Ask question

Ask question

All categories

2 years ago

14

State and prove Newton's second law of motion

1 answer:

LiRa [457]2 years ago

5 0

Answer:

HOPE IT HELP YOU A LOT :)

I prove it also .

You might be interested in

Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter

Aleonysh [2.5K]

We can solve the problem by using the first law of thermodynamics:

$\Delta U= Q-W$

where

$\Delta U$ is the variation of internal energy of the system

Q is the heat added to the system

W is the work done by the system

In this problem, the variation of internal energy of the system is

$\Delta U=U_f-U_i=134 J-87 J=47 J$

While the heat added to the system is

$Q=500 J$

therefore, the work done by the system is

$W=Q-\Delta U=500 J-47 J=453 J$

5 0

3 years ago

Read 2 more answers

What separates the terrestrial planets from the jovian or gas planets in space?

Flura [38]

In our solar system, terrestrial planets are separated from the gas giants by the asteroid belt. The asteroid belt is a region in the solar system between Mars and Jupiter where asteroids are located. Gas giants do not have a solid surface and possible a small rocky core. The gas giants are Jupiter, Saturn, Uranus and Neptune. The first four planets, Mercury, Venus, Earth and Mars.

8 0

2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

3 years ago

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a

zepelin [54]

Answer:

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.

Explanation:

Given Values:

L = 50 cm = 0.5 m

H = 170 j/s

To find the diameter of the rod, we have to find the area of the rod using the following formula.

Here Tc = 100.0° C

k = 50.2

H = k × A × $\frac{[T_{H -}T_{C} ] }{L}$

Solving for A

A = $\frac{H * L }{k * [ T_{H}- T_{C} ] }$

A = $\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}$

A = $\frac{85}{12550}$ = 6.77 × $10^{-3}$ m²

Now Area of cylinder is :

A = $\frac{\pi }{4}$ d²

solving for d:

d = $\sqrt{\frac{4 * 0.00677 }{\pi } }$

d = 9.28 cm

5 0

3 years ago

At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately

Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

$I = 2 kJ/s*m^2$

area on which photons are received is given as

$A = 4.80 cm^2 = 4.80 * 10^-4 m^2$

now we can find the power received due to sunlight

$P = I*A$

$P = 2* 10^3 * 4.80 * 10^-4$

$P = 0.96 Watt$

now we can say this power is due to photons that strikes on surface of earth

so here we can say

$P = N\frac{hc}{\lambda}$

given here that

$\lambda = 510 nm$

$0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}$

$0.96 = N * 3.88 * 10^{-19}$

$N = \frac{0.96}{3.88*10^{-19}}$

$N = 2.47 * 10^{18}$

so it will strike 2.47 * 10^18 photons on given area per second

3 0

3 years ago