Answer:
Part a)
Part b)
T = 4.68 s
Explanation:
Part a)
Shell is fired at speed of 40 m/s at angle of 35 degree
so here we have
since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero
so at the highest point the speed is given
Part b)
After completing the motion we know that the displacement of the object will be zero in Y direction
so we have
Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m
Assuming you mean temperature
Answer: The third law of thermodynamics
Answer:
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