During a car crash, energy is transferred from the vehicle to whatever it hits, be it another vehicle or a stationary object. ... The object that was struck will either absorb the energy thrust upon it or possibly transfer that energy back to the vehicle that struck it.
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Here, Initial momentum = mu = 6*2 = 12 Kg m/s
Final momentum = mv = 6*4 = 24 Kg m/s
In short, Your Answer would be Option C
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Answer:
A. Vx = 3.63 m/s
B. Vy = -45.73 m/s
C. |V| = 45.87 m/s
D. θ = -85.46°
Explanation:
Given that position, r, is given as:
r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk
Velocity is the derivative of position, r:
V = dr/dt = 3.63 - 11.46t^j
A. x component of velocity, Vx = 3.63 m/s
B. y component of velocity, Vy = -11.46t
t = 3.99 secs,
Vy = - 11.46 * 3.99 = -45.73 m/s
C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]
|V| = √(2091.2329 + 13.1769)
|V| = √(2104.4098)
|V| = 45.87 m/s
D. Angle of the velocity relative to the x axis, θ is given as:
tanθ = Vy/Vx
tanθ = -45.73/3.63
tanθ = -12.6
θ = -85.46°
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Answer:
Pressure = 5 x 10⁶ Pa
Explanation:
Given:
Height of building = 512 m
Find:
Pressure
Computation:
P2 = P1+dgh
P2 = 1 + (1000)(9.8)(512)
P2 = 51.2 atm
Pressure = 5 x 10⁶ Pa