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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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Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10
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Answer:

Explanation:

The formula for this is

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.

Filling in:

7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2} and moving things around to solve for r:

r=\sqrt{\frac{6.67*10^{-11}(90.0)(30.0)}{7.5*10^{-8}} } Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m

8 0
3 years ago
Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti
Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = m_1=6.50 kg- 1.55 kg=4.95 kg

Velocity achieved by squid = v_1=2.60 m/s

Momentum gained by squid = P=m_1v_1

Mass of water = m_2=1.55 kg

Velocity by which water was released by squid = v_2

Momentum gained by water but in opposite direction = P'=m_2v_2

P = P'

m_1v_1=m_2v_2

v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E  =\frac{1}{2}m_1v_1^{2}

Kinetic energy of water = K.E' = \frac{1}{2}m_2v_2^{2}

Total kinetic energy created by this maneuver:

K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}

=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules

4 0
3 years ago
A box is pulled up a rough ramp that makes an angle of 22 degrees with the horizontal surface. The surface of the ramp is the x-
kifflom [539]

Magnitude of the force  of tension: 139 N

Explanation:

The surface of the ramp here is assumed to be the positive x-direction.

To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.

There are three forces acting along the x-direction:

  • The force of tension, F_T, acting up along the plane
  • The force of friction, F_f=14.8 N, acting down along the plane
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We know that the magnitude of the weight is

F_g=70.0 N

So its x-component is

F_{gx}=F_g sin \theta =(70.0)(sin 22^{\circ})=26.2 N

The net force along the x-direction can be written as

F_x = F_T-F_f-F_{gx}

And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:

F_T=F_x+F_f+F_{gx}=98+14.8+26.2=139 N

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

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Answer:

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The vertical component is zero, however, it has a horizontal velocity, so its initial speed is not zero, therefore the secong ball has the greater speed at ground level.

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