All categories

3 years ago

Calculate the kinetic energy in joules of a 1,500 kg car that is moving at a speed of 42 km/h

1 answer:

4 0

Data:
KE (Kinetic Energy) = ? (Joule)
m (mass) = 1500 Kg
v (speed) = 42 Km/h
converting to m/s (42 / 3.6), we have: v (speed) = 11.6 m/s

Formula:
$K_{E} = \frac{1}{2} m*v^2$

Solving:
$K_{E} = \frac{1}{2} m*v^2$
$K_{E} = \frac{1}{2} *1500*(11.6)^2$
$K_{E} = \frac{1}{2} *1500*134.56$
$K_{E} = \frac{201840}{2}$
$\boxed{\boxed{K_{E} = 100920\:Joule}}\end{array}}\qquad\quad\checkmark$

You might be interested in

The increase of kinetic energy at the square of the speed of your vehicle has a major influence on all motor vehicles in three p

Viktor [21]

This implies that stopping distance and impact force grow as a function of speed. The best ways to improve manoeuvrability and lessen crash severity are to drive at an appropriate pace and to slow down as soon as you spot dangers in front of you.

Keep in mind that stopping distance increases with speed; at 50 mph, it is four times longer than at 25 mph, and at 75 mph, the force of impact is nine times greater.

<h3>What is the impact of speed on kinetic energy ?</h3>

When your car expends or absorbs energy to speed up or slow down, you may feel a pull or a jolt, called impulse. Impulse increases as the energy or force increases, and increases as the duration of the force decreases. You'll feel a harder jolt if you speed up or slow down suddenly.

Consider: coming to a stop from 60 mph in ten seconds doesn't hurt you or your vehicle because the force of this event is spread out over a long time. But if you hit a wall and come to a stop in just half a second, you'll feel twenty times the impulse, causing severe damage.

Learn more about Kinetic energy here:

brainly.com/question/25959744

#SPJ4

5 0

2 years ago

Help me please .Thank you for all

Nostrana [21]

Answer:

I can't understand the language....

7 0

2 years ago

A cart travels 4.00 meters east and then4.00 meters north. determine the magnitude ofthe cart’s resultant displacement.

gogolik [260]

^
|
|
|
----------->|

Square root of (4^2 + 4^2) = 4*squareRoot(2)

7 0

3 years ago

Which of the following statements explains how total time spent in the air is affected as a projectile's angle of launch is incr

charle [14.2K]

Answer:

Therefore letter <u>C is the correct answer.</u>

Explanation:

In a projectile motion the total time in the air can be calculated using the following equation:

We analyze the y-component motion.

$v_{fy}=v_{iy}-gt$

When the final velocity (v(f)) is equal to zero we calculate the upward time and multiplying it by 2 we find the total time in the air. So we will have:

$t_{tot}=2\frac{v_{iy}}{g}$

$t_{tot}=2\frac{v_{i}sin(\theta)}{g}$

We can see that the <u>total time is directly proportional to the angle</u>, then when <u>θ increase t increase.</u>

Therefore letter C is the correct answer.

I hope it helps you!

3 0

3 years ago

A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of

Rasek [7]

Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

8 0

3 years ago