Answer: 363 Ω.
Explanation:
In a series AC circuit excited by a sinusoidal voltage source, the magnitude of the impedance is found to be as follows:
Z = √((R^2 )+〖(XL-XC)〗^2) (1)
In order to find the values for the inductive and capacitive reactances, as they depend on the frequency, we need first to find the voltage source frequency.
We are told that it has been set to 5.6 times the resonance frequency.
At resonance, the inductive and capacitive reactances are equal each other in magnitude, so from this relationship, we can find out the resonance frequency fo as follows:
fo = 1/2π√LC = 286 Hz
So, we find f to be as follows:
f = 1,600 Hz
Replacing in the value of XL and Xc in (1), we can find the magnitude of the impedance Z at this frequency, as follows:
Z = 363 Ω
Answer:
Following are the answer to this question:
Explanation:
In option (a):
- The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
- Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.
In option (b):
- Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
- Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.
Explanation:
The first equation of motion in kinematics is given by :
.....(1)
u is initial speed
a is acceleration
v is final speed
t is time
Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.
Answer:
the horizontal distance is 4.355 meters
Explanation:
The computation of the horizontal distance while travelling in the air is shown below:
Data provided in the question is as follows
Velocity = u = 7.70 m/s
H = 1.60 m
R = horizontal direction
Based on the above information
As we know that
R = u × time
where,
Time = 
So,
= 
= 4.355 meters
hence, the horizontal distance is 4.355 meters