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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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Use the drop-down menus to complete each sentence.
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Answer:b

Explanation:

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Does earths Roation causes gravity?​
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he gravity of Earth, denoted by g, is the net acceleration that is imparted to objects due to the combined effect of gravitation (from mass distribution within Earth) and the centrifugal force (from the Earth's rotation )

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3 years ago
A "spherical capacitor" is constructed of two thin, concentric spherical shells of conducting material. Let a be the radius of t
Shalnov [3]

Answer:

C=\frac{ab}{k(b-a)}

Explanation:

We can assume this problem as two concentric spherical metals with opposite charges.

We have also to take into account the formulas for the electric field and the capacitance. Hence we have

C=\frac{Q}{V}\\\\E=k\frac{Q}{r^2}\\

Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating

dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]

Hence, the capacitance is

C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}

but R1=a and R2=b

C=\frac{ab}{k(b-a)}

HOPE THIS HELPS!!

3 0
3 years ago
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Which of the following affect friction (you can choose more than 1)
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A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows
jeka57 [31]

Answer:

(a) The impedance in the circuit is Z=183.33\Omega.

(b)The resistance is R=38.89\Omega.

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

Z=\frac{V_rms}{I_rms}

Here, V_rms is the rms voltage and I_rms is the rms current.

PutV_rms=110 V and I_rms=0.600 A.

Z=\frac{110}{0.600}

Z=183.33\Omega

Therefore, the impedance in the circuit is Z=183.33\Omega.

(b)

The expression for the average power is as follows;

P_{a}=I_{rms}^{2}R

Here, P_{a} is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

R=\frac{P_{a}}{I_{rms}^{2}}

Put P_{a}=14W and

R=\frac{14}{{0.600}^{2}}

R=38.89\Omega

Therefore, the resistance is R=38.89\Omega.

(c)

The expression for the impedance is as follows;

Z^{2}=R^{2}+X_{L}^{2}

Here,X_{L} is the inductive reactance.

Put Z=183.33\Omega and R=38.89\Omega.

(183.33)^{2}=(38.89)^{2}+X_{L}^{2}

X_{L}=179.16\Omega

The expression for the inductive reactance in terms of  frequency is as follows;

X_{L}=2\pi fL

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

L=\frac{X_{L}}{2\pi f}

Put X_{L}=179.16\Omega and f=50Hz.

L=\frac{179.16}{2\pi (50)}

L=0.57 H

Therefore, the inuctance is 0.57 H.

4 0
3 years ago
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