Answer:
The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N
Explanation:
The details of the given masses having gravitational attractive force between them are;
m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m
The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;
Where;
F = The gravitational force between m₁ and m₂
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r₂ = 0.1 m + 0.15 m = 0.25 m
Therefore, we have;
The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N
The velocity of B after elastic collision is 3.45m/s
This type of collision is an elastic collision and we can use a formula to solve this problem.
<h3>Elastic Collision</h3>
The data given are;
- m1 = 281kg
- u1 = 2.82m/s
- m2 = 209kg
- u2 = -1.72m/s
- v1 = ?
Let's substitute the values into the equation.
From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.
Learn more about elastic collision here;
brainly.com/question/7694106
Answer:
D= 1999.2 m
Explanation:
Given that
Average velocity ,v= 0.98 m/s
time ,t= 34 min
We know that
1 min = 60 s
That is why
t= 34 x 60 =2040 s
We know that
Displacement = Average velocity x time
D= v t
Now by putting the values in the above equation
D= 0.98 x 2040 m
D= 1999.2 m (eastward)
The direction of the displacement will be towards eastward.
That is why the displacement will be 1999.2 m or we can say that 1.9992 km.
