Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.
Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,
Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules
Just ignore the negative value for the final result because work is a scalar quantity.
Answer:
C) 40 N/m
Explanation:
If we ASSUME that the spring is un-stretched at the zero cm position
k = F/Δx = 10/0.25 = 40 N/m
Answer: William Thomson, better known as Lord Kelvin
Answer:
The acceleration is
Explanation:
From the question we are told that
The lift up speed is
The distance covered for the take off run is
Generally from kinematic equation we have that
Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest
So
=>
2.72 N
Explanation:
Step 1:
From the basic formula in electrostatics
F = E * q
where F = Force due to charges
E = Electric field strength
q = Charge
Step 2:
From the given question
q=
E =
F = N