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3 years ago

Why do you often freeze after a shower? Help mee

1 answer:

5 0

When water evaporates of your body it cools you down, hence why after a shower when the water droplets begin to evaporate you feel cold

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Another name for Newton’s 2nd law

monitta

According to Newton’s second law of motion, also know as the law of force and Accelerate , a force upon an object causes it to accelerate according to the formula net force = mass x acceleration

4 0

3 years ago

What is the potential energy of a 2500 g object suspended 5 kg above the earth's surface?

Alinara [238K]

Answer:

$potential \: energy = mgh \\ = ( \frac{2500}{1000} ) \times 10 \times 5 \\ = 125 \: newtons$

if height is 5 m

8 0

3 years ago

A car travelling at 15 m/s comes to rest in a distance of 14 m when the brakes are applied.

stira [4]

Answer:

-8.04 m/s2

Explanation:

To find the answer to this, you have to use the 4th kinematic equation:

$v^{2} = v^{2}_{0} + 2ax$

You plug into the equation to get:

$0 = 15^{2} + 2a(14)$

solve for a to get

-8.04 m/s2

3 0

3 years ago

A disk has a radius of 30 cm and a mass of 0.3 kg and is turning at 3.0 rev/s. A trickle of sand falls onto the disk at a distan

Pani-rosa [81]

Answer:

The mass of the sand that will fall on the disk to decrease the is 0.3375 kg

Explanation:

Moment before = Moment after

$I \omega_i = I \omega_f +mr^2 \omega_f\\\\mr^2 \omega_f = I \omega_i - I \omega_f \\\\m = \frac{ I \omega_i - I \omega_f}{r^2 \omega_f }$

where;

I is moment of inertia = Mr² = 0.3 x (0.3)² = 0.027 kg.m²

substitute this in the above equation;

$m = \frac{ 0.027[3(2 \pi) - 2(2 \pi)]} {0.2^2 * 6\pi } = \frac{ 0.027[6 \pi - 4\pi]} {0.2^2 * 4\pi }\\\\m = 0.3375kg$

Therefore, the mass of the sand that will fall on the disk to decrease the is 0.3375 kg

7 0

3 years ago

The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t

maxonik [38]

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s$

Work done is equal to the change in kinetic energy. It can be given as follows :

$W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J$

So, the required work done is 32.99 J.

3 0

2 years ago