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Arlecino [84]
4 years ago
9

Which statement relates most directly to the second law?

Physics
2 answers:
alukav5142 [94]4 years ago
7 0
Statement three i do believe

jok3333 [9.3K]4 years ago
6 0

Answer: Option (2) is the correct answer.

Explanation:

According to second law of thermodynamics, all the energy of a system cannot be converted into useful work because some amount of energy will be lost.

For example, a machine will never convert its energy into 100% work as some of the energy of machine will be lost into the atmosphere.

Therefore, we can conclude that the statement some thermal energy is lost and some is used to do work, relates most directly to the second law.

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A 7.77 kg mass is moving due west at 7.77 m/s. A second mass of 8.88 kg is moving due south at 8.88 m/s. What is the magnitude o
masya89 [10]

Answer:

8.362m/s

Explanation:

Given data

Mass m1= 7.77kg

Velocity v1= 7.77m/s

Mass m2= 8.88kg

Velocity v2= 8.88m/s

Apply the law of conservation of momentum for inelastic collision we have

m1v1+m2v2= (m+m2)V

7.77*7.77+ 8.88*8.88= (7.77+8.88)V

60.3729+78.8544= 16.65V

139.2273= 16.65V

Divide both sides by 16.65

V= 139.2273/16.65

V= 8.362m/s

Hence the final velocity is 8.362m/s

7 0
3 years ago
A 562 N trunk is on frictionless plane inclined at 30.0 degrees from the horizontal. What is the acceleration of the trunk down
Len [333]

Answer: 0m/s²

Explanation:

Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane

According newton's law of motion

Summation of forces along the plane = mass × acceleration

Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane

Ff = nR where

n is coefficient of friction = tan(theta)

R is normal reaction = Wcos(theta)

Fm = Wsin(theta)

Substituting in the formula of newton's first law we have;

Fm-Ff = ma

Wsin(theta) - nR = ma

Wsin(theta) - n(Wcos(theta)) = ma... 1

Given

W = 562N, theta = 30°, n = tan30°, m = 56.2kg

Substituting in eqn 1,

562sin30° - tan30°(562cos30°) = 56.2a

281 - 281 = 56.2a

0 = 56.2a

a = 0m/s²

This shows that the trunk is not accelerating

4 0
3 years ago
In first equation of motion at denotes​
RideAnS [48]

Answer:

Names of the Equations of Motion

1 The First Equation of Motion is, v=u+at is known as velocity – time relation. 2 The Second Equation of Motion is, s=ut+12at2 is known as position – time relation.

6 0
3 years ago
A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the ga
pashok25 [27]
A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
p_i V_i = nRT_i
where
p_i=1.0 atm=1.01 \cdot 10^5 Pa is the initial pressure of the gas
V_i is the initial volume of the gas
n=2.3 mol is the number of moles
R=8.31 J/K mol is the gas constant
T_i=240^{\circ}C=513 K is the initial temperature of the gas

By re-arranging this equation, we can find V_i:
V_i =  \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3

2) Now the gas cools down to a temperature of
T_f = 14^{\circ}C=287 K
while the pressure is kept constant: p_f = p_i = 1.01 \cdot 10^5 Pa, so we can use again the ideal gas law to find the new volume of the gas
V_f =  \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
W=p \Delta V= p(V_f -V_i)
by using the data we found at point 1) and 2), we find
W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J
where the negative sign means the work is done by the surrounding on the gas.
5 0
3 years ago
Which two conditions would result in the weakest electric force between objects?
Novay_Z [31]
The objects are far apart the weakest the result the two conditions
3 0
4 years ago
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