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:
A)
B)
Explanation:
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mass of car,
A)
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B)
given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:
Answer: d. 8.25 m/s
Explanation:
We are given that Current= 5 m/s in j direction
Velocity= 8 m/s i + 3 m/s j
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First we find Jada's velocity with respect to water
v= (8 i + 3 j) - (5 j)
v= 8i - 2 j
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|v|=
|v|= = 8.246 m/s
which comes out to be around = 8.25 m/s
So option d is correct.
If you draw the problem, it would look like that shown in the attached picture. The total length the ship will now travel can be solved using the Pythagorean theorem. The solution is as follows:
d = √(120)²+(100)²
d = 156.2 km
So, the ship will have to travel 156.2 km to the northwest direction.
d = √(120)²+(100)²
d = 156.2 km
So, the ship will have to travel 156.2 km to the northwest direction.