Answer:
the balls reached a height of 4.9985 m
Explanation:
Given the data in the question;
mass one m = 3.8 kg
mass two M = 2.1 kg
Initial velocities
u = 22 m/s
U = { moving downward} = 12 m/s
Now, using the law conservation of linear moment;
mu + MU = v( m + M )
we solve for "v" which is the velocity of the ball s after collision;
v = (mu + MU) / ( m + M )
so we substitute our given values into the equation
v = ( ( 3.8 × 22 ) + ( 2.1 × -12) ) / ( 3.8 + 2.1 )
v = ( 83.6 - 25.2 ) / 5.9
v = 58.4 / 5.9
v = 9.898 m/s
Now, we determine required height using the following relation;
v"² - v² = 2gh
where v" is the velocity at the top which is 0 m/s and g = -9.8 m/s²
0 - v² = 2gh
v² = -2gh
so we substitute
( 9.898 )² = -2 × -9.8 × h
97.97 = 19.6 × h
h = 97.97 / 19.6
h = 4.9985 m
Therefore, the balls reached a height of 4.9985 m
Answer:
112.5 J
Explanation:
I calculated it by K/G BY M/S TO POTENTIAL ENERGY.
C. Slamming on the brakes to come to a stop at a stop sign. Here you are Decelerating.
A radio telescope is simply a telescope that is designed to receive radio waves from space. In its simplest form it has three components:
1. One or more antennas to collect the incoming radio waves. Most antennas are parabolic dishes that reflect the radio waves to a receiver, in the same way as a curved mirror can focus visible light to a point.
2. A receiver and amplifier to boost the very weak radio signal to a measurable level. These days the amplifiers are extremely sensitive and are normally cooled to very low temperatures to minimise interference due to the noise generated by the movement of the atoms in the metal (called thermal noise).
3. A recorder to keep a record of the signal. Most radio telescopes nowadays record directly to some form of computer memory disk as astronomers use sophisticated software to process and analyse the data.
