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Ainat [17]
3 years ago
11

Tim's car breaks down just down the street from his house. he and his friends decide to try to push it to his house. using newto

n's three laws of motion, explain why this will be difficult.
Physics
2 answers:
daser333 [38]3 years ago
6 0
Pushing a broke down car, even done by more than one person, is difficult especially if the distance to be covered is quite far. A car is heavy and it requires a lot of force to start the car moving. This is because the inertia of the car to remain at rest is great. Additionally, the force applied in pushing the car must be greater than the frictional force to cause it to accelerate. The frictional force is dependent on the mass of the object which means that the frictional force acting on the car is also great. Finally, with every push of the car, the frictional force will always be present and acting on the opposite direction. The push that will be supplied must be sustained all throughout.
masha68 [24]3 years ago
3 0

The friends have to combine their forces to be greater than the force of the car pushing back on them (3rd law). The car will remain at rest until an outside force acts on it that it is greater than its own (1st). The speed in which the car is being moved is directly related to the amount of force and direction of the kids pushing the car (2nd).

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Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

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A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]   [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s   [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

8 0
3 years ago
Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circ
andreev551 [17]

Answer:

v = 15.8 m/s

Explanation:

Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement.  So the variation of mechanical energy is equal to the work of the fictional force

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Let's write the mechanical energy at each point

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Let's use Hooke's law to find compression

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Let's write the energy equation

    fr d = ½ m v² + mgy - ½ k x²

Let's clear the speed

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   v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50)   2/60.0

   v² = (160 + 8800 - 1470) / 30

   v = √ (229.66)

   v = 15.8 m/s

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