Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
Answer:
1.67 A
Explanation:
Given that,
→ Power (P) = 400 W
→ Potential difference (V) = 240 V
→ Current (I) = ?
The amount of current drawn will be,
→ P = V × I
→ I = P/V
→ I = 400/240
→ I = 1.66666666667
→ [ I = 1.67 A ]
Hence, the current drawn 1.67 A.
Answer: Could you please add the answer choices.
Explanation:
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Answer:
Explanation:
Position of charge 3q is x = 0
position of charge -2q is x = a
so here we know that
when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge
So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q
so here we can say
so we will have
so the x coordinate of this position is given as
