Answer:
(a) 0.833 j
(b) 2.497 j
(c) 4.1625 j
(d) 4.995 watt
Explanation:
We have given force F = 5 N
Mass of the body m = 15 kg
So acceleration 
As the body starts from rest so initial velocity u = 0 m/sec
(a) From second equation of motion 
For t = 1 sec

We know that work done W =force × distance = 5×0.1666 =0.833 j
(b) For t = 2 sec

We know that work done W =force × distance = 5×0.666 =3.33 j
So work done in second second = 3.33-0.833 = 2.497 j
(c) For t = 3 sec

We know that work done W =force × distance = 5×1.4985 =7.4925 j
So work done in third second = 7.4925 - 2.497 -0.833 = 4.1625 j
(d) Velocity at the end of third second v = u+at
So v = 0+0.333×3 = 0.999 m /sec
We know that power P = force × velocity
So power = 5× 0.999 = 4.995 watt
Answer:
V = (5.8cm/s)i, (4.7cm/s)j
Explanation:
Given :
r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^
To obtain the average velocity (V)
V = (r2 - r1) / (t2 - t1)
To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above
r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j
r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j
r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j
r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j
r2 = (16.1cm)i + (9.4cm)j
V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0
V = 11.6i / 2 ; 9.4j / 2
V = (5.8cm/s)i, (4.7cm/s)j
Answer:
If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.
Λ = 3*10^8 / 9*10^8 = 1/3 m
no. of wavelengths = 60/(1/3) = 180