An astronaut on the moon drops a feather from rest (v = 0). The acceleration due to gravity is 1.67 m/s 2 . If the feather beg
1 answer:
Given :
Initial velocity , u = 0 m/s .
Acceleration due to gravity on moon , .
Height , h = 2 m .
To Find :
Final position after falling for 1.5 seconds .
Solution :
We know , by equation of motion :
Here , .
So , equation will transform by :
Therefore , the height form moon's surface is 1.88 m .
Hence , this is the required solution .
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Answer:
Efriction = 768.23 [kJ]
Explanation:
In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.
We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.
The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.
where:
m = mass = 56 [kg]
h = elevation = 1400 [m]
v = velocity = 5.6 [m/s]
Answer:
The pressure on the ground is about 9779.5 Pascal.
The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side. See more details further below.
Explanation:
Start with the formula for pressure
(pressure P) = (Force F) / (Area A)
In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first
where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:
The density of water is 997 kg/m^3, so the mass of the barrel is:
and so the weight is
and so the pressure is
This answers the first part of the question.
The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.