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3 years ago

What is the distance that a car travels if it was brought to stop in 5 seconds and if it was traveling at 110 Km/h

Physics

1 answer:

Triss [41]3 years ago

6 0

Answer:

Suppose that the acceleration is a constant, a.

a(t) = a.

To write the velocity equation, we must integrate over time, and the constant of integration will be equal to the initial velocity, in this case is 110km/h.

v(t) = a*t + 110km/h

And we know that at t = 5s, the car was brought to stop, so the velocity must be zero.

v(5s) = 0 = a*5s + 110km/h.

a = (110km/h)*(1/5s)

now we have that:

1 hour = 3600 seconds.

1km = 1000m

then:

110km/h = (110*1000/3600)m/s = 30.56 m/s

Then we have:

a = (-30.55 m/s)/5s = -6.11 m/s^2

Now the velocity equation is:

v(t) = -6.11m/s^2*t + 30.56m/s

To write the positon equation we must integrate over time again, we can get:

p(t) = (1/2)*(-6.11m/s^2)*t^2 + (30.56m/s)*t + p0

Where p0 is the initial position, here i will assume that is zero, because it does no really mater.

The total displacement of the car will be equal to p(5s)

p(5s) = (1/2)*(-6.11m/s^2)*(5s)^2 + (30.56m/s)*(5s) = 76.425 meters.

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To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

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$I_0 =$ Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

$I = \frac{I_0}{2}$

Replacing we have that

$I = \frac{370}{2}$

$I = 185W/m^2$

Re-arrange the equation,

$I'= \frac{I_0}{2}cos^2\theta$

Re-arrange to find \theta

$cos^2\theta = \frac{2I'}{I_0}$

$cos^2\theta = \frac{2*138}{370}$

$\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})$

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The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

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3 years ago

A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti

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Use Newton's second law to determine the acceleration being applied to the sled. There are three forces at work on the sled (its weight, the force normal to the ground, and friction) but two of them cancel, leaving friction as the only effective force. This vector is pointed in the opposite direction of the sled's movement, so if we take the direction of its movement to be the positive axis, we would find the acceleration due to the friction to be

$\vec F_G+\vec F_N+\vec F_F=m\vec a\iff-12\,\mathrm N=(20\,\mathrm{kg})a\implies a=-0.6\,\dfrac{\rm m}{\mathrm s^2}$

Now we use the formula

${v_f}^2-{v_i}^2=2a(x_f-x_i)$

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3 years ago

A rock is thrown vertically upward with a speed of 18.0 m/s from the roof of a building that is 50.0 m above the ground. Assume

Andreyy89

Answer:

(a) 5.7 s

(b) 39 m/s

Explanation:

(a) u = 18 m/s

At the maximum height, the final velocity of ball is zero. lte teh time taken by the ball to go from 50 m height to maximum height is t.

use first equation of motion.

v = u + g t

0 = 18 - 10 x t

t = 1.8 s

Let the maximum height attained by the ball when it thrown from 50 m height is h'.

Use third equation of motion

v^2 = u^2 + 2 g h'

0 = 18^2 - 2 x 10 x h'

h' = 16.2 m

Total height from the ground H = h + h' = 50 + 16.2 = 76.2 m

Let t' be the time taken by the ball to hit the ground as it falls from maximum height.

use third equation of motion

H = ut + 1/2 x g t'^2

76.2 = 0 + 1/2 x 10 x t'^2

t' = 3.9 s

Total time taken by the ball to hit the ground = T = t + t' = 1.8 + 3.9 = 5.7 s

(b) Let v be the velocity with which the ball strikes the ground.

v^2 = u^2 + 2 g H

v^2 = 0 + 2 x 10 x 76.2

v = 39 m/s

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