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Novay_Z [31]
3 years ago
6

A pole-vaulter just clears the bar at 5.31 m and falls back to the ground. The change in the vaulter's potential energy during t

he fall is -3900 J. What is his weight?
Physics
1 answer:
irina [24]3 years ago
8 0

Answer:

Weight = 734.46 N

Explanation:

Given:

Initial height of the pole-vaulter is, h_i=5.31\ m

Final height of the pole-vaulter is, h_f=0\ m

Change in the vaulter's potential energy is, \Delta U=-3900\ J

We know that, change in potential energy is given as:

\Delta U=mg(h_f-h_i)

Where, 'm' is the mass of the object, 'g' is acceleration due to gravity and has a value of 9.8 m/s².

Now, weight of the object is given as the product of its mass and acceleration due to gravity. So, replace 'mg' by weight 'w'. So, the equation becomes,

\Delta U = w(h_f-h_i)

Now, rewriting in terms of 'w', we get:

w=\dfrac{\Delta U}{(h_f-h_i)}

Now, plug in all the given values and solve for 'w'. This gives,

w=\dfrac{-3900\ J}{(0-5.31)\ m}\\\\\\w=\dfrac{3900}{5.31}\ N\\\\\\w=734.46\ N

Therefore, weight of the vaulter is 734.46 N.

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An object's buoyant force and weight mean the same thing.<br><br>A. True<br><br>B. False​
german

Answer:

False

Explanation:

No. The buoyant force on an object is the portion of its weight that appears to vanish

when the object is in any fluid (could be either a liquid or a gas).

If the object happens to float in a particular fluid, then the buoyant force at that moment

is equal to the object's weight.

Notice that the buoyant force on an object will be different in different fluids.

6 0
2 years ago
W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively
MArishka [77]

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

 W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

      W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

    W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

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7 0
3 years ago
Which of these is not shared by Bohr's model and the modern atomic model?
Ainat [17]

"D. Both have electrons that orbit the atomic nucleus in a similar way ." is not shared by Bohr's model and the modern atomic model.

Hope this helps,

Davinia.

5 0
3 years ago
Zeb was lifting a box onto a moving truck. He lifted with a net force of 2000N and the box had a mass of 100 kg. What was the re
Aleksandr [31]

Answer:

<h2>20 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{2000}{100}  = 20 \\

We have the final answer as

<h3>20 m/s²</h3>

Hope this helps you

8 0
3 years ago
Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc
snow_lady [41]

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force F_g = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force

Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

5 0
3 years ago
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