Answer:
1.162 x 10^-7 Nm
Explanation:
length of wire, l = 25 cm
l = 2 π r
where, r is the radius of circular loop
25 = 2 x 3.14 x r
r = 3.98 cm
Magnetic field, B = 5.55 mT = 5.55 x 10^-3 T
Current, i = 4.121 mA = 4.21 x 10^-3 A
Torque, τ = i x A x B
τ = 4.21 x 10^-3 x 3.14 x 0.0398 x 0.0398 x 5.55 x 10^-3
τ = 1.162 x 10^-7 Nm
Thus, the maximum torque in the coil is 1.162 x 10^-7 Nm.
Answer:
(a) W=217 J
(b) Tc=378K
(c) e=0.39=39%
Explanation:
For part (a)
We to calculate the mechanical work W the engine does. By knowing QC and QH can obtain the work using equation
W = IQHI — IQcl .....................eq(1)
Put given values for QH and QC into equation (1) to get
the mechanical work of the engine
W = 550 - 335
W=217 J
For part (b)
We want to determine the temperature of low temperature reservoir which means Tc
IQc|/|Qh| =TC/TH
for Tc
Tc=(IQc|/|Qh|)*TH
Now we can put values
Tc= 620K (335/
550.1)
Tc=378K
For part (c)
Here we want to find the thermal efficiency (e) of the cycle
e=1-TC/TH
e=1-(378/620)
e=0.39=39%
Answer:
Option 2 is the correct answer.
Explanation:
We have equation of motion s = ut + 0.5at²
Here u = 0 m/s
So, s = 0.5at²
Distance traveled in first second = 0.5 x a x 1² = 0.5 a
Distance traveled in second second = 0.5 x a x 2² - 0.5 x a x 1²= 1.5 a
Distance traveled in third second = 0.5 x a x 3² - 0.5 x a x 2²= 2.5 a
Distance traveled in fourth second = 0.5 x a x 4² - 0.5 x a x 3²= 3.5 a
The percentage increase in its displacement during the 4th second compared to its displacement in the 3rd second
Option 2 is the correct answer.
That's pretty easy, have a look :
The only thing you need to do is to find t and the problem is solved.