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defon
3 years ago
8

Explain how sound travels through the air to observer

Physics
1 answer:
Harlamova29_29 [7]3 years ago
3 0
Sample:
As soon as the lightning is seen, sound travels in the air for several kilometers until it eventually reaches the observer.
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The blue mass is currently 4 meters away from the red mass.
umka21 [38]

Answer:

its a!

Explanation:

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6 0
3 years ago
A truck pulls a block 8 meters across a level surface at a force of 216 N over the course of 12 seconds. How much power did the
iren [92.7K]

Answer:

work = 1728

Power = 134

Explaination:

by using the formula,

Work(W)= Force(F)×Distance(D)

<h2> and</h2>

Power(P)= Work(W)/Time taken(T)

4 0
3 years ago
At the moment t = 0, a 24.0-v battery is connected to a 5.00-mh coil and a 6.00-ω resistor. (a) immediately thereafter, how does
agasfer [191]
At the moment the answer is Yeet
5 0
3 years ago
An apple hanging from a limb has potential energy because of its height . If it falls ,what becomes of this energy just before i
sineoko [7]

Answer: Just before its hits the ground it becomes kinetic energy and when it hits the ground it becomes in another form of energy (acoustic energy or thermal energy, for example)

Explanation:

Energy is the ability of matter to produce work in the form of movement, light, heat, among others.

In this sense, according to the Conservation of Energy principle: <em>"energy is not created or destroyed, it is transformed."</em>

So, in the case of the apple, its total energy is conserved.

When the apple is hanging from a limb, it has zero kinetic energy K_{o}=0 (because it is at rest) and has gravitational potential energy U_{o}, which depends on the mass m, the acceleration due gravity g and the height h:

U_{o}=mgh

When the apple falls, just before its hits the ground, this gravitational potential energy transforms in kinetic energy K_{1} (since the apple is moving), which depends on the mass and velocity V of the apple:

K_{1}=\frac{1}{2}mV^{2}

When the apple hits the ground, the gravitational potential energy is zero (h=0) and the kinetic energy transforms into some other form of energy (acoustic energy or thermal energy, for example).

3 0
3 years ago
The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

4 0
3 years ago
Read 2 more answers
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