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Maru [420]
3 years ago
9

The table below shows 4 examples of pairs of objects, their masses, and the distance between them. In which example is the gravi

tational force of attraction between the two objects the greatest.

Physics
1 answer:
svp [43]3 years ago
3 0

Answer:

example two

Explanation:

They have the greatest masses and close proximety relative to the rest, (If you have two black holes each with a solar mass only 1 mile away from one another, they will be highly atracted and probly

orbit each other once a second or so. But now lets try to put the earth and moon one half mile away from each other, they orbit each other much much slower then the two black holes, its becuase the gigantic mass of the black holes overwalms the closser distance between earth and the moon

Have a great day,

enjoy life.

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2225.6 grams =________ ft
AnnyKZ [126]

71606.562467

Is the complete answer

6 0
3 years ago
The figure above shows the net force exerted on an object as a function of the position of the object. The object starts from re
weqwewe [10]

Answer:

0.06 Kg

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Net Force (F) = 3 N

Mass (m) =?

Next, we shall determine the acceleration of the object. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Acceleration (a) =?

v² = u² + 2as

3² = 0² + (2 × a × 0.09)

9 = 0 + 0.18a

9 = 0.18a

Divide both side by 0.18

a = 9 / 0.18

a = 50 m/s²

Finally, we shall determine the mass of the object. This can be obtained as follow:

Net Force (F) = 3 N

Acceleration (a) = 50 N

Mass (m) =?

F = ma

3 = m × 50

Divide both side by 50

m = 3 / 50

m = 0.06 Kg

Therefore, the mass of the object is 0.06 Kg

5 0
2 years ago
For a projectile launched horizontally, which of the following best describes the downward component of a projectile's velocity?
Angelina_Jolie [31]

C. The downward component of the projectile's velocity continually increases

Explanation:

The motion of a projectile consists of two independent motions:  

- A uniform motion (with constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction  

Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:

v=u+at

where

u = 0 is the initial vertical velocity (zero since the projectile is fired horizontally)

a=g=9.8 m/s^2 downward is the acceleration of gravity

t is the time

So the equation becomes

v=gt

This means that

C. The downward component of the projectile's velocity continually increases

Because every second, it increases by 9.8 m/s in the downward direction.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0
4 years ago
I.Solve the following problems and answer the following questions. Show all your work and provide answers rounded off to the app
Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

  • initial velocity of the sprinter, u = 18 km/h
  • final velocity of the sprinter, v = 27 km/h
  • time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\

The time of motion is seconds;

t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s

The sprinter’s average acceleration is calculated as follows;

a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2

Thus, the sprinter’s average acceleration is 1.98 m/s²

Learn more here:brainly.com/question/17280180

6 0
3 years ago
A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge
Kisachek [45]

Answer:

\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

5 0
3 years ago
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