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3 years ago

12

a ball is dropped from a height of 120 meters. If it takes 2.00 seconds for a ball to fall 60 meters, how long will it take the

ball to reach the ground?

1 answer:

Phantasy [73]3 years ago

5 0

It will take the ball 4 seconds to reach the ground because (120/2= 60/2=0= 2+2=4)

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Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov

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Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

$K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}$

Final kinetic energy is :

$K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}$

Now, fraction of initial kinetic energy loss is :

$Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25$

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

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A 60-kg cheetah reaches a speed of 30 m/s as it chases its prey.What is the kinetic energy of the cheetah?

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Explanation:

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The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it i

andreyandreev [35.5K]

The apparent weight of a 1.1 g drop of water is 4.24084 N.

<h3>What is Apparent Weight?</h3>

According to physics, an object's perceived weight is a characteristic that describes how heavy it is. When the force of gravity acting on an object is not counterbalanced by a force of equal but opposite normality, the apparent weight of the object will differ from the actual weight of the thing.
By definition, an object's weight is equal to the strength of the gravitational force pulling on it. It follows that even a "weightless" astronaut in low Earth orbit, with an apparent weight of zero, has almost the same weight that he would have if he were standing on the ground; this is because the gravitational pull of low Earth orbit and the ground are nearly equal.

Solution:

N = Speed of rotation = 1250 rpm

D = Diameter = 45 cm

r = Radius = 22.5 cm

M = Mass of drop = 1.1 g

Angular speed of the water = $\omega = \frac{2\pi N}{60}$

$\omega = \frac{2\pi \times 1250}{60}$

$\omega = 130.89 rad/s$

Apparent weight is given by

$W _a = M\omega^{2}R$

$W_a = 1.1 \times 10^-^3\times (130.89)^2\times 0.225$

$W_a$ = 4.24084 N

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Question:

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?

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1 year ago