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Llana [10]
3 years ago
7

We can block light by placing obstacles in its path, but it’s much more difficult to block sound. Why?

Physics
2 answers:
olganol [36]3 years ago
8 0
It's harder to block sound because sound waves and transverse waves, causing the sound to bounce off objects. This isn't the same with light. Hope that helps. 
rjkz [21]3 years ago
5 0
All we have to do is focus on what you want to hear and you can tell your mind to block everything out  and its very easy to block light than noise all you have to do is focus
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A box rests on top of a flat bed truck. the box has a mass of m = 18 kg. the coefficient of static friction between the box and
konstantin123 [22]
For this case, the first thing you should do is write the kinematic motion equation of the block.
 We have then:
 vf = vo + a * t
 Where,
 vf: Final speed.
 vo: Initial speed.
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 t: time.
 Substituting the values:
 (16) = (0) + a * (16)
 Clearing the acceleration:
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5 0
3 years ago
A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5
svlad2 [7]
The magnetic force experienced by the proton is given by
F=qvB \sin \theta
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and \theta the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so \sin \theta=1 and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
F=ma

So we have
ma=qvB
from which we can find the magnitude of the field:
B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T
4 0
3 years ago
Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i
densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

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