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Anarel [89]
3 years ago
8

How will the electrostatic force be changed if the distance between electric charges is reduced one half of the original distanc

e?
A) more than original force
B) less than original force
C) impossible to determine
D) the same as original force
Physics
1 answer:
GuDViN [60]3 years ago
8 0
I think the answer to your question is A
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Suppose two point charges, Q1 and Q2, are separated by a distance d. Which correctly states Coulomb's Law for the electrical for
likoan [24]
The correct answer is (B)

Which is (kQ1Q2) / d^2
7 0
3 years ago
The student throws one cannonball directly upward at 5.0 m/s and simultaneously throws the other cannonball directly downward at
Mademuasel [1]
Let the cannonball be thrown at a height of h above ground.
Then  the potential energy of the ball is
V = m*g*h
where
m = the mass of the ball
g = 9.8 m/s²

Also, the kinetic energy of the ball is
K = (1/2)mu²
where
u = 5 m/s, the vertical launch velocity.
Ignore wind resistance.

Because the total energy is preserved, the total energy (n the form of only kinetic energy) when the ball strikes the ground is
(1/2)mV²
where V = vertical velocity when the ball strikes the ground.

Expressions for both the initial and final energy are equal regardless of whether the ball s thrown downward or upward.
Therefore there is no difference in the landing speed. 

Answer: There is no difference.
8 0
4 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
4 years ago
Jake drops a book out of a window from a height of 10 meters. At what velocity does the book hit the ground?.
Lelechka [254]
I think the best way to answer to this kind of question because some of the elements are missing specially the mass of the book is just states the variables and assigned as the mass of an object. So the initial velocity of an object starts at zero and increase rapidly by 9.8 m/s^2 because it is due to gravity
4 0
3 years ago
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PUBE: DATE: A particle is moving in a straight line with constant acceleration. it passes point p with speed of 2ms. ten seconds
ddd [48]

Answer:4m/s

Explanation:The car is moving with constant acceleration therefore average speed is equal to distance over time to find the distance covered we multiple the speed by the time which gives 20 metres meaning after every 20 metres it uses a speed of 2m/s so for 40 metres it's going to be 40 x2 over 20 giving us 4m/s

7 0
3 years ago
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