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3 years ago

which describes the attraction or repulsion caused by magnets? a. magnetic domain b. magnetic force c. magnetic field

1 answer:

5 0

B. Magnetic force-attraction or repulsion that arises between electrically charged particles because of their motion. It is the basic force responsible for such effects as the action of electric motors and the attraction of magnets for iron. Electric forces exist among stationary electric charges; both electric and magnetic forces exist among moving electric charges. The magnetic force between two moving charges may be described as the effect exerted upon either charge by a magnetic field created by the other.

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3 years ago

A pair of oppositely charged parallel plates is separated by 5.38 mm. A potential difference of 623 V exists between the plates.

Salsk061 [2.6K]

Answer:

115799.256V/m

Explanation:

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3 years ago

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2 years ago

Read 2 more answers

Three persons wants to push a wheel cart in the direction marked x in Fig. The two person push with horizontal forces F1 and F2

Svetllana [295]

Answer:

I had to search the Figure on Google to solve this question.

a) The magnitude of the force F₃ is:

$F_{3} = 87.47 N$

And the direction of F₃:

$\alpha = 79.04 ^{\circ}$ (with respect to the y-direction, in the third quadrant)

b) P = 4.22 N

Explanation:

I had to search the Figure on Google to solve this question.

a) We can find the force of the third person as follows:

$\Sigma F_{x} = F_{1x} + F_{2x} + F_{3x} = 0$

$\Sigma F_{y} = F_{1y} - F_{2y} + F_{3y} = 0$

So, in x-direction we have:

$\Sigma F_{x} = 45 N*cos(70) + 75 N*cos(20) + F_{3x} = 0$

$F_{3x} = -85.87 N$

In y-direction we have:

$\Sigma F_{y} = 45 N*sin(70) - 75 N*sin(20) + F_{3y} = 0$

$F_{3y} = -16.63 N$

The magnitude of the force F₃ is:

$F_{3} = \sqrt{F_{3x}^{2} + F_{3y}^{2}} = \sqrt{(-85.87 N)^{2} + (-16.63 N)^{2}} = 87.47 N$

To find the direction of F₃ we need to calculate its angle with respect to the y-direction (in the third quadrant):

$tan(\alpha) = \frac{|F_{3x}|}{|F_{3y}|} = \frac{85.87 N}{16.63 N}$

$\alpha = 79.04 ^{\circ}$

b) If the third person exerts the force found in part (a) the car will stop, so the only way for the cart to accelerate at 200 m/s² is that the third person does not exert the force found in a.

To find the weight of the cart when it accelerates at 200 m/s², we need to consider: F₃ = 0.

First, we need to find the cart's mass. Since the car is moving in the x-direction we have:

$\Sigma F_{x} = F_{1x} + F_{2x} = ma$

$45 N*cos(70) + 75 N*cos(20) = m*200 m/s^{2}$

$m = \frac{45 N*cos(70) + 75 N*cos(20)}{200 m/s^{2}} = 0.43 kg$

Now, the weight of the cart is:

$P = mg = 0.43 kg*9.81 m/s^{2} = 4.22 N$

I hope it helps you!

3 0

2 years ago