Answer: D
Reduced impact time will increase the impact force
Explanation: Collision occurs when two or more bodies collide and exert forces on each other within a short time.
If a body of mass M moving with a velocity V collide with another body, the kinetic energy of the body is equal to the work done by the body.
That is, K.E = 1/2mv^2 = F × s
Where workdone = Force × distance
Make F the subject of formula
Mv^2/2s = F
But V = distance s/time t
Substitute for V
Ms^2/2t^2s = F
Ms/2t^2 = F
From the equation above, we can deduce that F is inversely proportional to the square of time.
Therefore, the reduced impact time will increase the impact force
The faster car behind is catching up/closing the gap/gaining on
the slow truck in front at the rate of (90 - 50) = 40 km/hr.
At that rate, it takes (100 m) / (40,000 m/hr) = 1/400 of an hour
to reach the truck.
(1/400 hour) x (3,600 seconds/hour) = 3600/400 = <em>9 seconds</em>, exactly
The amplitude did not change when the recurrence was expanded on the grounds that the long headstrong time of the heart forestalls adjustment. It is the most extreme removal or separation moved by a point on a vibrating body or wave measured from its balance position. It is equivalent to the one-a large portion of the length of the vibration way.
Answer:
Your answer is: K.E = 8.3 J
Explanation:
If the height (h) = 169.2 meters (m) and the mass (m) is 0.005 kilograms (kg) the total energy will be kinetic energy which is equal to the potential energy.
K.E = P.E and also P.E equals to mgh
Then you substitute all the parameters into the formula ↓
P.E = 0.005 × 9.81 × 169.2
P.E = 8.2908 J
So your answer is 8.2908 but if you round it is K.E = 8.3