Ask question

Ask question

All categories

chubhunter [2.5K]

3 years ago

13

A race car is moving with a velocity of 144 kilometers/hour. The driver applies the brakes, and the car comes to a halt in 12.0

seconds. What is the average acceleration of the car during those 12.0 seconds?

1 answer:

anzhelika [568]3 years ago

8 0

-12 k/s/s (the number is right the abbreviation might not be)

You might be interested in

1. What is the acceleration of a car that goes from 20 Km/hr to 100 km/hr in 8 seconds?

san4es73 [151]

Answer:

10 :)

You have to divide the difference of speed and divide it by the time. So 100-20 would be 80, and if you divide that by 8 it would be 10.

Hope this helps.

7 0

3 years ago

Read 2 more answers

Astar begins as a blank a large cloud of gas and dust

crimeas [40]

This is a statement but yes a star forms inside nebulae which are gigantic clouds of gas. stars form inside as the gases own gravity pulls it together after which it becomes large enough to perform fusion and become a star.

3 0

3 years ago

A beam of light strikes a mirror reflects off at the angle in which it hit the mirror. which term does the diagram below illustr

zavuch27 [327]

Answer:

B.) reflection

Explanation:

3 0

3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago

A plane accelerates from 25 m/s to a velocity of 80 m/s in a time of 5 seconds. What

Natali [406]

Answer:

-7

Explanation:

-7 because -7 said so

6 0

3 years ago