Answer:
yes
Explanation:
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Answer:
The weight of the block on the moon is 15 kg.
Explanation:
It is given that,
The acceleration of the block, a = 7.5 m/s²
Force applied to the box, F = 70 N
The mass of the block will be, 
m = 9.34 kg
The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :
W = 15.13 N
or
W = 15 N
So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.
Answer:
1626.4 N
Explanation:
Given that a 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?
The parameters to be considered are:
Distance S = 3m
Time t = 0.55s
Since the man started from rest, initial velocity u = 0
Using second equation of motion
S = Ut + 1/2at^2
3 = 1/2 × a × 0.55^2
3 = 1/2 × a × 0.3025
a = 3/ 0.15125
a = 19.83 m/s^2
Force = mass × acceleration
Force = 82 × 19.83
Force = 1626.4 N
Therefore, the force that water exerted on him is 1626.4 N
Answer:
I am going to guess it shows that the balloon is going downwards because the speed of rise is in the negatives for the last 2.
Answer:
0.37 m/s to the left
Explanation:
Momentum is conserved. Initial momentum = final momentum.
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
Initially, both the fisherman/boat and the package are at rest.
0 = m₁ v₁ + m₂ v₂
Plugging in values and solving:
0 = (82 kg + 112 kg) v + (15 kg) (4.8 m/s)
v = -0.37 m/s
The boat's velocity is 0.37 m/s to the left.
