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3 years ago

Gravity is greater when there is

Physics

2 answers:

liberstina [14]3 years ago

7 0

Gravity is greater when there is more mass and less distance between two given objects.

Bess [88]3 years ago

7 0

when there is <span>more mass and less distance between two objects</span>

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If the distance between the center of two objects is quadrupled. The gravitational

Juliette [100K]

Answer:

F' = F/16

Explanation:

The gravitational force between masses is given by :

$F=G\dfrac{m_1m_2}{r^2}$

If the distance between the center of two objects is quadrupled, r' = 4r

New force will be :

$F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(4r)^2}\\\\F'=\dfrac{Gm_1m_2}{16r^2}\\\\F'=\dfrac{1}{16}\times \dfrac{Gm_1m_2}{r^2}\\\\F'=\dfrac{F}{16}$

So, the new force will change by a factor of 16.

6 0

3 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

3 years ago

A force F with arrow applied to an object of mass m1 produces an acceleration of 3.10 m/s2. The same force applied to a second o

Bas_tet [7]

Answer:

(a) The value of the ratio m₁/m₂ is 0.581

(b) the acceleration of the combined masses is 1.139 m/s²

Explanation:

Given;

The acceleration of force applied to M₁, a₁ = 3.10 m/s²

The same force applied to M₂ has acceleration, a₂ = 1.80 m/s²

Let this force = F

According Newton's second law of motion;

F = ma

(a) the value of the ratio m₁/m₂

since the applied force is same in both cases, M₁a₁ = M₂a₂

$\frac{m_1}{m_2} = \frac{a_2}{a_1} \\\\\frac{m_1}{m_2} = \frac{1.8}{3.1} \\\\\frac{m_1}{m_2} = 0.581$

(b) the acceleration of m₁ and m₂ combined as one object under the action force F

F = ma

$a = \frac{F}{M} \\\\a = \frac{F}{m_1 + m_2} \\\\a = \frac{F}{0.581m_2 + m_2}\\\\a = \frac{F}{1.581m_2}$

$But, m_2 = \frac{F}{a_2} \\\\a = \frac{F}{1.581m_2} = \frac{F*a_2}{1.581F} \\\\a = \frac{a_2}{1.581} \\\\a = \frac{1.8}{1.581} = 1.139 \ m/s^2$

Therefore, the acceleration of the combined masses is 1.139 m/s²

5 0

3 years ago

A wire perpendicular to the screen carries a current into the screen (away from you). What is the direction of the magnetic fiel

Llana [10]

Answer:

The direction is downwards. This answer to this question is found from the applicatikn of flemings right hand rule. This rule uses the thumb, index finger and the middle finger in the direction of current force and the magnetic field.

Explanation:

4 0

3 years ago

Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.

olganol [36]

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

$U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{x,1}$ , $K_{x,2}$ - Initial and final horizontal translational kinetic energy, measured in joules.

$K_{y,1}$ , $K_{y,2}$ - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

$m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})$

$y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}$

Where:

$y_{1}$ . $y_{2}$ - Initial and final height of the arrow, measured in meters.

$v_{y,1}$ , $v_{y,2}$ - Initial and final vertical speed of the arrow, measured in meters.

$g$ - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

$v_{y,1} = v_{1}\cdot \sin \theta$

Where:

$v_{1}$ - Magnitude of the initial velocity, measured in meters per second.

$\theta$ - Initial angle, measured in sexagesimal degrees.

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the initial vertical speed is:

$v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}$

$v_{y,1} \approx 33.352\,\frac{m}{s}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} \approx 33.352\,\frac{m}{s}$ and $v_{y,2} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{2} - y_{1} = 56.712\,m$

Second arrow

$U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}$

Where:

$U_{g,1}$ , $U_{g,3}$ - Initial and final gravitational potential energy, measured in joules.

$K_{y,1}$ , $K_{y,3}$ - Initial and final vertical translational kinetic energy, measured in joules.

$m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})$

$y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} = 82\,\frac{m}{s}$ and $v_{y,3} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{3} - y_{1} = 342.816\,m$

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

$E = U + K_{x}$

The expression is now expanded:

$E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}$

Where $v_{x}$ is the horizontal speed of the arrow, measured in meters per second.

$v_{x} = v_{1}\cdot \cos \theta$

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the horizontal speed is:

$v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}$

$v_{x} \approx 74.911\,\frac{m}{s}$

If $m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{max} = 56.712\,m$ and $v_{x} \approx 74.911\,\frac{m}{s}$ , the total mechanical energy is:

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}$

$E = 201.720\,J$

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

$E = m\cdot g \cdot y_{max}$

$m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $y_{max} = 342.816\,m$

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)$

$E = 201.720\,J$

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

7 0

3 years ago