All categories

3 years ago

Please help!!30 points and will give the brainliest answer for a clear explanation.I need it really fast. Thank you!

1 answer:

4 0

I would rather be hit by the deflated ball because it wouldn't hurt as bad because it wouldn't have a lot of weight to hurt me in anyway

You might be interested in

Select Light for the type of wave, adjust the wavelength so that the light is red, and increase the amplitude of the light to th

Sergeu [11.5K]

Answer:

here as we increase the distance the intensity will decrease and hence the amplitude of the electric field will decrease and vice-versa

Explanation:

As wee know that the amplitude of the wave will decide the energy of the wave

Here we know that energy density of electromagnetic wave is given as

$u = \frac{1}{2}\epsilon_0E_0^2$

now we have

$\frac{I}{c} = \frac{1}{2}\epsilon_0 E_0^2$

so here we can say that intensity of the wave at the given distance from the source is given by formula

$I = \frac{P}{4\pi r^2}$

so here as we increase the distance the intensity will decrease and hence the amplitude of the electric field will decrease and vice-versa.

5 0

3 years ago

Meg walks with a velocity of 0.9 m/s west. She does so while riding on a train that is traveling with a velocity of 2.7 m/s east

Shkiper50 [21]

<span>Velocities are vectors so we can add them!

Let's let +x be East and -x be West.

-0.9 + 2.7 = 1.8

Since our answer is positive that means East so the answer is C.</span>

6 0

3 years ago

The sun is part of which type of galaxy?

jeka94

I believe the answer is D. The milky way is a spiral galaxy, You can tell just by looking at it.

4 0

3 years ago

A 40W lamp wastes 34 J of energy every second by heating its surroundings.

Artemon [7]

Answer:

$15\%$ .

Explanation:

The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is $40\; \rm W$ (same as $40\; \rm J \cdot s^{-1}$ ,) meaning that $40\; \rm J$ of energy is supplied to this lamp every second.

The question states that $34\; \rm J$ out of that $40\; \rm J$ of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the $(40\; \rm J - 34\; \rm J) = 6\; \rm J$ of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive $40\; \rm J$ of energy input and would outputs $6\; \rm J$ of useful work. The efficiency of this lamp would be:

$\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}$ .

4 0

2 years ago

A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static frictio

lesya [120]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

$W = F\times d$

$KE = 0.5\times m\times v^2$

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

$W = F \times d = 2.62 N \times 100 m$