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Tcecarenko [31]
2 years ago
6

Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in this collisi

on? Explain why or why not.
Physics
2 answers:
mart [117]2 years ago
6 0
Conservation of momentum<span> is a fundamental law of physics. This law states that the </span>momentum<span> of a system is constant if there are </span>no external forces acting on the system. In a situation in which two balls, each with a mass of 0.5 kg, collide on a pool table<span> the law of conservation of momentum is not satisfied because there are external forces that moved the balls. </span>
swat322 years ago
4 0
Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s, and the total momentum after the collision is 
<span>1.5 kg • m/s. The momentum stays the same after the collision.</span>
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A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s,  u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest),  = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

6 0
2 years ago
calculate the work done to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it?
Charra [1.4K]
100N is how much work is needed 
4 0
2 years ago
The air surrounding an airplane in flight exerts a drag force that acts opposite to the airplane's motion. When an Airbus A380 i
Rainbow [258]

Answer:

4.32\cdot 10^5 hp, 3.22\cdot 10^8 W

Explanation:

The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.

Therefore, we can write:

4F_T - F_d = 0

where

F_T = 322,000 N is the thrust force generated by each engine of the jet

F_d is the drag force

Solving for Fd,

F_d = 4 F_T = 4(322,000)=1.288\cdot 10^6 N

The velocity of the jet is

v=250 m/s

So, the rate at which the drag force does work (which is the power) is

P=F_d v

and substituting

F_d = 1.288\cdot 10^6 N\\v = 250 m/s

we find

P=(1.288\cdot 10^6)(250)=3.22\cdot 10^8 W

Converting into horsepower,

P=\frac{3.22\cdot 10^8}{746}=4.32\cdot 10^5 hp

4 0
2 years ago
Astronomers were at first surprised to find complicated molecules in the interstellar medium. They thought ultra-violet light fr
jeka57 [31]

Answer:

The dust present in the clouds.

Explanation:

The complicated composition molecules that can be found in space are generally associated with clouds of dust. The significant amount of dust in these clouds provides protection not only for these molecules, but for any body that makes up or is associated with dust clouds.

It is exactly this dust that protects the molecules against the action of ultraviolet rays.

8 0
2 years ago
What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?
DIA [1.3K]

The elastic potential energy of the spring is 6.8 J

Explanation:

The elastic potential energy of a compressed/stretched spring is given by the equation:

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the elongation of the spring

The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):

F=kx

By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,

k=\frac{15.0-0}{10.0-0}=1.5 N/m

Therefore when the spring has a elongation of x=3.0 m, its potential energy is

E=\frac{1}{2}(1.5)(3.0)^2=6.8 J

Learn more about potential energy:

brainly.com/question/1198647

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#LearnwithBrainly

3 0
2 years ago
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