. If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular

1 answer:

7 0

<span>In this particular case, where car is moving through curvature, so it is moving in circular motion, force acting on car is centripetal force which holds car not to fly out. Centripetal force is always pointed in the middle of circle. Here frictional force has role of centripetal force. If frictional force is to weak, car would fly out of curvutare.</span>

You might be interested in

A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl

frosja888 [35]

Answer:

W ≅ 292.97 J

Explanation:

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;

W = (Fcosθ)d

Given that:

Tension of the force = 62 N

angle of incline θ = 34°

distance d =5.7 m.

Then;

W = 62 × cos(34) × 5.7

W = 353.4 cos(34)

W = 353.4 × 0.8290

W = 292.9686 J

W ≅ 292.97 J

Hence, the work done by tension before the block goes up the incline = 292.97 J

8 0

2 years ago

A baseball is thrown with an initial velocity of 45.4 m/s at an angle of 31.2 ∘ .

sveticcg [70]

Answer:

V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s

1/2 m V^2 = m g h conservation of energy

h = V^2 / (2 g) = 23.52^2 / 19.6 = 28.2 m max height

Check:

t = 28.2 / 9.8 = 2.88 sec time to reach max height

h = 23.52 * 2.88 - 1/2 g 2.88^2 = 27.1 m

8 0

2 years ago

When the force acting on the body equal to acceleration?

topjm [15]

Answer:

Acceleration and velocity Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force.

5 0

2 years ago

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$