Answer:
the magnitude of gravitational force is 6 x 10⁻⁸ N.
Explanation:
Given;
mass of the two people, m₁ and m₂ = 90 kg
distance between them, r = 3.0 m
The magnitude of gravitational force exerted by one person on another is calculated as;

where;
G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²

Therefore, the magnitude of gravitational force is 6 x 10⁻⁸ N.
Hi there!
Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.
We also know there was work done on the ball by air resistance that decreased the ball's total energy.
Let's do a summation using the equations:

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

And:

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.
Therefore:

Solving for the work done by air resistance:


If the trend changed toward traditional (pre-World War II) families, the women’s rights are employment in manufacturing sector.
<h3>What is World war?</h3>
The war between two countries to take over each other's kingdom using weapons to kill each other.
Before the world war, the army needs armor, weapons, guns and tanks. Their manufacturing is only possible with many workers to work for long hours. If the men are not enough, then women are given opportunities to work with them.
Thus, the women’s rights are employment in manufacturing sector when trend changed traditional families.
Learn more about world war.
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Answer:
a) θ = 2500 radians
b) α = 200 rad/s²
Explanation:
Using equations of motion,
θ = (w - w₀)t/2
θ = angle turned through = ?
w = final angular velocity = 1420 rad/s
w₀ = initial angular velocity = 420
t = time taken = 5s
θ = (1420 - 420) × 5/2 = 2500 rads
Again,
w = w₀ + αt
α = angular accelaration = ?
1420 = 420 + 5α
α = 1000/5 = 200 rad/s²
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.