Change in velocity of larger moose: (1/3)v - v = -(2/3)v
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2
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Answer:
Explanation:
a). Find the graph attached for the motion.
b). If a shopper walk 5.4 m westwards then 7.8 m eastwards,
Distance traveled by the shopper = Distance traveled in eastwards + Distance traveled westwards
= 5.4 + 7.8
= 13.2 m
c). Displacement of the shopper = Distance walked westwards - Distance traveled eastwards
= 5.4 - 7.8
= -2.4 m
Therefore, magnitude of the displacement of the shopper is = 2.4 m
And the direction of the displacement is eastwards.
M = 10.0 g, the mass of the iron sample
ΔT = 75 - 25.2 = 49.5°C, the decrease in temperature
c = 0.449 J/(g-°C), the specific heat of iron
The heat released is
Q = m*c*ΔT
= (10.0 g)*(0.449 J/(g-°C))*(49.5 C)
= 222.255 J
Answer: 222.3 J (nearest tenth)
Answer:
1.03 m/s
Explanation:
I'm too lazy to write the explanation down but my teacher graded this and it was right
