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lawyer [7]
3 years ago
14

A rubber ball thrown at a speed of 5 m/s hit a flat wall and

Physics
1 answer:
velikii [3]3 years ago
5 0

Answer:

See the explanion below

Explanation:

Linear momentum is defined as the product of mass by velocity and can be calculated by the following equation.

P =m*v

where:

P = momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

When the ball hits the wall it receives a force for a very short time, this time and force value is known as momentum and can be calculated by the following expression.

-(m_{1}*v_{1}) + (F*t)=(m_{1}*v_{2})

where:

m₁ = mass of the ball [kg]

v₁ = 5 [m/s]

v₂ = 5 [m/s]

v₁ = v₂

F*t = Imp

Imp = (m_{1}*v_{1})+ (m_{1}*v_{2})\\Imp = 2*m*v_{1}

It can be said that the impulse is equal to 2 times the speed of throwing by the mass.

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Lenses are described as convergent or divergent depending on how they refract light. What is the difference between these two ty
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Answer:

Explanation:

Convergent lens is the lens which converges the rays of light falling on it. It is thicker at the middle and thinner at the edges. It is also known as convex lens.

Divergent lens is the lens which diverge the rays of light falling on it. It is thinner at the middle and thicker at the edges. It is also called as concave lens.

5 0
3 years ago
If Mercury, moved two orbital paths closer to the sun: what would be different? List serval differences, ideas.......
elena-14-01-66 [18.8K]
Well first of all, when it comes to orbits of the planets around
the sun, there's no such thing as "orbital paths", in the sense
of definite ("quantized") distances that the planets can occupy
but not in between.  That's the case with the electrons in an atom,
but a planet's orbit can be any old distance from the sun at all. 

If Mercury, or any planet, were somehow moved to an orbit closer
to the sun, then ...

-- its speed in orbit would be greater,

-- the distance around its orbit would be shorter,

-- its orbital period ("year") would be shorter,

-- the temperature everywhere on its surface would be higher,

-- if it has an atmosphere now, then its atmosphere would become
less dense, and might soon disappear entirely,

-- the intensity of x-rays, charged particles, and other forms of
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5 0
3 years ago
Two objects are held close together. When they are released, they move toward one another. Which conclusion is supported by this
Liula [17]

Answer:

The objects have opposite charges.​

Explanation:

The object when released are moving towards each other simply because they attract each other.

  • According to coulombs law of charges, like charges repel and will move away from one another.
  • Unlike charges attracts and moves towards one another
  • Since the objects are coming towards each other, they are of opposite charges.  
6 0
3 years ago
Read 2 more answers
The first physicist gets a second physicist to help. They both push on the crate, parallel to the surface of the incline, and it
Romashka-Z-Leto [24]

Answer:

F_aplied = fr

Explanation:

Newton's second law states that the force is proportional to the acceleration of the system, as in this case they indicate that the body moves at constant speed, the acceleration is zero, therefore

                 F_applied - fr = 0

                 F_aplied = fr

therefore the force applied by people is equal to the friction force

3 0
3 years ago
You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice
Galina-37 [17]

The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

\sum F =ma

where

\sum F = F_f = -0.14 N is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)

m = 0.12 kg is the mass of the puck

a is the acceleration

Solving for a,

a=\frac{\sum F}{m}=\frac{-0.14}{0.12}=-1.17 m/s^2

The motion of the puck is a uniformly accelerated motion, therefore we can use the following suvat equation:

v^2-u^2=2as

where:

v = 0 is the final velocity (the puck comes to a stop)

u = 18.3 m/s is the initial velocity

a=-1.17 m/s^2 is the acceleration

s is the stopping distance

And solving for s, we find

s=\frac{v^2-u^2}{2a}=\frac{0-(18.3)^2}{2(-1.17)}=143.1 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

6 0
3 years ago
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