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2 years ago

What are the two factors that affect the gravitational pull between two objects?

Physics

1 answer:

elena-14-01-66 [18.8K]2 years ago

4 0

Answer:

hen dealing with the force of gravity between two objects, there are only two things that are important – mass, and distance.

Explanation:

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What important factor does the ozone layer serve in earth's atmosphere?

Dmitry_Shevchenko [17]

The Ozone layer is responsible for the absorption of harmful radiation from the sun before it hits the surface or exterior structure of our planet. It is known to be a belt of a known occurring gas, known as “ozone” which is sealed around the Earth’s atmosphere. It serves as a shield, which absorbs most of the sun’s ultraviolet radiation.

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3 years ago

Number of atoms of each element potassium chlorate?

11111nata11111 [884]

5 atoms is the answer

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2 years ago

If you double the velocity of a moving object, how is it's momentum affected?

Allushta [10]

Well momentum is = to Mass*Velocity so let's use an example to figure this out

If I weighed 50kg and I was jogging at 3m/s then I broke into a run at 6m/s how will me momentum be affected?
3m/s*50kg=150
6m/s*50kg=300

So as you can see by doubling the velocity you also double the momentum

8 0

3 years ago

A car travels along a straight line at a constant speed of 60.0 mi/h for a distance d and then another distance d in the same di

Makovka662 [10]

Answer:

velocity during second d = 20.0 mi/h

Explanation:

Total distance travelled is 2d, with an average velocity of 30.0 mi/h you can express the time travelled in terms of d:

distance = velocity * time

time = distance / velocity

time = 2d/30.0

The time needed for the first d at 60.0 is:

time = d/60.0

The time in the second d you can get it by substracting both times (total time - time for the first d)

second d time = 2d/30.0 - d/60.0

= 4d/60.0 - d/60.0

= 3d/60.0

and with the time (3d/60.0) and the distance travelled (d) you can get the velocity:

velocity = distance / time

velocity = d / (3d/60.0)

= 60.0/3 = 20.0 mi/h

8 0

2 years ago

A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c

kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately $20.3\; \rm m \cdot s^{-2}$ .

It would take approximately $3.9\; \rm s$ for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

Let $v$ denote the final velocity of the car.
Let $u$ denote the initial velocity of the car.
Let $a$ denote the acceleration of the car.
Let $x$ denote the distance that this car travelled.

$v^2 - u^2 = 2\, a\cdot x$ .

Given:

$u = 3\; \rm m \cdot s^{-1}$ .
$a = 4.5\; \rm m \cdot s^{-2}$ .
$x = 45\; \rm m$ .

Rearrange the equation $v^2 - u^2 = 2\, a\cdot x$ and solve for $v$ :

$\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}$ .

Calculate the time required for reaching this speed from $u = 3\; \rm m \cdot s^{-1}$ at $a = 4.5\; \rm m \cdot s^{-2}$ :

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}$ .

3 0

3 years ago