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3 years ago

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Two straight wires carry current of 5A opposite direction separated by a distance of 30cm. What is the magnitude of magnetic fie

ld at point P exactly the middle of them 15cm from each wire

1 answer:

jek_recluse [69]3 years ago

6 0

Answer:

The magnitude of magnetic field of both wires is β= 13. 33 $x10^{-6}$ T

Explanation:

$p=15cm*\frac{1m}{100cm}=0.15m$

$I=5A$

$u_{o} = 4*\pi x10^{-7}$

Using the right hand rule field magnetic as the current go in opposite direction the field in the point exactly in the middle have the same direction both so

$\beta _{t} =\beta _{1}+\beta _{2}$

They have the same direction and the P point is the middle as the same current field can be find:

$\beta _{t} =\beta _{1}*2$

$\beta = \frac{u_{o}*I }{2*\pi *p} \\\beta =\frac{4*\pi x10^{-7}*5A }{2*\pi *\frac{30}{2} } \\\beta =\frac{4x10^{-7}*5 }{0.30} \\\beta =6.6x10^{-6}$ T

$\beta _{t} =6.6x10^{-6}*2$ =13.33 $x10^{-6}$ T

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A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

W = $\int_{0}^{0.54} F dx$

= $\int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx$

= $[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}$

= 7560 + 1458 - 1364.69

= 7653.31 J

or, = 7.65 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

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2 years ago

A positive charge of 1 µC is taken from points A & B such that VA-VB = 100 V. Then the

Lunna [17]

Explanation:

It is given that,

Let Charge of $1\ \mu C$ is taken from points A & B such that $V_A-V_B=1000\ V$ .

We need to find the energy of charge. Electric potential is defined as the work done per unit of electric charge. So,

$W=(V_B-V_A)q\\\\W=-1000\times 10^{-6}\\\\W=E=-10^{-3}\ J$

So, the energy of charge decreases by $10^{-3}\ J$ . Hence, the correct option is (a).

4 0

2 years ago

A particle starts from the origin at t = 0 with an initial velocity of 4.8 m/s along the positive x axis.If the acceleration is

tia_tia [17]

Answer:

Part a)

Velocity = 6.9 m/s

Part b)

Position = (3.6 m, 5.175 m)

Explanation:

Initial position of the particle is ORIGIN

also it initial speed is along +X direction given as

$v_x = 4.8 m/s$

now the acceleration is given as

$\vec a = -3.2 \hat i + 4.6 \hat j$

when particle reaches to its maximum x coordinate then its velocity in x direction will become zero

so we will have

$v_f = v_i + at$

$0 = 4.8 - 3.2 t$

$t = 1.5 s$

Part a)

the velocity of the particle at this moment in Y direction is given as

$v_f_y = v_i + at$

$v_f_y = 0 + 4.6(1.5)$

$v_f_y = 6.9 m/s$

Part b)

X coordinate of the particle at this time

$x = v_x t + \frac{1}{2}a_x t^2$

$x = 4.8(1.5) - \frac{1}{2}(3.2)(1.5^2)$

$x = 3.6 m$

Y coordinate of the particle at this time

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 0(1.5) + \frac{1}{2}(4.6)(1.5^2)$

$y = +5.175 m$

so position is given as (3.6 m, 5.175 m)

5 0

3 years ago

What is the main difference between work power and energy

liubo4ka [24]

Answer:

Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.

Energy can also be defined as the ability to do work.

6 0

2 years ago

pentagon [3]

Answer:

6N

Explanation:

Given parameters:

Mass of object = 6kg

Initial velocity = 5m/s

Final velocity = 25m/s

Time = 30s

Unknown:

Net force acting on the object = ?

Solution:

From Newton's second law of motion:

Force = mass x acceleration

Acceleration is the rate of change of velocity with time

Acceleration = $\frac{Final velocity - Initial velocity }{time}$

Force = mass x $\frac{Final velocity - Initial velocity }{time}$

So;

Force = 6 x $\frac{25 - 5}{30}$ = 6N

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2 years ago