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4 years ago

If the current in each wire is the same, which wire produces the strongest magnetic field?

Physics

2 answers:

ASHA 777 [7]4 years ago

7 0

Answer:

the answer is D

Explanation:

lions [1.4K]4 years ago

3 0

Answer: option 4: A wire that is 2-mm thick and coiled.

Explanation:

The current in each wire is same. The magnetic field due to a current carrying wire increases if the wire is coiled with the more number of turns. A thick wire would cause low resistance to the current. Hence, a 2-mm thick wire which is coiled would produce the strongest magnetic field.

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A musician playing a string instrument may use a finger to push a string tightly against the fingerboard of her instrument, thus

Thepotemich [5.8K]

Answer:

a) increase

Explanation:

The frequency of sound produced due to vibration of a string is governed by the following formula

n = 1/2L√T/m ;

L is length of the string in vibration, T is tension in the string and m is mass of string per unit length.

So frequency is inversely proportional to length of the string. Hence shortning of length of string increases the frequency of vibration.

7 0

3 years ago

Heat naturally flows from an object that has a _______________ temperature to an object that has a _______________ temperature.

GaryK [48]

Answer:

higher, lower, external work, 100 %, work.

Explanation:

These paragraphs refer to the second law of thermodynamics. There are two statements for the second law of thermodynamics. They are as follows:

KELVIN STATEMENT:

All the heat from a source can never be transferred to the sink without the rejection of some heat.

CLAUSIUS STATEMENT:

Heat can not be transferred from a colder body to a hotter body without the application of som external work.

According to the statements the blanks can be filled as follows:

Heat naturally flows from an object that has a higher temperature to an object that has a Lower temperature. Heat can be made to flow in the reverse direction if external work is done. A machine can never have an efficiency of 100 %. This means that heat energy can never be fully converted into work energy.

7 0

3 years ago

Read 2 more answers

Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and

Lelu [443]

Answer:

The maximum shear stress in shaft AB, $T_{ABmax}$ is 15 MPa

The maximum shear stress in shaft CD, $T_{CDmax}$ is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by

J = $\pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}$

Therefore, we have

$J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}$

Where:

D $_{AB}$ = Diameter of shaft AB = 30 mm = 0.03 m

r $_{AB}$ = Radius of shaft AB = 15 mm = 0.015 m

∴ $J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2}$ = 7.95 × 10⁻⁸ m⁴

and

$J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}$

Where:

D $_{CD}$ = Diameter of shaft CD = 36 mm = 0.036 m

r $_{CD}$ = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

$J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2}$ = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= $1.58 \times \frac{2\pi }{360}$ rad = 2.76 × 10⁻² rad.

That is $\phi_r$ = 2.76 × 10⁻² rad.

However $\phi_r = \phi_{C/D} - \phi_{B/A}$

Where:

$\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$ and

$\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}$

$T_{AB}$ and $T_{CD}$ = Torque on shaft AB and CD respectively

$T_{AB}$ = Required

$T_{CD}$ = 500 N·m

$L_{AB}$ and $L_{CD}$ = Length of shafts AB an CD respectively

$L_{AB}$ = 600 mm = 0.6 m

$L_{CD}$ = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

$\phi_r = \phi_{C/D} - \phi_{B/A}$ = $\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

2.76 × 10⁻² rad = $\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

= $\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}$

Therefore;

T $_{AB}$ = 79.54 N.m

Where T = $T_{AB}$ + T $_{CD}$ =

Therefore T $_{CD total }$ = 500 - 79.54 = 420.46 N·m

τ $_{max}$ = $\frac{T\times R}{J}$

$\tau_{ABmax}$ = $\frac{T_{AB}\times R_{AB}}{J_{AB}}$ = $\frac{79.54\times 0.015}{7.95\times 10^{-8}}$ = 15 MPa

$\tau_{CDmax}$ = $\frac{T_{CD}\times R_{CD}}{J_{CD}}$ = $\frac{420.46\times 0.018}{1.65\times 10^{-7}}$ = 45.9 MPa

7 0

3 years ago

The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-

just olya [345]

Answer:

1.99 parsecs.

Explanation:

We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.

We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.

$6.5\text{ Light-years}=6.5\times 0.306601\text{ Parsecs}$