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3 years ago

Help me please i’ll pay you (paypal)

Physics

1 answer:

Arisa [49]3 years ago

8 0

If the award weighs 200 newtons and 200 newtons equals 44.96 pounds of force even though it is of such a force if it hits the ground the energy will either discharge into the air doing nothing but creating a loud sound or it will discharge into the ground altering the ground that it landed on.

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When conduction electrons pass through a device with resistance, which describes their energy?

Aleksandr-060686 [28]

Answer:

a) their potential energy increases.

Explanation:

Ohm's Law is

R= V/I

Where R= Resistance

V= potential difference or potential energy

I= current or conduction electron flow rate

Clearly R and V are directly proportional i-e Potential energy increases with resistance.

3 0

3 years ago

A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir

TEA [102]

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I=I_0 cos^2\theta$

Where,

$I_ {0}$ indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

$\theta$ indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

Since we have two objects the law would be,

$I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)$

Replacing the values,

$I=100*cos^2(20)*cos^2(40-20)$

$I=100*cos^4(20)$

$I=77.91W/m^2$

Therefore the intesity of the light after it has passes through both polarizers is $77.91W/m^2$

7 0

3 years ago

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What

Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

$\alpha = 6261 \ rev/minutes^2$

$\theta = 613 \ revolutions$

Explanation:

From the question we are told that

The initial angular speed is $w_i = 1120 \ rev/minutes$

The angular speed after $t = 13.8 s = \frac{13.8}{60 } = 0.23 \ minutes$ is $w_f = 2560 \ rev/minutes$

The time for revolution considered is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$

Generally the angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{2560 - 1120 }{0.23}$

=> $\alpha = 6261 \ rev/minutes^2$

Generally the number of revolution made is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$ is mathematically represented as

$\theta = \frac{1}{2} * (w_i + w_f)* t$

=> $\theta = \frac{1}{2} * (1120+ 2560 )* 0.333$

=> $\theta = 613 \ revolutions$

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