Answer:
12N
Explanation:
We are given that one of the forces are acting only in the horizontal x-direction. As a force must be applied on an object of mass in order to cause acceleration, the 6.0ms^-2 acceleration is due to the non-horizontal force acting on the 2.0kg object.
Using Newton's Second Law of motion; we know that for a constant mass, force is equal to mass times acceleration, F=ma.
Assuming the other force is acting only in the vertical direction (question doesn't specify, thus we are finding the minimum force to cause this acceleration):
F= 2.0kg * 6.0ms^-2
F=12.0 kgms^-2
F=12 N
Answer:
The magnitude of the induced voltage in the loop is 20 mV.
Explanation:
given;
length of loop, L = 0.43 m
width of loop,w = 0.43 m
velocity of moved loop, v = 0.15m/s
magnetic field strength,B = 0.31 T
To determine the magnitude of the induced voltage in the loop, we apply Faraday's law;
magnitude induced E.M.F = BLv
magnitude induced E.M.F = 0.31 x 0.43 x 0.15 = 0.02 V = 20 mV
Therefore, the magnitude of the induced voltage in the loop is 20 mV.
Answer:
The Troposphere is the lowermost portion of Earth’s atmosphere. It is the densest layer of the atmosphere and contains approximately 75 percent of the mass of the atmosphere and almost all the water vapor and aerosol. The troposphere extends from the Earth’s surface up to the tropopause where the stratosphere begins.
Explanation:
The planets revolving around the sun, The moon revolving around the Earth.
Answer:
1.25 m/s
Explanation:
Given,
Mass of first ball=0.3 kg
Its speed before collision=2.5 m/s
Its speed after collision=2 m/s
Mass of second ball=0.6 kg
Momentum of 1st ball=mass of the ball*velocity
=0.3kg*2.5m/s
=0.75 kg m/s
Momentum of 2nd ball=mass of the ball*velocity
=0.6 kg*velocity of 2nd ball
Since the first ball undergoes head on collision with the second ball,
momentum of first ball=momentum of second ball
0.75 kg m/s=0.6 kg*velocity of 2nd ball
Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg
=1.25 m/s