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HACTEHA [7]
3 years ago
6

A repelling force occurs between two charged objects when the charges are of Equal magnitude. Equal magnitude. a Unequal magnitu

de.b Unequal magnitude. c Like/same signs.d Like/same signs. e Unlike signs. help plz
Physics
1 answer:
Ahat [919]3 years ago
3 0

Answer:

c Like/same signs

Explanation:

A repelling force occurs between two or more charged objects with the charges are of like or same sign.

  • According to Coulombs law, like charges repel on another, unlike charges attracts on another.
  • If a positive charge comes into the vicinity of another positive charge, there will be repulsion.
  • When oppositely charge species are brought near each other, there is an attraction.

Therefore, repulsion occurs when like charges are brought close to each other.

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A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the
Vitek1552 [10]

Answer:

147 N

Explanation:

static friction = frictional force max / normal

normal force = mg = 50 kg × 9.8 m/s² = 490 N

0.3 = Frictional force / 490 N

0.3 × 490 N = 147 N

The body will not move because the frictional force is greater than the force applied.

4 0
3 years ago
Read 2 more answers
A 782-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Vadim26 [7]

Answer:

a) v = 5.59x10³ m/s

b) T = 4 h

c) F = 1.92x10³ N

Explanation:

a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:

F_{c} = F_{G}

\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}

v = \sqrt{\frac{gr^{2}}{r+h}          

Where:

g is the gravity = 9.81 m/s²        

r: is the Earth's radius = 6371 km

h: is the satellite's height = r = 6371 km      

v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s                                      

b) The period of its revolution is:

T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h

c) The gravitational force acting on it is given by:

F = \frac{GMm}{(r + h)^{2}}

Where:

M is the Earth's mass =  5.97x10²⁴ kg    

m is the satellite's mass = 782 kg

G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²

F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N

I hope it helps you!

3 0
3 years ago
assume that in an unpolished floor a lady us pushing a 40 kg box to the left with an applied force of 192 N. find the sum of for
son4ous [18]
The sum of force is 7,680...............
8 0
3 years ago
How does the gravitational force between two objects change if the distance between the objects doubles?
Ymorist [56]

if the distance between the objects is doubled the force is reduced by a factor of 4

<h3>Whats is gravitational force?</h3>

Gravitational force is the force of attraction between objects in the universe.

f = G * m1 * m2 / r^2

f = gravitational force

G = gravitational constant

m1 = mass of object 1

m2 = mass of object 2

r = distance between the objects

From the formular, the gravitational force and the distance is an inverse relationship so increasing the distance by a factor results to reduction of the force by the square of the factor. hence doubling the distance which is distance mutiplied by 2 will lead to reduction of the force by 2^2 = 4

Therefore: The force decreases by a factor of 4.

hope it helps

4 0
2 years ago
Difference between elastic and inelastic collisions
Eva8 [605]
An elastic collision<span> is where there is no loss of kinetic energy in the </span>collision<span>. Momentum is conserved saved in </span>inelastic collisions<span>, but cannot track the kinetic energy through the </span>collision<span> since some of it is changed into other forms of energy.</span>
3 0
3 years ago
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