I need more to answer this
Answer:
O The particles of the medium move more slowly and there are fewer chances to transfer energy.
Explanation:
Various media are made up of particles. These particles are in constant motion according to the kinetic theory of matter. Recall that temperature has been defined as the average kinetic energy of the particles in a medium. Hence, for any given medium, the velocity of particle motion increases or decreases linearly with temperature.
The speed of particles in any medium increases or decreases as the temperature of the medium increases or decreases as emphasised above. Hence, at low temperature, the velocity of waves set up by the motion of particles in a medium decreases and transfer the wave energy to neighbouring particles occurs more slowly than at high temperatures.
Answer:
The value of change in internal l energy of the gas = 1850 J
Explanation:
Work done on the gas (W) = - 1850 J
Negative sign is due to work done on the system.
From the first law we know that Q = Δ U + W ------------- (1)
Where Q = Heat transfer to the gas
Δ U = Change in internal energy of the gas
W = work done on the gas
Since it is adiabatic compression of the gas so heat transfer to the gas is zero.
⇒ Q = 0
So from equation (1)
⇒ Δ U = - W ----------------- (2)
⇒ W = - 1850 J (Given)
⇒ Δ U = - (- 1850)
⇒ Δ U = + 1850 J
This is the value of change in internal energy of the gas.
root mean square<span>= square root of ( 3RT/M)
R = 8.314 J/K/mole
T = 25 + 273 = 298 K
M = molecular mas of N2 in kg = 28 X 10^-3 kg
put values...
</span><span> root mean square</span> = square root of ( 3 X 8.314 X 298/28 X 10^-3)
= square root of ( 265454.143)
= 515.2 m/s
so option A is right
hope this helps
Use the Inverse square law, Intensity (I)<span> of a light </span>is inversely proportional to the square of the distance(d).
I=1/(d*d)
Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.
L1/L2=(D2*D2)/(D1*D1)
L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>
