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liraira [26]
3 years ago
5

an object is dropped from a height of 25 meters. at what velocity will it hit the ground? a 7.0 m/s b 11 m/s c 22 m/s d 49 m/s e

70 m/s
Physics
1 answer:
kipiarov [429]3 years ago
6 0
Assuming that the object starts at rest, we know the following values:

distance = 25m
acceleration = 9.81m/s^2 [down]
initial velocity = 0m/s

we want to find final velocity and we don't know the time it took, so we will use the kinematics equation without time in it:

Velocity final^2 = velocity initial^2 + 2 × acceleration × distance

Filling everythint in, we have:

Vf^2 = 0^2 + (2)(-9.81)(-25)
The reason why the values are negative is because they are going in the negative direction

Vf^2 = 490.5

Take the square root of that

Final velocity = 22.15m/s which is answer c
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A power plant running at 39 % efficiency generates 330 MW of electric power. Part A At what rate (in MW) is heat energy exhauste
marusya05 [52]

516.154 megawatts of heat are <em>exhausted</em> to the river that cools the plant.

By definition of energy efficiency, we derive an expression for the energy rate exhausted to the river (Q_{out}), in megawatts:

Q_{out} = Q_{in} - W

Q_{out} = \left(\frac{1}{\eta}-1 \right)\cdot W(1)

Where:

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  • W - Electric power, in megawatts.

If we know that \eta = 0.39 and W = 330\,MW, then the energy rate exhausted to the river is:

Q_{out} = \left(\frac{1}{0.39}-1 \right)\cdot (330\,MW)

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7 0
3 years ago
A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows
anastassius [24]

Answer:

In the vertical direction the acting forces are the normal force and the weight of the bobsleder plus the sled. In the horizontal direction the acting force is the friciton force.

Explanation:

Hi there!

Please, see the attached figure for a graphic representation of the forces acting on the sled after the bobsleder jumped in.

In the vertical direction, the acting forces are the normal force (N) and the weight of the sled plus the bobsledder (W).

Since the sled is not being accelerated in the vertical direction, the sum of forces in that direction is zero:

∑Fy = W + N = 0 ⇒ W = N

The weight is calculated as follows:

W = (mb + ms) · g

Where:

mb = mass of the bobsleder.

ms = mass of the sled.

g = acceleration due to gravity.

In the horizontal direction the only acting force is the friction force (Fr). The friction force is calculated a follows:

Fr = N · μ

Where:

N = normal force.

μ = kinetic friction coefficient.

Since N = W = (mb + ms) · g

Fr = (mb + ms) · g · μ

If we want to find the acceleration of the sled after the bobsleder jumps in, we can apply Newton's second law:

∑F = m · a

Where "a" is the acceleration and "m" is the mass of the object (in this case, the mass of bobsleder plus the mass of the sled).

∑F = Fr =  (mb + ms) · g · μ =  (mb + ms) · a

(mb + ms) · g · μ =  (mb + ms) · a

Solving for "a":

g · μ = a

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3 years ago
Please help me out !!
blagie [28]
I think the answer would be letter a
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