Answer:
The horizontal force that will be needed to move the pickup along the same road at the same speed is 230 N
Explanation:
given information:
horizontal force, F = 200 N
speed, v = 2.4 m/s
total weight increase 42%
coefficient of friction decrease by 19%
now, lets take a look of horizontal system
F - F(friction) = 0
F = F(friction)
F = μ N
F = μ m g
μ m g = 200 N
now find the force to move the pickup along the same road at the same speed(F(move))
total weight increase 42%, m g = 1+0.42 = 1.42
he coefficient of rolling friction decreases by 19%, μ = 1 - 0.19 = 0.81
F (move) = (0.81μ) (1.42 m g)
= (0.81) (1.42) (μ m g)
= (0.81) (1.42) (200)
= 230 N
We can't tell without more information. We know it will be higher than 40 and lower than 75, but we don't know exactly where it will settle. In order to work that out, we would need to know the volumes of the water and the cube, and WHAT metal the cube is made of.
Answer:
the equilibrium temperature Te = 19.9°C
Explanation:
Given;
specific heat for copper Cc is 390 J/kg⋅°C
for aluminun Ca is 900 J/kg⋅°C,
for water Cw is 4186 J/kg⋅°C
Mass of copper Mc= 265 g = 0.265kg
Temperature of copper Tc = 235°C
Mass of aluminium Ma = 135g = 0.135 kg
Temperature of aluminium Ta = 14.0°C
Mass of water Mw= 865 g = 0.865kg
Temperature of water Tw = 14.0°C
The equilibrium temperature can be derived by;
Te = (MaCaTa + McCcTc + MwCwTw)/(MaCa + McCc+ MwTw)
Substituting the values;
Te = ( 0.135×900×14 + 0.265×390×235 + 0.865×4186×14)/(0.135×900 + 0.265×390 + 0.865×4186)
Te = 19.939°C
Te = 19.9°C
Answer:3W
If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much work should it take to move three q point charges from infinity to a distance d apart from each other?
A) 2W
B) 3W
C) 4W
D) 6W
Explanation: calculating work done,W, in moving two positive q point charges from infinity to a valued distance d from each other is
W = k(+q)(+q)/ d
k is couloumb's constant
work done in moving 3 equal positive charges from infinity to a finite distance is given by
W₂=W₄=W₆=k(+q)(+q)/ d
Total work done, W' =k(+q)(+q)/ d + k(+q)(+q)/ d + k(+q)(+q)/ d
= W + W + W = 3W
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