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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Pnet = Po + dgh
<span>Density of saltwater = 1030 kg/m^3. </span>
<span>Disregard the thickness. Assuming it's a circular window, then the area is pi(r^2). </span>
<span>d = 20 cm = 0.2 m </span>
<span>r = d/2 = 0.1 m </span>
<span>A = pi(r^2) </span>
<span>A = 3.14159265(.1^2) </span>
<span>A = 0.0314159265 m^2 </span>
<span>p = F/A </span>
<span>p = (1.1 x 10^6) / (0.0314159265) </span>
<span>p = 35,014,087.5 Pa </span>
<span>1 atm = 101,325 Pa </span>
<span>P = Po + dgh </span>
<span>h = (P - Po) / dg </span>
<span>h = (35,014,087.5 - 101,325) / (1030 x 9.81) </span>
<span>h = 3 455.23812 m </span>
<span>h = 3.5 km</span>
The initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
CALCULATE INITIAL VELOCITY:
The initial velocity of the car can be calculated by using one of the equation of motion as follows:
V = u + at
Where;
- V = final velocity (m/s)
- u = initial velocity (m/s)
- a = acceleration due to gravity (m/s²)
- t = time (s)
According to this question, a car accelerates at a constant rate of 3 m/s² for 5 seconds. If it reaches a velocity of 27 m/s, its initial velocity is calculated as follows:
u = v - at
u = 27 - 3(5)
u = 27 - 15
u = 12m/s.
Therefore, the initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
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Answer:
7200 N/m
Explanation:
Metric unit conversion
100g = 0.1 kg
5 cm = 0.05 m
50 cm = 0.5 m
As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground
The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m
Let g = 10 m/s2. The change in potential energy can be calculated as the following:
Therefore, as elastic energy is converted to potential energy and work of friction:
Answer:
Explanation:
Assuming no friction between the roller coaster car and the hill, and neglecting air resistance, the kinetic energy the roller coaster car would have at the bottom of the hill would be equal to its gravitational potential energy at the top of the hill, by conservation of energy.
