Answer:
u = 23.68 m/s
Explanation:
given,
mass of the car, m = 1900 Kg
Force exerted on the wall, F = 150,000 N
time of contact, t = 0.3 s
final speed of the car = 0 m/s
initial speed of the car = ?
we know,
Impulse is equal to change in momentum
J = m v- mu
J = 1900 x 0 - 1900 x (-u)
J = 1900 u
impulse is also equal to force into time
J = F x t
equating both equation of impulse
1900 u = F x t
1900 u = 150000 x 0.3
1900 u = 45000
u = 23.68 m/s
Speed of the car before collision is equal to 23.68 m/s
Vo = 89 m/s
angle: 40°
=> Vox = Vo * cos 40° = 89 * cos 40°
=> Voy = Vo. sin 40° = 89 * sin 40°
x-movement: uniform => x =Vox * t = 89*cos(40)*t
x = 300 m => t = 300m / [89m/s*cos(40) = 4.4 s
y-movement: uniformly accelerated => y = Voy * t - g*t^2 /2
y = 89m/s * sin(40) * (4.4s) - 9.m/s^2 * (4.4)^2 / 2 = 156.9 m = height the ball hits the wall.
Answer:
v(7) = 52.915 m/s
Explanation:
First, find the value for acceleration.
F = ma
100 = .5 * a
a = 200 m/s²
Next find the velocity at x = 7 using kinematic equations.
v² = v₀² + 2a(Δx)
v² = (0)² + 2(200)(7)
v =
v = 52.915 m/s
I believe its air particles
Answer:
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