Answer:
Explanation:
Left block is on surface with higher inclination so it will go down . If T be tension
For motion of block A ,
net force = mgsin60 - (T + mg cos 60 x μ ) , μ is coefficient of friction .
ma = mgsin60 - T - mg cos 60 x .1
10a = 277.13 - T - 16
= 261.13 - T
T = 261.13 - 10a
For motion of block B
T - mg sin30 - mgcos30 x μ = ma
T- 160 - 27.71 = 10 a
261.13 - 10a - 160 - 27.71 = 10a
73.42 = 20a
a = 3.67 ft / s²
common acceleration = 3.67 ft / s²
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Answer:
P=4801.5
Explanation:
Given :
work done = W = 100,832 J
time = 21.0 sec
Find:
P = ?
Formula:
P = W/t
Solution:
P = W/t
P = 100,832/21.0
= 4801.52 J/s or Watts
Answer:
A. the magnitude of the velocity at which the two players move together immediately after the collision is 7.9m/s
B. The direction of this velocity is due north as the linebacker since he has obviously has more momentum
Explanation:
This problem bothers on the inelastic collision
Given data
Mass of linebacker m1= 110kg
Mass of halfbacker m2= 85kg
Velocity of linebacker v1= 8.8m/s
Velocity of halfbacker v2= 7.2m/s
Applying the principle of conservation of momentum for inelastic collision we have
m1v1 +m2v2= (m1+m2)v
Where v is the common velocity after impact
Substituting our data into the expression we have
110*8.5+85*7.2= (110+85)v
935+612=195v
1547=195v
v=1547/195
v=7.9m/s
Momentum of linebacker after impact = 110*7.9= 869Ns
Momentum of halfbacker after impact = 85*7.9= 671.5Ns
the direction after impact is due north since the linebacker has greater momentum
