Ai là người phát hiện trái đất hình cầu đầu tiên ?

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

A hydrogen atom in the n=7 state decays to the n=4 state. what is the wavelength of the photon that the hydrogen atom emits? use

frez [133]

A hydrogen atom in the n=7 state decays to the n=4 state. The wavelength of the photon that the hydrogen atom emits is 4592.59nm.

The Energy of photon is the energy possessed by a photon when it moves from a high energy level to a low energy level. It emits a photon of a certain wavelength. The following relation can be used to find out the relation between the energy levels and the energy possessed:

E = 13.6 × Z² (1/n₂² - 1/n₁²) eV

where, n₁ is the initial energy level i.e. n₁ =7

n₂ is the higher energy level i.e. n₂ = 4

E is the energy possessed

Z is the atomic number, Z = 1 for H-atom

Subsituting in above equation,

E = 13.6 (1/16 - 1/49) eV

E = 0.27 eV

We know that,

E = hc / λ

where, h is Planck constant

c is speed of light

λ is wavelength

On subsituting,

0.27 eV = 1240/ λ

⇒ λ = 4592.59 nm

Hence, the wavelength of photon emitted by Hydrogen atom is 4592.59nm.

Learn more about Energy of Photon here, brainly.com/question/2393994

#SPJ4

5 0

2 years ago

How many full length strands of hair are collected from the scalp to use as a sample?

laila [671]

50 strands is the standard procedure

4 0

3 years ago

A flute player hears four beats per second when she compares her note to a 523 HzHz tuning fork (the note C). She can match the

laiz [17]

Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, $\Delta f = 4\ beats/s$

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

$\Delta f = f - f_{i}$

$f_{i} = f \pm \Delta f$

when

$f_{i} = f + \Delta f = 523 + 4 = 527 Hz$

when

$f_{i} = f - \Delta f = 523 - 4 = 519 Hz$

But we know that as the length of the flute increases the frequency decreases

Hence, the initial frequency must be 527 Hz

7 0

3 years ago

Read 2 more answers

I need help on 18,19,20 and 21

Dmitriy789 [7]

When a relationship between two different things is shown in a fraction it is a ratio.

hope this helps :)

8 0

3 years ago