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mixer [17]
2 years ago
6

A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.30 T.

If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field?
Physics
1 answer:
Degger [83]2 years ago
5 0

Answer:

The angle the wire make with respect to the magnetic field is 30°

Explanation:

It is given that,

Length of straight wire, L = 0.6 m

Current carrying by the wire, I = 2 A

Magnetic field, B = 0.3 T

Force experienced by the wire, F = 0.18 N

Let θ be the angle the wire make with respect to the magnetic field. Magnetic force is given by :

F=ILB\ sin\theta

\theta=sin^{-1}(\dfrac{F}{ILB})

\theta=sin^{-1}(\dfrac{0.18\ N}{2\ A\times 0.6\times 0.3\ T})

\theta=30^{\circ}

So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.

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faust18 [17]
The answer is the suns gravity
5 0
3 years ago
Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c
Travka [436]

Answer:

22.1 V

Explanation:

We are given that

A_1=0.7 cm^2=0.7\times 10^{-4} m^2

A_2=2.5 cm^2=2.5\times 10^{-4} m^2

A_3=2.2 cm^2=2.2\times 10^{-4} m^2

A_4=3 cm^2=3\times 10^{-4} m^2

Using 1cm^2=10^{-4} m^2

We know that

R=\frac{\rho l}{A}

In series

R=R_1+R_2+R_3+R_4

R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}

R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}

Substitute the values

R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})

R=\rho l(2.62\times 10^4)

V=145 V

I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}

Voltage across the 2.5 square cm wire=IR=I\times \frac{\rho l}{A_2}

Voltage across the 2.5 square cm wire=\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V

Voltage across the 2.5 square cm wire=22.1 V

6 0
3 years ago
What is the quantity of heat energy required to convert 10g cube of ice at -30oC to steam at 120oC. also draw a graph of tempera
Nadusha1986 [10]

Answer:

The amount of energy required is 31.1692 kJ .

Explanation:

The transitions of water are as follows,

(243K ice to 273K ice to 273K water to 373k water to 373K steam to 393K steam)

The following required data is,

specific heat capacity of ice= 2.1 kJ/kg

specific heat capacity of water= 4.2 kJ/kg

specific heat capacity of ice= 1.996 kJ/kg

specific latent heat of fusion=334 kJ/kg

specific latent heat of vapourisation = 2260 kJ/kg

FORMULAS:-

temperature change=mc(T2-T1)

phase transistion = mL,

where, m=mass , c=specific heat capacity ,L = latent heat of fusion or vapourisation ,(T2-T1)= temparature change

Thus the total amount of heat is,

=mc(T4-T3) + mL + mc(T3-T2) + mL + mc(T2-T1)

=10(10^{-3} )(2.1)(30)+10(10^{-3} )(334)+10(10^{-3} )(100)(4.2)+10(10^{-3} )(2260) +10(10^{-3} )(20)(1.996)

=(0.3)(2.1) + (3.34) + (4.2) +  (22.6) + (0.2)(1.996)

=31.1692 kJ.

3 0
3 years ago
Make the following prefix conversions.<br> 0.001s =ms
Free_Kalibri [48]

Answer:

To convert a millisecond measurement to a second measurement, divide the time by the conversion ratio. The time in seconds is equal to the milliseconds divided by 1,000.

Explanation:

hope it helps

4 0
2 years ago
Which of the following relationships is correct?
ira [324]
1 N = 1 kg•m is the answer
5 0
3 years ago
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