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patriot [66]
3 years ago
6

7. An automobile with a radio antenna 1.0 m long travels at 100.0 km/h in a location where the Earth’s horizontal magnetic field

is 5.5×10−5T. What is the maximum possible emf induced in the antenna due to this motion?
Physics
1 answer:
VashaNatasha [74]3 years ago
3 0

Answer:

0.00479 volts

Explanation:

From Faraday's law of electromagnetic induction, the induced emf equals the change in magnitude flux (magnetic field strength multiplied by the area = BA) divided by the time change

Therefore we have the equation

EMF = BA÷t

Since area A = 2πr

EMF = Bπr²÷t

B = 5.5×10^(-5)

Velocity = 100km/h = 27.7778m/s

r = 1m, t = r÷V = 0.036

EMF = Bπr²÷t = (5.5×10^(-5) x π x (1)²)÷0.036 = 0.00479 Volts

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Answer:

Explanation:

General Equation of SHM is given by

x=A\cos \omega t

v=-A\omega \sin \omega t

where x=position of particle

A=maximum Amplitude

\omega =angular frequency

t=time

At any time Total Energy is the sum of kinetic Energy and Elastic potential Energy i.e. \frac{1}{2}kA^2

where k=spring constant

Potential Energy is given by U=\frac{1}{2}kx^2

also it is given that Potential Energy(U) is equal to Kinetic Energy(K)

Total Energy=K+U

Total=2U=2\times \frac{1}{2}kx^2

\frac{1}{2}kA^2=2\times \frac{1}{2}kx^2

x=\pm \frac{A}{\sqrt{2}}

at x=\frac{A}{\sqrt{2}}

velocity is v=\frac{A\omega}{\sqrt{2}}

6 0
3 years ago
What is the chemical formula for mercury(I) nitrate? Hgmc021-1.jpg(NOmc021-2.jpg) Hg(NOmc021-3.jpg)mc021-4.jpg Hgmc021-5.jpg(NOm
Anit [1.1K]
If you just type "<span>What is the chemical formula for mercury(I) nitrate?" into google you get the answer but HG(NO3)2 is the correct one.
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8 0
3 years ago
Read 2 more answers
Anyone know this lmk ASAP !!!
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Answer:1. Transparent

2. Transparent

3. Opaque

Explanation: In a transparent medium light can pass through. For opaque objects it does not allow to pass through instead it is reflected back.

8 0
2 years ago
A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
aivan3 [116]

Answer:

Part a)

F_n = 306 N

Part b)

v = 12.1 m/s

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

mg - F_n = ma_c

now we will have

mg - F_n = \frac{mv^2}{R}

F_n = mg - \frac{mv^2}{R}

now we have

F_n = 100(9.81) - \frac{100(9^2)}{12}

F_n = 981 - 675

F_n = 306 N

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

F_n + mg = ma_c

F_n = 0 at minimum speed

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = \sqrt{15 \times 9.81}

v = 12.1 m/s

So this speed is independent of the mass of the rider

5 0
3 years ago
The label has been scratched off a tuning fork and you need to know its frequency. From its size, you suspect that it is somewhe
bonufazy [111]

Answer:

255 Hz

Explanation:

With 5 beats per second with the 250 Hz fork, we know the unknown fork is either 250 - 5 = 245Hz or 250 + 5 = 255 Hz

With 15 beats per second with the 270 Hz fork, we know the unknown fork is either 270 - 15 = 255Hz or 270 + 15 = 285 Hz (most people would have a hard time discerning 15 beats per second... 5 per second is hard enough)

As 255 is the common frequency, it is the one selected.

3 0
2 years ago
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