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faust18 [17]
3 years ago
5

A cylinder of diameter 100 mm rolls from restdown a 5 m long ramp and its center of mass is moving with velocity 2 m/s at the bo

ttom of the ramp. (a) What is its (constant) acceleration down the ramp? (b) What is its angular acceleration?
Physics
1 answer:
RoseWind [281]3 years ago
3 0

Answer:

(a): a = 0.4m/s²

(b): α = 8 radians/s²

Explanation:

First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):

a= (Vf-Vi)/t = (2m/s)/t

a: linear acceleration.

Vf: speed at the end of the ramp.

Vi: speed at the beginning of the ramp (zero).

d= (1/2)×a×t² = 5m

d: distance of the ramp (5m).

We replace the first equation in the second to determine the travel time on the ramp:

d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s

And the linear acceleration will be:

a = (2m/s)/5s = 0.4m/s²

Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:

perimeter = π×diameter = π×0.1m = 0.3142m

To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:

α = (0.4m/s²)/(0.05m) = 8 radians/s²

α: angular aceleration.

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The maximum height reached by the ball is 99.2 m

Explanation:

When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration (g=9.8 m/s^2 towards the ground). Therefore, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

In this problem, we have:

u = 44.1 m/s is the initial vertical velocity of the ball

v = 0 is the final velocity when the ball reaches the maximum height

s is the maximum height

a=-g=-9.8 m/s^2 is the acceleration of gravity (downward, so negative)

Solving for s, we find the maximum height reached by the ball:

s=-\frac{u^2}{2a}=-\frac{44.1^2}{2(-9.8)}=99.2 m

Learn more about free fall:

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3 0
3 years ago
How long would it take a leopard, running at an average speed of 20 m/s to travel 500 m?
alexandr1967 [171]

Answer:

25 seconds

Explanation:

500/20

4 0
3 years ago
Two identical 0.400 kg masses are pressed against opposite ends of a light spring of force constant 1.75 N/cm, compressing the s
nlexa [21]

Answer:

0.853 m/s

Explanation:

Total energy stored in the spring = Total kinetic energy of the masses.

1/2ke² = 1/2m'v².................... Equation 1

Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.

make v the subject of the equation,

v = e[√(k/m')].................... Equation 2

Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.

Substitute into equation 2

v = 0.39[√(1.75/0.8)

v = 0.39[2.1875]

v = 0.853 m/s

Hence the speed of each mass = 0.853 m/s

7 0
3 years ago
Two stars have the same radius but have very different temperatures. the red star has a surface temperature of 3,000 k and the b
dangina [55]
I would say that insofar as the two stars temperatures are presumably closely related to their luminosity, that the blue star at 156,100 k compared to 3000k for the red star then the blue star has a luminosity of 52 times that of the red star.
4 0
3 years ago
A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?
katrin [286]

Answer:

<em>k = 25.18 N/m</em>

Explanation:

<u>Simple Harmonic Oscillator</u>

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}

If the period is known, we can find the value of the constant by solving for k:

\displaystyle k=m\left(\frac{2\pi}{T}\right)^2

Substituting the given values m=5 Kg and T=2.8 seconds:

\displaystyle k=5 \left(\frac{2\pi}{2.8}\right)^2

k = 25.18 N/m

5 0
3 years ago
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