A 2.0m long pendulum is released from rest when the support string is at an angle of 25 degrees with the vertical. What is the s
1 answer:
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Answer:
a)Initial speed of the projectile = 196.2 m/s
b)Maximum altitude = 490.5 m
c) Range of projectile = 3398.28 m
d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
Explanation:
Time of flight of a projectile is given by the expression,
Here θ = 30° and t = 20 s
a)
Initial speed of the projectile = 196.2 m/s
b) Maximum altitude is given by
Maximum altitude = 490.5 m
c) Range of projectile is given by
Range of projectile = 3398.28 m
d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s
Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s
We have equation of motion s = ut + 0.5 at²
Horizontal motion
u = 169.91 m/s
a = 0 m/s²
t = 15 s
Substituting
s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m
Vertical motion
u = 98.1 m/s
a = -9.81 m/s²
t = 15 s
Substituting
s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m
Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
The acceleration of the jet is
Explanation:
Since the motion of the jet is a uniformly accelerated motion, we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
For the jet in this problem, we have
u = 0
v = 75 m/s
s = 100 m
Solving for a, we find the acceleration:
Learn more about acceleration:
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Answer:
the ratio is
Explanation:
Given
The RMS velocity of molecules in a gas is given by
where T=temperature
For T = 387K
For T = 774
dividing eqn 1 and eqn 2
Thus,the ratio is