Answer:
a) a = -g = 9.8 m/s²
, b) a = 0 m/s² and c) t1 = 0.0213 s
Explanation:
a) At the moment the marble is released its velocity is zero, so it has no resistance force, the only force acting is its weight, so the acceleration is the acceleration of gravity
a = -g = 9.8 m / s²
b) When the marble goes its terminal velocity all forces have been equalized, therefore, the sum of them is zero and consequently if acceleration is also zero
a = 0 m / s²
c) We have to assume a specific type of resistive force, for liquid in general the resistive force is proportional to the speed of the body.
The expression of this situation is
v = mg / b (1 -
)
For a very long time the exponential is zero, so the terminal velocity is
= mg / b
b = mg /
b = 5 10-3 9.8 / 0.3
b = 0.163
We already have all the data to calculate the time for v = ½
½ = (1 -)
½ = 1- e (- 0.163 t1 / 5 10-3)
e (-32.6 t1) = 1-0.5 (by ln())
-32.6 t1 = ln 0.5
t1 = -1 / 32.6 (-0.693)
t1 = 0.0213 s
Explanation:
It is given that,
Speed of a wave, v = 251 m/s
Wavelength of the wave, λ = 5.1 cm = 0.051 m
(1) The frequency of the wave is given by :
(2) Angular frequency of the wave is given by :
(3) The period of oscillation is given by T as :
T = 0.000203 seconds
or
T = 0.203 milliseconds
Hence, this is the required solution.
Answer:
The magnitude of the average force exerted by the club on the ball during contact = mv/t
Explanation:
Impulse exerted on the ball = Momentum of the ball = mass * velocity = m*v
As we know,
m*v = Integration of F.dt with limits 0 to T
Ft = mv
F = mv/t
The magnitude of the average force exerted by the club on the ball during contact = mv/t
I think the right answer is the first one. If she stops moving her Position does not change any more-and the Graph Shows that after 6 seconds she stays at the Position of 5 m. If she Went Back to the start point the Graph would have Developed Back to 0m(decreased).
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is
and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:
