Answer:
23: Acceleration 24:1m/s^2
Explanation:
A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. a) Calculate the m
1 answer:
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Answer:
T = 2.4 + 2.4 = 4.8 [s]
Explanation:
In order to solve this problem, we must use the following kinematics equation and calculate the acceleration value.
Vo = inital velocity = 0
x - xo = 15 [m]
t = time = 2.4 [s]
15 = 0.5*a*(2.4)^2
a = 5.208 [m/s^2]
We can use the same equation to find the time.
30 = 15 + 0.5*(5.208)*t^2
t = 2.4 [s]
T = 2.4 + 2.4 = 4.8 [s]
Answer:
Explanation:
Yes , the horizontal distance travelled by the ball will depend upon the height of release .
When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play . But the horizontal range covered by the body thrown
depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .
Horizontal distance covered = t x horizontal velocity
If V be the velocity of throw and Vx be its horizontal component
Horizontal distance covered = t x Vx
Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .
If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity
from the relation
s = ut + 1/2 at²
h = - Vy t + 1/2 at²
As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only horizontal velocity initially , Vy = 0
h = 1/2 gt²
Horizontal distance covered = t x Vx
=
From this expression also
Horizontal distance covered is proportional to .