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qaws [65]
2 years ago
8

A small car is pushing a large truck. They are speeding up.

Physics
1 answer:
Gnom [1K]2 years ago
3 0

The force of the truck on the car is equal to the force of the car on the truck.

<h3>What is Newton's third law of motion?</h3>
  • Newton's third law of motion states that action and reaction are equal and opposite.

Let the mass of the small car = m

Let the mass of the truck = M

The force exerted by the small car is calculated as;

F₁ = ma

The force exerted by the large truck is calculated as;

F₂ = Ma

According to Newton's third law, the magnitude of the two forces are equal but opposite in direction.

|F₁| = |F₂|

Thus, we can conclude that, the force of the truck on the car is equal to the force of the car on the truck.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

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3 years ago
There is a 12 v potential difference between the positive and negative ends of the jumper cables, which are a short distance apa
steposvetlana [31]
The electric potential energy of the electron depends on the potential difference applied between the two ends of the cable. Indeed, the electric potential energy of a charge is given by
U=q \Delta V
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4 0
3 years ago
f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If
Luden [163]

Answer:

The angular velocity is   w= \sqrt[4]{\frac{a_{max}^2}{r^2}  - \alpha ^2}      

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

          The Radial acceleration which is mathematically represented as

                              a_r = \frac{v^2}{r}  = w^2r

And the Tangential  acceleration which is mathematically represented as

                                a_r = \alpha r

  The net acceleration is evaluated as

                      a = \sqrt{a_r^2 + a_t^2}

       

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

                 

So at maximum angular acceleration we a have

             a_{max} = \sqrt{a_r^2 + a_t^2}

                     a_{max}^2 = a_r^2 + a_t^2

                    a_{max}^2 = (w^2r)^2 + (\alpha r)^2

                 a_{max}^2 =  r^2 w^4 + r^2 \alpha ^2

                  a_{max}^2 = r^2 (w^4 + \alpha^2 )

                w^4 +\alpha ^2 = \frac{a_{max}^2}{r^2}

                         w^4 = \frac{a_{max}^2}{r^2}  - \alpha ^2

                         w= \sqrt[4]{\frac{a_{max}^2}{r^2}  - \alpha ^2}        

                     

3 0
3 years ago
A skater with a mass of 72 kg is traveling east at 5.8 m/s when he collides with another skater of mass 45 kg heading 60° south
Akimi4 [234]

The final velocity is 5.87 m/s

<u>Explanation:</u>

Given-

mass, m_{1} = 72 kg

speed, v_{1} = 5.8 m/s

Mass_{2},m_{2}  = 45 kg

speed_{2},v_{2}  = 12 m/s

Θ = 60°

Final velocity, v = ?

Applying the conservation of momentum:

m_{1} X v_{1} + m_{2} X v_{2} = (m_{1} +m_{2} ) v

72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v

v = 417.6 + 540 X \frac{0.5}{117}

v = 417.6 + \frac{270}{117}

v = 5.87 m/s

The final velocity is 5.87 m/s

8 0
3 years ago
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