The second stone hits the ground exactly one second after the first.
The distance traveled by each stone down the cliff is calculated using second kinematic equation;
where;
- <em>t is the time of motion </em>
- <em />
<em> is the initial vertical velocity of the stone = 0</em>
The time taken by the first stone to hit the ground is calculated as;
When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as
Thus, we can conclude that the second stone hits the ground exactly one second after the first.
"<em>Your question is not complete, it seems be missing the following information;"</em>
A. The second stone hits the ground exactly one second after the first.
B. The second stone hits the ground less than one second after the first
C. The second stone hits the ground more than one second after the first.
D. The second stone hits the ground at the same time as the first.
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Answer:
0.000025s
Explanation:
Period it’s. : T(s)= 1/f(Hz)=1/40000Hz=0.000025s
Answer:
-39.2m/s
Explanation:
Using the equation of motion;
v = u + at
Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g
v = u - gt
Since g = 9.8m/s²
t = 4.0s
u = 0m/s
v = 0 + (-9.8)(4)
v = 0 + (-9.8)(4)
v = -39.2m/s
Hence the speed of the ball before release is -39.2m/s
Answer:
Explanation:
The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.
Let Vector 
is the tidal current velocity as shown in the diagram.
In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, 
must be in the north direction.
Let 
is the speed of the kayaker having angle \theta measured north of east as shown in the figure.
For the resultant velocity in the north direction, the tail of the vector 
and head of the vector must lie on the north-south line.
Now, for this condition, from the triangle OAB
Hence, the kayaker must paddle in the direction of in the north of east direction.
Answer:
4.45×10¯¹¹ N
Explanation:
From the question given above, the following data were obtained:
Mass of ball (M₁) = 4 Kg
Mass of bowling pin (M₂) = 1.5 Kg
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Distance apart (r) = 3 m
Force of attraction (F) =?
The force of attraction between the ball and the bowling pin can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 4 × 1.5 / 3²
F = 4.002×10¯¹⁰ / 9
F = 4.45×10¯¹¹ N
Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N
