Ask question

Ask question

All categories

2 years ago

13

Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?

2 answers:

sashaice [31]2 years ago

8 0

<h2>Question :- </h2>

Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?

<h2>Answer :-</h2>

<h3>Gravitational Force :- </h3>

It is the attractive force between the two body having some mass.
Example of gravitational force :- attraction between the human and earth . earth pulls the body towards it with the help of this
This force helps the celestial bodies to revolve in there orbits

<h3>Factors effecting Gravitational force:- </h3>

Mass :- gravitational force directly proportional on the masses of the body between which it is acting $gravitational \: force ∝m{ \tiny1}m{ \tiny2}$
Distance between the masses :- Gravitational force is inversely proportional to the square of the distance between the masses $gravitational \: force∝ \frac{1}{ {r}^{2} }$

<h3>Formula of the Gravitational force :- </h3>

$F = \frac{G m{ \tiny1}m{ \tiny2} }{ {r}^{2} }$

where

F is gravitational force
G is gravitational constant ( universal constant )
m 1 is mass of first body
m2 is mass of second body
r is the distance between the two bodies

MakcuM [25]2 years ago

5 0

Answer:

Gravitational force is the force that attracts objects towards each other. Two factors that affect the gravitational force between objects are the mass of the two objects and the distance between

Explanation:

Gravity is what pulls us towards the Earth if we were to jump into the air, so it is the force that pulls things towards other things. The bigger the objects are the more gravity they have, so a planet has more gravity than say, an apple. Distance between objects also makes their gravity change, so the Earth's pull on the moon is different than the Earth's pull on the sun.

Hopefully this helps- let me know if you have any questions!

You might be interested in

A solar collector is a device that absorbs ___________from the sun ?

Gekata [30.6K]

Answer:

SUNLIGHT OR SUNRAYS OR ENERGY OR HEAT

Explanation:

5 0

3 years ago

Read 2 more answers

The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration

a_sh-v [17]

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>

Forces along x axis direction are as follows

4N along +x axis, so it's taken as +4 N
2N along -x axis , so it's taken as -2N.

Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>

As per Newton's second law, Force = mass × acceleration of the object
Force along x axis= mass × acceleration along x axis= 2N
Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

#SPJ1

3 0

2 years ago

Why are unit of length mass and time independent with each other?

mariarad [96]

Масса (килограмм), длина (метр) и время (секунда) - это 3 из 7 основных единиц в системе СИ.

Базовые единицы произвольно определены как то, что они есть, поэтому они не могут быть получены.

Все другие единицы в системе СИ являются производными от 7 основных единиц, таких как скорость (метр / секунда).

Остальные 4 - это температура (Кельвин), количество вещества (моль), электрический ток (ампер) и светимость (кандела).

4 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

While walking past a construction site, a person notices a pipe sticking out of a second floor window with water pouring out. As

tia_tia [17]

Answer:

Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.

P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0

Explanation:

P1 = P2 = atmospheric pressure so, P1 - P2 = 0

h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²

From the continuity equation for fluids

A1v1 = A2v2

v2 = A1v1/A2

Substituting into the equation above

(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho

Making A2² the subject of the formula,

A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}

The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.

Thank you for reading.

4 0

3 years ago