Answer: 330 m
Explanation:
The speed of sound
is defined as the distance traveled by the sound wave
in a especific time
:
Where:
is the speed of sound
is half the time the sound wave travels since the person speaks, the sound wave hits the hill and then returns to the person again as an echo
is the distance between the person and the hill
So, we have to find
:


Finally:

Answer:
0.074 V
Explanation:
Parameters given:
Number of turns, N = 121
Radius of coil, r = 2.85 cm = 0.0285 m
Time interval, dt = 0.179 s
Initial magnetic field strength, Bin = 55.1 mT = 0.0551 T
Final magnetic field strength, Bfin = 97.9 mT = 0.0979 T
Change in magnetic field strength,
dB = Bfin - Bin
= 0.0979 - 0.0551
dB = 0.0428 T
The magnitude of the average induced EMF in the coil is given as:
|Eavg| = |-N * A * dB/dt|
Where A is the area of the coil = pi * r² = 3.142 * 0.0285² = 0.00255 m²
Therefore:
|Eavg| = |-121 * 0.00255 * (0.0428/0.179)|
|Eavg| = |-0.074| V
|Eavg| = 0.074 V
Answer:
v = 15.8 m/s
Explanation:
Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force
= ΔEm =
-Em₀
Let's write the mechanical energy at each point
Initial
Em₀ = Ke = ½ k x²
Final
= K + U = ½ m v² + mg y
Let's use Hooke's law to find compression
F = - k x
x = -F / k
x = 4400/1100
x = - 4 m
Let's write the energy equation
fr d = ½ m v² + mgy - ½ k x²
Let's clear the speed
v² = (fr d + ½ kx² - mg y) 2 / m
v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0
v² = (160 + 8800 - 1470) / 30
v = √ (229.66)
v = 15.8 m/s
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