A golf club hits a 0.04551 kg golf ball off a golf tee. The club is in contact with the ball for 0.020 s, and the force applied
1 answer:
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Answer:
The farther star will appear 4 times fainter than the star that is near to the observer.
Explanation:
Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time
Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)
This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by
For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by
Hence we sense it as 4 times fainter than the nearer star.
Answer:
The angular magnification is
Explanation:
From the question we are told
The focal length is
The near point is
The angular magnification is mathematically represented as
Substituting values
Answer:
The work done will be 57.15 J
Explanation:
Given that,
Mass = 12 kg
Distance = 3 m
Force = 22 N
Angle = 30°
We need to calculate the work done
The work done is defined as,
Where, F = force
d = displacement
Put the value into the formula
Hence, The work done will be 57.15 J
Answer:
Remains same
Explanation:
= Time period of oscillation
= mass
= spring constant
Time period of oscillation is given as
we know that as we move from earth to moon, the value of spring constant "k" and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.
Time period depends on spring constant inversely and directly on the mass.
hence the time period remains the same.