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maxonik [38]
1 year ago
11

Lamar writes several equations trying to better understand potential energy. H = d with an arrow to the equation W = F d and P E

Subscript g Baseline = m g h. F Subscript g Baseline = mg with arrows to the F in W = F d and to P E Subscript g Baseline = m g h. What conclusion is best supported by Lamar's work? The elastic potential energy is the same for any distance from a reference point. The gravitational potential energy equals the work needed to lift the object. The gravitational potential energy is the same for any distance from a reference point. The elastic potential energy equals the work needed to stretch the object.
Physics
2 answers:
Marina86 [1]1 year ago
4 0

Answer:

The correct answer is B.

B. The gravitational potential energy equals the work needed to lift the object.

Explanation:

it is correct

Debora [2.8K]1 year ago
3 0

Answer:

The gravitational potential energy equals the work needed to lift the object.

Explanation:

here we know that

H = \vec d

work done is given as

W = \vec F . \vec d

Potential energy is given as

PE_g = mgh

force due to gravity is given as

\vec F_g = mg

now here if we plug in the value of distance and force in the formula of work done then we will have

W = (mg)(h)

so here we got

W = PE_g

so we can concluded that

The gravitational potential energy equals the work needed to lift the object.

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John runs 1.0 m/s at first, and then accelerates to 1.6 m/s during
erastova [34]

Answer: 0.13m/s^2

Explanation:

Formula: a=\frac{V_2-V_1}{t}

Where;

a = acceleration

V2 = final velocity

V1 = initial velocity

t = time

If John runs 1.0 m/s first, we assume this is V1. He accelerates to 1.6 m/s; this is V2.

a=\frac{1.6m/s-1.0m/s}{4.5s}

a=\frac{0.6m/s}{4.5s}

a=0.13m/s^2

7 0
1 year ago
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Assume that the mass has been moving along its circular path for some time. You start timing its motion with a stopwatch when it
lesya692 [45]

Answer:

v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)

Explanation:

As we know that the mass is revolving with constant angular speed in the circle of radius R

So we will have

\theta = \omega t

now the position vector at a given time is

r = Rcos\theta \hat i + R sin\theta \hat j

now the linear velocity is given as

v = \frac{dr}{dt}

v = (-R sin\theta \hat i + R cos\theta \hat j)\frac{d\theta}{dt}

v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)

6 0
1 year ago
The image shows positions of the earth and the moon in which region would an astronaut feel the lightest
trapecia [35]

Answer:

The moon region

Explanation:

This is because there is little to no gravity on the moon. That is where the astronaut would feel the lightest.

5 0
1 year ago
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In a charge-free region of space, a closed container is placed in an electric field. Which of the following is a requirement for
galina1969 [7]

Answer:

D. The requirement does not exist -the total electric flux is zero no matter what.

Explanation:

According to Gauss's law , total electric flux over a closed surface is equal to 1 / ε₀ times charge inside.

If charge inside is zero , total electric flux over a closed surface is equal to

zero . It has nothing to do with whether external field is uniform or not. For any external field , lines entering surface will be equal to flux going out.

8 0
2 years ago
An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o
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Answer:

The gplanet is 0.193 m/s^2

Explanation:

The speed of the pulse is:

v=\frac{lengthofthewipe}{traveltime} =\frac{1.6}{0.0656} =15.24m/s

v=\sqrt{\frac{MgL}{m} } \\v^{2} =\frac{MgL}{m} \\g=\frac{mv^{2} }{ML}

where

m=mass of the wire=4 g= 4x10^-3 kg

M=mass of the object= 3 kg

Replacing values:

g=\frac{4x10^{-3}*15.24^{2}  }{3*1.6} =0.193 m/s^{2}

7 0
2 years ago
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