A 2.0 kg mass moving at 5.0 m/s suddenly collides head-on with a 3.0 kg mass at rest. If the collision is perfectly inelastic, h
1 answer:
You might be interested in
Explanation:
Given that,
2 strings both vibrate at exactly 220 Hz. The frequency of sound wave depends on the tension in the strings.
The tension in one of them is then decreased sightly, then will decrese.
Beat frequency,
So, the new frequency of the string is 217 Hz. Hence, this is the required solution.
Consider, please, this solution:
The final heat is:
Q=Q₁+Q₂+Q₃≈726.116 kJ
All the details are in the attachment.
Answer:
a = 3.27 m/s²
T = 275 N
Explanation:
Given that:
Mass m₁ = 42.p0 kg
Mass m₂ = 21.0 kg
Consider both masses to be in a whole system, then:
The acceleration can be determined as:
Making acceleration the subject in the above formula;
a = 3.27 m/s²
in the string, the tension is calculated using the formula:
T = 274.68 N
T ≅ 275 N
Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.
Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.
On dividing equation (1) by (2), we have
Now, putting this value in equation (2), we have
Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.