Answer:
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Ec₁=mv₁²/2=450*26²/2=225*676=152100 J
Ec₂=mv₂²/2=450*30²/2=225*900=202500 J
ΔEc=Ec₂-Ec₁=202500-152100=50.4 kJ
Answer:
3.73m
Explanation:
What we are asked to find is the range covered by the fish. This is given by the following equation (1). Range can simply be defined as the horizontal distance covered by a body whose motion freely under gravity is in two dimensions. The motion of the fish is in two dimensions, the vertical dimension and the horizontal dimension.
where u = 6.5m/s is the initial velocity, g is acceleration due to gravity which is taken as and .
Substituting these values into equation (1), we obtain the following;
Answer:
Acceleration due to gravity will be .
Explanation:
We can use the gravitational force equation:
The F is equal to the weight of the astronaut, so we will have:
- M(e) is the mass of the earth
- R is the radius of the earth
- G is the gravitational constant
But the distance between the astronaut and the center of the earth is 2R, then we have:
Therefore the acceleration due to gravity will be .
I hope it helps you!