25 newtons.
Since the force of gravity varies inversely with the square of the distance. At distance R, the force is 100 Newtons. So we now have a distance 2R. So
(1) 100 = M/R^2
(2) X = M/(2R)^2
Solve equation (1) for M
100 = M/R^2
100R^2 = M
Substitute the value for M into equation (2) and solve
X = M/(2R)^2
X = (100R^2)/(2R)^2
X = (100R^2)/(4R^2)
X = (100)/(4)
X = 25
So the object at the higher altitude will weight 25 newtons.
<h2>
Answer:</h2><h3><u>QUESTION①)</u></h3>
<em>✔ First step : calculate the kinetic energy that this car requires to reach 95 km/h</em>
95/ 3,6 ≈ 26,4 m/s
<em>Ec = ½ m x V² </em>
With Ec in J; m in kg; and V in m/s
- Ec = ½ 1750 x 26,4²
- Ec ≈ 610 000 J
<em>✔ Knowing that the car has a p power of 215,000 W, so :
</em>
T = E/P
- T = 610 000/215 000
- T ≈ 2.8 s
<h3>
The car takes 2.8 s to reach 95 km/h </h3>
<h3><u>QUESTION②)</u></h3>
N = 2,8/6,5 x 100 = 43.07
<h3>The car efficiency is 43 % </h3>
The average velocity of an object is given by:
Average velocity = total displacement / total time
First, we calculate the time taken to reach the maximum height of the ball. This occurs when the final velocity is 0.
Vf = Vi + at
0 = 19.6 - 9.81 * t
t = 2 seconds
The total trip will be of 4 seconds (2 seconds up, 2 seconds down)
The total displacement is given by:
s = ut + 1/2 * at²
s = 19.6 * 2 - 0.5 * 9.81 * 2²
s = 19.6 meters
This is the distance maximum height, so the total height is
19.6 * 2 = 39.2 meters
Average velocity = 39.2 / 4
Average velocity = 9.8 m/s
Answer:
Explanation:
In a velocity/time (aka acceleration) graph, the slope of a line indicates the value of the acceleration in m/s/s. Acceleration is the change in velocity over the change in time. From 0 - 2 seconds, there is no change in velocity, so the acceleration during this interval is 0 (which is the same as the slope of the line). From 2 - 4 seconds, the slope of the line is -2, so the acceleration during the time interval from 2 to 4 seconds is -2 (negative because David is slowing down but is still going the same direction: to the right).
