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Stells [14]
3 years ago
10

The angular speed of an automobile engine is increased at a constant rate from 1200 rev/min to 3000 rev/min in12 s.(a) What is i

ts angular acceleration in revolutions per minute­squared?(b) How many revolutions does theengine make during this 12 s interval?
Physics
1 answer:
Agata [3.3K]3 years ago
8 0

Answer:

The angular acceleration is  

=

15.71

r a d s −  2  and the number of revolutions is  = 419.9

Explanation:

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Please answer the following question.
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A child in a boat throws a 5.10-kg package out horizontally with a speed of 10.0 m/s. the mass of the child is 27.6 kg and the m
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The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

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dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

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Answer:

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