A frictionless inclined plane is 8.0 m long and rests on a wall that is 2.0 m high. How much force is needed to push a block of
ice weighing 300.0 N up the plane
1 answer:
Given Information:
Inclined plane length = 8 m
Inclined plane height = 2 m
Weight of ice block = 300 N
Required Information:
Force required to push ice block = F = ?
Answer:
Force required to push ice block = 75 N
Explanation:
The force required to push this block of ice on a inclined plane is given by
F = Wsinθ
Where W is the weight of the ice block and θ is the angle as shown in the attached image.
Recall from trigonometry ratios,
sinθ = opposite/hypotenuse
Where opposite is height of the inclined plane and hypotenuse in the length of the inclined plane.
sinθ = 2/8
θ = sin⁻¹(2/8)
θ = 14.48°
F = 300*sin(14.48)
F = 75 N
Therefore, a force of 75 N is required to push this ice block on the given inclined plane.
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Answer:
The initial velocity of the water from the tank is 5.42 m/s
Explanation:
By applying Bernoulli equation between point 1 and 2
At the point 1
P₁=0 ( Gauge pressure)
V₁= 0 m/s
Z₁=3 m
At point 2
P₂=0 ( Gauge pressure)
Z₂= 0 m/s
Now by putting the values
V₂= 5.42 m/s
The initial velocity of the water from the tank is 5.42 m/s
The heat released by the water when it cools down by a temperature difference 
is
where
m=432 g is the mass of the water
is the specific heat capacity of water
is the decrease of temperature of the water
Plugging the numbers into the equation, we find
and this is the amount of heat released by the water.
where
m=432 g is the mass of the water
Plugging the numbers into the equation, we find
and this is the amount of heat released by the water.