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3 years ago

How can electricty be used to create a magnetic field

2 answers:

5 0

A magnetic field pulls and pushes electrons in some objects near them to make them move metals like copper have electrons that are moved easily and can be readily moved from their orbits if a magnet is moved quickly through coil of copper wire electrons move and electricity is made u can use this but maybe u will want to shorten it down

Strike441 [17]3 years ago

4 0

When a charged particle such as an electron, proton or ion is in motion, magnetic lines of force rotate around the particle. Since electrical current moving through a wire consists of electrons in motion, there is a magnetic field around the wire. i hope u have smileing day

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What objects has large volume and small mass

viva [34]

Answer:

A hot air balloon

7 0

3 years ago

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

ollegr [7]

<h2>Answer:</h2>

(a) standard deviation = σ = 4.9996

(b) variance = σ² = 24.996

<h2>Explanation:</h2><h2 />

<em>Given frequency table (find attached as Table 1);</em>

<u></u>

(a) To find the sample standard deviation and sample variance, follow these steps;

<em>i. Calculate the mid-point c for each group by using the mid-point formula;</em>

c = (lower bound + upper bound) / 2

=> c = (6.51 + 8.50) / 2 = 7.505

=> c = (8.51 + 10.50) / 2 = 9.505

=> c = (10.51 + 12.50) / 2 = 11.505

=> c = (12.51 + 14.50) / 2 = 13.505

=> c = (14.51 + 16.50) / 2 = 15.505

<em>So the new table becomes (find attached as Table 2);</em>

<em>ii. Calculate the total number of samples (n) which is the sum of all the frequencies.</em>

n = 50+18+42+20+46

n = 176

<em>iii. Calculate the mean (M)</em>

This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).

M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176

M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176

M = [2012.88] / 176

M = 11.44

<em>iv. Find the variance (σ²);</em>

The variance is calculated using the following formula

σ² = [Σ(f x c²) - (n x M²)] / (n - 1) ------------(i)

Where;

f = frequency of each boundary data point

<em>=> Let's first calculate </em>Σ(f x c²).

This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)

Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]

Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]

Σ(f x c²) = 24708.1244

<em>=> Now calculate (n x M²)</em>

n x M² = 176 x 11.44²

n x M² = 23033.7536

<em>=> Now substitute these values into equation (i) to calculate the variance</em>

σ² = [Σ(f x c²) - (n x M²)] / (n - 1)

σ² = [24708.1244 - 23033.7536] / (176 - 1)

σ² = [4374.3708] / (175)

σ² = 24.996

Therefore, the variance is 24.996

<em>v. Find the standard deviation (σ)</em>

The standard deviation is the square root of the variance. i.e

σ = √σ²

σ = √24.996

σ = 4.9996

Therefore, the standard deviation is 4.9996

4 0

3 years ago

A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while

dem82 [27]

12.8 rad

Explanation:

The angular displacement $\theta$ through which the wheel turned can be determined from the equation below:

$\omega^2 = \omega_0^2 + 2\alpha\theta$ (1)

where

$\omega_0 = 0$

$\omega = 34.7\:\text{rad/s}$

$\alpha = 47.0\:\text{rad/s}^2$

Using these values, we can solve for $\theta$ from Eqn(1) as follows:

$2\alpha\theta = \omega^2 - \omega_0^2$

$\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}$

$\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}$

$\:\:\:\:= 12.8\:\text{rad}$

7 0

2 years ago

When measuring speed, you will measure the

qaws [65]

Answer:

Speed is the rate at which an object's position changes, measured in meters per second. The equation for speed is simple: distance divided by time

Explanation:

5 0

3 years ago

A +12 Î¼C charge and -8 Î¼C charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t

Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle $q_1=+12\ \mu C$

Charge of second Particle $q_2=-8\ \mu C$

distance between them $d=4\ cm$

$k=9\times 10^{9}$

magnetic field due to first charge at mid-way between two charged particles is

$E_1=\frac{kq_1}{r^2}$

$r=\frac{d}{2}=\frac{4}{2}=2\ cm$

$E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_1=27\times 10^7\ N/C$ (away from it)

Electric field due to $q_2=-8\ \mu C$

$E_2=\frac{kq_2}{r^2}$

$E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_2=-18\times 10^7\ N/C$ (towards it)

$E_{net}=E_1+E_2$

$E_{net}=9\times 10^7\ N/C$ (away from first charge)

8 0

3 years ago