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Aleks04 [339]
3 years ago
14

A car drives straight down toward the bottom of a valley and up the other side on a road whose bottom has a radius of curvature

of 115m. At the very bottom, the normal force on the driver is twice his weight. At what speed was the car traveling?
Physics
1 answer:
mina [271]3 years ago
6 0

Answer:33.58794 m/s

Explanation : fCenrtipetal force = mv2/r

m = Mass of car

r = Radius of curvature = 115 m

g = Acceleration due to gravity = 9.81 m/s²

The normal force is given by

N=Fc+mg

At the bottom N = 2N

2mg=fc+mg

⇒2mg=mv2

The car was traveling at the speed of 33.58794 m/s

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The tangential acceleration of a cart moving at a constant speed in a horizontal circle is:
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a=0                                  ∵ \alpha=0\ rad.s^{-1}

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The tangential acceleration of a cart moving at a constant speed in a circle is:

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We know that the (instantaneous) tangential velocity of such object is given by:

v=r.\omega

Now for angular acceleration we have a constant angular speed:

\alpha=0\ rad.s^{-1}

And angular acceleration is related to tangential acceleration as:

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3 years ago
Why does conduction occur more slowly in gases than in solids?​
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3 years ago
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8 0
3 years ago
A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio
Fed [463]

Answer:

Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

m = 100 g = 0.10 kg

g = acceleration due to gravity = 9.8 m/s²

θ = 15°

N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

N = normal reaction = 0.9466 N

Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

8 0
4 years ago
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