Answer:
The bowling ball would
Explanation:
Because it contains more weight! And that will make it fall down quicker.
Hope it helped
Short Answer
3: C
4: D
Problem Three
Remark
Somewhere we ought to be told that this is the Doppler Effect. I have never done a problem using this formula, so I think I'm doing it correctly, but no guarantees. My guess is that the frequency increases as it comes towards you and decreases as it moves away from you. I think that is correct.
Formula
<em><u>Givens</u></em>
- f' = observed frequency
- f = actual frequency
- v = velocity of sound or light waves.
- vo = velocity of observer (in both cases 0)
- vs = velocity of source.
f' = (v + vo) * f / (v - vs)
Solution
- v = 3*10^8 m/s
- f' = 1.1 f
- f = f
- vo = 0 We are standing still while all this is going on.
- vs = ???
f'/f = 1.1
1.1 = (3*10^8 + 0 ) / (3*10^8 - vs)
3.3*10^8 - 1.1*vs = 3*10^8
3.3*10^8 - 3*10^8= 1.1 vs
0.3 * 10^8 = 1.1 vs
2.73 * 10^7 = vs
The closest answer is 3.00 * 10^7 which is C
Problem Four
Here what is happening is that you are looking for the frequency resulting from a wave moving towards you at 1/2 the speed of sound. You are not moving.
<em><u>Givens</u></em>
- v = v
- vs = 1/2 v
- f ' = ?
- f = 1000 hz
- vo =0
f' = v/(v - 1/2v) * 1000
f' = v/ (1/2 v) * 1000
f' = 2 * 1000
f' = 2000 which is D
Answer:
a = 0.5 m/s²
Explanation:
Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:
ωf = ω₀ + α*t
⇒ ωf = α*t ⇒ α = 
= 
The angular velocity, and the linear speed, are related by the following expression:
v = ω*r
Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:
a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²
The three types of magnets are temporary, permanent, and electromagnets.
I believe it was famous quantum physicist <span>Erwin Schrödinger
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