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3 years ago

Why does Newton's law of universal gravitation apply to all objects in the universe? Explain

2 answers:

5 0

Newton's law of Universal Gravitation tells that Force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

As there is mass for every object, and they are separated by certain distance (from 1 cm to more than hundreds of light years), they exerts Gravitational force on each other

Hope this helps!

SCORPION-xisa [38]3 years ago

3 0

Because all objects attract in proportion to the product of their masses.Gravitational interactions exist between all objects with an intensity that is directly proportional to the product of their masses.

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An 800-g block of ice at 0.00°C is resting in a large bath of water at 0.00°C insulated from the environment. After an entropy c

Allisa [31]

Answer:

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

ΔH is enthalpy change

T is absolute temperature

ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

Entropy change,ΔS = 600 J/K

Latent heat of fusion of water = 333 J/g

∴ΔH = TΔS

∴ΔH = 273 x 600

= 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

∴ΔH = mass of ice melted x latent heat of fusion of water

Mass of ice melted = ΔH / latent heat of fusion of water

= 163800 / 333

= 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

= 800-491.891

= 308.109 g

Amount of ice remained after melting = 308.109 g

8 0

3 years ago

In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point

Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ X volume of the oil drop

volume of the oil drop (spherical) = (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

= (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0

2 years ago

A person with mass mp = 76 kg stands on a spinning platform disk with a radius of R = 1.98 m and mass md = 191 kg. The disk is i

nalin [4]

<span>1.7 rad/s The key thing here is conservation of angular momentum. The system as a whole will retain the same angular momentum. The initial velocity is 1.7 rad/s. As the person walks closer to the center of the spinning disk, the speed will increase. But I'm not going to bother calculating by how much. Just remember the speed will increase. And then as the person walks back out to the rim to the same distance that the person originally started, the speed will decrease. But during the entire walk, the total angular momentum remained constant. And since the initial mass distribution matches the final mass distribution, the final angular speed will match the initial angular speed.</span>

3 0

3 years ago

The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24°C, and a wet-bulb temperature of 17°C. Using the psychro

Licemer1 [7]

Answer:

Given that

Dry-bulb temperature(T) =24°C

Wet-bulb temperature(Tw) = 17°C

Pressure ,P = 1 atm

As we know that psychrometric chart are drawn at constant pressure.

From the diagram

ω= specific humidity

Lets take these two lines Dry-bulb temperature(T) line and Wet-bulb temperature(Tw) cut at point P

From chart at point P

Specific humidity,ω = 0.00922 kg/kg

The enthalpy ( h)

h=47.59 KJ/kg

The relative humidity, RH

RH= 49.58 %

Specific volume ,

v= 0.853 m³/kg

5 0

2 years ago

I've had several mental breakdowns please help me

sesenic [268]

Answer:

The efficiency of the system is 63.7 %

Explanation:

Given;

input power of the motor, = 1.5 kW = 1,500 W

mass of the car lifted, m = 1300 kg

height through which the car was lifted, h = 1.8 m

time, t = 24 s

The output power of the motor is calculated as;

Output Power = F x v

= (mg) x (d/t)

= (1300 x 9.8) x (1.8 / 24)

= 12,740 x 0.075

= 955.5 W

The efficiency of the system is calculated as;

$E = \frac{0utput \ power}{1nput \ power} \times \ 100\%\\\\E = \frac{955.5}{1500} \times \ 100\% \\\\E = 63.7 \ \%$

The correct answer is 63.7%

4 0

2 years ago