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.
I believe the answer is (4) The reason that is, is because if the exponents are the same like 10^2 and 10^3, you can add them. Then you would get 10^5. You can go ahead though and multiply 5.0 and 1.0. Now remember that with decimals you don't need the zeros behind the decimal point. So that simplifies it with just 5 x 1. Leaving you with 5.0 x 10^5.
When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:
F = mg
Where,
m = mass
g = Gravitational acceleration
F = 5*9.8
F = 49N
Therefore the correct answer is E.
Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
= 13.625 V
<span>A decrease in the overall volume of gases namely hydrogen would prevent nuclear fusion in a nebula.</span>
