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3 years ago

3 write the three laws given by kepler.How did they help Newton to arrive at the inverse square law of gravity?

Physics

1 answer:

stiv31 [10]3 years ago

5 0

Answer:

Kepler's laws apply: First Law: Planetary orbits are elliptical with the sun at a focus. Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times. Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semimajor axis is the same for all planets.

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A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.

Luba_88 [7]

Answer:

The final equilibrium temperature of the system is $T = 12.48^oC$

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice $M_{i}$ = 129 g = 0.129 kg

Mass of the steam $M_s$ = 19 g = 0.019 kg

Initial temperature is $T_i$ = 0°C

Temperature of steam $T_s$ = 100°C

Following the change of state of water in the question

The energy required by ice to change to water is mathematically given as

$Q_A = M_iL_f$

Where $L_f$ is a constant known as heat of fusion and the value is $334*10^3 J/kg$

$Q_A = 0.129 *334 *10^3 = 43086 J$

The energy been released when the steam changes to water is mathematically given as

$Q_B = M_s * L_v$

Where $L_v$ is a constant known as heat of vaporization and the value is $2256*10^3J/kg$

$Q_B = 0.019 * 2256*10^3 = 42864J$

The energy released when the temperature of water decrease from 100°C to 0°C is

$Q_C = M_s *C_water ($ 100°C)

Where $C_{water}$ is the specific heat of water which has a value $4186J/kg \cdot K$

$Q_C = 0.019 *4186*100 = 7953.4$

Looking at the values we obtained we noticed that ]

$Q_B + Q_C > Q_A$

What this means is that the ice will melt

bearing in mind the conservation of energy

looking the way at which water at different temperature were mixed according to the question

Heat lossed by the vapor = heat gained by ice

$Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T$

$T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}$

$T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}$

$T = 12.48^oC$

3 0

3 years ago

A 60 kg person sits in a chair. How much does the earth pull on the person? (Acceleration due to gravity is -10m/s2) F = mxa → F

MA_775_DIABLO [31]

Answer:

600N(downwards)

Explanations

600N(downwards)

Mas of the person = 60kg

Acceleration due to gravity = -10m/s²

To get the earths pull on the person, we will use the Newton second law of motion;

Force = mass * acceleration;

Force = 60 * -10

Force- -600N

Hence the earth gravitational pull on the person is 600N(downwards). It is downwards due to the negative sign.

7 0

2 years ago

A sample of helium behaves as an ideal gas as energy is

alisha [4.7K]

Answer:

0.0321 g

Explanation:

Let helium specific heat $c_h = 5.193 J/g K$

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

$E_h = m_hc_h \Delta T = 20 J$

where $m_h$ is the mass of helium, which we are looking for:

$m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g$

4 0

3 years ago

We want to design a cylindrical vacuum capacitor, with a given radius a for the outer cylindrical shell, that will be able to st

anastassius [24]

Solution :

a). Using Gauss's law :

$E=\frac{Q}{4 \pi \epsilon_0r^2}$ , $$b$ .........(1)

Let $E=E_0,\ r=b$ in equation (1)

Therefore, $Q=4 \pi \epsilon_0b^2E_0$ .............(2)

$V_b-V_a = \int^a_b \vec E. d\vec l$

$=\int^a_b E \ dx$

$=\frac{Q}{4 \pi \epsilon_0} \int^a_b \frac{1}{x^2} \ dx$

$=\frac{Q(a-b)}{4 \pi \epsilon_0 a b}$ ....................(3)

Therefore, $U=\frac{1}{2}Q \Delta V$

$=\frac{1}{2}(4 \pi \epsilon_0 b^2 E_0)\left(\frac{Q(a-b)}{4\pi \epsilon_0 a b}\right)$

$=\frac{4 \pi \epsilon_0}{2a} \ E^2_0 b^3(a-b)$ .............(4)

Now differentiating the equation (4) w.r.t. 'b', we get

$b=\frac{3}{4}a$

Thus the radius for the inner cylinder conductor is $b=\frac{3}{4}a$

b). For the energy storage, substitute the radius in (4), we get

$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$

This is the amount of energy stored in the conductor.

8 0

3 years ago

a rescue line is to be thrown horizontally from the bridge of a cruise ship at 40.0 m above sea level to a life boat 50.0 m away

Gemiola [76]

There is a 40m drop from the bridge, so we can work out the time of the vertical component first (how long it will take to reach the water)

do this by using a SUVAT equation s=ut+0.5at^2
this rearranges to t= sqrt(2s/a)
- note that u has been removed since there is no initial velocity.

sqrt((2*40)/9.81)= 2.86 seconds

this means that the horizontal component of 50m has so be covered in 2.86s
and speed = distance/ time

50m/2.86s=17.48ms^-1

the line has to be thrown at 17.48ms^-1 or more to reach the life boat.

8 0

3 years ago

3 write the three laws given by kepler.How did they help Newton to arrive at the inverse square law of gravity?​

3 write the three laws given by kepler.How did they help Newton to arrive at the inverse square law of gravity?