The answer is D. As the ambulance gets closer, the sound waves are compressed relative to the person; so the frequency increases.
Answer:
A car
Explanation:
A car can travel 100 m in 5 seconds
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Answer:
Explanation:
a. The equation of Lorentz transformations is given by:
x = γ(x' + ut')
x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.
x' = 0
t' = 5.00 s
u =0.800 c,
c is the speed of light = 3×10⁸ m/s
Then,
γ = 1 / √ (1 - (u/c)²)
γ = 1 / √ (1 - (0.8c/c)²)
γ = 1 / √ (1 - (0.8)²)
γ = 1 / √ (1 - 0.64)
γ = 1 / √0.36
γ = 1 / 0.6
γ = 1.67
Therefore, x = γ(x' + ut')
x = 1.67(0 + 0.8c×5)
x = 1.67 × (0+4c)
x = 1.67 × 4c
x = 1.67 × 4 × 3×10⁸
x = 2.004 × 10^9 m
x ≈ 2 × 10^9 m
Now, to find t we apply the same analysis:
but as x'=0 we just have:
t = γ(t' + ux'/c²)
t = γ•t'
t = 1.67 × 5
t = 8.35 seconds
b. Mavis reads 5 s on her watch which is the proper time.
Stanley measured the events at a time interval longer than ∆to by γ,
such that
∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)
c. According to Stanley,
dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m
which is the same as in part (a)
Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;
Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
The period of the block's mass is changed by a factor of √2 when the mass of the block was doubled.
The time period T of the block with mass M attached to a spring of spring constant K is given by,
T = 2π(√M/K).
Let us say that, when we increased the mass to 2M, the time periods of the block became T', the spring constant is not changed, so, we can write,
T' = 2π(√2M/K)
Putting T = 2π(√M/K) above,
T' =√2T
So, here we can see, if the mass is doubled from it's initial value. The time period of the mass will be changed by a factor of √2.
To know more about time period of mass, visit,
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