Which object would have the most momentum? A. a 500-pound car moving at 20 miles per hour B. a 200-pound car moving at 60 miles
2 answers:
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Hi there,
for this question we have:
Signal 2.0 MHz = Emitted so we can call it
and we need the Reflected =
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted).
we also know: ΔBeat frequency(max) = 560 Hz =
so we have:
- =
so frequency of Reflected is:
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz =
now you know that Lambda = v/f
so if we find the lambda with our Emitted then we can find v with the Reflected:
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m
=>
= (lambda)(
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s
so the is equal to 1532 m/s :)))
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day
for this question we have:
Signal 2.0 MHz = Emitted so we can call it
and we need the Reflected =
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted).
we also know: ΔBeat frequency(max) = 560 Hz =
so we have:
so frequency of Reflected is:
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz =
now you know that Lambda = v/f
so if we find the lambda with our Emitted then we can find v with the Reflected:
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m
=>
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s
so the
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day
Answer:
a) v = 0.4799 m / s, b) K₀ = 1600.92 J, K_f = 5.46 J
Explanation:
a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.
Initial instant. Before the crash
p₀ = m v₁ + M v₂
where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive
Final moment. After the crash
p_f = (m + M) v
as the system is isolated, the moment is preserved
p₀ = p_f
m v₁ + M v₂ = (m + M) v
v =
let's calculate
v =
v = 0.4799 m / s
b) let's find the initial kinetic energy of the system
K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2
K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2
K₀ = 1600.92 J
the final kinetic energy
K_f = ½ (m + M) v ^ 2
k_f = ½ (95 + 111) 0.4799 ^ 2
K_f = 5.46 J