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3 years ago

75 points

Physics

1 answer:

Vikki [24]3 years ago

5 0

Answer:

Explanation:

The height to which a ball will bounce depends on the height from which it is dropped, what the ball is made out of (and if it is inflated, what the pressure is), and what the surface it bounces from is made out of. The radius of the ball doesn't really matter, if you are measuring the height of the ball from the bottom of the ball to the ground.

A ball's gravitational potential energy is proportional to its height. At the bottom, just before the bounce, this energy is now all in the form of kinetic energy. After the bounce, the ball and the ground or floor have absorbed some of that energy and have become warmer and have made a noise. This energy lost in the bounce is a more or less constant fraction of the energy of the ball before the bounce. As the ball goes back up, kinetic energy (now a bit less) gets traded back for gravitational potential energy, and it will rise back to a height that is the original height times (1-fraction of energy lost). We'll call this number f. For a superball, f may be around 90% (0.9) or perhaps even bigger. For a steel ball on a thick steel plate, f is >0.95. For a properly inflated basketball, f is about 0.75. For a squash ball, f might be less than 0.5 or 0.25 - squash balls are not very bouncy. The steel ball on an unvarnished pine wood floor may not bounce at all, but rather make a dent, and so what the floor is made out of makes quite a lot of difference.

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A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra

skelet666 [1.2K]

Answer:

So the acceleration of the child will be $8.05m/sec^2$

Explanation:

We have given angular speed of the child $\omega =1.25rad/sec$

Radius r = 4.65 m

Angular acceleration $\alpha =0.745rad/sec^2$

We know that linear velocity is given by $v=\omega r=1.25\times 4.65=5.815m/sec$

We know that radial acceleration is given by $a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2$

Tangential acceleration is given by

$a_t=\alpha r=0.745\times 4.65=3.464m/sec^$

So total acceleration will be $a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2$

7 0

3 years ago

Of the solar radiation entering the atmosphere, about ___% reaches the surface directly as direct radiation.

strojnjashka [21]

Of the solar radiation entering the atmosphere, about 22.5% reaches the surface directly as direct radiation and is being absorbed. Thirty five percent of it is reflected back to the space. While 10.5% and 14.5% are scattered to the earth from the blue sky and from clouds, respectively. Also, 17.5 percent was being absorbed by the atmosphere. Thus leaving only 22.5% to be given directly to the Earth's surface. Also, another factor of this low value would be the transparency of the atmosphere. The atmosphere of the Earth has only an effective transparent to radiation from the sun of about .34 to .7 micrometer.

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3 years ago

Is the strange solid/liquid changing behavior of Oobleck the SAME as an ice cube melting?

Dmitrij [34]

I don’t think so because an ice cube melting needs heat and relies on temp while ooblecks transition from solid to quickly depends on force and speed

6 0

3 years ago

A single-turn circular loop of wire that has a radius of 3.5 cm lies in the plane perpendicular to a spatially uniform magnetic

Lena [83]

Explanation:

Given that,

Radius of the circular loop, r = 3.5 cm = 0.035 m

(a) During a 0.12-s time interval, the magnitude of the field increases uniformly from 0.2 T to 0.5 T. Due to the change in the magnetic field, an emf will induced in it. The magnitude of induced emf is given by :

$\epsilon=-\dfrac{d(BA)}{dt}\\\\\epsilon=-\pi r^2\dfrac{B_f-B_i}{dt}\\\\\epsilon=-\pi (0.035)^2\dfrac{0.5-0.2}{0.12}\\\\\epsilon=-9.62\times 10^{-3}\ volts$

So, the magnitude of the emf induced in the loop during the time interval is $9.62\times 10^{-3}\ V$ .

(b) The negative sign shows that the direction of induced emf in the loop is in anitclockwise direction.

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3 years ago

A point source of light is 1.24 m below the surface of a pool. What is the diameter of the circle of light that a person above t

Ksivusya [100]

Answer:

D = 2.828 m

Explanation:

given,

distance of source of light = 1.24 m below surface of pool

refractive index of the water = n₁ = 1.33

refractive index of air = n₂ = 1

refraction angle be = 90°

let C be the critical angle

Radius = d tan C

d is the depth of the source

Using Snell's law

n₁ sin C = n₂ sin R

1.33 x sin C = 1 x sin 90°

$sin C = \dfrac{1}{1.33}$

$C = sin^{-1}(0.752)$

C = 48.75°

hence,

R = 1.24 x tan 48.75°

R = 1.414 m

Diameter = 2 x R

D = 2 x 1.414

D = 2.828 m

5 0

3 years ago