A pulley is used to raise a heavy crate. With the pulley it only takes an effort force of 223 N to
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Answer:
<em>The magnitude of the force is 10 N</em>
Explanation:
<u>Coulomb's Law</u>
The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.
Written as a formula:
Where:
q1, q2 = the objects' charge
d= The distance between the objects
We have two identical charges of q1=q2=1 c separated by d=30000 m, thus the magnitude of the force is:
F = 10 N
The magnitude of the force is 10 N
Refer to the diagram shown below.
When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.
The time, t, for the bolt to fall a known distance obeys the equation
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.
Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s
The total time, T, to fall 94 m is given by
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s
The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s
The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s
The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s
Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).
When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.
The time, t, for the bolt to fall a known distance obeys the equation
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.
Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s
The total time, T, to fall 94 m is given by
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s
The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s
The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s
The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s
Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).