If the force on an object is negative, what is known about the change in velocity?

O The velocity will decrease.
O The velocity will increase.
O The velocity will not change.
O The object will change direction.

Answer:

a)  U = 506 J, b)  U = 37.11 J, c) U = 0

Explanation:

The gravitational power energy is given by the expression

         U = m g (y -y₀)

In general, a reference system is set that allows the expression to be simplified, in this case let's assume the reference system at the child's lowest point, therefore y₀ = 0

Let's use trigonometry to find the child's height

          h = y = L - L cos θ

we substitute

           U = m g L (1 - cos θ)

a) when the chain is horizontal θ = 90 and cos 90 = 0

           U = mg L

weight and mass are related

            W = mg

            m = W / g

           U = 230 2.20

           U = 506 J

b) θ = 33.0º

           cos 33 = 0.83867

           U = 230 (1 - 0.83867)

           U = 37.11 J

c) in this case θ = 0 cos 0 = 1

            U = 0

Answer:

[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]

Explanation:

We are given that

[tex]PV^{1.4}=C[/tex]

Where C=Constant

[tex]\frac{dP}{dt}=-7KPa/minute[/tex]

V=420 cubic cm and P=99KPa

We have to find the rate at which the  volume increasing at this instant.

Differentiate w.r.t t

[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]

Substitute the values

[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]

[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]

[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]

[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]

Answer:

[tex]\dot V=2786.52~cm^3/min[/tex]

Explanation:

Given:

initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]

initial volume during the process, [tex]V_1=420~cm^3[/tex]

The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.

Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]

Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:

[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]

[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]

[tex]\dot V=2786.52~cm^3/min[/tex]

[tex]\because P\propto\frac{1}{V}[/tex]

[tex]\therefore[/tex] The rate of change in volume will be increasing.

I see six (6) loops.

I attached a drawing to show where I get six loops from.

Answer:

I think it would be (-7 C )..

Answer:

hmmmmmmmmmmmmmmmmmmmmmmmm y

Answer:

6.3 m/s

Explanation:

From the given information:

The displacement (x) = 15 m

time (t) = 7.0 s

initial velocity = -2.0 m/s (since it is moving in the opposite direction)

We need to determine the acceleration then find the final velocity.

By applying the kinematics equation:

[tex]x = ut + \dfrac{1}{2}at^2[/tex]

[tex]15 = (-2.0)(7.0) + \dfrac{1}{2}a(7.0)^2\\ \\ 15 = -14.0 + \dfrac{49}{2}a \\ \\ 29= 24.5a \\ \\ a= \dfrac{29}{24.5} \\ \\ a = 1.184 \ m/s^2[/tex]

Now, to determine the final velocity by using the equation:

v = u + at

v = -2 + 1.184(7.0)

v = 6.288 m/s

v ≅ 6.3 m/s

Answer:

9.25 m/s

Explanation:

Answer:Diagram A

Explanation:

Since the air resistance is to be neglected, only the gravitational force acts on the ball ( and has acted all the way from the throw upward). Diagram A reflects this fact correctly indicating the gravity acting on the ball downward.

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:

[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)

Where:

[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.

[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.

[tex]\Delta P[/tex] - Pressure change, in pascals.

If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:

[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]

[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]

[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Answer:

0.113 J

Explanation:

Applying,

w = ke²/2................. Equation 1

Where w = workdone in stretching the spring, k = spring constant, e = extension

make k the subject of the equation

k = 2w/e²................ Equation 2

From the question,

Given: w = 3 J, e = 49-32 = 17 cm = 0.17 m

Substitute these values into equation 2

k = (2×3)/0.17²

k = 6/0.17

k = 35.29 N/m

(a) if the spring from 37 cm to 45 cm,

Then,

w = ke²/2

Given: e = 45-37 = 8 cm = 0.08

w = 35.29(0.08²)/2

w = 0.113 J

Answer:

17.8cm

Explanation:

1.3kg --> 4cm

1kg --> 3, 1/13cm

5.8kg --> 18.8cm

Answer:

The magnitude of the maximum possible torque exerted on the coil is 5.73 x 10⁻³ Nm

Explanation:

Given;

number of turns of the circular coil, N = 49.5 turns

radius of the coil, r = 5.10 cm = 0.051 m

magnitude of the magnetic field, B = 0.535 T

current in the coil, I = 26.5 mA = 0.0265 A

The magnitude of the maximum possible torque exerted on the coil is calculated as;

τ = NIAB

where;

A is the area of the coil

A = πr² = π(0.051)² = 0.00817 m²

Substitute the given values and solve for the maximum torque

τ = (49.5) x (0.0265) x (0.00817) x (0.535)

τ = 0.00573 Nm

τ = 5.73 x 10⁻³ Nm

It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

The sagging sheet gives the impact with the egg additional time, which prevents the egg from breaking when it is hurled against it. This lessens the force the egg would have applied to the wall had it been flung at it.

It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.

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Answer:

1140 J, 6840 J, 10260 J

Explanation:

Lo x 2 Lo x 3 Lo, Lo = 0.2 m,  K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,

time, t = 6 s

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]

The heat conducted is

[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]

Answer:

 T = 5.45 10⁻¹⁰   [tex]\sqrt{(R_e + h)^3}[/tex]

Explanation:

Let's use Newton's second law

          F = ma

force is the universal force of attraction and acceleration is centripetal

          G m M / r² = m v² / r

          G M / r = v²

as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion

          v = d / T

the length of a circle is

          d = 2π r

we substitute

        G M / r = 4π² r² / T²

        T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]

the distance r is measured from the center of the Earth (Re), therefore

        r = Re + h

where h is the height from the planet's surface

let's calculate

         T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex]   (Re + h) ³

         T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]

         T = 5.45 10⁻¹⁰   [tex]\sqrt{(R_e + h)^3}[/tex]

Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Explanation:

Let us assume that

[tex]q_{1} = q_{2} = +q[/tex]

[tex]q_{3} = -q[/tex]

As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).

Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r

Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.

[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]

where,

k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]

Substitute the values into above formula as follows.

[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)

Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.

[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex]   ... (2)

As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.

Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Let m be the mass of the second car, so the first car's mass is 2m.

Let K be the kinetic energy of the second car, so the first car's kinetic energy would be K/2.

Let u and v be the speeds of the first car and the second car, respectively. At the start,

• the first car has kinetic energy

K/2 = 1/2 (2m) u ² = mu ²   ==>   K = 2mu ²

• the second car starts with kinetic energy

K = 1/2 mv ²

It follows that

2mu ² = 1/2 mv ²

==>   4u ² = v ²

When their speeds are both increased by 2.76 m/s,

• the first car now has kinetic energy

1/2 (2m) (u + 2.76 m/s)² = m (u + 2.76 m/s)²

• the second car now has kinetic energy

1/2 m (v + 2.76 m/s)²

These two kinetic energies are equal, so

m (u + 2.76 m/s)² = 1/2 m (v + 2.76 m/s)²

==>   2 (u + 2.76 m/s)² = (v + 2.76 m/s)²

Solving the equations in bold gives u ≈ 1.95 m/s and v ≈ 3.90 m/s.

Answer:

Archimedes was born about 287 BCE in Syracuse on the island of Sicily. He died in that same city when the Romans captured it following a siege that ended in either 212 or 211 BCE. One story told about Archimedes' death is that he was killed by a Roman soldier after he refused to leave his mathematical work.

Answer:

A = 26.875 rad/s²

Explanation:

Given the following data;

Initial angular speed, Uw = 150 rads/s.

Final angular speed, Vw = 580 rads/s.

Time = 16 seconds.

To calculate the angular acceleration;

From kinematics equation;

At = Vw - Uw

Where;

Substituting into the formula, we have;

A*16 = 580 - 150

16A = 430

A = 430/16

A = 26.875 rad/s²

(D)

Explanation:

The more massive an object is, the greater is the curvature that they produce on the space-time around it.

The theory of relativity explain the gravity exerted by massive objects is

more massive objects create larger curves of space-time (option-d).

The term "gravitational force" refers to the attraction between masses. The gravitational force increases in size as the masses get bigger (also called the gravity force). As the distance between masses grows, the gravitational force progressively lessens.

Greater gravitational forces will be used to attract heavier things since the gravitational force is directly proportional to the mass of both interacting objects. Therefore, when two things' respective masses increase, so does their gravitational pull to one another.

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P = F v

where P is power, F is the magnitude of force, and v is speed. So

50.0 hp = 37,280 W = F (24.6 m/s)

==>   F = (37,280 W) / (24.6 m/s) ≈ 1520 N

Answer:

The current across the resistance is 0.011 A.

Explanation:

Total resistance, R = 25 ohms

Total current, I = 100 mA = 0.1 A

Let the voltage is V.

By the Ohm's law

V = I R

V = 0.1 x 25 = 2.5 V

Now the resistance is R' = 220 ohm

As they are in parallel so the voltage is same. Let the current is I'.

V = I' x R'

2.5 = I' x 220

I' = 0.011 A

Answer:

1568J

Explanation:

Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.

Use conservation of Energy

ΔUg+ΔKE=0

ΔUg= mgΔh=2*9.8*(20-100)=-1568J

ΔKE-1568J=0

ΔKE=1568J

since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J

Complete question:

A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.

Answer:

The tension on the rope is 1.95 N

Explanation:

The general equation of a progressive wave is given as;

[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]

Compare the given equation with the general equation of wave, the following parameters will be deduced.

A = 0.75 cm

k = 0.400π cm⁻¹

ω = 250π s⁻¹

The frequency of the wave is calculated as;

ω = 2πf

2πf = 250π

2f = 250

f = 250/2

f = 125 Hz

The wavelength of the wave is calculated as;

[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]

The velocity of the wave is calculated as;

v = fλ

v = 125 x 0.05

v = 6.25 m/s

The tension on the rope is calculated as;

[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]

Therefore, the tension on the rope is 1.95 N

Answer:

t=0.417s

Explanation:

After the ball hits the racket it is in freefall(assume air resistance as negligible)

so a=-g

use

x-x0=v0t+1/2at^2

Plug in givens

20=50t-4.9t^2

Solve quadratic equation using quadratic formula

t= 0.417 seconds, (the other answer is extraneous because it is too big because in 1 second, the ball travels 50 meters)

Answer:

The mount saved is $ 0.105.

Explanation:

Concerned about the increase in the electricity bill, a person decides to save by reducing bathing time from 20 to 15 minutes. Your shower has the following specifications: 4200 W and 220V. Knowing that the kWh costs R$0.30, the savings made in 10 days were approximately​.

The electrical energy is given by

E = P x t

where, P is the electrical power and t is the time.

When he is using the shower for 20 minutes, the energy consumed is

E = 4200 x 20 x 60 = 5040,000 J = 1.4 kWh

When he is using the shower for 15 minutes, the energy consumed is

E' = 4200 x 15 x 60 = 3780000 J = 1.05 kWh

The difference in energy is

E'' = E - E' = 1.4 - 1.05 = 0.35 kWh

The money saved is

= 4 0.3 x 0.35 = $ 0.105

The object's kinetic energy changes according to

dK/dt = 15 J/s

If v is the object's initial speed, then its initial kinetic energy is

K (0) = 1/2 (5 kg) v ²

Use the fundamental theorem of calculus to solve for K as a function of time t :

[tex]K(t) = K(0) + \displaystyle\int_0^t \left(15\frac{\rm J}{\rm s}\right)\,\mathrm du = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)t[/tex]

After t = 13 s, the object's kinetic energy is

K (13 s) = 1/2 (5 kg) (13 m/s)² = 422.5 J

Put this as the left side in the equation above for K(t) and solve for v :

[tex]422.5\,\mathrm J = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)(13\,\mathrm s)[/tex]

==>   v ≈ 9.5 m/s

We have that  the distance to Betelgeuse, and the uncertainty in that measurement is

From the Question we are told that

Betelgeuse (in Orion)  has a parallax of 0.00451 + 0.00080

Generally

[tex]Distance\ in\ parsecs =\frac{ 1}{(parallax\ measured\ in\ arcseconds}[/tex]

Where

Parallax [tex]P =0.00451[/tex]

Uncertainty [tex]U = 0.00080[/tex]

Generally the equation for the distance  is mathematically given as

[tex]d=(\frac{1}{P}pc\pm(\frac{U}{P}*100\%))[/tex]

Therefore

[tex]d=(\frac{1}{0.00451}pc\pm(\frac{0.00080}{0.00451}*100\%))[/tex]

[tex]d=(221.7\pm39.33)pc[/tex]

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6.07 m

Explanation:

Given:

[tex]v_0=24.5\:\text{m/s}[/tex]

[tex]\theta_0 = 35.5°[/tex]

First, we need to find the amount of time it takes to travel a horizontal distance of 25.8 m. We know that

[tex]x = v_{0x}t \Rightarrow t = \dfrac{x}{v_0 \cos \theta_0}[/tex]

or

[tex]t = 1.29\:\text{s}[/tex]

To find the vertical height where the ball hit the wall, we use

[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]

[tex]\:\:\:\:=(24.5\:\text{m/s})\sin 35.5(1.29\:\text{s}) \\ - \frac{1}{2}(9.8\:\text{m/s}^2)(1.29\:\text{s})^2[/tex]

[tex]\:\:\:\:=6.07\:\text{m}[/tex]