Answer:
246.28 K
Explanation:
The total energy of one mole of gas molecules can be calculated by the formula given below
E = 
Where R is gas constant and T is absolute temperature.
Put the value of R as 8.314 and temperature as 245 , we get
E = 
= 3055.4 J
Add 16 j to it
Total energy of gas molecules = 3055.4 + 16 = 3071.4 J.
If T be the temperature after addition of energy then
= 3071.4
T =
T = 246.28 K
Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.
However, photons are considered to have a rest mass zero.
However note the point carefully "rest mass". A body in motion cannot have mass to be zero.
<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>
Answer:
A. It must be zero
Explanation:
A spacecraft leaves the solar system at a velocity of 1,500 m/s. The net force on this spacecraft is zero. What can we say about the spacecraft's acceleration?
According to Newton's second law
Force = Mass × acceleration
If the net force is zero
0 = mass × acceleration
0 = ma
a = 0/m
a = 0m/s²
this shows that the acceleration will be zero If the net force is zero
Answer:
Explanation:
Given that
The mass of the body is 0.04kg
M=0.04kg
The radius of the paths is 0.6m
r=0.6m
The normal force exerted at A is 3.9N
Fa=3.9N
The normal force exerted at B is 0.69N
Fb=0.69N
Then work done by friction from point A to B will be the change in K.E
W=∆K.E+P.E
So we need to know the velocity at both point A and B
Then since the centripetal force is given as
Ft=mv²/r
Then,
For point A
Fa=mv²/r
3.9=0.04v²/0.6
3.9=0.0667v²
v²=3.9/0.0667
v²=58.5
v=√58.5
v=7.65m/s
Va=7.65m/s
Now at point B
Fb=mv²/r
0.69=0.04v²/0.6
0.69=0.0667v²
v²=0.69/0.0667
v²=10.35
v=√10.35
v=3.22m/s
Vb=3.22m/s
Then, the work done is
W=∆K.E+P.E
P.E is given as mgh
The height will be 2R =1.2m
P.E=mgh
P.E=0.04×9.81×1.2
P.E=0.471J
Final kinetic energy at B minus initial kinetic energy at A
W=K.Eb-K.Ea
K.E is given as 1/2mv²
W=1/2m(Vb²-Va²) +P.E
W=0.5×0.04(3.22²-7.65²) +0.471
W=0.5×0.04×(-48.1541) +0.471
W=-0.96+0.471
W=-0.49J
work was done on the block by friction during the motion of the block from point A to point B is 0.49J.
Friction opposes motions and that is why the work done is negative
