In an automobile, the system voltage varies from about 12 V when the car is off to about 13.8 V when the car is on and the charging system is in operation, a difference of 15%. By what percentage does the power delivered to the headlights vary as the voltage changes from 12 V to 13.8 V? Assume the headlight resistance remains constant

Answers

Answer 1

The power delivered to the headlights varies by approximately 32.25% as the voltage changes from 12 V to 13.8 V, assuming the headlight resistance remains constant.

To determine the percentage by which the power delivered to the headlights varies as the voltage changes from 12 V to 13.8 V, we can use the formula for power:

Power = (Voltage²) / Resistance

Given that the headlight resistance remains constant, we can compare the powers at the two different voltages.

At 12 V:

Power_12V = (12^2) / Resistance = 144 / Resistance

At 13.8 V:

Power_13.8V = (13.8^2) / Resistance = 190.44 / Resistance

To calculate the percentage change, we can use the following formula:

Percentage Change = (New Value - Old Value) / Old Value × 100

Percentage Change = (Power_13.8V - Power_12V) / Power_12V × 100

Substituting the values:

Percentage Change = (190.44 / Resistance - 144 / Resistance) / (144 / Resistance) × 100

Simplifying:

Percentage Change = (190.44 - 144) / 144 * 100

Percentage Change = 46.44 / 144 * 100

Percentage Change ≈ 32.25%

Therefore, the power delivered to the headlights varies by approximately 32.25% as the voltage changes from 12 V to 13.8 V, assuming the headlight resistance remains constant.

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Related Questions

A skier starts from rest at the top of a hill that is inclined at 9.2° with respect to the horizontal. The hillside is 235 m long, and the coefficient of friction between snow and skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?

Answers

The skier glides around 133.8 meters along the level portion of the snow before stopping.

To find the distance the skier glides along the horizontal portion of the snow before coming to rest, we need to consider the forces acting on the skier. Initially, the skier is subject to the force of gravity, which can be decomposed into two components: the parallel force along the slope and the perpendicular force normal to the slope. The parallel force contributes to the acceleration down the hill, while the normal force counteracts the force of gravity.

Using trigonometry, we can find that the component of the force of gravity parallel to the slope is given by mg * sin(9.2°), where m is the mass of the skier and g is the acceleration due to gravity. The force of friction opposing the skier's motion is then μ * (mg * cos(9.2°) - mg * sin(9.2°)), where μ is the coefficient of friction.

The net force acting on the skier along the slope is equal to the parallel force minus the force of friction. Using Newton's second law (F = ma), we can determine the acceleration of the skier down the hill.

Next, we can find the time it takes for the skier to reach the bottom of the hill using the kinematic equation: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity (which is zero), a is the acceleration, and t is the time.

After finding the time, we can calculate the distance the skier glides along the horizontal portion of the snow using the equation: d = ut + (1/2)at^2, where d is the distance, u is the final velocity (which is zero), a is the acceleration, and t is the time.

The skier glides approximately 133.8 meters along the horizontal portion of the snow before coming to rest.

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Two 20 W resistances are connected in series. Find the value of
a single resistor that could be used to replace both 20 W resistors
without changing the current in the circuit.

Answers

The single resistor that could be used to replace both 20 W resistors without changing the current in the circuit is a 40 W resistor.

When two resistors are connected in series, their resistances add up. In this case, we have two 20 W resistors connected in series. Therefore, the total resistance in the circuit is:

Two 20 W resistors are connected in series, resulting in a total resistance of 40 W.

To replace these two resistors with a single resistor without changing the current in the circuit, the equivalent resistance should also be 40 W.

Therefore, a single 40 W resistor can be used to replace the two 20 W resistors.

This single resistor will have the same effect on the circuit's current flow as the original configuration of two resistors in series.

R_total = R1 + R2 = 20 W + 20 W = 40 W

To replace these two 20 W resistors with a single resistor, we need to find a resistor with an equivalent resistance of 40 W.

Therefore, the single resistor that could be used to replace both 20 W resistors without changing the current in the circuit is a 40 W resistor.

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The low-frequency speaker of a stereo set has a surface area of 0.06 m and produces 1.83 W of acoustical power. What is the intensity at the speaker (in W/m)? W/m2 If the speaker projects sound uniformly in all directions, at what distance (in m) from the speaker is the intensity 0.204 W/m2

Answers

The intensity at the speaker is 30.5 W/m², and the distance from the speaker at which the intensity is 0.204 W/m² is 6.33 m.

Given data:

Surface area of low-frequency speaker, A = 0.06 m²

Acoustical power produced, P = 1.83 W

The intensity at the speaker is given by I = P/A. Thus, I = 1.83 W/0.06 m² = 30.5 W/m².

Intensity is inversely proportional to the square of the distance. The formula used for finding the distance from the speaker is:

I₁r₁² = I₂r₂²

Where:

I₁ = intensity at a distance r₁ from the speaker

I₂ = intensity at a distance r₂ from the speaker

Putting the given data into the formula, we get:

0.204 × r₁² = 30.5 × r₂²

The distance from the speaker at which the intensity is 0.204 W/m² is given by r₂. Substituting r₂ = 1 m in the above equation, we can find r₁.

r₁ = sqrt(30.5/0.204) × r₂ = 6.33 m × 1 m = 6.33 m

Therefore, the intensity at the speaker is 30.5 W/m², and the distance from the speaker at which the intensity is 0.204 W/m² is 6.33 m.

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equipotentials. In a region (xy plane), the potential between x=0 and x=6.00 m satisfies the equation V =a+bx where a=10.0 V and b=+7.00 V/m. Determine:
a) the electric potential at x=0, x=3.00 m and x=6.00 m.
b) the magnitude and direction of the electric field at x=0, x=3.00 m and x=6.00 m. Use the relationship ⃗ E=−∇⃗ V which in one dimension corresponds to Ex=−dV/dx.
c) Make a drawing of some equipotentials in the xy plane and of the field lines in the xy plane in the region between x=0 and x=6.00 m.
d) If a positive charge of value 1.0 μC and mass 4.0 g is released from rest at x=3.00, calculate the speed it attains in advancing a distance of 3.00 m. Between which points does it move?

Answers

The electric potential is  - 7.00 V/m. the magnitude of the electric field at x = 0, x = 3.00 m, and x = 6.00 m is 7.00 V/m.The change in its potential energy is  2.10 × 10-5 J.The charged particle moves between x = 3.00 m and x = 6.00 m.

To determine the electric potential at x = 0, x = 3.00 m and x = 6.00 m, substitute the given values of a, b, and x in the equation V = a + bx. Here's how to compute it:

For x = 0, V =  10.0 V,For x = 3.00 m, V = a + bx

10.0 + (7.00 V/m)(3.00 m) = 31.0 V.

For x = 6.00 m, V = a + bx

10.0 + (7.00 V/m)(6.00 m) = 52.0 V

To determine the magnitude and direction of the electric field at x = 0, x = 3.00 m, and x = 6.00 m, use the relationship ⃗E = −V, which in one dimension corresponds to Ex=−dV/dx. Thus:For x = 0,E = - dV/dx|0

- (7.00 V/m) = - 7.00 V/m,

pointing in the negative x-direction.

For x = 3.00 m,E = - dV/dx|3

- (7.00 V/m) = - 7.00 V/m ,

pointing in the negative x-directionFor x = 6.00 m,E = - dV/dx|6 = - (7.00 V/m) = - 7.00 V/m pointing in the negative x-direction.

Therefore, the magnitude of the electric field at x = 0, x = 3.00 m, and x = 6.00 m is 7.00 V/m, and it points in the negative x-direction.

The equipotentials in the xy-plane and field lines in the xy-plane in the region between x = 0 and x = 6.00 m are illustrated in the following figure.

The contour lines in the figure represent the equipotentials, which are perpendicular to the electric field lines. They are uniformly spaced, indicating that the electric field is constant and uniform. Since the electric field is uniform, the electric field lines are also uniformly spaced and parallel. Since the electric field is directed from positive to negative, the electric field lines are directed from positive to negative in the x-direction.

The potential energy of the charged particle at x = 3.00 m is Ep = qV

(1.0 × 10⁻⁶ C)(31.0 V) = 3.10 × 10⁻⁵ J.

Therefore, the kinetic energy of the particle at x = 0 is equal to its potential energy at x = 3.00 m, or KE = 3.10 × 10⁻⁵ J. The total energy of the particle is conserved, so at x = 6.00 m, the sum of the kinetic and potential energy of the particle is equal to its total energy. Thus, KE + Ep = ET. or KE = ET - Ep.

The velocity of the charged particle at x = 6.00 m is v = sqrt(2KE/m), where m is the mass of the particle. Substituting the given values of KE, m, and v, the speed is calculated as:

v = √[(2KE)/(m)]

√[(2(ET - Ep))/(m)] = √[(2[(4.0 × 10⁻³ kg)(7.00 V/m)(3.00 m)] - (3.10 × 10⁻⁵J))/(4.0 × 10⁻³ kg)]

√[(2[(4.0 × 10⁻³ kg)(7.00 V/m)(3.00 m)] - (3.10 × 10⁻⁵ J))/(4.0 × 10⁻³ kg)] = 0.60 m/s.

The charged particle moves between x = 3.00 m and x = 6.00 m.

Therefore, the change in its potential energy is ΔEp = qΔV

(1.0 × 10⁻⁶ C)(52.0 V - 31.0 V) = 2.10 × 10⁻⁵ J.

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(a) A wire that is 1.50 m long at 20.0°C is found to increase in length by 1.90 cm when warmed t 420.0'C. Compute its average coefficient of linear expansion for this temperature range. (b) The wire i stretched just taut (zero tension) at 420.0*C. Find the stress in the wire if it is cooled to 20.0°C withou being allowed to contract. Young's modulus for the wire is 2.0 x 10^11 Pa.

Answers

(a) Thee average coefficient of linear expansion for this temperature range is approximately 3.17 x 10^(-5) / °C. (b) The stress in the wire, when cooled to 20.0°C without being allowed to contract, is approximately 2.54 x 10^3 Pa.

(a) The average coefficient of linear expansion (α) can be calculated using the formula:

α = (ΔL / L₀) / ΔT

Where ΔL is the change in length, L₀ is the initial length, and ΔT is the change in temperature.

Given that the initial length (L₀) is 1.50 m, the change in length (ΔL) is 1.90 cm (which is 0.019 m), and the change in temperature (ΔT) is 420.0°C - 20.0°C = 400.0°C, we can substitute these values into the formula:

α = (0.019 m / 1.50 m) / 400.0°C

= 0.01267 / 400.0°C

= 3.17 x 10^(-5) / °C

(b) The stress (σ) in the wire can be calculated using the formula:

σ = E * α * ΔT

Where E is the Young's modulus, α is the coefficient of linear expansion, and ΔT is the change in temperature.

Given that the Young's modulus (E) is 2.0 x 10^11 Pa, the coefficient of linear expansion (α) is 3.17 x 10^(-5) / °C, and the change in temperature (ΔT) is 420.0°C - 20.0°C = 400.0°C, we can substitute these values into the formula:

σ = (2.0 x 10^11 Pa) * (3.17 x 10^(-5) / °C) * 400.0°C

= 2.0 x 10^11 Pa * 3.17 x 10^(-5) * 400.0

= 2.54 x 10^3 Pa.

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1) What is the energy transformed in a 10.0 Ohm resistor when 100.0 V DC is applied for 5.00-minutes. 2) A 12.0 V DC supply is connected to two resistors in series. The first is 0.400 ks and the second is 0.800 kg. What is the current through and the potential difference across each resistor? 3) An 18.0 V source is connected to three resistors in parallel. These are 3.00 £2, 6.00 2 and 9.00 22 what are the currents through each resistor and the power converted in each resistor? Show that the sum of these currents is equal to the current through a single equivalent resistor of 1.64 22 (to 3 s.f.) connected to an 18.0 V source. What is the power converted in this resistor? 4) An AC current with amplitude 2.00 A flows through a 10 Ohm resistor. What is the average power transformed in the resistor?

Answers

The average power transformed in the 10 Ω resistor is 20 W.

1. The energy transformed in a 10.0 Ohm resistor when 100.0 V DC is applied for 5.00-minutes is 30,000 J.

2. The current through the first resistor is 30 A and the potential difference across it is 12 V.

The current through the second resistor is 15 A and the potential difference across it is 12 V.

3. The current through the 3.00 Ω resistor is 6 A, the current through the 6.00 Ω resistor is 3 A, and the current through the 9.00 Ω resistor is 2 A.

The power converted in the 3.00 Ω resistor is 108 W, the power converted in the 6.00 Ω resistor is 54 W, and the power converted in the 9.00 Ω resistor is 32 W.

The sum of these currents is 11 A, which is equal to the current through a single equivalent resistor of 1.64 Ω (to 3 s.f.) connected to an 18.0 V source.

The power converted in this resistor is 356 W.4.

The average power transformed in the 10 Ω resistor is 20 W.

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The rms current flowing through an RLC series circuit increases as the capacitive reactance is decreased. Select one: True O False

Answers

The rms current flowing through an RLC series circuit increases as the capacitive reactance is decreased. - False

The rms (root mean square) current flowing through an RLC series circuit does not increase as the capacitive reactance is decreased. In fact, as the capacitive reactance (XC) decreases, the impedance of the circuit decreases, which results in an increase in the current magnitude.

In an RLC series circuit, the impedance (Z) is given by the formula:

Z = √(R^2 + (XL - XC)^2)

Where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance.

As XC decreases, the term (XL - XC) in the above formula becomes larger, resulting in a larger overall impedance. According to Ohm's Law (V = I * Z), for a given voltage (V), a larger impedance leads to a smaller current (I).

Therefore, as the capacitive reactance is decreased in an RLC series circuit, the rms current actually increases.

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Hollow flywheel system a 400kg hollow steel flywheel energy storage with 2m outer diameter and a thickness of 225mm spins at 6000rpm. with 80fficiency, how long it will support 100kw load?

Answers

The hollow steel flywheel system will support a 100 kW load for approximately 1.77 hours.

To determine the duration for which the flywheel system can support a 100 kW load, we need to calculate the energy stored in the flywheel and then divide it by the power required by the load.

1. Calculate the moment of inertia of the hollow flywheel:

The moment of inertia (I) of a hollow cylinder can be calculated using the formula:

I = (1/2) * m * (r1^2 + r2^2)

Given:

Mass of the flywheel (m) = 400 kg

Outer radius (r2) = 1 m (diameter = 2 m)

Inner radius (r1) = r2 - thickness = 0.875 m (225 mm)

Plugging in the values:

I = (1/2) * 400 * (0.875^2 + 1^2)

I = 225 kg*m^2

2. Calculate the energy stored in the flywheel:

The energy stored in a rotating flywheel can be calculated using the formula:

E = (1/2) * I * ω^2

Given:

Angular velocity (ω) = 6000 rpm = 6000 * 2π / 60 rad/s

Plugging in the values:

E = (1/2) * 225 * (6000 * 2π / 60)^2

E = 1,413,716 J (Joules)

3. Calculate the duration of support:

The duration can be calculated by dividing the energy stored by the power required by the load:

Duration = E / (Power * Efficiency)

Given:

Power of the load = 100 kW

Efficiency = 80% = 0.8

Plugging in the values:

Duration = 1,413,716 / (100,000 * 0.8)

Duration ≈ 1.77 hours

The hollow steel flywheel system will support a 100 kW load for approximately 1.77 hours.

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An electron has an energy of 2.4 eV. It is incident on a single slit which has a width of 0.1 microns (10-6 m). What is the angle at which the first diffraction minimum is found?
Enter your answer in radians to four decimal places but do not enter the units.
If you could not determine the wavelength of the electron in the previous question, you may use a wavelength of 1 nm.

Answers

The angle at which the first diffraction minimum is found can be calculated using the formula for single-slit diffraction. Given an electron with an energy of 2.4 eV incident on a single slit with a width of 0.1 microns, we can determine the angle by considering the wavelength of the electron.

The formula for the angle of the first diffraction minimum in single-slit diffraction is given by:

sin(θ) = λ / (w),

where θ is the angle, λ is the wavelength of the incident wave, and w is the width of the slit.

To calculate the angle, we need to determine the wavelength of the electron. If the wavelength is not provided, we can assume a value of 1 nm (10^(-9) m) for the electron wavelength.

Using the given width of the slit (0.1 microns = 10^(-7) m) and the assumed wavelength (1 nm = 10^(-9) m), we can substitute these values into the formula:

sin(θ) = (10^(-9) m) / (10^(-7) m) = 10^(-2).

To find the angle θ, we take the inverse sine of 10^(-2):

θ = sin^(-1)(10^(-2)) ≈ 0.01 radians.

Therefore, the angle at which the first diffraction minimum is found is approximately 0.01 radians.

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Write a x; in a form that includes the Kronecker delta. Now show that V.r=3.

Answers

x; = Σn=1 to ∞ δn,x vn,
where δn,x is the Kronecker delta and vn is a vector in the basis of x.


Kronecker delta is a mathematical symbol that is named after Leopold Kronecker. It is also known as the Kronecker's delta or Kronecker's symbol. It is represented by the symbol δ and is defined as δij = 1 when i = j, and 0 otherwise. Here, i and j can be any two indices in the vector x. The vector x can be expressed as a sum of vectors in the basis of x as follows: x = Σn=1 to ∞ vn, where vn is a vector in the basis of x.

Using the Kronecker delta, we can express this sum in the following form:

x; = Σn=1 to ∞ δn,x vn, where δn,x is the Kronecker delta. Now, if we take the dot product of the vector V and x, we get the following:

V·x = V·(Σn=1 to ∞ vn) = Σn=1 to ∞ (V·vn)

Since V is a 3-dimensional vector, the dot product V·vn will be zero for all but the third term, where it will be equal to 3. So, V·x = Σn=1 to ∞ (V·vn) = 3, which proves that V·x = 3.

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a skateboarder uses an incline to jump over a wall. the skateboarder reaches their maximum height at the wall barely making it over. the height of the wall is h=.86 m. the ramp makes an angle of 35 degrees with respect to the ground. Assume the height of the ramp is negligible so that it can be ignored.
Write the known kinematic variables for the horizontal and vertical motion.
What initial speed does the skateboarded need to make the jump?
How far is the wall from the ramp?

Answers

Known kinematic variables:

Vertical motion: Maximum height (h = 0.86 m), angle of incline (θ = 35 degrees), vertical acceleration (ay = -9.8 m/s^2).

Horizontal motion: Distance to the wall (unknown), horizontal velocity (unknown), horizontal acceleration (ax = 0 m/s^2).

To calculate the initial speed (vi) needed to make the jump, we can use the vertical motion equation:

h = (vi^2 * sin^2(θ)) / (2 * |ay|)

Plugging in the given values:

h = 0.86 m

θ = 35 degrees

ay = -9.8 m/s^2

We can rearrange the equation to solve for vi:

vi = √((2 * |ay| * h) / sin^2(θ))

Substituting the values and calculating:

vi = √((2 * 9.8 m/s^2 * 0.86 m) / sin^2(35 degrees))

vi ≈ 7.12 m/s

Therefore, the skateboarder needs an initial speed of approximately 7.12 m/s to make the jump.

To find the distance to the wall (d), we can use the horizontal motion equation:

d = vi * cos(θ) * t

Since the height of the ramp is negligible, the time of flight (t) can be determined solely by the vertical motion. We can use the equation:

h = (vi * sin(θ) * t) + (0.5 * |ay| * t^2)

We can rearrange this equation to solve for t:

t = (vi * sin(θ) + √((vi * sin(θ))^2 + 2 * |ay| * h)) / |ay|

Substituting the values and calculating:

t = (7.12 m/s * sin(35 degrees) + √((7.12 m/s * sin(35 degrees))^2 + 2 * 9.8 m/s^2 * 0.86 m)) / 9.8 m/s^2

t ≈ 0.823 s

Finally, we can substitute the time value back into the horizontal motion equation to find the distance to the wall (d):

d = 7.12 m/s * cos(35 degrees) * 0.823 s

d ≈ 4.41 m

Therefore, the wall is approximately 4.41 meters away from the ramp.

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"Two 4.0 cmcm ××4.0 cmcm square aluminum electrodes, spaced 0.50
mmmm apart, are connected to a 200 VV battery.What is the
capacitance?What is the charge on the positive electrode?

Answers

The system's capacitance is approximately 2.83 nanofarads (nF) and the charge on the positive electrode is about 5.66 micro coulombs (μC).

To find the capacitance (C) of the system, we can use the formula:

C = ε₀ × (A / d)

where:

C = capacitance

ε₀ = permittivity of free space (constant value)

A = area of overlap between the electrodes

d = separation distance between the electrodes

The area of overlap between the electrodes can be calculated as follows:

A = a × a

Plugging in the values, we get:

A = 0.04 m × 0.04 m = 0.0016 m²

The permittivity of free space (ε₀) is a constant value of approximately 8.85 x 10^-12 F/m.

Now, let's calculate the capacitance (C):

C = (8.85 x 10⁻¹² F/m) * (0.0016 m² / 0.0005 m)

C ≈ 2.83 x 10⁻⁹ F

Therefore, the system's capacitance is approximately 2.83 nanofarads (nF).

To find the charge on the positive electrode, we can use the formula:

Q = C × V

where:

Q = charge

C = capacitance

V = voltage

Substituting in the values, we get:

Q = (2.83 x 10⁻⁹ F) × (200 V)

Q ≈ 5.66 x 10⁻⁷ C

Therefore, the charge on the positive electrode is approximately 5.66 micro coulombs (μC).

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A steel rule is calibrated for measuring lengths at 20.00°C. The rule is used to measure the length of a Vycor glass brick; when both are at 20.00°C, the brick is found to be 23.90 cm long. If the rule and the brick are both at 57.00°C, what would be the length of the brick as measured by the rule? Coefficient of linear expansion α for steel is 12.0 × 10−6 K−1 and for glass (Vycor) is 0.750 × 10−6 K−1. answer in cm

Answers

The length of the brick measured by the rule is 0.011926cm at 57°C.

The change in length due to thermal expansion is given by:

ΔL = α × L × ΔT

Where:

ΔL is the change in length,

α is the coefficient of linear expansion,

L is the initial length, and

ΔT is the change in temperature.

Coefficient of linear expansion, α(steel) = 12.0 × 10⁻⁶ K⁻¹

Coefficient of linear expansion, α(vycor) = 0.750 × 10⁻⁶ K⁻¹

Initial length, L(steel) = 23.90 cm

Initial temperature, T₁(steel) = 20.00°C = 293K

Final temperature, T₂(steel) = 57.00°C = 330K

ΔT(steel) = T₂(steel) - T₁(steel) = 37K

ΔL(steel) = α(steel) × L(steel) × ΔT(steel) = 0.0106cm

Similarly,

ΔL(vycor) = 6.63 × 10⁻⁴

ΔL(total) = ΔL(steel) + ΔL(vycor)

ΔL(total) = 0.0112cm

Length at 57.00°C = L(vycor) + ΔL(total) = 0.011926cm.

Hence, the length of the brick measured by the rule is 0.011926cm at 57°C.

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(14.8) In the figure, a particle moves along a circle in a region of uniform magnetic field of magnitude B = 4.6 mT. The particle is either a proton or an electron (you must decide which). It experiences a magnetic force of magnitude 3.0 × 10-15 N. What are (a) the particle's speed, (b) the radius of the circle, and (c) the period of the motion?

Answers

(a) Since the force is given, we can equate it to qvB and solve for the velocity (v). By knowing the charge of the particle, we can determine if it's a proton or an electron.

The particle in the uniform magnetic field experiences a magnetic force, which is given by the equation F = qvB, where q is the charge of the particle, v is its velocity, and B is the magnetic field strength.

(b) The radius of the circle can be determined using the centripetal force equation, F = mv²/r, where m is the mass of the particle, v is its velocity, and r is the radius of the circle. By rearranging the equation, we can solve for the radius (r).

(c) The period of the motion is the time it takes for the particle to complete one full revolution around the circle. It can be calculated using the equation T = 2πr/v, where T is the period, r is the radius, and v is the velocity.

(a) To determine the particle's speed, we need to know whether it is a proton or an electron since their charges differ. Once we know the charge, we can rearrange the equation F = qvB to solve for the velocity (v) by dividing both sides of the equation by qB. The resulting velocity will represent the speed of the particle.

(b) The centripetal force experienced by the particle is responsible for its circular motion. By equating the magnetic force (given) to the centripetal force (mv²/r), we can rearrange the equation to solve for the radius (r). The mass of the particle can be obtained based on whether it is a proton or an electron.

(c) The period of the motion represents the time taken for the particle to complete one full revolution around the circle. It can be calculated using the equation T = 2πr/v, where r is the radius and v is the velocity. Substituting the known values will give us the period of the motion.

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A pendulum on Earth is released from rest at an angular displacement of 8.9 degrees to the right, and is at an angular displacemet of -4.76886 degrees when measured 1.12131 s after it is released. Assume the positive angular displacement direction is to the right. Help on how to format answers: units a. What is the length of the pendulum?

Answers

The positive angular displacement direction is to the right. The length of the pendulum is approximately 0.288 meters.

To determine the length of the pendulum, we can use the equation 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.

First, we need to find the period of the pendulum. The time it takes for the pendulum to complete one full oscillation can be calculated using the given angular displacements.

The difference in angular displacement between the two measurements is:

Δθ = final angular displacement - initial angular displacement

    = (-4.76886 degrees) - (8.9 degrees)

    = -13.66886 degrees

To convert the angular displacement to radians:

Δθ_rad = Δθ * (π/180)

       = -13.66886 degrees * (π/180)

       = -0.2384767 radians

Next, we can find the period using the formula for the period of a pendulum:

T = (time for one oscillation) / (number of oscillations)

Since the pendulum is released from rest, it takes one oscillation for the given time interval of 1.12131 s. Therefore, the period is equal to the time interval:

T = 1.12131 s

Now, we can rearrange the equation for the period of a pendulum to solve for the length:

L = (T^2 * g) / (4π^2)

Substituting the values:

L = (1.12131 s)^2 * g / (4π^2)

To find the length of the pendulum, we need to know the value of acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.

L = (1.12131 s)^2 * (9.8 m/s^2) / (4π^2)

L ≈ 0.288 m

Therefore, the length of the pendulum is approximately 0.288 meters.

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A ball thrown horizontally from the top of a building 0.2km high. The ball hits the ground at a point 47m horizontally away from and below the launch point. What is the speed of the ball (m/s) just before it hits the ground?
Give your answer in whole numbers

Answers

A ball thrown horizontally from the top of a building 0.2km high.  the speed of the ball just before it hits the ground is approximately 7 m/s.

To find the speed of the ball just before it hits the ground, we can use the equations of motion. Since the ball is thrown horizontally, there is no vertical acceleration acting on it.

Given:

Height of the building (h) = 0.2 km = 200 m

Horizontal distance (d) = 47 m

We need to find the speed (v) of the ball just before it hits the ground.

Using the equation of motion for vertical displacement:

h = (1/2) * g * t^2

Where g is the acceleration due to gravity and t is the time of flight. Since the initial vertical velocity is zero, the time of flight can be determined using the equation:

t = sqrt((2h) / g)

Substituting the values, we have:

t = sqrt((2 * 200) / 9.8) ≈ 6.42 s

Now, we can use the equation for horizontal distance traveled:

d = v * t

Rearranging the equation, we can solve for v:

v = d / t

Substituting the values, we have:

v = 47 / 6.42 ≈ 7.32 m/s

Therefore, the speed of the ball just before it hits the ground is approximately 7 m/s.

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The resistive force that occurs when the two surfaces do side across each other is known as _____

Answers

The resistive force that occurs when two surfaces slide across each other is known as friction.

Friction is the resistive force that opposes the relative motion or tendency of motion between two surfaces in contact. When one surface slides over another, the irregularities or microscopically rough surfaces of the materials interact and create resistance.

This resistance is known as friction. Friction occurs due to the intermolecular forces between the atoms or molecules of the surfaces in contact.

The magnitude of friction depends on factors such as the nature of the materials, the roughness of the surfaces, and the normal force pressing the surfaces together. Friction plays a crucial role in everyday life, affecting the motion of objects, enabling us to walk, drive vehicles, and control the speed of various mechanical systems.

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X-rays of wavelength 0.116 nm reflect off a crystal and a second-order maximum is recorded at a Bragg angle of 22.1°. What is the spacing between the scattering planes in this crystal?

Answers

To determine the spacing between the scattering planes in the crystal, we can use Bragg's Law.

Bragg's Law relates the wavelength of X-rays, the angle of incidence (Bragg angle), and the spacing between the scattering planes.

The formula for Bragg's Law is: nλ = 2d sinθ

In this case, we are dealing with second-order diffraction (n = 2), and the wavelength of the X-rays is given as 0.116 nm. The Bragg angle is 22.1°.

We need to rearrange the equation to solve for the spacing between the scattering planes (d):

d = nλ / (2sinθ)

Plugging in the values:

d = (2 * 0.116 nm) / (2 * sin(22.1°))

 ≈ 0.172 nm

Therefore, the spacing between the scattering planes in the crystal is approximately 0.172 nm.

when X-rays with a wavelength of 0.116 nm are incident on the crystal, and a second-order maximum is observed at a Bragg angle of 22.1°, the spacing between the scattering planes in the crystal is approximately 0.172 nm.

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An interference pattern is formed on a screen when light of
wavelength 500 nm is incident on two parallel slits 60
μmapart.
Find the angle of the third order bright fringe.

Answers

The angle of the third-order bright fringe in the interference pattern formed by light of wavelength 500 nm incident on two parallel slits spaced 60 μm apart is approximately 0.18 degrees.

In the double-slit interference pattern, the bright fringes are formed at specific angles due to constructive interference of the light waves. The formula for calculating the angle of the bright fringes is given by the equation

dsinθ = mλ,

where d is the slit spacing, θ is the angle of the bright fringe, m is the order of the fringe, and λ is the wavelength of light.

For the third-order bright fringe (m = 3), we can rearrange the formula to solve for θ: θ = arcsin(mλ/d).

Substituting the values, we have θ = arcsin((3 * 500 nm) / 60 μm). Converting the units to be consistent, we get θ ≈ arcsin(0.015) ≈ 0.18 degrees.

Therefore, the angle of the third-order bright fringe in the interference pattern is approximately 0.18 degrees.

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In a parallel circuit
Select one or more than one:
a. The currents of all the resistors, added together, is the current of the battery
b. The voltage across each resistor is the same.
c. The current through each resistor is the same.
d. The voltages of all the resistors, added together, is the battery voltage.

Answers

In a parallel circuit:b. The voltage across each resistor is the same.c. The current through each resistor is the same.

Both options b and c are correct for a parallel circuit. In a parallel circuit, the voltage across each resistor is the same because all the resistors are connected directly across the voltage source. Additionally, the current through each resistor is the same because the total current entering the parallel circuit is divided among the individual branches, with each resistor experiencing the same amount of current.Option a is incorrect for a parallel circuit because in a parallel circuit, the currents of all the resistors are not added together to give the current of the battery. The total current entering the parallel circuit is the sum of the currents through each individual branch.Option d is incorrect for a parallel circuit because the voltages of the resistors in a parallel circuit do not add up to the battery voltage. The voltage across each resistor is the same and equal to the battery voltage.

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18-5 (a) Calculate the number of photons in equilibrium in a cavity of volume 1 m? held at a temperature T = 273 K. (b) Compare this number with the number of molecules the same volume of an ideal gas contains at STP.

Answers

For a cavity of volume 1 m³ held at a temperature of 273 K (equivalent to 0 degrees Celsius or 32 degrees Fahrenheit), the number of photons in equilibrium can be determined.

The number of photons in equilibrium can be obtained by integrating the Planck radiation law over all possible photon energies. The calculation involves considering the photon energy levels and their respective probabilities according to the temperature. The result yields a value for the number of photons in equilibrium.

In comparison, the number of molecules in the same volume of an ideal gas at standard temperature and pressure (STP) can be calculated using the ideal gas law, which relates the pressure, volume, and temperature of a gas. At STP, which is defined as a temperature of 273.15 K (0 degrees Celsius) and a pressure of 1 atmosphere (atm), the number of molecules in a given volume can be determined.

By comparing the number of photons in equilibrium in the cavity to the number of molecules in the same volume of an ideal gas at STP, we can observe the significant difference between the two quantities.

The number of photons in equilibrium depends on the temperature and is determined by the Planck radiation law, while the number of molecules in an ideal gas at STP is governed by the ideal gas law. These calculations provide insights into the vastly different nature and behavior of photons and gas molecules in equilibrium systems.

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Please help ASAP! Physics Final is depending on this.

Answers

Answer:

Explanation:

Given:

The image shows a diagram of a circuit with electrical components.There are 3 resistors with resistance values 10 ohms, 4 ohms and 6 ohms. They are labeled R1, R2 and R3 respectively.There is a 20V battery supplying power to the circuit. It is labeled VThere is a capacitor with capacitance 5 microfarads. It is labeled C.There are 2 switches in the circuit labeled S1 and S2. Both are in the open position in this diagram.There are nodes or junctions labeled a, b, c, d and e.Wires connect the different components together to form the circuit.

D Question 4 6.25 pts A space station shaped like a ring rotates in order to generate an acceleration of 12 m/s2. If the station has a radius of 151 m, what is the speed of the outer edge of the ring

Answers

The speed of the outer edge of the ring is approximately 42.62 m/s.

To find the speed of the outer edge of the ring, we can use the formula for centripetal acceleration:

a = v^2 / r

Where:

a is the acceleration (given as 12 m/s^2)v is the velocity (speed) of the outer edge of the ring (what we're trying to find)r is the radius of the ring (given as 151 m)

Rearranging the formula, we get:

v = √(a * r)

Substituting the given values:

v = √(12 m/s^2 * 151 m)

v ≈ √(1812 m^2/s^2)

v ≈ 42.62 m/s

Therefore, the speed of the outer edge of the ring is approximately 42.62 m/s.

The complete question should be:

A space station shaped like a ring rotates in order to generate an acceleration of 12 m/s2. If the station has a radius of 151 m, what is the speed of the outer edge of the ring in m/s?

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A single slit experiment forms a diffraction pattern with the fourth minima 0 =2.1° when the wavelength is X. Determine the angle of the m =6 minima in this diffraction pattern (in degrees).

Answers

A single slit experiment forms a diffraction pattern with the fourth minima 0 =2.1°, the angle of the m = 6 minima in this diffraction pattern is approximately 14.85°.

The position of the minima in a single slit diffraction pattern is defined by the equation:

sin(θ) = m * λ / b

sin(2.1°) = 4 * X / b

sin(θ6) = 6 * X / b

θ6 = arcsin(6 * X / b)

θ6 = arcsin(6 * (sin(2.1°) * b) / b)

Since the width of the slit (b) is a common factor, it cancels out, and we are left with:

θ6 = arcsin(6 * sin(2.1°))

θ6 ≈ 14.85°

Thus, the angle of the m = 6 minima in this diffraction pattern is approximately 14.85°.

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A tower cranc has a hoist motor rated at 155 hp. If the cranc is limited to using 69.0% of its maximum hoisting power for safety reasons, what is the shortest time in which the crane can lift a 5550 kg load over a distance of 87.0 m? Assume the
load is lifted at a constant velocity.

Answers

The shortest time in which the crane can lift the 5550 kg load over a distance of 87.0 m under constant velocity is approximately 58.74 seconds.

To find the numerical value of the shortest time, we need to calculate the maximum hoisting power (P_max) and substitute it into the equation.

Hoist motor rated power: 155 hp

Load mass: 5550 kg

Distance lifted: 87.0 m

Percentage of maximum hoisting power used: 69.0%

First, let's calculate the maximum hoisting power in watts:

P_max = 155 hp * 746 W/hp

P_max ≈ 115630 W

Next, let's calculate the actual hoisting power (P_actual):

P_actual = 0.69 * P_max

P_actual ≈ 0.69 * 115630 W

P_actual ≈ 79869 W

Now, let's calculate the work done by the crane:

W = mg * d

W = 5550 kg * 9.8 m/s^2 * 87.0 m

W ≈ 4689930 J

Finally, let's calculate the shortest time (t):

t = W / P_actual

t ≈ 4689930 J / 79869 W

t ≈ 58.74 seconds

Therefore, the shortest time in which the crane can lift the 5550 kg load over a distance of 87.0 m is approximately 58.74 seconds.

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If it takes a ball dropped from rest 2.417 s to fall to the ground, from what height h was it released?

Answers

To find the height from which the ball was released, we can use the formula for the distance fallen by an object under free fall: d = 0.5 g t 2. In this formula, d represents the distance fallen, g is the acceleration due to gravity (approximately 9.8 m/s 2), and t is the time taken to fall.

Given that the time taken to fall is 2.417 s, we can plug in these values into the formula:

d = 0.5 * 9.8 * (2.417)^2

Simplifying this equation, we get:

d = 0.5 9.8  5.855489

d ≈ 28.672 m

Therefore, the ball was released from a height of approximately 28.672 meters. This is the main answer.

The formula used to calculate the distance fallen by an object under free fall is derived from the equations of motion. In this case, we assumed that the ball was dropped from rest, which means it started with an initial velocity of zero. If the ball had an initial velocity, we would need to use a different formula, such as d = where v_0 represents the initial velocity. However, since the question states that the ball was dropped from rest, we can use the simplified formula.

In conclusion, the ball was released from a height of approximately 28.672 meters.

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A man holds a double-sided spherical mirror so that he is looking directly into its convex surface, 45 cm from his face. The magnification of the image of his face is +0.25. What will be the image distance when he reverses the mirror (looking into its concave surface), maintaining the same distance between the mirror and his face?

Answers

Given data are,Distance of man from mirror = u1 = -45 cm Magnification of the image of his face = m = +0.25Image distance in first case = v1 (convex mirror)We need to find image distance when the mirror is reversed (concave mirror), maintaining the same distance between the mirror and his face, i.e.,v2 = ?

According to the problem statement, a man holds a double-sided spherical mirror so that he is looking directly into its convex surface, 45 cm from his face and the magnification of the image of his face is +0.25. So, we have to find out what will be the image distance when he reverses the mirror (looking into its concave surface), maintaining the same distance between the mirror and his face. Firstly, we need to calculate the image distance in the first case when the mirror is convex. So, the distance of the man from the mirror is -45 cm.

As given, the magnification of the image of his face is +0.25. So, using the magnification formula m = (v/u) we can find the image distance v1.v1 = m × u1v1 = 0.25 × (-45)v1 = -11.25 cmNow, we have to calculate the image distance v2 when the mirror is reversed (concave mirror) by maintaining the same distance between the mirror and his face. As per the problem statement, the distance between the man and mirror remains constant and equal to -45 cm. Now, we have to find the image distance v2. As the mirror is now concave, the image is real, and hence, v2 is negative.

Therefore, we can write the magnification formula asm = -v2/u1Here, m = +0.25 and u1 = -45 cmSo, the image distance isv2 = m × u1v2 = 0.25 × (-45)v2 = -11.25 cm. Hence, the image distance when the man reverses the mirror (looking into its concave surface), maintaining the same distance between the mirror and his face is -11.25 cm.

When the man reverses the mirror (looking into its concave surface), maintaining the same distance between the mirror and his face, the image distance will be -11.25 cm.

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What is the y component of vector ? Cy-3 Suppose C - A4 B where vector Ä nas components A = 5, A, 2 and vector i nas components B, 3B, 5 Previous Answers Correct Important: If you use this answer in later parts, use the full unrounded value in your calculations Y Part C What is the magnitude of vector VAXD OBI? Submit Request Answer Part What is the deection of vector ca Express your answer in degrees VOAS ?

Answers

The y-component of a vector is denoted as the second element of the vector when using the standard Cartesian coordinate system. The y-component of vector C is A + 12B.

To find the y-component of vector C, we look at the given information: C = A + 4B, where vector A has components A = (5, A, 2) and vector B has components B = (B, 3B, 5).

To find the y-component of C, we focus on the y-component of each vector and add them together: C_y = A_y + 4B_y

Since A = (5, A, 2), A_y = A.

Similarly, B = (B, 3B, 5), so B_y = 3B.

Substituting these values into the equation, we have:

C_y = A + 4(3B)

C_y = A + 12B

Therefore, the y-component of vector C is A + 12B.

To find the magnitude of vector VAXB, we need to calculate the cross product of vectors A and B. The cross product of two vectors is a vector perpendicular to both vectors, and its magnitude represents the area of the parallelogram formed by the two vectors.

The magnitude of the cross product can be calculated using the formula:

|VAXB| = |A| * |B| * sin(theta)

Where |A| and |B| are the magnitudes of vectors A and B, and theta is the angle between them.

Since the magnitudes of vectors A and B are not provided, we cannot calculate the magnitude of vector VAXB without this information.

To find the deflection of vector CA, we need to determine the angle between vectors C and A.

Using the dot product of vectors C and A, we can find the angle theta between them:

C · A = |C| * |A| * cos(theta)

The dot product can also be calculated as:

C · A = C_x * A_x + C_y * A_y + C_z * A_z

Since only the y-components of vectors C and A are given, we can focus on those:

C_y * A_y = |C| * |A| * cos(theta)

Substituting the given values:

(C - 3) * 5 = |C| * |A| * cos(theta)

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A figure skating couple change their configuration so that they
rotate faster, from 15 rpm to 30 rpm. What is the ratio of the
couples moment of Inertia before the deformation to the moment of
inertia

Answers

The ratio of the couples moment of Inertia before the deformation to the moment of inertia is 2 by applying conservation of angular momentum.

The couples moment of inertia can be defined as a measure of the amount of energy needed to move an object rotating on an axis. On the other hand, angular speed is a measure of how fast an object is rotating on an axis.  Let us now solve the given problem. A figure skating couple changed their configuration so that they rotate faster, from 15 rpm to 30 rpm. The ratio of the couples moment of Inertia before the deformation to the moment of inertia is calculated as follows: Since the figure skating couple rotates faster, the initial angular speed is 15 rpm, while the final angular speed is 30 rpm. Therefore, the ratio of the couples moment of Inertia before the deformation to the moment of inertia is given by: I1/I2 = ω2/ω1

Where I1 is the moment of inertia before deformation, I2 is the moment of inertia after deformation, ω1 is the initial angular speed, and ω2 is the final angular speed. Substituting the given values, we get:

I1/I2 = (30 rpm)/(15 rpm)

I1/I2 = 2

Therefore, the ratio of the couples moment of Inertia before the deformation to the moment of inertia is 2.

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The mass of 1 mol of 13C (carbon-13) is 13.003 g.
a. What is the mass in u of one 13C atom? answer in
u
b. What is the mass in kilograms of one 13C atom?
____ *10^-26 kg

Answers

The mass of one 13C atom is 13.009 u. The mass in kilograms of one 13C atom is 2.160 × 10⁻²⁶ kg.

a. To calculate the mass in u (atomic mass units) of one 13C atom, we need to divide the molar mass of 13C by Avogadro's number (6.022 × 10²³). The molar mass of 13C is given as 13.003 g/mol.

Mass of one 13C atom

= (13.003 g/mol) / (6.022 × 10²³) = 2.160 × 10⁻²³ g

To convert the mass from grams to atomic mass units (u), we need to divide it by the atomic mass constant. The atomic mass constant is defined as 1/12th the mass of a carbon-12 atom, which is approximately 1.66 × 10⁻²⁴ g.

Mass of one 13C atom =[tex](2.160 \times 10^{(-23)} g) / (1.66 \times 10^{(-24)} g) = 13.009 u[/tex]

b. To convert the mass of one 13C atom from grams to kilograms, we divide it by 1000 since there are 1000 grams in a kilogram.

Mass of one 13C atom =  [tex](2.160 \times 10^{(-23)} g) / (1000) = 2.160 \times 10^{(-26)} kg[/tex]

Therefore, the mass of one 13C atom is 13.009 u, and its mass in kilograms is [tex]2.160 \times 10^{(-26)} kg[/tex].

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Final answer:

The mass of one 13C atom is 13.003 u and 2.161 x 10^-26 kg.

Explanation:

a. The mass in u of one 13C atom is 13.003 u.
b. To convert this to kilograms, we need to convert u to kg using the conversion factor:
1 u = 1.66054 * 10-27 kg
Therefore, the mass in kilograms of one 13C atom is 13.003 * (1.66054 * 10-27) kg = 2.161 x 10-26 kg.

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