Answer:The speed if hailstone dependly largely on its size. A hailstone with a diameter of 0.39 inches,falls wit a speed of 20mph while a hailstone with 3.1 inches in diameter falls at a speed of 110mph.
No speed does not depend on the distance that the hailstone falls.
Explanation: There are other factors that affect the speed of the falling hailstone apart from its size.They are:
1. Friction between the air and the hailstone
2. Wind condition( windy or moist air)
3. The rate at which it melts falling.
Many materials produce static charge
Velocity, because if an object is in motion with no direction we will consider it as speed, but if it has direction we will consider it as Velocity. Hope it helps
Answer:
Explanation:
Given
length of window ![h=2.9\ m](https://tex.z-dn.net/?f=h%3D2.9%5C%20m)
time Frame for which rock can be seen is ![\Delta t=0.134\ s](https://tex.z-dn.net/?f=%5CDelta%20t%3D0.134%5C%20s)
Suppose h is height above which rock is dropped
Time taken to cover ![h+2.9 is t_1](https://tex.z-dn.net/?f=h%2B2.9%20is%20t_1)
so using equation of motion
![y=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=y%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where y=displacement
u=initial velocity
a=acceleration
t=time
time taken to travel h is
![h=0+0.5\times g\times (t_2)^2---2](https://tex.z-dn.net/?f=h%3D0%2B0.5%5Ctimes%20g%5Ctimes%20%28t_2%29%5E2---2)
Subtract 1 and 2 we get
![2.9=0.5g(t_1^2-t_2^2)](https://tex.z-dn.net/?f=2.9%3D0.5g%28t_1%5E2-t_2%5E2%29)
![5.8=g(t_1+t_2)(t_1-t_2))](https://tex.z-dn.net/?f=5.8%3Dg%28t_1%2Bt_2%29%28t_1-t_2%29%29)
and from equation ![t_1-t_2=0.134\ s](https://tex.z-dn.net/?f=t_1-t_2%3D0.134%5C%20s)
so ![t_1+t_2=\frac{5.8}{9.8\times 0.134}](https://tex.z-dn.net/?f=t_1%2Bt_2%3D%5Cfrac%7B5.8%7D%7B9.8%5Ctimes%200.134%7D)
![t_1+t_2=4.416\ s](https://tex.z-dn.net/?f=t_1%2Bt_2%3D4.416%5C%20s)
and ![t_1=t_2+\Delta t](https://tex.z-dn.net/?f=t_1%3Dt_2%2B%5CDelta%20t)
so ![t_2+\Delta t+t_2=4.416](https://tex.z-dn.net/?f=t_2%2B%5CDelta%20t%2Bt_2%3D4.416)
![2t_2+0.134=4.416](https://tex.z-dn.net/?f=2t_2%2B0.134%3D4.416)
![t_2=0.5\times 4.282](https://tex.z-dn.net/?f=t_2%3D0.5%5Ctimes%204.282)
![t_2=2.141\ s](https://tex.z-dn.net/?f=t_2%3D2.141%5C%20s)
substitute the value of
in equation 2
![h=0.5\times 9.8\times (2.141)^2](https://tex.z-dn.net/?f=h%3D0.5%5Ctimes%209.8%5Ctimes%20%282.141%29%5E2)
![h=22.46\ m](https://tex.z-dn.net/?f=h%3D22.46%5C%20m)
Answer:
ξ = 0.00845020162 V or 8.4 mV
Explanation:
Magnetic flux measures the total magnetic field that passes through a known area. Magnetic flux describe the effect of magnetic field in a given area. Mathematically,
magnetic flux (Ф) = BA cos ∅
where
A = test area
B = magnetic field
before the flip
Ф = Bπr²N
N = number of turn
magnitude of induced emf = N |ΔФ/Δt|
ξ = 2Nπr²B/dt
ξ = 2 × 22 × π × (1.02/2)² × 0.000047/0.2
ξ = 44 × π × 0.51² × 0.000047/0.2
ξ = 44 × π × 0.2601 × 0.000047/0.2
ξ = 0.0005378868 × 3.142/0.2
ξ = 0.00169004032/0.2
ξ = 0.00845020162 V or 8.4 mV