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seropon [69]
3 years ago
11

A small ball with a mass of 0.6 kg and a velocity of 12 m/s hits another ball with the same mass. The firs forward and hits a th

ird ball with a mass of 0.2 kg. If the system is closed, what is the velocity of the third ball?
4 m/s

36 m/s

30 m/s

1.44 m/s​
Physics
1 answer:
Ket [755]3 years ago
8 0
I think is 30m/s I am not sure
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If a 20 N object has been lifted 5 meters above the ground, how much gravitational potential energy does it have?
Andreyy89

The gravitational potential energy of the object is 100 J.

Gravitational potential energy stored in an object is the work done in raising the object to a height <em>h</em> against the gravitational force acting on it.

The gravitational force acting on a body is its weight mg, where m is its mass and g, the acceleration due to gravity.

Work done by a force is equal to the product of the force and the displacement made by the point of application of the force.

GPE = mgh

The weight of the object is given as 20 J and it is raised to a height of 5 m.

GPE =(mg)*h = (20 N)*(5m)=100 J

The gravitational potential energy of the object is 100 J.

5 0
3 years ago
Read 2 more answers
The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2
viktelen [127]

Answer:

Distance, d = 778.05 m                          

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2

Let d is the distance covered by car. Using second equation of motion as :

d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m

So, the car will cover a distance of 778.05 meters.

5 0
3 years ago
A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.
kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

F_k = \mu_k N = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0
3 years ago
Find f (the force) in relation to moments.​
Masja [62]

Answer:

B

Explanation:

i7tigp79ug978uyioh;

7 0
2 years ago
Rank the following in order of increased size? A. Proton B. Nucleus C. Electron D. Atom
Basile [38]
From largest to smallest- atom, nucleus, proton and electron
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