Suppose the Earth's magnetic field at the equator has magnitude 0.50×10−4T and a northerly direction at all points. Part A Estim
1 answer:
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Average force on the window: 0.32 N
Explanation:
The average force exerted on the window is given by Newton's second law
where
is the net change in momentum of the hailstones in a time interval of
In order to find the change in momentum, we have to consider only the component of the hailstone's momentum perpendicular to the window, therefore:
is the initial momentum of one hailstone, with
m = 7 g = 0.007 kg is the mass
is the initial speed
is the angle with the window
The final momentum is
where
v = 4.5 m/s is the final speed (the collision is elastic so the speed is equal, while the direction changes)
(after the rebound, the direction has changed)
So the change in momentum of 1 hailstone is
We are interested only in the magnitude, so
There are 595 hailstones hitting the window in 49 s, so the total change in momentum is
And therefore, the average force on the window is
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Answer:
wavelength = 0.968 m
frequency = 39.02 Hz
Explanation:
given data
mass = 0.0127 kg
force = 9.33 N
length = 1.93 m
to find out
wavelength and Frequency
solution
we know here linear density that is
linear density = .........1
linear density =
linear density = 6.5803 × kg/m
so
wavelength will be here
wavelength = ..............2
here n = 4 for forth harmonic
wavelength =
wavelength = 0.968 m
and
frequency will be for 4th normal mode of vibration is
frequency = ..........3
frequency =
frequency = 1.036269 × 37.654594
frequency = 39.02 Hz