I need a Science EXPERT some one who is really good at Science

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

Explanation:

Given that,

Positive charge = 24.00 μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

$\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})$

$\theta=\sin^{-1}(\dfrac{1}{4})$

$\theta=14.47^{\circ}$

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

$E=\dfrac{2k\lambda}{r}\times2\cos\theat$

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}$

$E=408094.00\ N/C$

$E=4.08\times10^{5}\ N/C$

Hence, The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

6 0

3 years ago

What is the efficiency of a machine?

Julli [10]

Answer:

Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work).

Explanation:

It is a measure of how well a machine reduces friction.

7 0

2 years ago

The magnetic field at the centre of a toroid is 2.2-mT. If the toroid carries a current of 9.6 A and has 6.000 turns, what is th

ziro4ka [17]

Answer:

Radius, r = 0.00523 meters

Explanation:

It is given that,

Magnetic field, $B=2\ mT=2.2\times 10^{-3}\ T$

Current in the toroid, I = 9.6 A

Number of turns, N = 6

We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :

$B=\dfrac{\mu_oNI}{2\pi r}$

$r=\dfrac{\mu_oNI}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 6\times 9.6}{2.2\pi \times 2\times 10^{-3}}$

r = 0.00523 m

or

$r=5.23\times 10^{-3}\ m$

So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.

7 0

2 years ago

Read 2 more answers

The current in a single-loop circuit with one resistance R is 6.3 A. When an additional resistance of 3.4 Ω is inserted in serie

Dmitry [639]

Answer:

10.15Ω

Explanation:

From ohm's law,

V = IR...................... Equation 1

Where V = Voltage, I = current, R = resistance.

Assume the voltage across the resistance = V,

Given: I = 6.3 A

Substitute into equation 1

V = 6.3R.................. Equation 2

When an additional resistance of 3.4 Ω is inserted in series with R,

The voltage remain the same, but the current changes

Total Resistance(Rt) = (R+3.4)Ω, I' = 4.72 A

Also from ohm' law,

V = I'Rt............... Equation 3

Substitute the value of I' and Rt into equation 3

V = 4.72(R+3.4)............... Equation 5.

Divide equation 2 by equation 5

V/V = 6.3R/4.72(R+3.4)

1 = 1.335R/(R+3.4)

R+3.4 = 1.335R

3.4 = 1.335R-R

3.4 = 0.335R

R = 3.4/0.335

R = 10.15Ω

6 0

2 years ago

Read 2 more answers

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago