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Lina20 [59]
3 years ago
9

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

st after the collision.
Physics
1 answer:
iren [92.7K]3 years ago
5 0
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.
therefore; 
(1060× 16) + (1830 ×0) = (1060 +1830) V
 16960 = 2890 V
      V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s
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Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

<u>h' = 55.3 m</u>

4 0
2 years ago
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Answer:

id say D

Explanation:

because its the only answer that has to do with colors and visable light

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