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3 years ago

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The force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If th

e charge on particle A is
7.2 × 10-8 coulombs, what is the charge of particle B?
(k = 9.0 × 109 newton·meters2/coulomb2)

A 2.4 × 10-2 coulombs

B 4.6 × 10-11 coulombs

C 5.2 × 10-11 coulombs

D 7.2 × 10-8 coulombs

1 answer:

sergejj [24]3 years ago

6 0

$F_e = \frac{q_1q_2k}{ r^{2} }$

$q_1 = \frac{F_e r^{2} }{q_2k}$

$q_1 = 4,6 * 10^{-11}$

I hope.

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A sound wave is moving through air. The wave has a wavelength of 0.65 m and a frequency of 512 Hz. Calculate the speed of the wa

nexus9112 [7]

Answer:

"332.8 m/s" is the right solution.

Explanation:

The given values are:

Wavelength,

$\lambda=0.65 \ m$

Frequency,

$f = 512 \ Hz$

As we know,

⇒ $Speed=Frequency\times Wavelength$

or,

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3 years ago

A uniform rod is attached to a wall by a hinge at its base. The rod has a mass of 6.5 kg, a length of 1.3 m, is at an angle of 1

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Answer:

tension wire 104 N, horizontal force hinge 104 N, vertical force hinge 63.7 N

Explanation:

The system described is in static equilibrium under the action of several forces, which are shown in the attached diagram, where T is the tension of the wire, W is the weight of the bar, Fx is the horizontal reaction of the wall and Fy It is the vertical reaction of the wall.

The directions of the forces are indicated by the arrows and are marked intuitively, but when solving the problem if one gives a negative value this indicates that the direction is wrong, but does not alter the results of the problem

For the resolution we use Newton's second law, both in translational and rotational equilibrium, if necessary.

We must establish a reference system to assign the positive meaning, we place it with the origin in the hinge, and the positive directions to the right and up. The location of the coordinate system allows us to eliminate the reaction of the hinge by having zero distance to origin. We write the equilibrium equations for each axis

∑ Fx = 0

Fx -T =0

∑ Fy =0

Fy -W =O W = m g

∑τ =0

-T dy + W dx =0

For rotational equilibrium we take the positive direction as counterclockwise rotation and distance is the perpendicular distance of the force to the axis of the coordinate system

dy = L sin 17

dx = (L/2) cos 17

Where L is the length of the bar and the weight is applied at the center of it, we write and simplify the equation

-T L sin 17 + mg (L / 2) cos 17 = 0

- T sin 17 + mg/2 Cos 17 =0

We write the three equations together

Fx- T = 0

Fy -W = 0

-T sin 17 + (m g / 2) cos 17 = 0

a) With the third equation we can find the wire tension

T = m g / 2 cos 17 / sin17

T = 6.5 9.8/2 cotan 17

T = 104 N

b) We use the first equation to find Fx

Fx- T =0

Fx = T = 104 N

c) We use the second equation to find Fy

Fy = W = m g

Fy = 6.5 9.8

Fy = 63.7 N

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3 years ago

The 6 kg block is then released and accelerates to the right, toward the 4 kg block. The surface is rough and the coefficient of

KatRina [158]

Answer:

v =3.41 m/s

Explanation:

given,

mass of block 1 = 6 Kg

mass of another block 2 = 4 Kg

coefficient of friction = 0.3

Assuming 6 Kg block is attached to the spring of spring constant 350 N/m

and distance between the two block is equal to 0.5 m

using formula

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 350 \times 0.5^2$

U = 43.75 J

using conservation of energy

KE = U - f.d

where f is the frictional force acting

$\dfrac{1}{2}mv^2 = 43.75- \mu m g d$

$\dfrac{1}{2}\times 6 \times v^2 = 43.75- 0.3\times 6 \times 9.8 \times 0.5$

$v= \sqrt{11.643}$

v =3.41 m/s

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4 years ago

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Diano4ka-milaya [45]

Answer:

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Explanation:

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3 years ago

Consider the following four objects: a hoop, a flat disk, a solid sphere, and a hollow sphere. Each of the objects has mass M an

Mumz [18]

Answer:

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Explanation:

We need to define the moment of inertia of the different objects, that is,

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HOOP:

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SOLID SPHERE:

$I_{ss} = \frac{2}{5}mR^2$

HOLLOW SPHERE

$I_{hs} = \frac{2}{3}mR^2$

If we have the same acceleration for a Torque applied, then

$mR^2>\frac{2}{3}mR^2>\frac{1}{2} mR^2>\frac{2}{5}mR^2$

$I_{hoop}>I_{hs} >I_{disk}>I_{ss}$

The greatest momement of inertia is for the hoop, therefore will require the largest torque to give the same acceleration

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3 years ago