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4 years ago

Why do you think it would be more practical to use an electromagnet to move scrap metal than to use a permanent magnet?

Physics

1 answer:

Natalija [7]4 years ago

8 0

Answer:

#See solution for details.

Explanation:

-An electromagnet's field can be rapidly changed by just controlling the amount of current. The magnetic strength can be varied over time.

-This way, only a sufficient amount of current will be supplied to meet the needs.

-Unlike the electromagnet, the permanent magnet has a fixed magnetic strength and can only lift a specified weight.

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A force F₁ of magnitude 6.50 units acts at the origin in a direction 49.0° above the positive x axis. A second force F₂ of magni

Sergeu [11.5K]

Found this hope it helps

4 0

4 years ago

A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla

Fiesta28 [93]

Answer:

a) $P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

b) For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

Explanation:

For this case we have the following data given:

$D= 1mm = 0.001 m$ represent the diameter

$r = D/2= 0.0005m$ represent the radius

$T= 5.7 N$ represent the tension

$f = 57 Hz$ represent the frequency of the oscillator

$A= 0.54 cm = 0.0054 m$ represent the amplitude of the wave

Part a

For this case we can assume that the power transmitted to the wave is the same power of the oscillator. and we have the following formula for the power:

$P= 2 \pi^2 \rho S v f^3 A^2$

This expression can be written in different ways:

$P= 2 \pi^2 \rho S \sqrt{\frac{T}{\mu}} f^2 A^2$

$P= 2\pi^2 \rho S \sqrt{\frac{T}{\rho S}} f^2 A^2$

$P= 2 \pi^2 f^2 A^2 \sqrt{S \rho T}$

Where f is the frequency , T the tension $rho= 7800 \frac{kg}{m^3}$ the density of the steel, A the amplitude and $S= \pi r^2$ the area, so then we have everuthing in order to replace and we got:

$P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

Part b

For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

8 0

4 years ago

Your little sister (mass 25 kg) is sitting in her little red wagon (mass

Zielflug [23.3K]

Answer:

p = 60.6N*s

Explanation:

v_f = v_0+a*t

a = (v_f-v_0)/t

a = (1.8m/s)/2.35s

a = 0.77m/s²

F = m*a

F = (25kg+8.5kg)*0.77m/s²

F = 25.8N

^p = F*t

p = 25.8N*2.35s

p = 60.6N*s

5 0

3 years ago

The magnitude of perpendicular vectors can be determined using ....

maxonik [38]

Determined by cross product or ( vector product )

8 0

3 years ago

What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0 m?

Alex_Xolod [135]

Answer:

(a) 69.52 x 10⁻⁹ C

(b) 6.25 N/C

Explanation:

(a) The electric field (E) due to a point charge is directly proportional to the magnitude of the point charge (Q) and inversely proportional to the square of the distance (r) between the point charge and the point where the electric field is. This can be represented as follows;

E = k Q / r² ---------------------------(i)

Where;

k = constant of proportionality called electric constant = 8.99 x 10⁹Nm²/C²

From the question, the following are given;

E = 10000N/C

r = 0.250m

Substitute these values into equation (i) as follows;

=> 10000 = 8.99 x 10⁹ x Q / 0.250²

=> 10000 = 8.99 x 10⁹ x Q / 0.0625

=> 10000 = 143.84 x 10⁹ x Q

Solve for Q;

Q = 10000 / (143.84 x 10⁹)

Q = 69.52 x 10⁻⁹ C

Therefore, the magnitude of the point charge is 69.52 x 10⁻⁹ C.

(b) By the same token, to calculate the magnitude of the electric field at 10.0m, substitute the values of Q = 69.52 x 10⁻⁹ C, k = 8.99 x 10⁹ and r = 10.0m into equation (i) as follows;

=> E = 8.99 x 10⁹ x 69.52 x 10⁻⁹ / 10.0²

=> E = 8.99 x 69.52 / 100.0

=> E = 6.25 N/C

Therefore, the electric field at 10.0m is as large as 6.25 N/C

5 0

3 years ago