Answer:
The velocity of the plane at take off is 160 m/s.
The distance travel by the plane in that time is 3200 meter.
Explanation:
Given:
Acceleration, a = 4 m/s²
Time, t = 40 s
u = 0 i .e initial velocity
To Find:
velocity , v = ?
distance , s =?
Solution:
we have first Kinematic equation
v = u + at
∴ v = 0 + 4×40
∴ v = 160 m/s
Now by Third Kinematic equation

∴ s = 0 + 0.5 × 4× 40²
∴ s = 3200 meter
Answer:
48kg
Explanation:
i could be wrong if i am srry
Answer:
0.243 m/s
Explanation:
From law of conservation of motion,
mu+m'u' = V(m+m')................. Equation 1
Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.
make V the subject of the equation
V = (mu+m'u')/(m+m')................. Equation 2
Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s
Substitute into equation 2
V = (260000×0.32+52500×(-0.14))/(260000+52500)
V = (83200-7350)/312500
V = 75850/312500
V = 0.243 m/s