John walks four kilometers east, two kilometers north as shown in the diagram below. What is John’s displacement

What effect would decreasing the distance between objects have on their gravitational attraction to each other?

Lady_Fox [76]

Decreasing the distance between two objects having a considerable mass would increase the attraction on gravitation. The reverse is true that if you separate or inrease the objects distance would substantially decrease their gravitational attraction. Most object in our planet is held by its gravitational force towards it's center.

5 0

3 years ago

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Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20Â°C. Container A is rigid. Container B has a 100 cm^2 pis

Vikki [24]

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) / 0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

7 0

2 years ago

water has an index of refraction of 1.33. What is the critical angle for light leaving a pool of water into air? 0 37 90 49

vesna_86 [32]

When light travels from a medium with higher refractive index into a medium with lower refractive index, there is a maximum angle (called critical angle) for which all the light is reflected, so there is no refraction.

The value of the critical angle is given by:
$\theta = \arcsin ( \frac{n_2}{n_1} )$
when n1 is the refractive index of the first medium, while n2 is the refractive index of the second medium. In our case, n1=1.33 (the water) and n1=1.00 (the air). Putting numbers in, we get
$\theta = \arcsin ( \frac{1.00}{1.33} )=49^{\circ}$

6 0

3 years ago

3. The shrillness of sound is known as_________

swat32

Answer:

pitched sound

Explanation:

The shrillness of sound is known as Pitched Sound

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3 0

2 years ago

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Please someone help, I’m very confused and it’s due soon, thanks

Anit [1.1K]

Answer:

1 s
19.6 m
2 s
0.8 m/s^2
28 m/s
79 m/s
0.37 s
26 m/s
242 m/s
19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

PE = Mgh

At velocity v, the kinetic energy of the object is ...

KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

PE = KE

Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

2gh = v^2

This can be solved for height:

h = v^2/(2g) . . . . [eq1]

or for velocity:

v = √(2gh) . . . . [eq2]

__

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

t = 2(v^2/(2g))/v = v/g . . . . [eq3]

t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

g = 2h/t^2 . . . . [eq6]

__

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

__

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

__

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

__

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

__

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

__

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

__

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

2(9/49 s) ≈ 0.37 s

__

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

__

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

__

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

<em>Additional comment</em>

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

5 0

3 years ago