We have that the instantaneous velocity of the
shuttlecock when it hits the ground is
From the question we are told
Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
shuttlecock when it hits the ground? Show your work below.
Generally the equation for acceleration is mathematically given as
Where
acceleration is still -9.81 m/s2,
Hence,
Therefore
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Answer:
An object on the moon would weigh the LEAST among these. So correct answer is B.
Explanation:
- Weight of an object on any place is given by:
W = Mass * Acceleration due to gravity(g)
- It means when masses of different objects those are in different places are same, the weight of those objects depends upon the 'g' of that particular place.
- As we know, acceleration due to gravity on surface of moon (g') is 6 times weaker than the acceleration on surface of earth (g), which is due to the large M/R^2 of the earth than the moon.
i.e. g' = g/6 so W' = W/6
- And in the space between the two, the object is weightless.
Light waves travel in straight lines when they are travelling in a uniform medium. This is because the waves are travelling at the same speed.
Answer: 0.817A
Explanation:
Assuming , that one coulomb per second of negative charge alone flow through a conductor and no positive charges flow. I.e Q=It
It means a current of one A flow in the opposite direction.
This is similar to one coulomb per second of positive charge flowing through and there is no negative charge,
In addition, the one coulomb per second of positive charge flows. This is flowing in the current direction of the previous one. Then, the total current is 2 A. Since 2 coulomb of positive charges flow through one due to real positive charge and another due to the negative charge flowing in opposite direction.The charges cannot cancel each other, because even before the current flow the conductor was neutral.
According to this, the current in the given problem is
[2.7 + 2.4] x 10 ^ 18 * 1.602 x 10^ [-19] C/s
= 0.817 A
Answer:
Pressure = 20 MPa
Explanation:
Given:
Force acting on the shoe is, 
Area of shoe on which the force acts is, 
Now, first we convert the area into its standard unit of m².
We have the conversion factor as:
1 cm² = 
Therefore, the area of shoe in square meters is given as:
Now, pressure on the shoe is given as:
Plug in 100 N for 'F', 
for 'A' and solve for 'P'. This gives,
Now, we know that,
Therefore, the pressure acting on the shoe is 20 MPa.
