The mass of the quarterback is 61.2 kg.
Explanation:
mass of the football player = m1 = 102 kg
mass of the quarterback = m2 = ?
velocity of the football player = v1 = 8 m/s
According to the law of conservation of momentum:
The total momentum of a system before and after the collision remains constant. Assuming the situation as an isolated system which is not affected by any external factors, we have:
m₁v₁ + m₂v₂ = (m₁+m₂)V
Here, we need to find m₂.
We assume that the quarterback is standing still when he is attacked by the football player so v₂ = 0 m/s
After the collision both of them fall to the ground with a velocity of 5 m/s so V = 5 m/s
Keywords: momentum, velocity, law of conservation of momentum
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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
Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
-- The unit of frequency is "per second" (Hz), which is [reciprocal time].
-- The unit of period is "second", which is [time].
Do you see where this is going ?
'Frequency' and 'period' are reciprocals of each other.
For any wave ...
Period = (1) / (frequency) .
Frequency = (1) / (period) .
Answer:
The surface gravity g of the planet is 1/4 of the surface gravity on earth.
Explanation:
Surface gravity is given by the following formula:
So the gravity of both the earth and the planet is written in terms of their own radius, so we get:
The problem tells us the radius of the planet is twice that of the radius on earth, so:
If we substituted that into the gravity of the planet equation we would end up with the following formula:
Which yields:
So we can now compare the two gravities:
When simplifying the ratio we end up with:
So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.
