All categories

3 years ago

If a car has a kinetic energy of 110250j its mass is 180kg, how fast is the car travelling

Physics

1 answer:

ANTONII [103]3 years ago

5 0

Answer:

Velocity = v = 35 m/s

Explanation:

Kinetic energy of an object is defined as the energy possess by an object due to its motion. Kinetic energy K.E of an object is equal to the half of the mass of that object multiplied by square of the object's velocity.

Mathematically,

v = 35 m/s

You might be interested in

I'm new to brainy, but here is my question.

skad [1K]

680 is the right answer

8 0

3 years ago

Read 2 more answers

What effect does an unbalanced force have on an object?

Mars2501 [29]

<span>it either changes the direction of the object or it stops the object in motion.</span>

5 0

3 years ago

Read 2 more answers

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

2 years ago

A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. What is the vertical compon

elixir [45]

Answer:

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity = 670 (0.7071)

Vertical component of the cannonball’s velocity = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

6 0

2 years ago

a goalkeeper catches a 491 g soccer ball traveling horizontally at 29.4 m/s. if it took 2,218 n of force to stop the ball, how m

yarga [219]

The ball will take 2.551 seconds to reach its peak position.

We must know how long the balls are in the air before we can predict where they will fall. It will take 2 seconds for both balls to touch the ground.

<h3>How quickly does a ball drop?</h3>

The falling ball travels a distance of d = 12 9.8 (m/s2) t2, with a speed of v = 9.8 (m/s2) t as a function of time. The ball travels 4.9 m in a second. The falling ball's velocity is v = -9.8 (m/s2) t j, and its position is r = (4.9 m - 12 9.8 (m/s2) t2) j as a function of time.

To know more about balls visit:-

brainly.com/question/19930452

#SPJ4

6 0

7 months ago