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NemiM [27]
2 years ago
13

Can someone solve it?

Physics
1 answer:
True [87]2 years ago
3 0

Answer:A

Explanation:

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A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

Emf = -N\frac{\Delta \phi}{\Delta t}

Where N is the number of turns

\Delta \phi is the change in magnetic  flux

\Delta t is the time interval

The change in magnetic flux is given by

\Delta \phi = \phi _{f} - \phi _{i}

Where \phi _{f} is the final magnetic flux

and \phi _{i} is the initial magnetic flux

Magnetic flux is given by the formula

\phi = BAcos(\theta)

Where B is the magnetic field

A is the area

and \theta is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, \theta = 0^{o}

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

A = \pi r^{2}

∴ A = \pi (0.15)^{2}

A = 0.0225\pi

∴\phi_{i}  = (1.5)(0.0225\pi)cos(0^{o} )

\phi_{i}  = (1.5)(0.0225\pi)

( NOTE: cos (0^{o}) = 1 )

\phi_{i}  = 0.03375\pi Wb

For \phi_{f}

The field pointed upwards, that is \theta = 90^{o}. Since cos (90^{o}) = 0

Then

\phi_{f} = 0

Hence,

\Delta \phi = 0- 0.03375\pi

\Delta \phi = - 0.03375\pi

From the question

\Delta t = 2.8 ms = 2.8 \times 10^{-3} s

Here, N = 1

Hence,

Emf = -N\frac{\Delta \phi}{\Delta t} becomes

Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }

Emf = 37.9 V

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0
3 years ago
In the sum →A+→B=→C, vector →A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the +x direction, and vector
icang [17]

Answer:

Explanation: Ok, first caracterize the two vectors that we know.

A = ax + ay = (12*cos(40°)*i + 12*sin(40°)*j) m

now, see that C is angled 20° from -x, -x is angled 180° counterclockwise from +x, so C is angled 200° counterclockwise from +x

C = cx + cy = (15*cos(200°)*i + 15*sin(200°)*j) m

where i and j refers to the versors associated to te x axis and the y axis respectively.

in a sum of vectors, we must decompose in components, so: ax + bx  = cx and ay + by = cy. From this two equations we can obtain B.

bx= (15*cos(200°) - 12*cos(40°)) m = -23.288 m

by = (15*sin(200°) - 12*sin(40°)) m = -12.843 m

Now with te value of both components of B, we proceed to see his magnitude an angle relative to +x.

Lets call a to the angle between -x and B, from trigonometry we know that tg(a) = by/bx, that means a = arctg(12.843/23.288) = 28.8°

So the total angle will be 180° + 28.8° = 208.8°.

For the magnitude of B, lets call it B', we can use the angle that we just obtained.

bx = B'*cos(208.8°) so B' = (-23.288 m)/cos(208.8°) =  26.58 m.

So the magnitude of B is 26.58 m.

7 0
3 years ago
The venn diagram shown below compares the nuclear reactions in the sun and nuclear power plants.
BARSIC [14]

Answer:

Formation of new elements

Explanation:

3 0
3 years ago
SYW A force of 175 N is needed to keep a stationary engine of weight 640 N on wooden skids from
grigory [225]

Answer:

phle follow karo yrr tab hee bat u ga

4 0
3 years ago
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A child in a boat throws a 5.80-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 24.6kg and the ma
Sergio [31]

Answer:

-0.912 m/s

Explanation:

When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

(m_c + m_b)v_b + m_pv_p = 0

where m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg are the mass of the child, the boat and the package, respectively. , v_p = 10m/s, v_b are the velocity of the package and the boat after throwing.

(24.6 + 39)v_b + 5.8*10 = 0

63.6v_b + 58 = 0

v_b = -58/63.6 = -0.912 m/s

3 0
3 years ago
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