Answer:
The change in momentum of the car is 16380.8 kg.m/s
Explanation:
Given;
mass of the car, m = 1600 kg
time of motion, t = 4.20s
force of friction on the car, F = 3900 N
final velocity of the car after the brakes were applied, v = 0
The initial velocity of the car during the motion is calculated as;
The change in momentum of the car is calculated as;
ΔP = mu
ΔP = 1600 x 10.238
ΔP = 16380.8 kg.m/s
Answer:
I dont have any context, but my best guess is it will be all of the above.
<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
Answer:
D. 12.4 m
Explanation:
Given that,
The initial velocity of the ball, u = 18 m/s
The angle at which the ball is projected, θ = 60°
The maximum height of the ball is given by the formula
h = u² sin²θ/2g m
Where,
g - acceleration due to gravity. (9.8 m/s)
Substituting the values in the above equation
h = 18² · sin²60 / 2 x 9.8
= 18² x 0.75 / 2 x 9.8
= 12.4 m
Hence, the maximum height of the ball attained, h = 12.4 m
