Answer:
6s
Explanation:
Assume it is dropped from rest and the gravitational acceleration is 10
By the equation of motion under constant acceleration:

180 = (0)t+10(t^2)/2
t = 6 or -6 (rejected)
t = 6 s
Answer:
b) 5 J
Explanation:
Work is the energy transferred by an object when acted by a force along a displacement. Work is the product of force and displacement. The SI unit of work is the joules (J)
To calculate the work done by the force, we have to first get the displacement (D) of the object. Hence:
Displacement (D) = Q(3, 8) - P(1, 3) = (3 - 1, 8 - 3) = (2, 5) = 2i + 5j
The work done is the dot product of the force and the displacement. Force = 5i - j. Hence:
Work done = (5i - j)(2i + 5j) = 10 - 5 = 5 J
Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds
Answer:
it's false coz
Speed is a scalar quantity.
Answer:
no material that were once part of living things