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An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the moment
1 answer:
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Answer:
the frequency clock would be 1262.85 Hz
Explanation:
Given data;
height of building h = 25 m
from the third equation of motion;
v² = u² + 2as
Since the Alarm clock falls with an acceleration equal to the acceleration due to gravity; a = g = 9.81 m/s²
initial velocity u = 0
so we substitute our values into the kinematic equation
v² = (0)² + 2 × 9.81 × 25
v² = 490.5
v = √490.5
v = 22.1472 m/s
Now, since the alarm clock is moving both I am stationary;
my velocity will be zero.
so Frequency of the alarm clock will be;
f' = [ (v - ) / ( v +
) ] × f
we know that; speed of sound is 343 m/s, so v = 343 m/s, is 22.1472 m/s, f is 1350 Hz,
is 0 m/s
so we substitute the values into the equation
f' = [ (343 - 22.142 ) / ( 343 + 0 ) ] × 1350
f' = [ 320.858 / 343 ] × 1350
f' = 0.935446 × 1350
f' = 1262.85 Hz
Therefore, the frequency clock would be 1262.85 Hz
To solve the problem it is necessary to take into account the concepts related to the Magnetic Force, which is given by,
Where
F= Magnetic force
q= charge of proton
v= velocity
B Magnetic field
Angle between the velocity and the magnetic field.
Re-arrange the equation to find the angle we have,
Replacing our values we have,
Then,
The angle between 0 to 180 degrees would be,
Therefore the two angles required are 13.63° and 166.36°
Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity = 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity is proportional to 1/(distance)²
i.e
∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e ₂ =
₁/2
Hence,
₂/
₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m