Answer:
Force of 37.8 × 10^(6) N attracts the two charges
Explanation:
The force between two charges is given by
F = k*q1*q2/r²
Where q1 and q2 are 0.06 C and 0.07 C.
r is the distance between q1 and q2 which is equal to 3 m
k is a constant = 9 × 10^(9) N.m²/C²
F = (9 × 10^(9) × 0.06 × 0.07)/3²
F = 37.8 × 10^(6) N
Answer:
t = 300.3 seconds
Explanation:
Given that,
The mass of a freight train, 
Force applied on the tracks, 
Initial speed, u = 0
Final speed, v = 80 km/h = 22.3 m/s
We need to find the time taken by it to increase the speed of the train from rest.
The force acting on it is given by :
F = ma
or

So, the required time is 300.3 seconds.
Answer : The angle between the string and the horizontal is 30 degrees
Explanation: Imagine this a a triangle where the length of the string (200m) is the hypotenuse and the height of the kite is the opposite side (100m) .
Let the angle between the string and the horizontal be theta.
Now sin (Theta) = opposite side/hypotenuse
= 100/200 = 1/2
Therefore Theta = Sin ⁻¹ ( 1/2 )
Theta = 30 degrees
Ciara is winging....etc
The answer is : 0.60 N, toward the center of the circle
A satellite....etc
The Answer is : 7400 m/s
What is the .....etc
The Answer is : 2.60 m/s
Answer:
The radius of the loop is 20.66 km
Explanation:
let the radius of the loop be r
mass of airplane is m
At the top, the pilot experiences two radial forces, which are
1) Gravitational force is mg
2) Centrifugal forces mv²/r out of the center
When the pilot experiences no weight,
then, mg = mv²/r
r = v² / g
= 450² / 9.8
= 20.66 x 10³3
= 20.66 km