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Alexxandr [17]
1 year ago
15

What is the work done by the 200.-N tension shown if it is used to drag the 150-N crate 25 m across the floor at a constant spee

d?
Physics
1 answer:
WINSTONCH [101]1 year ago
8 0

Answer:

0 J

Explanation:

Work equals force times distance, but the force is zero because the crate being dragged will have zero acceleration. Force equals mass times acceleration and since acceleration is zero, force has to equal zero as well. Since the force is zero, the work required also has to be zero.

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A block of mass m is pushed up against a spring with spring constant k until the spring has been compressed a distance x from eq
Snowcat [4.5K]

Answer:d

Explanation:

Spring is compressed to a distance of x from its equilibrium position

Work done by block on the spring is equal to change in elastic potential energy

i.e. Work done by block W=\frac{1}{2}kx^2

therefore spring will also done an equal opposite amount of work on the block in the absence of external force

Thus work done by spring on the block W=-\frac{1}{2}kx^2

Thus option d is correct

6 0
2 years ago
Do as physics instructor fred cauthen does and place a tennis ball close to and above the top of a basketball. drop the balls to
Ksivusya [100]

Answer:

after shock

creating a system for the conservation of the energy of the basketball ball and creating a system for the tennis ball only, the conservation of energy should be applied to each system independently

Explanation:

When the two balls fall they acquire the same speed since they are accelerated by the same force, their weight and the acceleration of the acceleration of gravity. When reaching the floor the mechanical energy of the system is conserved.

Upon reaching the floor, the first ball (basketball) collides with the floor, this process is very fast, at the end of the process the basketball comes out with a velicad up and collides with the much lighter tennis ball that is still descending .

we assume that the shocks are elastic, when solving the momentary and kinetic energy findings, we find the velocities after each shock

     

In this clash the tennis ball acquires a high kinetic speed with an upward direction that makes a very high height high. Again this shock is very fast and the tennis ball almost does not move.

Here we must separate the system, creating a system for the conservation of the energy of the basketball ball and another system for the tennis ball only, the conservation of energy should be applied to each system independently

Em₀ =K = 1/2 m v²

                    Em_{f} = U = m g h

As in the elastic shock the final speed of the tennis ball is approximately 2 vo, we can calculate the maximum height

                 m g h = 1/2 m (2v₀)²

                 h = 2 v₀²/g

To reconcile this with the conservation of energy we must calculate the energy for the tennis ball at two points, the first when the crash with the tennis ball ends and at the end point at its maximum height.

6 0
2 years ago
Nathan accelerates his skateboard uniformly along a straight path from rest to 12.5 m/s in 2.5 s.
kicyunya [14]

Answer:

<h2>a) Nathan's acceleration is 5 m/s² </h2><h2>b) Nathan's displacement during this time interval is 15.625 m</h2><h2>c) Nathan's average velocity during this time interval is 6.25 m/s</h2>

Explanation:

a) We have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = 12.5 m/s    

     Time, t = 2.5 s

     Substituting

                      v = u + at  

                      1.25 = 0 + a x 2.5

                      a = 5 m/s²

     Nathan's acceleration is 5 m/s²

b) We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 5 m/s²  

        Time, t = 2.5 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 2.5 + 0.5 x 5 x 2.5²

                      s = 15.625 m

      Nathan's displacement during this time interval is 15.625 m

c) Displacement = 15.625 m

   Time = 2.5 s

  We have

           Displacement = Time x Average velocity

           15.625 = 2.5 x  Average velocity

           Average velocity = 6.25 m/s

     Nathan's average velocity during this time interval is 6.25 m/s

5 0
2 years ago
A 1,160 kg satellite orbits Earth with a tangential speed of 7,446 m/s. If the satellite experiences a centripetal force of 8,95
Juli2301 [7.4K]

Answer:

8.02×10⁵ m

Explanation:

Equation for centripetal force:

F = mv²/r

Solving for r:

r = mv²/F

Given:

F = 8955 N

m = 1160 kg

v = 7446 m/s

r = (1160 kg) (7446 m/s)² / 8955 N

r = 7.182×10⁶ m

The height above the surface is:

h = 7.182×10⁶ m − 6.38×10⁶ m

h = 0.802×10⁶ m

h = 8.02×10⁵ m

4 0
2 years ago
Wonder Woman and Superman fly to an altitude of 1510 km, carrying between them a chest full of jewels that they intend to put in
blagie [28]

Answer:

10061.56 m/s

Explanation:

Gravitation potential of a body in orbit from the center of the earth is given as

Pg = -GM/R

Where G is the gravitational constant 6.67x10^-11 N-m^2kg^-2

M is the mass of the earth = 5.98x10^24 kg

R is the distance from that point to the center of the earth = r + Re

Where r is the distance above earth surface, Re is the earth's radius.

R = 1510 km + 6370 km = 7880 km

Pg = -(6.67x10^-11 x 5.98x10^24)/7880x10^3

Pg = -50617512.69 J/kg

The negative sign means that the gravitational potential is higher away from earth than it is at the earth's surface (it shows convention).

This indicates the kinetic energy per kilogram that the chest of jewel will fall with to earth.

For the jewel chest, the velocity V will be

0.5v^2 = 50617512.69

V^2 = 101235025.4

V = 10061.56 m/s

4 0
2 years ago
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