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Dvinal [7]
2 years ago
12

Three cars (car F, car G, and car H) are moving with the same velocity when the driver suddenly slams on the brakes, locking the

wheels. The most massive car is car F, the least massive is car H, and all three cars have identical tires.
Which car travels the longest distance to skid to a stop?
Physics
1 answer:
Nastasia [14]2 years ago
5 0

Answer:

Car H

Explanation:

Frictional force is a resistant force. It is given as:

F = u*m*g

Where u = coefficient of friction

m = mass

g = acceleration due to gravity

From the formula above, we see that frictional force is dependent on the mass of object and the coefficient of friction.

Since they all have the same tires, the coefficient of friction between the tire and the floor is the same for each car. Acceleration due to gravity, g, is constant.

The only factor that determines the frictional force of each car is the mass. Hence, the more the mass, the more the frictional force.

So, the most massive car will have the most frictional force and hence, will come to a stop quicker than the others. The least massive car will have the least frictional force and so, will take a longer time to stop.

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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
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Answer:

magnitude of force on charge 2Q  = \frac{KQ^{2} }{I^{2} }

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q  i.e x-component of the net force and the y-component of the net force

║F║ = \sqrt{f_{x}^{2} + f_{y}^{2}    }  =  after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = \frac{KQ^{2} }{I^{2} }

The direction of the force on charge 2Q is calculated as

tan ∅ = \frac{f_{y} }{f_{x} } = 1.8284

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3 0
3 years ago
3. In 1989, Michel Menin of France walked on a tightrope suspended under a
Tamiku [17]

Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

Let the distance of Menin's pocket to the ground be x

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Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

3070m

Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

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I personally think the answer is B.
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