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3 years ago

Newton's law of universal gravitation works well in ordinary situations on earth, but it does not work well

Physics

2 answers:

IRINA_888 [86]3 years ago

6 0

Answer:

For large masses,It does not work well

Explanation:

Newton's law of universal gravitation :

In the universe every particle attract the other particle by a force and this force is directly proportional to product of their masses and inversely proportional to distance between these two masses.This law given as follows

$F=G\dfrac{m_1m_2}{r^2}$

Where G is the constant.

Th one limitation of this law that the masses should not be too large.So this law does not work well for large bodies.

USPshnik [31]3 years ago

3 0

In large stars because they are too massive

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Explain velocity ratio. Explain and define the efficiency of a machine

Norma-Jean [14]

Velocity ratio is also defined as the ratio of a distance through which any part of a machine moves, to that which the driving part moves during the same time. An object has a mechanical advantage if it exerts a force higher than the velocity ratio.

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3 years ago

What is the net force

nadya68 [22]

Answer:

20 N in West direction.

Explanation:

opposite forces cancel each other. so 20 N in north and 20N in south cancel each other. In west and east direction...

70N in west-50N in east= 20N in west

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3 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago

A spring stretches 1.68cm vertically when a 2.50kg object is suspended from it.Find the distance (in cm) the spring stretches if

Llana [10]

Answer:

2.96 cm

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extension(d)

F = k × d

Force is the weight of the object, F = W = mg

So we get, mg = kd ⇒ m ∝ d

2.5 ∝ 1.68 --------------(1)

4.4 ∝ d' --------------(2)

From (1) & (2), 4.4/2.5 = d'/1.68

d' = 2.96 cm ⇒ the required extension.

7 0

3 years ago

Calculate the density of a tin of mass 100g whose dimensions are 2cmx5cmx

Luda [366]

Answer:

your question in not complete.

you need to the high too.

7 0

3 years ago