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schepotkina [342]
3 years ago
14

Find the speed vfinal of the joined cars after the collision. mastering physics

Physics
1 answer:
Tanya [424]3 years ago
6 0
<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>
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An object is falling from a height of 7.5 meters. At what height will it’s velocity be 7 meters/second?
Nady [450]
One of the equations of gravity is this:
{v}^{2} = {u}^{2} + 2gh
Where v = final velocity which is 7m/s
u = initial velocity which is 0 for objects falling from a height
g = acceleration due to gravity and it is approximately 10m/s^2. It's a constant so pretty much remember this number. It's positive since the work being done is caused by gravity (in other words, it's falling down). It can also be negative if the work being down is against gravity (in other words, it's going up)
h = height of object

Substitute for the values and you should have something like this
{7}^{2} = {0}^{2} + 2 \times 10 \times h
49 = 0 + 20h
h = \frac{49}{20}
h = 2.5m
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3 years ago
A 1060-kg car moving west at 16 m/s collides with and locks onto a 1830-kg stationary car. determine the velocity of the cars ju
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M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
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The velocity of the cars after collision will be 5.689 m/s
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