We have to calculate the impulse of a hockey puck.
Imp = m * ( v 1 - v 2 ) = m * Δ v
v 1 = - 10 i m/s,
v 2 = ( 20 * cos 40° ) i + ( 20 * sin 40° ) j =
= ( 20 * 0.766 ) i + ( 20 * 0.64278 ) j = ( 15.32 i + 12.855 j ) m/s
Δ v = ( 15.32 i + 12.855 j ) - ( - 10 i ) =
= 15.32 i + 12.855 j + 10 i = 25.32 i + 12.855 j
| Δv | = √ ( 25.32² + 12.855²) = √806.35 = 28.4 m/s
Imp = 0.2 kg * 28.4 m/s = 5.68 N-s
Answer: D ) 5.68 N-s.
Answer:
Explanation:
Work done in lifting the weight once = mgh
= 20 x 9.8 x (1.9+1.7)
= 705.6 J
= 705.6 / 4.2 calorie
= 168 cals
Total energy to be spent = 600 x 10³ cals
No of times weight is required to be lifted
= 600 x 10³ / 168
= 3.57 x 10³ times
Total time to be taken = 2 x 3.57 x 10³
= 7.14 x 10³ s
=119 minutes .
Answer:
25N
Explanation:
Assuming the lab is on earth:
w = mg = 2.5 (9.81) = 25N
Answer:
2,500 watts
Explanation:
I got this answer right on a test. I hope it works for you to.
Answer:
450 joules ; 450 joules ; 45.9 m
Explanation:
Given that :
Initial Velocity, u = 30m/s
Mass, m = 1 kg
Kinetic Energy of ball (KE) = 0.5mu²
K. E = 0.5 * 1 * 30^2
K.E = 0.5 * 900
K.E = 450 Joules
B.) Potential Energy (P. E)
P. E = mgh
At the highest point, all kinetic energy has would have become potential energy, hence
K. E = P. E = 450 Joules
C) Height of the ball :
From ; P. E = mgh
Where ; g = acceleration due to gravity = 9.8m/s² ; h = height
450 = 1 * 9.8 * h
450 = 9.8h
h = 450 / 9.8
h = 45.918
h = 45.9 m
