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Alina [70]
3 years ago
10

A single conducting loop of wire has an area of 7.26E-2 m2 and a resistance of 117 Ω. Perpendicular to the plane of the loop is

a magnetic field of strength 0.289 T. At what rate (in T/s) must this field change if the induced current in the loop is to be 0.367 A?
Physics
1 answer:
cricket20 [7]3 years ago
3 0

Answer:

\frac{dB}{dt} = 591.45 T/s

Explanation:

i = induced current in the loop = 0.367 A

R = Resistance of the loop = 117 Ω

E = Induced voltage

Induced voltage is given as

E = i R

E = (0.367) (117)

E = 42.939 volts

\frac{dB}{dt} = rate of change of magnetic field

A = area of loop = 7.26 x 10⁻² m²

Induced emf is given as

E = A\frac{dB}{dt}

42.939 = (7.26\times 10^{-2})\frac{dB}{dt}

\frac{dB}{dt} = 591.45 T/s

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The diagram shows a ballistic pendulum. A 200 g bullet is fired into the suspended 4 kg block of wood and remains embedded insid
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Question

What was the initial momentum of the bullet before collision?

Answer:

10 Kg.m/s

Explanation:

Momentum is a product of velocity of an object in m/s and its mass in kgs hence numerically expressed as p=mv where p is momentum, v is velocity and m is mass. Substituting m for 0.2 kg and v for 50 m/s then p=0.2*50=10 kg.m/s

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A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv
sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

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Using formula of energy

K_{rel}=(\gamma-1)mc^2

\dfrac{K_{rel}}{mc^2}=\gamma-1

Put the value into the formula

\gamma=\dfrac{105.7}{0.511}+1

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Using formula of velocity

\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}

\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}

\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1

v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2

Put the value into the formula

v^2=\dfrac{1-(208)^2}{-208^2}\times c^2

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6 0
3 years ago
Hi, I need your help with this Physics exercise, I hope you can help me A pulse moving to the right along the x axis is represen
igomit [66]

Answer:

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y(x,t) = 2/ (x - 3t)² + 1

At t =0

y(x,0) = 2/ ((x - 3(0))² + 1)

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At t = 1

y(x,t) = 2/ ((x - 3(1))² + 1)

= 2 /(( x - 3)² + 1)

At t = 2

y(x,t) = 2/ ((x - 3(2))² + 1)

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3 0
3 years ago
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