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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Explanation:
2.
4.
In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of
You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).
In both cases, the net force applied to the object must equal zero.
Answer: d = 4750n/3.1+95n
Explanation:
Using the principle of moment to solve the question.
Sum of clockwise moments = sum of anti clockwise moments
Since there are n identical coins with mass 3.1g placed at point 0cm, 1 coin will have mass of 3.1/n grams
Taking moment about the pivot,
Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.
Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).
To get 'd'
We have 3.1/n × d = 95 × (50-d)
3.1d/n = 4750-95d
3.1d = 4750n-95dn
3.1d+95dn=4750n
d(3.1+95n) = 4750n
d = 4750n/3.1+95n
Answer:
see below for the truth table
Explanation:
<u>Truth Table</u>
As we will see from the description of operation, any input low causes the output to be high. This is the logic of a NAND gate. The truth table is attached.
<u>Working Principle</u>
Pulling any of A, B, or C low will saturate transistor Q1, depriving Q2 of any base current, cutting it off. Then Q5 is also deprived of base current and is cut off. Meanwhile, the current through R2 supplies base current to Q4, allowing it to pull the output high.
If all of A, B, and C are high (or open), base current is supplied to Q2 through the base-collector junction of Q1. Then Q2 saturates, supplying base current to Q3. Diode D1 ensures that the voltage across Q2 will be insufficient to supply any base current to Q4, so it stays cut off.
