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2 years ago

5

What is the correct name for the ionic compound, MgCl2? A) manganese chloride B) magnesium chloride C) magnesium dichloride D) m

anganese dichloride

2 answers:

VladimirAG [237]2 years ago

7 0

The corret answer is B. I just had this question on USATestPrep.

Aloiza [94]2 years ago

6 0

Magnesium dichloride correct name for the ionic compound, MgCl2

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yarga [219]

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2 years ago

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This same car gets pulled over for speeding, and goes from 68 m/s to 0 m/s in 14

Harrizon [31]

Answer:

the acceleration of the car is -4.9m/s2.

the direction is opposite to the actual direction, since the acceleration is negative.

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2 years ago

A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the

Aleksandr-060686 [28]

Answer:

(a): The magnitude of the electric force on the small sphere = $\dfrac{q\sigma}{2\epsilon_o}.$

(b): Shown below.

Explanation:

<u>Given:</u>

m = mass of the small sphere.
q = charge on the small sphere.
L = length of the silk fiber.
$\sigma$ = surface charge density of the large vertical insulating sheet.

<h2>(a):</h2>

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

$\rm E = \dfrac{\sigma}{2\epsilon_o}.$

<em>where,</em>

$\epsilon_o$ is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

$\rm F_e=qE.$

Thus, the magnitude of the electric force on the small sphere is given by

$\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.$

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

<h2>(b):</h2>

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

$\rm \tan \theta = \dfrac{F_e}{W}.$

<em>where,</em>

$\rm F_e$ = electric force on the sphere, acting along left.
$\rm W$ = weight of the sphere, acting vertically downwards.

<em />

$\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .$

g is the acceleration due to gravity.

6 0

2 years ago

A point charge of +3 C is located at the origin of a coordinate system and a second point charge of -6 C is at x = 1.0 m. At w

Butoxors [25]

Answer:

The point at which the electrical potential is zero is x = +0.33 m.

Explanation:

By definition the electrical potential is:

$V_{E} = \frac{K*q}{r}$

Where:

K: is Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

r: is the distance

The point at which the electrical potential is zero can be calculated as follows:

$V_{1} + V_{2} = 0$

$K(\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}) = 0$ (1)

q₁ is the first charge = +3 mC

r₁ is the distance from the point to the first charge

q₂ is the first charge = -6 mC

r₂ is the distance from the point to the second charge

By replacing r₁ = 1 - r₂ into equation (1) we have:

$K(\frac{q_{1}}{1 - r_{2}} + \frac{q_{2}}{r_{2}}) = 0$ (2)

By solving equation (2) for r₂:

$r_{2} = \frac{q_{1}}{q_{1} - q_{2}} = \frac{3 mC}{3 mC - (-6 mC)} = +0.33 m$

Therefore, the point at which the electrical potential is zero is x = +0.33 m.

I hope it helps you!

8 0

2 years ago

A 7kg block slides down a 30 degree slope at a constant velocity of 2 m/s. How big is the force of friction acting on the block?

lisabon 2012 [21]

The force of friction is <u>34.3 N.</u>

A block of mass m slides down a plane inclined at an angle θ to the horizontal with a constant velocity. According to Newton's first law of motion, every body continues in its state of rest or a state of uniform motion in a straight line, unless acted upon, by an external unbalanced force. This means that when balanced forces act on a body, the body moves with a constant velocity.

The free body diagram of the sliding block is shown in the attached diagram. Resolve the weight mg of the block into two components mg sinθ along the direction of the plane and mg cosθ perpendicular to the plane . The force of friction F acts upwards along the plane and the normal reaction acts perpendicular to the plane.

Since the block moves down with a constant velocity, the downward force mg sinθ must be equal to the upward frictional force.

$F = mg sin\theta$

Substitute 7 kg for m, 9.8 m/s² for g and 30° for θ.

$F = mg sin\theta = (7 kg)(9.8 m/s^2)(sin30^o) =34.3 N$

The force of friction is <u>34.3 N</u> up the plane.

6 0

3 years ago