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Irina18 [472]
4 years ago
13

A 78.1 kg ice skater is moving at 2.93 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole.

she then moves in a circle of radius 0.556 m around the pole. find the force exerted by the rope on her arms. the acceleration of gravity is 9.8 m/s 2 . answer in units of kn.
Physics
1 answer:
Rina8888 [55]4 years ago
3 0
That would be 8.9m that is because.............,...nxjdjdbndjenxjendndndnnenxbbejxnbdnzjdnkNdj
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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}

Our values are given as

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

3 0
3 years ago
As a way of determining the inductance of a coil used in a research project, a student first connects the coil to a 5.62 V batte
Reptile [31]

Answer:

Its inductance L = 166 mH

Explanation:

Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR

R = V/I = 5.62/0.698 = 8.052 Ω

Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from

V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'

Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω

WE now find the reactance X of the coil from

Z² = X² + R²

X = √(Z² - R²)

= √(97.5² - 8.05²)

= √(9506.25 - 64.8025)

= √9441.4475

= 97.17 Ω

Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.

L = X/2πf

= 97.17/2π(93.1 Hz)

= 97.17 Ω/584.965 rad/s

= 0.166 H

= 166 mH

Its inductance L = 166 mH

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3 years ago
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Answer:

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Where is Jesus located, I would like to pay him a visit ;)
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