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WITCHER [35]
3 years ago
6

what should be done in order to increase the gravitational force between two objects? Decrease the mass of both of the objects.

Decrease the mass of one of the objects. Decrease the distance between the two objects. Increase the distance between the two objects.
Physics
1 answer:
sesenic [268]3 years ago
8 0

Answer:

Decrease the distance between the two objects.

Explanation:

The force (F) of attraction between two masses (M₁ and M₂) separated by a distance (r) is given by:

F = GM₁M₂ / r²

NOTE: G is the gravitational force constant.

From the equation:

F = GM₁M₂ / r²

We can say that the force is directly proportional to the masses of the object and inversely proportional to the square of the distance between them. This implies that an increase in any of the masses will increase the force of attraction and likewise, a decrease in any of the masses will lead to a decrease in the force of attraction.

Also, an increase in the distance between the masses will result in a decrease in the force of attraction and a decrease in the distance between the masses, will result in an increase in the force of attraction.

Considering the question given above,

To increase the gravitational force between the two objects, we must decrease the distance between the two objects as explained above.

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Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in
sattari [20]

Answer:

49.3 N

Explanation:

Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?

The weight of the bucket of water = mg.

Weight = 4.25 × 9.8

Weight = 41.65 N

The tension and the weight will be opposite in direction.

Total force = ma

T - mg = ma

Make tension T the subject of formula

T = ma + mg

T = m ( a + g )

Substitutes all the parameters into the formula

T = 4.25 ( 1.8 + 9.8 )

T = 4.25 ( 11.6 )

T = 49.3 N

Therefore, the tension in the rope is 49.3 N approximately.

8 0
3 years ago
A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie
Paul [167]

The range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x 10^{-7}

90°              0

180°        - 375 π x 10^{-7}

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $\theta

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $\theta

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $\theta

U = Iπr^{2}B cos $\theta

For $\theta = 0° →

U(Max) = MB cos(0) = MB =  Iπr^{2}B = 5 × π × ( 0.05 ) ^{2} × 3 × 10^{-3} =

375 π x 10^{-7}.

For $\theta = 90° →

U = MB cos (90) = 0

For $\theta = 180° →

U(Min) = MB cos(0) = - MB =  - Iπr^{2}B = - 5 × π × ( 0.05 ) ^{2} × 3 × 10^{-3} =

- 375 π x 10^{-7}.

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x 10^{-7}

90°              0

180°        - 375 π x 10^{-7}

To solve more questions on Magnetic potential energy, visit the link below-

brainly.com/question/13708277

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3 0
1 year ago
Why should flexibility exercises be done in conjunction with strength-building exercises?
bixtya [17]
<span>Flexibility exercises should be done in conjunction with strength-building exercises because: 

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I hope my answer has come to your help. God bless and have a nice day ahead!
7 0
3 years ago
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From the top of the engineering building, you throw a ball vertically upward. the ball strikes the ground 4.00 s later. the engi
anastassius [24]
Equations of the vertical launch:

Vf = Vo - gt

y = yo + Vo*t - gt^2 / 2

Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
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=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s

The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.

Answer: 11.25 m/s


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