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ladessa [460]
2 years ago
9

What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road, (b) on an uphill road?

Physics
1 answer:
xeze [42]2 years ago
6 0

The net force on the car for either case will be zero, since the car is moving at a constant velocity.

<h3>Net force on the car</h3>

The net force on the car is the sum of all the forces acting on the car.

∑F = ma

<h3>When the car is on a level road</h3>

When the car is on a level road, the only two forces acting on the car is the applied force and frictional force of the road.

F - Ff = 0

The net force will be zero, since the car is moving at a constant velocity.

<h3>When the car is on uphill</h3>

The forces acting on the car at the uphill is the applied force, weight of the car acting downwards and the frictional force acting opposite direction.

F - Wsinθ - μFₙcosθ = 0

The net force will be zero, since the car is moving at a constant velocity.

Learn more about net force at constant velocity here: brainly.com/question/14392124

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Answer:

a

   F_A  =425.42 \ N

b

  F_A_H  = 358.58 \ N

Explanation:

From the question we are told that

    The diameter of the Ferris wheel is  d  =  80 \ ft =  \frac{80}{3.281} =  24.383

    The  period of the Ferris wheel is  T  =  24 \ s

     The  mass of the passenger is  m_g  =  40 \ kg

The  apparent weight of the passenger at the lowest point is mathematically represented as

           F_A_L  =  F_c  + W

Where  F_c is the centripetal force on the passenger,  which is mathematically represented as

         F_c  =m *  r *  w^2

Where w is the angular velocity which is mathematically represented as

         w =  \frac{2* \pi   }{T}

substituting values

         w =  \frac{2* 3.142 }{24}

         w =  0.2618 \ rad/s

and  r  is the radius which is evaluated as r =  \frac{d}{2}

   substituting values

         r =  \frac{24.383}{2}

         r = 12.19 \ ft

So

          F_c  = 40 * 12.19* (0.2618)^2

          F_c  =  33.42 \ N

W is the weight which is mathematically represented as

           W =  40 * 9.8

           W =  392 \ N

So

         F_A    =  33.42 + 392

         F_A  =425.42 \ N

The  apparent weight of the passenger at the highest point is mathematically represented as

          F_A_H  =  W- F_c

substituting values

         F_A_H  = 392 -  33.42

         F_A_H  = 358.58 \ N

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3 years ago
How could two waves on a rope interfere so the rope does not move at all?
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Answer:

If a crest formed by one wave interferes with a trough formed by the other wave then the rope will not move at all.

Explanation:

Assume a straight rope tied to both ends is at rest. When a wave is created at one end of the rope, it travels to the other end of the rope through formation of alternative crest and trough. Due to these crest and trough the rope shifts up and down.

But when there are two waves travelling through the rope and both have opposite direction (directed towards one another) in such a way that crest formed by one wave is interfering with the trough formed by the other wave then due to this interference the waves will cancel the effects of each other on the rope and rope will be stable.

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Ch 27 HW Exercise 27.12 10 of 20 Constants A horizontal rectangular surface has dimensions 2.80 cm by 3.15 cm and is in a unifor
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Answer:

Magnetic field, B = 0.88 T

Explanation:

It is given that,

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Let B is the magnitude of magnetic field in which the rectangular surface is placed. It is given by :

\phi=BA\ cos\theta

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Here, \theta=90-31=59^{\circ}

B=\dfrac{\phi}{A\ cos\theta}

B=\dfrac{4\times 10^{-4}}{0.000882\times cos(59)}

B = 0.88 T

So, the magnitude of magnetic field is 0.88 T. Hence, this is the required solution.

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