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3 years ago

A point charge of 6.30 uC is lecated 15.0 cm from a second point charge of-4.80 yC. The force on the 6.30 juC charge is

Physics

1 answer:

faltersainse [42]3 years ago

7 0

Answer:

12.096 N

Explanation:

Q = 6.30 micro coulomb = 6.3 x 10^-6 C

q = - 4.8 micro coulomb = - 4.8 x 10^-6 C

d = 15 cm = 0.15 m

According to the Coulomb's law, the force between the two charges is given by

$F = \frac{KQq}{d^{2}}$

Where, k be the Coulombic constant which is equal to 9 x 10^9 Nm^2 / C^2. This value is for free space or vacuum be the medium between two charges.

By substituting the values

$F = \frac{9\times 10^{9}\times 6.3 \times10^{-6}\times 4.8\times 10^{-6}}{0.15\times 0.15}$

F = 12.096 N

This force is attractive in nature as the charges as unlike in nature and by the properties of charges, if there are two unlike charges, then the force between the charges is attractive in nature.

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bixtya [17]

Answer:

Since the ball becomes positively charged, it will repel as like charges repel.

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3 years ago

What is the least force you could exert on the refrigerator to move it?

Ulleksa [173]

<span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365

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Fsf = (0.49)(1365N)</span><span>
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3 years ago

A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N.

kaheart [24]

A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be 372.77 m/s

<h3>What is a sound wave?</h3>

It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.

For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula

V= √F/μ

where v is the wave velocity of the wave travel on the string

F is the tension in the string of piano

μ is the mass per unit length of the string

As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.

The μ is the mass per unit length of the string would be

μ = 4.80/(1.150×1000)

μ = 0.0041739 kg/m

By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity

V= √F/μ

V=√(580/0.0041739)

V = 372.77 m/s

Thus, the speed of transverse waves on the wire comes out to be 372.77 m/s

Learn more about sound waves from here

brainly.com/question/11797560

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6 0

1 year ago

Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude

DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, $v_p_i$ = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

$W =K.E_f - K.E_i$

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d = (EQ) x d = EQd

$K.E_f =EQd + \frac{1}{2}m_pv_p_i^2$

$m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C$

$K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5 \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\$

$K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J$

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0

3 years ago

For a storm to be a blizzard, the wind must be at least

Anton [14]

For a storm to be a blizzard, the wind must be at least 35 miles per hour. This is just one criteria for considering a storm to be a blizzard. The wind speed of 35 miles per hour should reduce the visibility to less than 400 meters. The last criteria is that the storm must continue for a time frame of at least 3 hours.

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3 years ago