Answer:
All the given options will result in an induced emf in the loop.
Explanation:
The induced emf in a conductor is directly proportional to the rate of change of flux.
where;
A is the area of the loop
B is the strength of the magnetic field
θ is the angle between the loop and the magnetic field
<em>Considering option </em><em>A</em>, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.
<em>Considering option </em><em>B</em>, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.
Option C has a similar effect with option A, thus both will result in an induced emf.
Finally, <em>considering option</em> D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will<em> </em>change the angle<em> </em>between the loop and the magnetic field. This effect will also result in an induced emf.
Therefore, all the given options will result in an induced emf in the loop.
Explanation:
The given data is as follows.
Voltage = 2.50 V
Hence, calculate the equivalence capacitor as follows.
= 
C = 
Now, we will calculate the charge across each capacitance as follows.
Q = CV
= 
= 
= 
Thus, we can conclude that is the charge stored on each given capacitor.
From the information in the question, the acceleration of the cart is 2.55 ms-2.
We can find the acceleration using the following information from the question;
u = 0.9 m/s
v = 6 m/s
a = 2.0 seconds
But;
a = v - u/t
v = final velocity
u = initial velocity
t = time
a = 6 - 0.9/2 = 2.55 ms-2
Force= ma
m = mass
a = acceleration
F = 6 kg × 2.55 ms-2
F = 15.3 N
Learn more about acceleration: brainly.com/question/12134554
All of the observations except "powerful gravitational field" are consistent with the current theory of black holes.
The gavitational field of a black hole is thought to be no different than that of an ordinary star with the same mass.
Answer:
1302 K or 1029 C
Explanation:
Air at atmospheric pressure has pressure of 1 atm
20 C = 20 + 273 = 293 K
Assume ideal gas, according to the ideal gas law:
Where P1, V1 and T1 are the pressure, volume and temperature of the gas before the compression and P2, V2 and T2 are the pressure, volume and temperature of the gas after the compression
Since the gas is compressed to 1/9 of its original volume, V2/V1 = 1/9:
or 1029 C
