The velocity of the target and arrow after collision is 6.67m/s
<u>Explanation:</u>
Given:
Mass of arrow, mₐ = 415g
Speed of arrow, vₐ = 68.5m/s
Mass of the target, mₓ = 3.3kg = 3300g
speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction
Velocity of the target and arrow after collision, vₙ = ?
Applying the conservation of momentum,
mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ
415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ
28427.5 - 3630 = 3715 X vₙ
24797.5 = 3715 X vₙ
vₙ = 6.67m/s
Therefore, the velocity of the target and arrow after collision is 6.67m/s
Answer:
This is because it has ns1 electron configuration like the alkali metals.
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A )
t 1 = 2 h, t 2 = 6 h
Δ t = t 2 - t 1 = 6 - 2 = 4 h
54 = 50 + a Δ t
54 = 50 + 4 a
4 a = 54 - 4
4 a = 4
a = 4 : 4
a = 1 km/h²
v o = 48 km/h
An equation that can be used to describe the velocity of the car at the different times is:
v = 48 + t
B ) The graph is in the attachment.
Answer:
Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.
Explanation:
Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.