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maxonik [38]
2 years 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]2 years 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]2 years 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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The speed of a bus increases uniformly from 15 ms per second to 60 ms per second in 20 seconds. calculate 1. the average speed 2
joja [24]
Vf= 60m/s
vs = 15m/s
t=20s
a = vf-vs/t
a = 45m/s / 20s
a = 2,25 m/s²
S = vs *t + 1/2 a*t²
S = 15 m/s*20s +1/2 * 2,25m/s² * 400 s²
S = 300m +  450m
S = 750m
v average = 750m/20s
v average = 37,5 m/s
3 0
2 years ago
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A 10kg rocket is traveling at 80 m/s when the booster engine applies a constant forward force of 60 N for 3.0 seconds. What impu
Lina20 [59]

Answer:

Impulse = 90

Resulting Velocity = 89

Explanation:

Use F * change in time = m * change in velocity.

For the first part of the question, the left side of the equation is the impulse. Plug it in.

60 * (3.0 - 0) = 90.

For the second half. we use all parts of the equation. I'm gonna use vf for the final velocity.

60 * (3.0 - 0) = 10 * (vf - 80). Simplify.

90 = 10vf - 800. Simplify again.

890 = 10vf. Divide to simplify and get the answer.

The resulting velocity is 89.

4 0
2 years ago
1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra
UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.  

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

F_r = mgsin\theta

F_r = \mu N

F_r = \mu mg cos\theta

Matching the two equation we have,

\mu N = \mu mg cos\theta

\mu = tan\theta

Applying energy conservation,

\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1

h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})

Frictionless surface

\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2

\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2

h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between h_1 and h_2 is given by

\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}

\frac{h_1}{h_2} = \frac{1}{2}

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D. Reduces the force of air resistance

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Answer:

last answer

Explanation:

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