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3 years ago

a car accelerates from rest at 3 m / s^2 along a straight road. how far has the car travelled after 4 s

Physics

1 answer:

evablogger [386]3 years ago

4 0

Answer:

24 m

Explanation:

The motion of the car is a uniformly accelerated motion (=at constant acceleration), therefore we can find the distance covered by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time elapsed

a is the acceleration

For the car in this problem:

u = 0, since the car starts from rest

$a=3 m/s^2$ is the acceleration

t = 4 s is the time elapsed

Therefore, the distance covered is:

$s=0+\frac{1}{2}(3)(4)^2=24 m$

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a 700k/g race car slowed down from 30m/s to 15m/s what is the impulse and what was the average force applied by the brakes

posledela

So impulse is a change in momentum.

Mass*(final velocity - initial velocity)

I dont think you will be able to find the average force with the given info because you need to know the time it takes for the car to slow down.

5 0

3 years ago

A shovel is the third class lever

Nookie1986 [14]

Answer:

Yes

Explanation:

In a third-class lever, the effort force lies between the resistance force and the fulcrum. Some kinds of garden tools are examples of third-class levers. When you use a shovel, for example, you hold one end steady to act as the fulcrum, and you use your other hand to pull up on a load of dirt.

8 0

3 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

$y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , so,

$y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0

3 years ago

2. Explain brightness of light using the wave model of light.

Dafna11 [192]

The amplitude of a wave tells us about the intensity or brightness of the light relative to other light waves of the same wavelength.

5 0

2 years ago

Read 2 more answers

Match the indices of refraction with the corresponding effects on the waves.

Yanka [14]

Answer:

B) waves speed up

C) waves bend away from the normal

Explanation:

The index of refraction of a material is the ratio between the speed of light in a vacuum and the speed of light in that medium:

$n=\frac{c}{v}$

where

c is the speed of light in a vacuum

v is the speed of light in the medium

We can re-arrange this equation as:

$v=\frac{c}{n}$

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is

B) waves speed up

Moreover, when light enters a medium bends according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1 ,n_2$ are the index of refraction of the 1st and 2nd medium

$\theta_1,\theta_2$ are the angles made by the incident ray and refracted ray with the normal to the interface

We can rewrite the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

So we see that if the index of refraction of the second medium is lower ( $n_2$ ), then the ratio $\frac{n_1}{n_2}$ is larger than 1, so the angle of refraction is larger than the angle of incidence:

$\theta_2>\theta_1$

This means that the wave will bend away from the normal. So the other correct option is

C) waves bend away from the normal

3 0

3 years ago