initial speed of the ball by which it is kicked is 8 m/s
now the angle at which it is kicked is 20 degree
here we will have
now we will have
so if the ball again land on the ground at same level then we have
so total time will be 0.56 s
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We/Wm = ge/gm = 120N/1.2N
or
gm = ge/100 = 0.1 m/s^2
density = mass/volume = 3M/(4pir^3)
Re-arranging this equation, we get
M/r^2 = (4/3)×pi×(density)×r
From Newton's universal law of gravitation, the acceleration due to gravity on the moon gm is
gm = G(M/r^2) = G×(4/3)×pi×(density)×r
Solving for density, we get the expression
density = 3gm/(4×pi×G×r)
= 3(0.1)/(4×3.14×6.67×10^-11×2.74×10^6)
= 130.6 kg/m^3
Answer:
the ball's velocity was approximately 0.66 m/s
Explanation:
Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.
Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.
Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:
0.7 = (1/2) g t^2
solve for t:
t^2 = 1.4 / g
t = 0.3779 sec
which we can round to about 0.38 seconds
No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:
horizontal distance covered = vi * t
0.25 = vi * (0.38)
solve for vi:
vi = 0.25/0.38 m/s
vi = 0.65798 m/s
Then the ball's velocity was approximately 0.66 m/s
Answer:
A - Crest, B - amplitude, C - wavelength, D - trough
Explanation:
