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3 years ago

The diagram shows two charged objects, X and Y.

Physics

2 answers:

Vanyuwa [196]3 years ago

7 0

X: positive

Y: positive

Explanation:

To answer correctly the question, we need to remind the difference between the electric field produced by a positive charge and by a negative charge:

- For a positive charge, the direction of the field lines is away from the charge (because they represent the direction in which a positive test charge would move, if placed in this field)

- For a negative charge, the direction of the field lines is toward the charge (because the positive test charge would move toward the negative charge, if placed in this field)

In the figure, we see that both charges have electric field lines that go away from the charges: therefore, according to the previous statements, they both have to be positive charges.

Scilla [17]3 years ago

3 0

Answer: both charges are positive.

Explanation :

The imaginary lines around a bar magnet are called the electric field lines. The direction of field lines is from positive charge to negative charge.

The electric field lines for the positive charge is in outward direction while for negative charge the field lines are inwards.

In the given figure, the electric field lines are away from both of the charges. So, both charges X and Y must be positive.

Hence, the correct option is (D) " X: positive and Y: positive ".

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What is the net force on the object represented by the FBD below?

ololo11 [35]

Explanation:

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3 years ago

a 5.0 kg block is at rest on a horizontal floor. if you push horizontally on the 5.0 kg block with a force of 15.0 n, it just st

zheka24 [161]

For the 5.0Kg block with the force of 15 Newton, the coefficient of static friction (μ) comes out to be 0.306. Applying the formula F = μmg.

Static friction acts on stationary body/body at rest. Let μ be the coefficient of static friction between the block and the horizontal floor.

Using the formula: F = μmg

Where,

F = Force , m= mass of the block and g = gravity.

and values of: m= 5.0Kg, F= 15.0N, g= 9.8m/s²

We can get: μ = F/(mg)

μ = 15/49

μ =0.306

Therefore, coefficient of static friction (μ) = 0.306.

To learn more about static friction visit the link- brainly.com/question/13680415

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2 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

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4 years ago

What are the dimensions of B

joja [24]

Well i think the answer is impossible to find because there is no picture

4 0

4 years ago

A force of 1.200×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 615 N. If he starts

AleksAgata [21]

Answer:

15.45 m/s

Explanation:

F = 1.2 x 10^3 N, Fk = 615 N, u = 0, s = 25 m

Let the speed after traveling 25 m is v.

mass = F / g = 1200 / 9.8 = 122.45 kg

Net force, Fnet = 1.2 x 1000 - 615 = 1200 - 615 = 585 N

acceleration = Fnet / mass = 585 / 122.45 = 4.78 m/s^2

Use third equation of motion

v^2 = u^2 + 2 a s

v^2 = 0 + 2 x 4.78 x 25 = 238.88

v = 15.45 m/s

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3 years ago