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4 years ago

Car mass of 1167 kg accelerates on a flat highway from 10.0 m/s to 28.0 m/s. how much work does thr car's engine do on the car ?

1 answer:

6 0

For the work-energy theorem, the work done by the engine on the car is equal to the variation of kinetic energy of the car:
$W= K_f-K_i=\frac{1}{2}mv_f^2 - \frac{1}{2} mv_i^2$
where m=1167 kg is the mass of the car, $v_i=10.0 m/s$ is the initial speed and $v_f=28.0 m/s$ is the final speed.

Putting numbers into the equation, we find:
$W= \frac{1}{2}(1176 kg)(28.0m/s)^2- \frac{1}{2}(1167 kg)(10.0 m/s)^2=4.0 \cdot 10^5 J$

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In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a

SOVA2 [1]

Answer: $0.084 m/s^2$

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance $d_1=8.13 km$

time taken=25 min =1500 s

initial speed $v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s$

after 8.13 km mark steve started to accelerate

speed after 60 s

$v_2=v_1+at$

$v_2=5.6+a\times 60$

distance traveled in 60 sec

$d_2=v_1\times 60+\frac{a60^2}{2}$

$d_2=336+1800 a$

time taken in last part of journey

$t_3=1663.6-1560=103.6 s$

distance traveled in this time

$d_3=v_2\times t_3$

$d_3=\left ( 5.6+a\times 60\right )103.6$

and total distance $=d_1+d_2+d_3$

$10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$a=0.084 m/s^2$

5 0

4 years ago

Write a hypothesis why the moon has very little liquid water.

Llana [10]

Because the Moon has a very small surface area compared to other spacial geo-bodies, it has cooled down much faster than Earth. Any water on the moon would freeze.

6 0

4 years ago

Hey what is you answering streak and what fact dose it say

Valentin [98]

Answer:

It’s 53:) Or - quit

Explanation:

That’s my strong streak

5 0

3 years ago

Read 2 more answers

Please do number 25! Explain how you got your answer with detail to get Brainliest! Thank you!

Ad libitum [116K]

John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.

The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds) = 4,000 foot-pounds of work.

If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds) = 645.2 foot-pounds per second.

The rate of doing work is called "power".

(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)

So back to our problem.

John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.

Well, 550 foot-pounds per second is called 1 "horsepower".

So as John runs up the steps to the balcony, he's doing the work
at the rate of

(645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)

= 1.173 Horsepower. GO JOHN !

(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________

Oh my gosh ! Look at #26 ! There are the metric units I was talking about.

Do you need #26 ?

I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.

a). 5
b). 750 Joules
c). 800 Joules
d). 93.75%

You're welcome.

And #27 is 0.667 m/s .

7 0

3 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago