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3 years ago

How does decreasing the distance between the two objects affect the gravitational force?

Physics

1 answer:

Ann [662]3 years ago

6 0

Answer:

Decreasing the distance between the two objects affect the gravitational force by strengthening it.

Explanation:

Gravity is stronger when the distance is shorter.

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The energy of motion or the energy an object has as a result of its motion.

nirvana33 [79]

Answer:

Kinetic energy is what i think the answer is

Explanation:

I also looked it up

8 0

3 years ago

A wave can be described as

rewona [7]

A wave can be described as the disturbance of particles in an area. Think about it this way: particles (matter) carry energy. For all the laws of physics to work, this energy must be "traded" somehow. This happens by miniscule vibrations in the particles, which are apparent disturbances. This creates a wave, and therefore a wave is, indeed, a disturbance.

4 0

4 years ago

A 0.100-kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, travel- ing horizontally at 350 m>s, s

mariarad [96]

Answer:

Magnitude is 25.8 m/s and direction is $35.5^{o}$

Explanation:

From the law of conservation of linear momentum

$m_{b}v_{ib}+ m_{s}v_{is}= m_{b}v_{fb}+ m_{b}v_{fs}$ where $m_{b}$ and $m_{s}$ are masses of bullet and stones respectively, $v_{ib}$ and $v_{is}$ are the initial velocities of bullet and stone respectively, $v_{fb}$ and $v_{fs}$ are the final velocities of bullet and stone respectively

Substituting 6g=0.006 Kg for mass of bullet, 0.1Kg for mass of stone, 350 m/s for initial velocity of bullet and 250 m/s as final velocity for stone but initial velocity of stone is zero

$0.006 Kg *350 m/s (\hat i)+0.1 Kg*0=0.006 Kg* 250 m/s (\hat j)+0.1*v_{fs}$

$2.1 Kg.m/s(\hat i)=1.5 Kg.m/s(\hat j)+ 0.1*v_{fs}$

$0.1*v_{fs}=2.1 Kg.m/s(\hat i)- 1.5 Kg.m/s(\hat j)$

$v_{fs}=21 Kg.m/s(\hat i)- 15 Kg.m/s(\hat j)$

$|v_{fs}|=\sqrt {(v_{x}^{2}+v_{y}^{2})}$

Substituting 21 for $v_{x}$ and 15 for $v_{y}$

$|v_{fs}|=\sqrt {(21^{2}+15^{2})}=25.8 m/s$

To find direction

$tan\theta=\frac {v_{y}}{v_{x}}$

$\theta=tan^{-1}(\frac {15}{21})=35.5^{o}$

Therefore, magnitude is 25.8 m/s and direction is $35.5^{o}$

8 0

4 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$