × 10^8 m/s)) * (200 V/m) * [sin ((0.5 m^−1)x − (5 × 10^6 rad/s)t)]ˆj.

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Calculate the resistance of a wire which has a uniform diameter 10.74mm and a length of 70.63cm If the resistivity is known to be 0.00092 ohm m Give your answer in units of Ohms up to 3 decimals. Take it as 3.1416 Answer:

Answers

The resistance of the wire is approximately 0.007 ohms.

To calculate the resistance of the wire, we can use the formula: R = (ρ * L) / A where R is the resistance, ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire. The cross-sectional area of the wire can be calculated using the formula:

A = π * r^2

where r is the radius of the wire.

Given that the diameter of the wire is 10.74 mm, we can calculate the radius as:

r = (10.74 mm) / 2 = 5.37 mm = 0.00537 m

Substituting the values into the formulas, we have:

A = π * (0.00537 m)^2 = 0.00009075 m^2

R = (0.00092 ohm m * 0.7063 m) / 0.00009075 m^2 ≈ 0.007168 ohms

Therefore, the resistance of the wire is approximately 0.007 ohms.

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Specific heat of salt solution measured are given with the following table. a) Determine the most suitable method and write your reasons to find the. value cp at 40°C b) Calculate the approximated value of specific heat at 40°C T(oC) 21 24 31 37 42 (x) CP 0351 0.453 0956 0.958 0.36 (fx).

Answers

Interpolation is the most suitable method, and the approximated value can be calculated by weighting the specific heat values based on their proximity to 40°C and summing them up.

What is the most suitable method to determine the specific heat at 40°C, and how can the approximated value be calculated using interpolation?

a) To determine the specific heat (cp) at 40°C, the most suitable method would be interpolation. Interpolation is a technique used to estimate values within a given set of data points. In this case, since we have specific heat values at nearby temperatures (21°C, 37°C, and 42°C), we can use interpolation to estimate the specific heat at 40°C.

Interpolation is suitable because it allows us to make a reasonable estimate based on the trend observed in the data.

b) Using the given data, we can calculate the approximated value of specific heat at 40°C using linear interpolation. We can calculate the weightage (fx) for each specific heat value based on the proximity of the corresponding temperature to 40°C.

Then, we multiply each specific heat value (CP) with its weightage (fx). The sum of these values will give us the approximated specific heat at 40°C.

For example, for the specific heat value at 37°C, the weightage (fx) would be calculated as (40 - 37) / (42 - 37) = 0.6. Multiplying this weightage with the specific heat value at 37°C (0.958) gives us the contribution to the overall approximated specific heat value.

Similarly, we calculate the contributions from other specific heat values and sum them up to obtain the approximated specific heat at 40°C.

The specific heat value at 31°C seems to be missing from the given data. If it is available, it can be included in the calculation using the appropriate weightage.

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An ideal gas expands isothermally, performing 5.00×10 3
J of work in the process. Calculate the change in internal energy of the gas. Express your answer with the appropriate units. Calculate the heat absorbed during this expansion. Express your answer with the appropriate units.

Answers

For an isothermal expansion of an ideal gas, the change in internal energy is zero. In this case, the gas performs 5.00×10^3 J of work, and the heat absorbed during the expansion is also 5.00×10^3 J.

An isothermal process involves a change in a system while maintaining a constant temperature. In this case, an ideal gas is expanding isothermally and performing work. We need to calculate the change in internal energy of the gas and the heat absorbed during the expansion.

To calculate the change in internal energy (ΔU) of the gas, we can use the first law of thermodynamics, which states that the change in internal energy is equal to the heat (Q) absorbed or released by the system minus the work (W) done on or by the system. Mathematically, it can be represented as:

ΔU = Q - W

Since the process is isothermal, the temperature remains constant, and the change in internal energy is zero. Therefore, we can rewrite the equation as:

0 = Q - W

Given that the work done by the gas is 5.00×10^3 J, we can substitute this value into the equation:

0 = Q - 5.00×10^3 J

Solving for Q, we find that the heat absorbed during this expansion is 5.00×10^3 J.

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A 5.0 μFμF capacitor, a 11 μFμF capacitor, and a 17 μFμF
capacitor are connected in parallel.
What is their equivalent capacitance?

Answers

The question involves finding the equivalent capacitance when three capacitors, with capacitance values of 5.0 μF, 11 μF, and 17 μF, are connected in parallel. The objective is to determine the combined capacitance of the parallel arrangement.

When capacitors are connected in parallel, their capacitances add up to give the equivalent capacitance. In this case, the three capacitors with capacitance values of 5.0 μF, 11 μF, and 17 μF are connected in parallel. To find the equivalent capacitance, we simply add up the individual capacitances.

Adding the capacitance values, we get:

5.0 μF + 11 μF + 17 μF = 33 μF

Therefore, the equivalent capacitance of the three capacitors connected in parallel is 33 μF. This means that when these capacitors are connected in parallel, they behave as a single capacitor with a capacitance of 33 μF.

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4. ARL circuit as shown below has a battery with ε = 10 V, resistors R. = 1000 R2 = 200 n, and an inductor L = 500 mH. R w celell R₂ w A. Find the time constant for energizing this circuit (switch is in position a). B. Find the current through the inductor when the switch has been in position a for a long time. C. With the inductor initially energized (switch has been at a for a long time) find the time necessary when de-energizing (switch moved to b at time t=0) to reduce the current to half of its initial value.

Answers

we need to fine the de-energizing time needs to half the current to its initial value. The problem mentioned above is related to an ARL circuit with certain components and conditions. Here is the solution to the problem:

Given, ε = 10 V,
R1 = 1000 Ω,
R2 = 200 Ω,
L = 500 mH

The time constant for energizing this circuit (switch is in position a):The formula for time constant (τ) is given as:

τ = L/R1

The value of L is given as 500 mH or 0.5 H, and R1 is 1000 Ω.

τ = L/R1

τ = 0.5 H/1000 Ω

τ = 0.0005 sb

The current through the inductor when the switch has been in position a for a long time: For t = ∞, the switch is in position a, and the circuit is energized. Thus, the current through the inductor would be maximum. The current (I) through the inductor (L) is given as:

I = ε/R1I = 10/1000= 0.01 Ac

With the inductor initially energized (switch has been at a for a long time) find the time necessary when de-energizing (switch moved to b at time t = 0) to reduce the current to half of its initial value:
The formula for current is given as:

I = I0e-t/τ

At half of its initial value, I = I0/2
The formula for the time taken to reach half of the initial value of current is given as:

t = τln2

The value of τ is already calculated, which is 0.0005 s.
Substitute the value of τ in the above formula:
tau = 0.0005 s

Therefore,
t = τ ln2

t = 0.0005 × ln2

t = 0.00035 s (approximately).

Hence, the main answer to the problem is: A. The time constant for energizing this circuit (switch is in position a) is 0.0005 s. B. The current through the inductor when the switch has been in position a for a long time is 0.01 A.C. The time necessary when de-energizing (switch moved to b at time t = 0) to reduce the current to half of its initial value is 0.00035 s. Hence, the conclusion to the problem is that the inductor in the circuit has certain properties and conditions, as calculated through the above solution.

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An impulse internal to the system will not change the momentum of
that system ( True or False)

Answers

False. An impulse internal to the system can change the momentum of that system.

According to Newton's third law of motion, every action has an equal and opposite reaction. When an impulse occurs within a system, it involves the application of an internal force for a certain period of time, resulting in a change in momentum. The impulse-momentum principle states that the change in momentum of an object is equal to the impulse applied to it. Therefore, an impulse internal to the system can indeed cause a change in the momentum of the system.

For example, in a collision between two objects, such as billiard balls on a pool table, the impulses exerted between the balls during the collision will cause their momenta to change. The change in momentum is a result of the internal forces between the objects during the collision. This demonstrates that an impulse internal to the system can alter the momentum of the system as a whole.

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The area of a pipeline system at a factory is 5 m 2
. An incompressible fluid with velocity of 40 m/s. After some distance, the pipe has another opening as shown in Figure 2 . The output of this opening is 20 m/s. Calculate the area of this opening if the velocity of the flow at the other end is 30 m/s Figure 2 (6 marks)

Answers

Given that the area of a pipeline system at a factory is 5 m2, an incompressible fluid with a velocity of 40 m/s. After some distance.

The output of this opening is 20 m/s. We need to calculate the area of this opening if the velocity of the flow at the other end is 30 m/s.

Let us apply the principle of the continuity of mass. The mass of a fluid that enters a section of a pipe must be equal to the mass of fluid that leaves the tube per unit of time (assuming that there is no fluid accumulation in the line). Mathematically, we have; A1V1 = A2V2Where; A1 = area of the first section of the pipeV1 = velocity of the liquid at the first sectionA2 = area of the second section of the pipeV2 = velocity of the fluid at the second section given that the area of the first section of the pipe is 5 m2 and the velocity of the liquid at the first section is 40 m/s; A1V1 = 5 × 40A1V1 = 200 .................(1)

Also, given that the velocity of the liquid at the second section of the pipe is 30 m/s and the area of the first section is 5 m2;A2 × 30 = 200A2 = 200/30A2 = 6.67 m2Therefore, the area of the opening of the second section of the pipe is 6.67 m2. Answer: 6.67

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Calculate the energies of the scattered photon and the Compton electron when incident gamma radiation of 167 keV (from Tl-210) is scattered through an angle of 23o. Assume the Compton electron originated from the L1 shell of oxygen (binding energy = 37.3 eV). (Draw a simple diagram). (10 points)
Hint: Eo = ESC + ECE + B.E.
Where Eo is the energy of the incident photon
ECE is the energy of the Compton electron
B.E. is the binding energy of the electron

Answers

The energy of the scattered photon is 157.9 keV, and the energy of the Compton electron is 9.12 keV.

The energy of the scattered photon, we use the Compton scattering formula: λ' - λ = (h / mc) * (1 - cosθ), where λ' is the wavelength of the scattered photon, λ is the wavelength of the incident photon, h is the Planck's constant, m is the electron mass, c is the speed of light, and θ is the scattering angle.

First, we convert the energy of the incident photon to its wavelength using the equation E = hc / λ. Rearranging the equation, we get λ = hc / E.

Substituting the given values, we have λ = (6.63 x 10⁻³⁴ J·s * 3.0 x 10⁸ m/s) / (167 x 10³ eV * 1.6 x 10⁻¹⁹ J/eV) ≈ 7.42 x 10⁻¹² m.

Next, we use the Compton scattering formula to calculate the wavelength shift: Δλ = (h / mc) * (1 - cosθ).

Substituting the known values, we find Δλ ≈ 2.43 x 10⁻¹² m.

Now, we can calculate the wavelength of the scattered photon: λ' = λ + Δλ ≈ 7.42 x 10⁻¹² m + 2.43 x 10⁻¹² m ≈ 9.85 x 10⁻¹² m.

Finally, we convert the wavelength of the scattered photon back to energy using the equation E = hc / λ'. Substituting the values, we find E ≈ (6.63 x 10⁻³⁴ J·s * 3.0 x 10⁸ m/s) / (9.85 x 10⁻¹² m) ≈ 157.9 keV.

To calculate the energy of the Compton electron, we use the equation ECE = Eo - ESC - B.E., where ECE is the energy of the Compton electron, Eo is the energy of the incident photon, ESC is the energy of the scattered photon, and B.E. is the binding energy of the electron.

Substituting the known values, we have ECE = 167 keV - 157.9 keV - 37.3 eV ≈ 9.12 keV.

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(10%) Problem 8: A detailed graph of acceleration versus time is shown. 10.0 (s/w)v +5.0- -5.0 5.0 15.0 te: 5/19/2022 11:59:00 PM 20.0 25.0 t(s) 20% Part (a) What is the instantaneous acceleration at time 14.25 s? a = 1 m/s² sin() tan() Л () 7 8 9 HOME cotan() acos() E 4 5 6 atan() sinh() 7 1 2 3 cosh() cotanh() + END . 0 VO BACKSPACE 1 Degrees CLEAR Submit Hint Feedback I give up! Hints: 5% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback. 20% Part (b) What is the change in velocity during the time interval from 3.75 s to 7.75 s? A 20% Part (c) What is the change in velocity during the time interval from 7.75 s to 14.25 s? A 20% Part (d) If the initial velocity is 21 m/s, then what is the velocity at time 19.25 s? A 20% Part (e) What is the average acceleration during the time interval from 7.75 s to 26 s? All content 2022 Expert TA, LLC. cos() asin() acotan() tanh() Radians

Answers

Part(a) The instantaneous acceleration at time 14.25 s is 1 m/s².

Part (b) The change in velocity during the time interval from 3.75 s to 7.75 s is 40 m/s.

Part (c) The change in velocity during the time interval from 7.75 s to 14.25 s is 0 m/s.

Part (d) The velocity at time 19.25 s is 211.5 m/s.

Part (e) The average acceleration during the time interval from 7.75 s to 26 s is 10 m/s².

Part (a)

Instantaneous acceleration is the derivative of velocity with respect to time. So, a = dv/dt. The instantaneous acceleration at time t = 14.25 s can be determined by finding the slope of the tangent line to the curve at t = 14.25 s. Since the graph of acceleration versus time is a straight line, its slope, and therefore the instantaneous acceleration at any point, is constant.

Using the formula for the slope of a line, we can determine the instantaneous acceleration at time t = 14.25 s as follows:

slope = (change in y-coordinate)/(change in x-coordinate)

slope = (5 m/s² - (-5 m/s²))/(15 s - 5 s)

slope = 10 m/s² / 10 s

slope=1 m/s²

Therefore, the instantaneous acceleration at time 14.25 s is 1 m/s².

Part (b)

The change in velocity from 3.75 s to 7.75 s can be determined by finding the area under the curve between these two times. Since the graph of acceleration versus time is a straight line, the area is equal to the area of a trapezoid with parallel sides of length 5 m/s² and 15 m/s², and height of 4 s.

area = (1/2)(5 + 15)(4) = 40 m/s

Therefore, the change in velocity during the time interval from 3.75 s to 7.75 s is 40 m/s.

Part (c)

The change in velocity from 7.75 s to 14.25 s can be determined in the same way as in part (b). The area of the trapezoid is given by:

area = (1/2)(-5 + 5)(14.25 - 7.75) = 0 m/s

Therefore, the change in velocity during the time interval from 7.75 s to 14.25 s is 0 m/s.

Part (d)

The velocity at time t = 19.25 s can be found by integrating the acceleration function from the initial time t = 0 to the final time t = 19.25 s and adding the result to the initial velocity of 21 m/s. Since the acceleration is constant over this interval,

we can use the formula:

v = v0 + at where v0 is the initial velocity, a is the constant acceleration, and t is the time interval. The velocity at time 19.25 s is therefore:

v = 21 m/s + (10 m/s²)(19.25 s - 0 s)

= 211.5 m/s

Therefore, the velocity at time 19.25 s is 211.5 m/s.

Part (e)

The average acceleration during the time interval from 7.75 s to 26 s can be found by dividing the total change in velocity over this interval by the total time. The total change in velocity can be found by subtracting the final velocity from the initial velocity:

v = v1 - v0v = (10 m/s²)(26 s - 7.75 s)

= 182.5 m/s

The total time is:

t = 26 s - 7.75 s

=18.25 s

Therefore, the average acceleration during the time interval from 7.75 s to 26 s is:

a = (v1 - v0)/t

= 182.5 m/s / 18.25 s

10 m/s².

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Final answer:

This question about acceleration, velocity, and time can be resolved using principles in physics. Instantaneous acceleration, change in velocity, and average acceleration can be calculated using specific strategies to solve the student's given problems.

Explanation:

The problems mentioned are about the relationship of acceleration, velocity, and time, which are fundamental concepts in Physics. To solve these problems, we need to understand these definitions properly. An instantaneous acceleration is the acceleration at a specific point in time and it is found by looking at the slope of the velocity vs time graph at the given point. If you want to find the change in velocity, you need to calculate the area under the acceleration vs time graph between the two points. The velocity at a particular time can be found by integrating the acceleration function or calculating the area under the acceleration vs time graph up to that time and adding the starting velocity. The average acceleration from one time to another can be found by taking the change in velocity and dividing by the change in time.

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D) light only 8) Kathryn is diving from a tower 10.0 m above the water. When she is 5.0 m above the surface of the water, her. A) momentum and kinetic energy are about equal B) velocity is half of the

Answers

Kathryn's velocity is greater than when she is at the top of the tower because she has lost some potential energy by coming down 5.0 m. So, the option is (D) light only which is the answer. Hence, the correct answer is (D) light only.

When Kathryn is 5.0 m above the surface of the water, her kinetic energy is greater than her potential energy. When she falls to the water surface, her potential energy becomes zero, and her kinetic energy is maximum, according to the law of conservation of energy. The kinetic energy of Kathryn is converted into thermal energy, sound energy, and a small amount of potential energy due to the splashing of water.As per the given problem, Kathryn is diving from a tower 10.0 m above the water and when she is 5.0 m above the surface of the water, her kinetic energy is greater than her potential energy.

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By what factor does the force between two protons change if each of the following occurs:One of the protons is replaced with an electron.
One of the protons is replaced with 3 electrons.

Answers

The force between two protons can be calculated using Coulomb's law,

which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

By what factor does the force between two protons change if each of the following occurs:

1. One of the protons is replaced with an electron:

Electrons have a negative charge, which is equal in magnitude to the positive charge on a proton. Therefore, if one of the protons is replaced with an electron, the net charge on the pair of particles becomes zero. .

2. One of the protons is replaced with 3 electrons:

If one of the protons is replaced with 3 electrons, the net charge on the system becomes negative. In this case, the force between the particles is attractive as opposite charges attract each other

Since the force between the particles increases by a factor of more than 3.

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a 190-lb man carries a 20-lb can of paint up a helical staircase that encircles a silo with radius 15 ft. if the silo is 80 ft high and the man makes exactly four complete revolutions, how much work is done by the man against gravity in climbing to the top?

Answers

The work done by the man against gravity in climbing to the top is 9,480 foot-pounds.

To calculate the work done by the man, we need to determine the total change in potential energy as he climbs up the helical staircase that encircles the silo. The potential energy can be calculated using the formula PE = mgh, where m represents the mass, g represents the acceleration due to gravity, and h represents the height.

In this case, the mass of the man is 190 lb, and the height of the silo is 80 ft. Since the man makes exactly four complete revolutions around the silo, we can calculate the circumference of the helical staircase. The circumference of a circle is given by the formula C = 2πr, where r represents the radius. In this case, the radius of the silo is 15 ft.

To find the work done against gravity, we need to multiply the change in potential energy by the number of revolutions. The change in potential energy is obtained by multiplying the mass, the acceleration due to gravity (32.2 ft/s²), and the height. The number of revolutions is four.

Therefore, the work done by the man against gravity in climbing to the top can be calculated as follows:

Work = 4 * m * g * h

    = 4 * 190 lb * 32.2 ft/s² * 80 ft

    = 9,480 foot-pounds.

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Suppose that we start a major scale on concert B, which is defined to have a frequency of 495 Hz. If we call this frequency do, what is the ideal-ratio frequency of (a) re (b) la (c) fa

Answers

The ideal-ratio frequencies of the notes in the major scale starting on concert B (do) are approximate:

(a) Frequency of re ≈ 556.875 Hz

(b) Frequency of la ≈ 743.4375 Hz

(c) Frequency of fa ≈ 660 Hz

In a major scale, the ideal ratio frequencies of the notes are determined by the specific intervals between them. The intervals in a major scale follow the pattern of whole steps (W) and half steps (H) between adjacent notes.

(a) Re:

In a major scale, the interval between do and re is a whole step (W). A whole step corresponds to a frequency ratio of 9/8.

Therefore, the ideal-ratio frequency of re can be calculated as:

Frequency of re = Frequency of do * (9/8)

Substituting the frequency of do as 495 Hz:

Frequency of re = 495 Hz * (9/8)

Frequency of re ≈ 556.875 Hz

(b) La:

In a major scale, the interval between do and la is a perfect fifth, which consists of seven half steps (H). A perfect fifth corresponds to a frequency ratio of 3/2.

Therefore, the ideal-ratio frequency of la can be calculated as:

Frequency of la = Frequency of do * (3/2)^7

Substituting the frequency of do as 495 Hz:

Frequency of la = 495 Hz * (3/2)^7

Frequency of la ≈ 743.4375 Hz

(c) Fa:

In a major scale, the interval between do and fa is a perfect fourth, which consists of five half steps (H). A perfect fourth corresponds to a frequency ratio of 4/3.

Therefore, the ideal-ratio frequency of fa can be calculated as:

Frequency of fa = Frequency of do * (4/3)^5

Substituting the frequency of do as 495 Hz:

Frequency of fa = 495 Hz * (4/3)^5

Frequency of fa ≈ 660 Hz

Therefore, the ideal-ratio frequencies of the notes in the major scale starting on concert B (do) are approximate:

(a) Frequency of re ≈ 556.875 Hz

(b) Frequency of la ≈ 743.4375 Hz

(c) Frequency of fa ≈ 660 Hz

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Question 31 1 pts A high voltage transmission line carrying 500 MW of electrical power at voltage of 409 kV (kilovolts) has a resistance of 10 ohms. What is the power lost in the transmission line? Give your answer in megawatts (MW).

Answers

The power lost in the transmission line is approximately 14.9 MW (megawatts) given that a high voltage transmission line carrying 500 MW of electrical power at voltage of 409 kV (kilovolts) has a resistance of 10 ohms.

Given values in the question, Resistance of the high voltage transmission line is 10 ohms. Power carried by the high voltage transmission line is 500 MW. Voltage of the high voltage transmission line is 409 kV (kilovolts).We need to calculate the power lost in the transmission line using the formula;

Power loss = I²RWhere,I = Current (Ampere)R = Resistance (Ohms)

For that we need to calculate the Current by using the formula;

Power = Voltage × Current

Where, Power = 500 MW

Voltage = 409 kV (kilovolts)Current = ?

Now we can substitute the given values to the formula;

Power = Voltage × Current500 MW = 409 kV × Current

Current = 500 MW / 409 kV ≈ 1.22 A (approx)

Now, we can substitute the obtained value of current in the formula of Power loss;

Power loss = I²R= (1.22 A)² × 10 Ω≈ 14.9 MW

Therefore, the power lost in the transmission line is approximately 14.9 MW (megawatts).

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Simple Rotational Variables Problem Points:40 The angular position of a point on the rim of a rotating wheel is given by 0 = 2.2t + 4.2t² + 1.9t3, where 0 is in radians if t is given in seconds. What is the angular speed at t = 3.0 s? 95.7rad/s Submit Answer Incorrect. Tries 1/40 Previous Tries What is the angular speed at t = 5.0 s? 353.5rad/s Submit Answer What is the Incorrect. Tries 2/40 Previous Tries average angular acceleration for the time interval that begins at t = 3.0 s and ends at t = 5.0 s? Submit Answer Tries 0/40 What is the instantaneous acceleration at t = 5.0 s? Submit Answer Tries 0/40 Post Discussion Send Feedback

Answers

The angular position of a point on the rim of a rotating wheel is given by θ = 2.2t + 4.2t² + 1.9t³, θ where  is in radians if t is given in seconds.

The angular speed at t = 3.0 s is 78.7 rad/s.

The angular speed at t = 5.0 s is 186.7 rad/s.

The average angular acceleration for the time interval that begins at t = 3.0 s and ends at t = 5.0 s is 54.0 rad/s².

The instantaneous acceleration at t = 5.0 s is 65.4 rad/s².

To find the angular speed at t = 3.0 s, we need to differentiate the given equation for angular position (θ) with respect to time (t):

ω = dθ/dt

Given that the equation for angular position is θ = 2.2t + 4.2t² + 1.9t³, we can differentiate it to find the angular speed:

ω = dθ/dt = 2.2 + 8.4t + 5.7t²

Now we can substitute t = 3.0 s into the equation to find the angular speed at t = 3.0 s:

ω = 2.2 + 8.4(3.0) + 5.7(3.0)²

= 2.2 + 25.2 + 51.3

= 78.7 rad/s

Therefore, the angular speed at t = 3.0 s is 78.7 rad/s.

To find the average angular acceleration for the time interval from t = 3.0 s to t = 5.0 s, we can use the formula:

Average angular acceleration (αₐ) = (ω₂ - ω₁) / (t₂ - t₁)

Given that t₁ = 3.0 s, t₂ = 5.0 s, and ω₁ = 78.7 rad/s (from the previous calculation), we need to find ω₂ at t = 5.0 s. Following the same process as before, we differentiate the equation for angular position:

ω = 2.2 + 8.4t + 5.7t²

ω₂ = 2.2 + 8.4(5.0) + 5.7(5.0)²

= 2.2 + 42 + 142.5

= 186.7 rad/s

Substituting the values into the average angular acceleration formula:

αₐ = (ω₂ - ω₁) / (t₂ - t₁)

= (186.7 - 78.7) / (5.0 - 3.0)

= 108.0 / 2.0

= 54.0 rad/s²

Therefore, the average angular acceleration for the time interval from t = 3.0 s to t = 5.0 s is 54.0 rad/s².

Finally, to find the instantaneous acceleration at t = 5.0 s, we need to differentiate the angular speed equation:

ω = 2.2 + 8.4t + 5.7t²

Differentiating with respect to time:

α = dω/dt = 8.4 + 11.4t

Substituting t = 5.0 s:

α = 8.4 + 11.4(5.0)

= 8.4 + 57

= 65.4 rad/s²

Therefore, the instantaneous acceleration at t = 5.0 s is 65.4 rad/s².

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The angular position of a point on the rim of a rotating wheel is given by θ = 2.2t + 4.2t² + 1.9t³, θ where  is in radians if t is given in seconds. What is the angular speed at t = 3.0 s? What is the angular speed at t = 5.0 s?  What is the average angular acceleration for the time interval that begins at t = 3.0 s and ends at t = 5.0 s? What is the instantaneous acceleration at t = 5.0 s?

The angular speed at t = 3.0 s can be found by taking the derivative of the given equation with respect to time and evaluating it at t = 3.0 s. Differentiating the equation [tex]0 = 2.2t + 4.2t^2 + 1.9t^3[/tex] with respect to t gives us the angular speed as the coefficient of the first-order term.

By differentiating the equation, we obtain [tex]0 = 2.2 + 8.4t + 5.7t^2[/tex]. Substituting t = 3.0 s into the equation, we can find the angular speed at t = 3.0 s.

The average angular acceleration for the time interval that begins at t = 3.0 s and ends at t = 5.0 s can be calculated by finding the change in angular speed over the given time interval and dividing it by the duration of the interval.

To find the instantaneous acceleration at t = 5.0 s, we need to take the derivative of the angular speed equation with respect to time and evaluate it at t = 5.0 s. The derivative of the angular speed equation will give us the angular acceleration at any given time.

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how
many joules of energy must be added to an ice cube of mass 45.0 g
at -19 Celsius in order to fully converted to water with a
temperature of 65°C?

Answers

The amount of energy needed to convert an ice cube of mass 45.0 g at -19 Celsius to water with a temperature of 65°C is 30,825.27 joules.

To calculate the amount of energy needed, we can use the following equation:

               Q = m * L + m * c * ΔT

where:

Q is the amount of energy needed in joules

m is the mass of the ice cube in grams

L is the latent heat of fusion for water, which is 333.55 joules per gram

c is the specific heat capacity of water, which is 4.184 joules per gram per degree Celsius

ΔT is the change in temperature, which is 65°C - (-19°C) = 84°C

Plugging in the values, we get:

Q = 45.0 g * 333.55 J/g + 45.0 g * 4.184 J/g/°C * 84°C

= 30,825.27 J

Therefore, 30,825.27 joules of energy must be added to an ice cube of mass 45.0 g at -19 Celsius to fully convert it to water with a temperature of 65°C.

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When a pendulum with a period of 2.00000 s in one location ( = 9.80 m/s) is moved to a new location from one where the period is now 1.00710. What is the change in acceleration in my due to gravity at its new location?

Answers

The change in acceleration due to gravity at the new location is 0 m/s². The acceleration due to gravity remains the same regardless of the change in the period of the pendulum.

To calculate the change in acceleration due to gravity at the new location, we can use the formula for the period of a simple pendulum:

T = 2π * √(L / g)

where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

The change in acceleration due to gravity at the new location is 0 m/s². The acceleration due to gravity remains the same regardless of the change in the period of the pendulum.

Let's denote the initial period as T1, the final period as T2, and the initial acceleration due to gravity as g1.

From the given information:

T1 = 2.00000 s

T2 = 1.00710 s

g1 = 9.80 m/s²

We can rearrange the formula for the period to solve for the acceleration due to gravity:

g = (4π² * L) / T²

First, we need to calculate the length of the pendulum at the new location. We can do this by rearranging the formula for the period:

L = (T² * g1) / (4π²)

Substituting the values:

L = (1.00710 s)² * (9.80 m/s²) / (4π²)

Now, we can calculate the new acceleration due to gravity (g2) using the length at the new location:

g2 = (4π² * L) / T2²

Substituting the values:

g2 = (4π² * [(1.00710 s)² * (9.80 m/s²) / (4π²)]) / (1.00710 s)²

Simplifying the equation:

g2 = (9.80 m/s²)

Therefore, the change in acceleration due to gravity at the new location is 0 m/s². The acceleration due to gravity remains the same regardless of the change in the period of the pendulum.

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A 250-lb man supports all of his weight on a snowshoe with an area of 200 in2. What pressure is exerted on the snow (in pounds per square inch)?I know the answer is: 1.25 lb/in.2
But how do I get this answer? Please explain and show work in legible writing, thank you.

Answers

The pressure exerted on the snow is 1.25 lb/in². Pressure is defined as the force applied per unit area.

To calculate the pressure exerted on the snow, we divide the force (weight) by the area of the snowshoe.

Given that the man's weight is 250 lb and the snowshoe's area is 200 in², we can calculate the pressure as follows:

Pressure = Force / Area

Pressure = 250 lb / 200 in²

To simplify the calculation, we convert the units to pounds per square inch (lb/in²):

Pressure = (250 lb / 200 in²) * (1 in² / 1 in²)

Pressure = 1.25 lb/in²

Therefore, the pressure exerted on the snow is 1.25 lb/in².

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A boat motor is rated at 56000W. What is the fastest it can do
42000J of work?
0.75s
1.33s
8.93s
14000s

Answers

The boat motor, rated at 56,000 watts, can perform 42,000 joules of work in approximately 0.75 seconds. Therefore, the correct option is (a).

In order to determine the time it takes for the motor to do a certain amount of work, we can use the formula:

Work = Power × Time

Given that the work is 42,000 joules and the power is 56,000 watts, we can rearrange the formula to solve for time:

Time = Work / Power

Plugging in the values, we get:

Time = 42,000 J / 56,000 W = 0.75 s

Therefore, the fastest the boat motor can perform 42,000 joules of work is approximately 0.75 seconds.

The power rating of a motor represents the rate at which work can be done. In this case, the boat motor has a power rating of 56,000 watts. This means that it can deliver 56,000 joules of energy per second. When we divide the work (42,000 joules) by the power rating (56,000 watts), we get the time it takes for the motor to perform the given amount of work. In this scenario, the boat motor can complete 42,000 joules of work in approximately 0.75 seconds. It's important to note that this calculation assumes that the motor is operating at its maximum power continuously.

Hence, the correct option is (a) 0.75 seconds.

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A 957-g empty iron kettle is put on a stove. How much heat. in joules. must it absorb to raise its temperature from 15.0°C to 37.0°C? (The specific heat for iron is 113 cal/kg•C°, 1 cal = 4.190 J) 10,900 J 9950 J 2380 J 16,700 J A monatomic ideal gas undergoes an isothermal expansion at 300 K, as the volume increased from 0.02 m2 to 0.14 m3. The final pressure of the gas is 140 kPa. The ideal gas constant is R = 8.314 J/mol · K. The change in the internal (thermal) energy of the gas is closest to 0.00 kJ. -38 kJ. -19 kJ. 19 kJ. 38 kJ.

Answers

1. The heat absorbed by the iron kettle is approximately 10,900 J.

2. The change in the internal energy of the gas is closest to 0.00 kJ.

1. To calculate the heat absorbed by the iron kettle, we can use the formula:

Q = m × c × ΔT

where Q is the heat, m is the mass of the iron kettle, c is the specific heat of iron, and ΔT is the change in temperature.

Given:

m = 957 g = 0.957 kg (converting to kilograms)

c = 113 cal/kg·°C = 113 × 4.190 J/kg·°C (converting to joules)

ΔT = (37.0°C - 15.0°C)

Substituting the values into the formula:

Q = 0.957 kg × (113 × 4.190 J/kg·°C) × (37.0°C - 15.0°C)

Q ≈ 10900 J

Therefore, the heat absorbed by the iron kettle is approximately 10900 J.

2. For an isothermal process, the change in internal (thermal) energy of the gas is zero. Therefore, the change in internal energy is closest to 0.00 kJ.

Therefore, the answer is 0.00 kJ.

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6. Design an absorption packed tower that is used to reduce NH; in air from a concentration of 0.10 kg'm' to a concentration of 0.0005 kg/ml Given: Column diameter - 3.00 m Operating temperature - 20.0°C Air density at 20.0°C -1.205 kg/m Operating pressure 101.325 kPa For 15 kg NH, per 100 kg H:0 measured partial pressure of NH3 – 15.199 kPa Q-01-10.0 kg's H-0.438 m H=0.250 m Incoming liquid is water free of NH3(x2) = 0 GMW of NH) - 17.030 GMW of Air - 28.970 GMW of H20 - 18.015 Determine 6.1 Mole fraction of pollutant in the gas phase at inlet of tower (y) 6.2 Equilibrium mole fraction of pollutant in the liquid phase (3) 6.3 Slope of equilibrium curve (m) 6.4 Absorption factor (A) 6.5 Height of an absorption packed tower (2)

Answers

The objective is to design an absorption packed tower to reduce NH3 concentration in air, and the parameters to be determined are the mole fraction of pollutant in the gas phase at the tower inlet (y), the equilibrium mole fraction of pollutant in the liquid phase (x), the slope of the equilibrium curve (m), the absorption factor (A), and the height of the tower.

What is the objective of the given problem and what parameters are to be determined for designing the absorption packed tower for NH3 removal?

The given problem involves the design of an absorption packed tower for removing NH3 from air. The tower should reduce the NH3 concentration from 0.10 kg/m3 to 0.0005 kg/m3.

The operating conditions include a column diameter of 3.00 m, operating temperature of 20.0°C, air density at 20.0°C of 1.205 kg/m3, and operating pressure of 101.325 kPa. The relevant data includes the measured partial pressure of NH3, the flow rate of H2O, and the molecular weights of NH3, air, and H2O.

The objectives are to determine the mole fraction of the pollutant in the gas phase at the inlet of the tower (y), the equilibrium mole fraction of the pollutant in the liquid phase (x), the slope of the equilibrium curve (m), the absorption factor (A), and the height of the absorption packed tower.

These parameters will help in designing an effective tower for NH3 removal.

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Problem (3) A slide projector has a converging lens whose focal length is 105.mm. (a) How far (in meters) from the lens must the screen be located if a slide is placed 108. mm from the lens? (b) If the slide measures 24.0 mm×36.0 mm, what are the dimensions (in mm ) of its image?

Answers

a) To determine the distance of the screen from the slide projector, we can use the lens formula. Let's recall the lens formula:Object distance (u) + Image distance (v) = Focal length (f)Given that the focal length of the converging lens is 105mm, the object distance is 108mm.Substituting the given values in the lens formula;u + v

= foru = 108mm, f

= 105mmTherefore, 108mm + v

= 105mmv

= - 3mmSince the image is on the other side of the lens, it is a virtual image. Thus, the screen must be placed 3mm from the lens. To convert mm to meters, we divide by 1000; hence, the screen is located at 0.003m.b) To determine the dimensions of the slide image, we use the thin lens equation:magnification (m) = image height (h')/object height (h)h = 24.0 mm (width), h

= 36.0 mm (height), image height (h')

= v * tan θIn part a, we determined that the image distance is -3 mm. We will use this value to determine the image height. To do so, we must first determine the angle of the image formed by the lens, θ. Recall the formula;tan θ = (h')/v, thus θ

= tan-1 (h'/v). Let's find the value of θ by substituting the value of v.tan θ

= (h')/v, where v

= - 3mm, h

= 36.0mm, and h

= 24.0mmθ

= tan-1(h'/v)θ

= tan-1 (24.0 / (- 3.0))θ

= tan-1 (- 8)θ

= - 83.66°Now we can calculate the image height. We can use trigonometry to calculate the height since we have the angle. Thus,h'

= v * tan θh'

= (- 3mm) * tan (- 83.66°)h'

= - 106.67mmSince the image is virtual, the dimensions of the slide image are 106.67mm × 160.0mm

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A point charge q1 = 4.10 nC is placed at the origin, and a second point charge q2 = -2.95 nC is placed on the x- axis at x = +20.0 cm. A third point charge 93 = 2.10 nC is to be placed on the x-axis between qi and 92. (Take as zero the potential energy of the three charges when they are infinitely far apart.) ▾ Part B Where should qs be placed between qi and q2 to make the potential energy of the system equal to zero? Express your answer in centimeters. [5] ΑΣΦ I ? H= cm.

Answers

The third point charge should be placed at approximately 6.77 cm from q1 towards q2 to make the potential-energy of the system equal to zero.

To determine the position at which the third point charge (qs) should be placed on the x-axis between q1 and q2 to make the potential energy of the system equal to zero, we can utilize the principle of superposition and the concept of potential energy.

The potential energy (U) of a system of point charges is given by the equation:

U = k * (q1 * q2) / r12 + k * (q1 * qs) / r1s + k * (q2 * qs) / r2s

where k is Coulomb's constant (k = 8.99 * 10^9 N m^2/C^2), q1, q2, and qs are the charges of q1, q2, and qs respectively, r12 is the distance between q1 and q2, r1s is the distance between q1 and qs, and r2s is the distance between q2 and qs.

Given that we want the potential energy of the system to be zero, we can set U = 0 and solve for the unknown distance r1s. By rearranging the equation, we get:

r1s = (-(q2 * r12) + (q2 * r2s) + (q1 * r2s)) / (q1)

Substituting the given values: q1 = 4.10 nC, q2 = -2.95 nC, r12 = 20.0 cm, and r2s = r1s - 20.0 cm, we can calculate the value of r1s. After solving the equation, we find that r1s is approximately 6.77 cm. Therefore, the third point charge (qs) should be placed at approximately 6.77 cm from q1 towards q2 to make the potential energy of the system equal to zero.

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A spring is stretched to a length of 5.40 m. You generate a standing wave using a frequency of 4.75 Hz If there are 5 antinodes along the spring, calculate the speed of the wave. Be sure to draw a picture of this standing wave.

Answers

The speed of the wave in the given scenario is approximately 12.83 m/s.

To calculate the speed of the wave in the given scenario, we need to use the formula that relates wave speed, frequency, and wavelength.

Given:

Frequency of the wave (f) = 4.75 Hz

Number of antinodes (n) = 5

Length of the stretched spring (L) = 5.40 m

The wavelength (λ) of the standing wave can be determined by considering the distance between adjacent antinodes. In a standing wave, the distance between adjacent nodes or antinodes is half the wavelength.

Since there are 5 antinodes along the spring, there are 4 intervals between them, which correspond to 4 half-wavelengths. Therefore, the total length of 5.40 m is equal to 4 times the half-wavelength.

Let's denote the wavelength as λ:

4 × (λ/2) = L

2λ = L

λ = L/2

Now, we can calculate the wavelength of the wave:

λ = 5.40 m / 2 = 2.70 m

The speed (v) of the wave can be calculated using the formula v = f × λ, where v is the speed, f is the frequency, and λ is the wavelength:

v = 4.75 Hz × 2.70 m

v ≈ 12.83 m/s

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Final answer:

To calculate the speed of the wave in the stretched spring, use the formula speed = frequency × wavelength. Find the wavelength by multiplying the length between two adjacent antinodes by 2. Substitute the frequency and wavelength into the formula to find the speed.

Explanation:

To calculate the speed of the wave, we can use the formula:

speed = frequency × wavelength

First, we need to find the wavelength of the wave. Since there are 5 antinodes along the spring, the distance between two adjacent antinodes is equal to half the wavelength:

wavelength = 2 × length between two adjacent antinodes

Next, we can substitute the frequency and wavelength into the formula to find the speed of the wave.

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