Answer:
521 nm
Explanation:
Given the values and units we are given, I'm assuming 5.76*10^14 Hz is frequency.
The formula to use here is λ * υ = c, where λ is wavelength, υ is frequency, and c is the speed of light.
λ =
Which of the following statements are true concerning electromagnetic induction? Check all that apply. It is possible to induce
a current in a closed loop of wire located in a uniform magnetic field by either increasing or decreasing the area enclosed by the loop. It is possible to induce a current in a closed loop of wire without the aid of a power supply or battery. It is possible to induce a current in a closed loop of wire by changing the strength of a magnetic field enclosed by the wire. It is possible to induce a current in a closed loop of wire by change the orientation of a magnetic field enclosed by the wire.
2 answers:
Answer:
<u>Option: </u> All of them are correct.
Explanation:
- <u>Electromagnetic induction:</u>
The electromagnetic induction is the phenomenon or process in which the electromotive force e.m.f is generated inside the coil or the conductor through the movement of the magnetic field,E.
As due to the position of the conductor inside the electromagnetic field,E of any magnet can results in the e.m.f generation inside that very conductor.Leading to potential difference inside the circuit, as it leads to the polarity difference created between any two points inside the circuit or any closed loop.
- It is possible to induce a current in a closed loop of wire located in a uniform magnetic field by either increasing or decreasing the area enclosed by the loop.
- It is possible to induce a current in a closed loop of wire without the aid of a power supply or battery.
- It is possible to induce a current in a closed loop of wire by changing the strength of a magnetic field enclosed by the wire.
- It is possible to induce a current in a closed loop of wire by change the orientation of a magnetic field enclosed by the wire.
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2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
Answer:
Explanation:
Given two mass on an incline code and
and an angle of inclination
.
. Assume that
is the weight being pulled up and
the hanging weight.
-The equations of motion from Newton's Second Law are:
where a is the acceleration.
#Substituting for (tension) gives:
#and solving for
which is the system's acceleration.