Answer:
2) twice the amplitude and half the wavelength
Explanation:
Answer:
3.8 secs
Explanation:
Parameters given:
Acceleration due to gravity, g = 9.8 
Initial velocity, u = 11.76 m/s
Final velocity, v = 49 m/s
Using one of Newton's equations of linear motion, we have that:

where t = time of flight of arrow
The sign is positive because the arrow is moving downward, in the same direction as gravitational force.
Therefore:

The arrow was in flight for 3.8 secs
Answer:
B=9.1397*10^-4 Tesla
Explanation:
To find the velocity first we put kinetic energy og electron is equal to potential energy of electron
K.E=P.E

where :
m is the mass of electron
v is the velocity
V is the potential difference
eq 1
Radius of electron moving in magnetic field is given by:
eq 2
where:
m is the mass of electron
v is the velocity
q=e=charge of electron
B is the magnitude of magnetic field
Put v from eq 1 into eq 2



B=9.1397*10^-4 Tesla
Yeah yeah I just got a hold of you and I saw that you were doing a good job and I thought you were doing a good job and I thought you were doing a good job and I thought you were doing a good job and I thought you were doing a good job and I thought you were doing a good job and I thought you were doing a good job and I thought you were doing a good job and I thought you were doing a good job and I thought you were doing a good job and I thought you were doing a good job and I thought you were doing a good job.
Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

v is the speed of cat, 

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.