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3 years ago

Find the magnitude of the emf induced in the rod when it is moving toward the right with a speed 7.50 m/s.

Physics

1 answer:

Flauer [41]3 years ago

6 0

Answer:

3 volts

Explanation:

It is given that,

Magnetic field, B = 0.8 T

Length of a conducting rod, l = 50 cm = 0.5 m

Velocity of the conducting rod, v = 7.5 m/s

We need to find the magnitude of the emf induced in the rod when it is moving toward the right. When a rod is moved in a magnetic field, an emf is induced in it and it is given by :

$\epsilon=Blv$

Putting all the values,

$\epsilon=0.8\times 0.5\times 7.5\\\\\epsilon=3\ V$

So, the magnitude of the emf induced in the rod is 3 volts.

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C. The pressure decreases, and we can precisely predict by how much.

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Which of the following represents an element?

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3 years ago

The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge

Eva8 [605]

Answer:

a) $v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) $v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) $v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) $v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) $v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

$KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)$ (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

$\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}$

We can write the mass of a proton in MeV/c².

$m_{p}=938.28 MeV/c^{2}$

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.04$

$v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) Linac (400 MeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.71$

$v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) Booster (8 GeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.994$

$v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) Main ring or injector (150 Gev)

$\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.999$

$v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) Tevatron (1 TeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.9999$

$v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Have a nice day!

4 0

4 years ago

You have developed a method in which a paint shaker is used to measure the coefficient of static friction between various object

Paladinen [302]

Answer:

0.62

Explanation:

A = Amplitude

f = Frequency = 1.85 Hz

$\mu$ = Coefficient of static

g = Acceleration due to gravity = 9.81 m/s²

Angular frequency

$\omega=2\pi f\\\Rightarrow \omega=2\pi\times 1.85$

Maximum acceleration

$a_{max}=A\omega^2\\\Rightarrow a_{max}=0.045\times (2\pi\times 1.85)^2$

$a_{max}=\mu g\\\Rightarrow \mu=\frac{a_{max}}{g}\\\Rightarrow \mu=\frac{0.045\times (2\pi\times 1.85)^2}{9.81}\\\Rightarrow \mu=0.62$

Coefficient of static friction between penny and tabletop is 0.62

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4 years ago

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Nimfa-mama [501]

The answer is true. The table does show an object moving with changing speed.

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3 years ago

Read 2 more answers