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ZanzabumX [31]
2 years ago
10

Two gravitational forces act on a particle, in perpendicular directions. to find the net force, can we add the magnitudes of tho

se two forces? no yes
Physics
1 answer:
Anna71 [15]2 years ago
6 0

No, we can not add the magnitudes of those two forces.

<h3>What is Gravitational Force?</h3>

An attractive force between masses is a gravitational force. According to Isaac Newton's second law, F = ma, a gravitational force generates an acceleration just like all other forces do. Remember that according to Newton's second law, a body will accelerate if there is a net force acting on it in an inertial frame of reference (a coordinate system travelling at a constant speed). According to Newton's Universal Law of Gravitation, when two bodies, such as the Earth and the Sun, are in close proximity to one another, an attractive force naturally attracts them. This attraction results in an acceleration of the two objects. 

To learn more about Gravitational Force, visit:

brainly.com/question/12528243

#SPJ4

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A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth
andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

F_{D}=\frac{kq^{2}}{DC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

F_{B}=\frac{kq^{2}}{BC^{2}}

F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

F_{A}=\frac{kq^{2}}{AC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N

The direction of FA is along A to C.

The net force along +X axis

F_{x}=F_{A}Cos45-F_{D}

F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N

The net force along +Y axis

F_{y}=F_{B}-F_{A}Sin45

F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N

The resultant force is given by

F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}

F = 4.03\times10^{7}N

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0
3 years ago
An object has a fixed volume and a variable shape before it changes state.
Wittaler [7]

Answer:

solid to liquid :)

Explanation:

APEX i just took it ;)

4 0
3 years ago
Read 2 more answers
You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?
In-s [12.5K]
-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is  14 cos(</span><span>51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.


As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
4 0
3 years ago
Read 2 more answers
What is the momentum of a 1550kg car that is traveling 38.0 m/s?
konstantin123 [22]

Answer:

p = 58,900 kg m/s

Explanation:

p = m × v

p = 1,550 × 38.0

p = 58,900 kg m/s

5 0
2 years ago
Need answers quick !! will give brainliest !!
stiv31 [10]

Answer:

use the bowl of water because Earth's magnetic field is relatively weak. Allowing it to float freely on the water, allows the magnetized needle to freely react to Earth's magnetic field, causing it to align North to South. If you watched closely, the same end of the needle should always point to the North

Explanation:

6 0
3 years ago
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