this is due to the existence of other forces called the strong nuclear forces that overcomes the repulsion forces between the protons and keeps the nucleons holding to each other also there is a type of energy that is called the nuclear binding energy and this energy also works on binding the components of the nucleus together
False because your deltoids are in your shoulders not your back
Answer:
Given that
V2/V1= 0.25
And we know that in adiabatic process
TV^န-1= constant
So
T1/T2=( V1 /V2)^ န-1
So = ( 1/0.25)^ 0.66= 2.5
Also PV^န= constant
So P1/P2= (V2/V1)^န
= (1/0.25)^1.66 = 9.98
A. RMS speed is
Vrms= √ 3RT/M
But this is also
Vrms 2/Vrms1= (√T2/T1)
Vrms2=√2.5= 1.6vrms1
B.
Lambda=V/4π√2πr²N
So
Lambda 2/lambda 1= V2/V1 = 0.25
So the mean free path can be inferred to be 0.25 times the first mean free path
C. Using
Eth= 3/2KT
So Eth2/Eth1= T2/T1
So
Eth2= 2.5Eth1
D.
Using CV= 3/2R
Cvf= Cvi
So molar specific heat constant does not change
Answer:
a = 1.764m/s^2
Explanation:
By Newton's second law, the net force is F = ma.
The equation for friction is F(k) = F(n) * μ.
In this case, the normal force is simply F(n) = mg due to no other external forces being specified
F(n) = mg = 15kg * 9.8 m/s^2 = 147N.
F(k) = F(n) * μ = 147N * 0.18 = 26.46N.
Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.
Thus, F(net) = F(k) = ma
26.46N = 15kg * a
a = 1.764m/s^2
Explanation:
Work done is the force applied to move a body through a specific or particular direction.
It is also the difference in the amount of energy expended in using an effort.
Work done is given as;
Work done = F x d CosФ
F is the force applied
d is the displacement
Ф is the angle
The unit of work done is in Joules.
