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8090 [49]
2 years ago
5

A circular loop of wire with radius r=0.0250 m and resistance r=0.390 ohms is in a region of spatially uniform magnetic field. t

he magnetic field is directed into the plane. at t=0, b=0. the magnetic field then begins increasing, with b(t)=(0.380t/s^3)t^3. what is the current in the loop (magnitude and direction) at the instant when b=1.33 t.
Physics
1 answer:
slavikrds [6]2 years ago
3 0

Answer:

0.0133 A

Explanation:

The time at which B=1.33 T is given by  

1.33 = 0.38*t^3  

t = (1.33/0.38)^(1/3) = 1.52 s  

Using Faraday's Law, we have  

emf = - dΦ/dt = - A dB/dt = - A d/dt ( 0.380 t^3 )  

Area A = pi * r² = 3.141 *(0.025 *0.025) = 0.00196 m²

emf = - A*(3*0.38)*t^2  

thus, the emf at t=1.52 s is  

emf = - 0.00196*(3*0.38)*(1.52)^2 = -0.0052 V  

if the resistance is 0.390 ohms, then the current is given by  

I = V/R = 0.0052/0.390 = 0.0133 A

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6 0
2 years ago
Find the net force and acceleration. 15 points. Will give brainliest!
gladu [14]

Answer:

\boxed{F_{net} = 28.7 \ N}

\boxed{a = 2.1 \ m/s^2}

Explanation:

<u><em>Finding the net force:</em></u>

<u><em>Firstly , we'll find force of Friction:</em></u>

F_{k} = (micro)_{k}mg

Where (micro)_{k} is the coefficient of friction and m = 13.6 kg

F_{k} = (0.16)(13.6)(9.8)\\

F_{k} = 21.32 \ N

<u><em>Now, Finding the net force:</em></u>

F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\

F_{net} = 28.7 \ N

<u><em>Finding Acceleration:</em></u>

a = \frac{F_{net}}{m}

a = \frac{28.7}{13.6}

a = 2.1 \ m/s^2

8 0
3 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
2 years ago
1. A soccer ball is kicked horizontally off a cliff with an initial speed of 8 m/s and lands 16 m from the base of
lana66690 [7]

Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

Initial velocity = 8 m/s

Distance s = 16 m

Starting acceleration (a) = 0

Computation:

s = ut + 1/2a(t)²

16 = 8t

t = 2 sec

Height of cliff = S

Gravitational acceleration = 10 m/s

S = 1/2a(t)²

S = 1/2(10)(2)²

Height of cliff = S = 20 m (Approx)

3 0
3 years ago
Please help me
melamori03 [73]
<h2>Question:</h2>

An automobile is driving uphill. Which form of energy is not involved in this process?

<h2>Choosing:</h2>

electromagnetic

potential

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<h2>Answer:</h2>

<u>Electromagnetic</u><u> </u>

<h3><u>#READINGHELPSWITHLEARNING</u><u> </u></h3><h3><u>#CARRYONLEARNING</u><u> </u></h3><h3><u>#STUDYWELL</u><u> </u></h3>
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