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Vanyuwa [196]
3 years ago
11

A charge of +1.4×10^6 coulombs moves from point a to b. What is the potential difference between the two points if the work done

on the charge is +3.4×10^7 joules?
A:10 volts
B:14 volts
C:24 volts
D:34 volts
Physics
1 answer:
Rama09 [41]3 years ago
8 0
The potential difference between the two points if the work done on the charge is +3.4×10^7 joules would be A. 10volts.
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Overload refers to: A. Performing a weight-lifting exercise with the resistance (load) held overhead B. Using a demand (load) ab
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Answer:

E. All of the answers are correct

Explanation:

Overload principle in fitness training is associated with a gradual development of an athlete's abilities by progressively increasing the athlete's load and training.

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If a circle was flattened by pushing down on it, it would most likely form into the shape of ----------- which has ----------- f
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A vessel is filled with a liquid of density 1900 kg/m3 . There are two holes (one above the other) in the side of the vessel. Li
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Answer:

110 meters is the distance where they will intersect

Explanation:

given,

liquid density = 1900 kg/m³

distance of upper hole = 19 m

distance of lower hole = 117 m

acceleration due to gravity  = 9.8 m/s²

the speed at each point

v = \sqrt{2gh}  

for upper hole  v = \sqrt{2\times 9.8 \times 19}  

                         v  = 19.29 m/s

lower hole    v = \sqrt{2\times 9.8 \times 117}  

                          v = 47.88 m/s

The path for each is parabolic

x = v t

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y = \dfrac{1}{2}g(\dfrac{x}{v})^2  

y = \dfrac{gx^2}{2v^2}  

we get

upper hole

y = \dfrac{9.8\times x^2}{2\times 19.29^2}= 0.0132 x^2  lower hole

y= \dfrac{9.8\times x^2}{2\times 47.88^2}=0.00214x^2  y for upper hole = 80 + y for lower hole

0.0132 x^2= 98 + 0.00214 x^2

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4 0
3 years ago
A rock is sitting at the edge of a flat merry-go-round at a distance of 1.6 meters from the center. The coefficient of static fr
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Answer:

ω = 2.1 rad/sec

Explanation:

  • As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
  • This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
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  • In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
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       f_{frs} = \mu_{s} * F_{n} (1)

      where μs = coefficient of static friction between the rock and the merry-

      go-round surface = 0.7, and Fn = normal force.

  • In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
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  • The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

       F_{c} = m* \omega^{2}*r (3)

  • Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

       \omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec

  • This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.
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3 years ago
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