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3 years ago

Select the correct answer.

Physics

2 answers:

Liono4ka [1.6K]3 years ago

7 0

Answer: The electromagnetic waves reach Earth, while the mechanical waves do not.

Explanation:

yaroslaw [1]3 years ago

4 0

Answer:

(B) The electromagnetic waves reach Earth, while the mechanical waves do not.

Explanation:

Electromagnetic waves do not require material medium for their propagation, so it will reach the Earth. Mechanical waves require material medium for their propagation, so mechanical waves will not reach Earth.

Therefore, the correct option is B "The electromagnetic waves reach Earth, while the mechanical waves do not".

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object x and y fall from a same height and object x is heavier than y which object would fall faster qnd y

hjlf

Answer: They’d fall at the same speed, because air resistance is the only thing that makes an object fall faster than another. There’s a video somewhere on the internet of a bowling ball and a feather falling at the same speed in a vacuum, if you look for it. Hope this helps!

4 0

3 years ago

What is the electric field strength at a distance of .50 m from a 1.00x10^-6 c charge?

ss7ja [257]

Answer:

e = k q / r² then,

e = 9×10^9 * 1×10^-6 / 50² = 3.6 N/C

hope this helps ❤.

7 0

2 years ago

Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!

Bas_tet [7]

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

$\theta_8 = 1.12^{\circ}$ = angle to 8th max (note how m = 8 since we're comparing this to the form $\theta_m$ )

$x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}$ (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

$d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}$

Make sure your calculator is in degree mode.

-----------------

Method 2

$\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\$

-----------------

Method 3

$\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\$

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

7 0

3 years ago

when a person is making cardiovascular gains through aerobic activity, they are increasing what to the lungs and heart?

gladu [14]

When a person is making cardiovascular gains through aerobic activity they are increasing the amount of oxygen to the heart and lungs. This keeps the heart and lungs of a person very healthy. It decreases the chance of heart diseases.

8 0

3 years ago

The springs of a 1500 kg car compress 5.00 mm when its 68 kg driver gets into the driver's seat. Part A If the car goes over a b

elena-s [515]

Answer:

the frequency of the oscillation is 1.5 Hz

Explanation:

Given;

mass of the spring, m = 1500 kg

extention of the spring, x = 5 mm = 5 x 10⁻³ m

mass of the driver = 68 kg

The weight of the driver is calculated as;

F = mg

F = 68 x 9.8 = 666.4 N

The spring constant, k, is calculated as;

k = F/m

k = (666.4 N) / (5 x 10⁻³ m)

k = 133,280 N/m

The angular speed of the spring is calculated;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{133280}{1500} } = 9.426 \ rad/s$

The frequency of the oscillation is calculated as;

ω = 2πf

f = ω / 2π

f = (9.426) / (2π)

f = 1.5 Hz

Therefore, the frequency of the oscillation is 1.5 Hz

6 0

3 years ago