Answer:
I = 27kg.mi/h
Explanation:
In order to calculate the impulse of the ball, you use the following formula:
(1)
m: mass of the ball = 0.3kg
v: speed of the ball after the bat hit it = 60mi/h
vo: speed of the ball before the bat hit it = 30mi/h
You replace the values of all parameters in the equation (1):
where the minus sign of the initial velocity means that the motion of the ball is opposite to the final direction of such a motion.
The imulpse of the ball is 27 kg.miles/hour
Answer:
T = 19.75 N
Explanation:
given,
mass of ball = 0.25 Kg
radius = 0.5 m
frequency = 2 s⁻¹
tension in the string = ?
angular velocity
ω = 2 π f
ω = 2 π x 2
ω = 12.57 rad/s
tension on the string is equal to the centripetal force
T = m ω² r
T = 0.25 x 12.57² x 0.5
T = 19.75 N
Tension in the string is equal to T = 19.75 N
Answer:
Transferred material is in the same relative position on the disk as on the original sample
Explanation:
The usefulness of blotting techniques in molecular biology is that transferred material is in the same relative position on the disk as on the original sample
Answer:
k = 49 N/m
Explanation:
Given that,
Mass, m = 250 g = 0.25 kg
When the mass is attached to the end of the spring, it elongates 5 cm or 0.05 m. We need to find the spring constant. Let it is k.
The force due to mass is balanced by its weight as follows :
mg=kx
So, the spring constant of the spring is 49 N/m.
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