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tangare [24]
2 years ago
14

You are driving to the grocery store at 20 m/s. You are 110m from an intersection when the traffic light turns red. Assume that

your reaction time is 0.50s and that your car brakes with constant acceleration. How far are you from the intersection when you begin to apply the brakes? What acceleration will bring you to rest right at the intersection? How long does it take you to stop after the light changes? the known values: vi=20m/s xi=110m t=.50s vf=0m/s xf=0m a=?
Physics
1 answer:
oksano4ka [1.4K]2 years ago
3 0

As we know that reaction time will be

t = 0.50 s

so the distance moved by car in reaction time

d = vt

d = 20 \times 0.50

d = 10 m

now the distance remain after that from intersection point is given by

d = 110 - 10 = 100 m

So our distance from the intersection will be 100 m when we apply brakes

now this distance should be covered till the car will stop

so here we will have

v_f = 0

v_i = 20 m/s

now from kinematics equation we will have

v_f^2 - v_i^2 = 2 a d

0 - 20^2 = 2(a)100

a = \frac{-400}{200} = -2 m/s^2

so the acceleration required by brakes is -2 m/s/s

Now total time taken to stop the car after applying brakes will be given as

v_f - v_i = at

0 - 20 = -2 (t)

t = 10 s

total time to stop the car is given as

T = 10 s + 0.5 s = 10.5 s

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Answer:

    The increase in the internal energy = 350 J

Explanation:

Given that

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Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

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Lets take change in the  internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

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A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic frict
Harlamova29_29 [7]

Answer:

The mass of the crate is 5kg.

We know that the force of friction can be obtained by:

F = N*k

where k is the coefficient of friction, where we use the static one if the object is at rest, and the kinetic one if the object os moving. N is the normal force

If we tilt the base making an angle of 30° with the horizontal, now the normal force against the plank will be equal to the fraction of the weight in the direction normal to the surface of the plank.

Knowing that the angle is 30°, then the fraction of the weight that pushes against the normal is Cos(30°)*W = cos(30°)*5kg*9.8m/s^2 = 42.4N

The fraction of the force in the parallel direction to the plank (the force that would accelerate the crate downwards) is:

F = sin(30°)*5k*9,8m/s = 24.5N

now, the statical friction force is:

Fs = 42.4N*0.5 = 21.2N

The statical force is less than the 24.5N, so the crate will move downwards, then the force that acts on the crate is the kinetic force of friction:

Fk = 42.4N*0.4 = 16.96N

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