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pychu [463]
4 years ago
13

A current of 1.53 a flows through a 63.5 resistor for 6.4 min. How much heat was generated by the resistor

Physics
1 answer:
Ira Lisetskai [31]4 years ago
8 0
I am not sure exactly maybe u should ask a friend
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A square coil, enclosing an area with sides 2.0 cm long, is wrapped with 2 500 turns of wire. A uniform magnetic field perpendic
Delvig [45]

Answer:

The induced voltage in the coil is 0.25 V.

Explanation:

It is given that,

Area of a square coil is 2 cm or 0.02 m

Number of turns in the wire is 2500

A uniform magnetic field perpendicular to its plane is turned on and increases to 0.25 T during an interval of 1.0 s.

We need to find the induced voltage in the coil. According to Faraday's law, the induced emf in the coil is given by the rate of change on magnetic flux. So,

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(NBA)}{dt}\\\\\epsilon=NA\dfrac{-dB}{dt}\\\\\epsilon=-2500\times (0.02)^2\times \dfrac{0.25}{1}\\\\\epsilon=-0.25\ V

So, the induced voltage in the coil is 0.25 V.

3 0
3 years ago
A block of gelatin is 120mm by 120mm by 40mm whrn unstressed. A force of 49N is applied tangentially to the upper surface causin
Inessa05 [86]

Answer:

The shearing stress is 10208.3333 Pa

The shearing strain is 0.25

The shear modulus is 40833.3332 Pa

Explanation:

Given:

Block of gelatin of 120 mm x 120 mm by 40 mm

F = force = 49 N

Displacement = 10 mm

Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain​, SS = ?

The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:

A = 40*120 = 4800 mm² = 0.0048 m²

The shearing stress:

Ss=\frac{F}{A} =\frac{49}{0.0048} =10208.3333Pa

The shearing strain is defined as the tangent of the displacement that the block over its length:

SS=tan\theta =\frac{Displacement}{L}  =\frac{10}{40} =0.25

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

Sm=\frac{10208.3333}{0.25} =40833.3332Pa

6 0
4 years ago
What is the resultant of vectors shown
Ilya [14]

Answer:

adding to or more vectors together . When displacement vectors are added, the result is a resultant displacement. But any two vectors can be added as long as they are the same vector quantity.

Explanation:

4 0
4 years ago
two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of m
alexdok [17]
The electrostatic force is directly proportional to the product of the charges, by Coulomb's law.

F α Qq

If the charges are now half the initial charges: 

<span>F α (1/2)Q *(1/2)q
</span>
F α (1/4)Q<span>q

The new force when the charges are each halved is (1/4) the first initial force experienced at full charge.</span>
4 0
3 years ago
Read 2 more answers
A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to
White raven [17]

Answer:

1) \Delta s=1000\ ft

2)  \Delta s'=998.11\ ft.s^{-1}

3) t\approx125\ s

t'\approx463.733\ s

Explanation:

Given:

width of river, w=500\ ft

speed of stream with respect to the ground, v_s=8\ ft.s^{-1}

speed of the swimmer with respect to water, v=4\ ft.s^{-1}

<u>Now the resultant of the two velocities perpendicular to each other:</u>

v_r=\sqrt{v^2+v_s^2}

v_r=\sqrt{4^2+8^2}

v_r=8.9442\ ft.s^{-1}

<u>Now the angle of the resultant velocity form the vertical:</u>

\tan\beta=\frac{v_s}{v}

\tan\beta=\frac{8}{4}

\beta=63.43^{\circ}

  • Now the distance swam by the swimmer in this direction be d.

so,

d.\cos\beta=w

d\times \cos\ 63.43=500

d=1118.034\ ft

Now the distance swept downward:

\Delta s=\sqrt{d^2-w^2}

\Delta s=\sqrt{1118.034^2-500^2}

\Delta s=1000\ ft

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

v'=v.\cos37

v'=4\times \cos37

v'=3.1945\ ft.s^{-1}

<u>Now the net effective speed of stream sweeping the swimmer:</u>

v_n=v_s-v'

v_n=8-3.1945

v_n=4.8055\ ft.s^{-1}

<u>The  component of swimmer's velocity heading directly towards the opposite bank:</u>

v'_r=v.\sin37

v'_r=4\sin37

v'_r=2.4073\ ft.s^{-1}

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

\tan\phi=\frac{v_n}{v'_r}

\tan\phi=\frac{4.8055}{2.4073}

\phi=63.39^{\circ}

  • Now let the distance swam in this direction be d'.

d'\times \cos\phi=w

d'=\frac{500}{\cos63.39}

d'=1116.344\ ft

<u>Now the distance swept downstream:</u>

\Delta s'=\sqrt{d'^2-w^2}

\Delta s'=\sqrt{1116.344^2-500^2}

\Delta s'=998.11\ ft.s^{-1}

3)

Time taken in crossing the rive in case 1:

t=\frac{d}{v_r}

t=\frac{1118.034}{8.9442}

t\approx125\ s

Time taken in crossing the rive in case 2:

t'=\frac{d'}{v'_r}

t'=\frac{1116.344}{2.4073}

t'\approx463.733\ s

7 0
4 years ago
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