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3 years ago

Please any one need help on this

Physics

1 answer:

Romashka [77]3 years ago

3 0

-- The wavelength and the amplitude were described in my answer to your previous question.

-- A "compression" is a place where the wave is <em>compressed</em>. It's the darker section of the wave in the picture, where the wavelength is temporarily shorter, so several waves are all bunched up (compressed) in a small time.

-- A "rarefaction" is exactly the opposite of a "compression". It's a place where the wave gets more "<em>rare</em>" ... the wavelength temporarily gets longer, so that several waves get stretched out, and there are fewer of them in some period of time. The arrow in the picture points to a rarefaction.

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In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single

FinnZ [79.3K]

Answer:

The ratio of electric force to the gravitational force is $2.27\times 10^{39}$

Explanation:

It is given that,

Distance between electron and proton, $r=4.53\ A=4.53\times 10^{-10}\ m$

Electric force is given by :

$F_e=k\dfrac{q_1q_2}{r^2}$

Gravitational force is given by :

$F_g=G\dfrac{m_1m_2}{r^2}$

Where

$m_1$ is mass of electron, $m_1=9.1\times 10^{-31}\ kg$

$m_2$ is mass of proton, $m_2=1.67\times 10^{-27}\ kg$

$q_1$ is charge on electron, $q_1=-1.6\times 10^{-19}\ kg$

$q_2$ is charge on proton, $q_2=1.6\times 10^{-19}\ kg$

$\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}$

$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}$

$\dfrac{F_e}{F_g}=2.27\times 10^{39}$

So, the ratio of electric force to the gravitational force is $2.27\times 10^{39}$ . Hence, this is the required solution.

3 0

3 years ago

A hydraulic lift has pistons with diameter 28cm and 70cm, respectively. If a force of 500 N is exerted at the input piston. What

Crazy boy [7]

Answer:

3125 N

Explanation:

diameter /2 =radius

so r1 =14cm , r2 =35cm

f1/A1 =f2/A2.

f2 = f1 × A2 / A1

=500×1225 pi cm² / 96 pi cm²

f2 =3125N

4 0

3 years ago

If you had 8 balls and 7 of them were a certain weight, and 1 of them was heavier, how could you find the heaviest ball. All the

dybincka [34]

Answer:

There are two method of comparing the balls 1) using a balance 2) by only 2 weighings.

Explanation:

There are two method of comparing the balls 1) using a balance 2) by only 2 weighings.

Make the following groups - --- (1,2,3),(4,5,6),(7,8)

Step 1. compare the Weigh (1,2,3) and (4,5,6)

there are 2 possible outcomes:

1---both the group are of same weight. and named as (Case A)

2--- one of the group is heavier than other and named as (Case B)

Step 2. Let examine both case

In Case A --in this case, now compare the weight of 7th and 8th ball. By this you have recognize the heavier ball by 2 weighing method.

In Case B -- considered the heaviest group (assume group (1,2,3) is heavy), from this group take randomly two ball and compare the their weight. out of these two ball, one is heavy else the third ball is.

7 0

3 years ago

For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci

MAXImum [283]

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

$z = \frac{V_o}{i_o}$

now we will have

$z = \frac{125}{3.20} = 39.06 ohm$

Part b)

Now we also know that

$\frac{R}{z} = cos\theta$

$\theta = 0.982 rad = 56.3 degree$

now we have

$\frac{R}{39.06} = cos56.3$

$R = 21.7 ohm$

5 0

3 years ago

A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteri

fgiga [73]

Answer:

4.29×10⁵ C

Explanation:

From the question,

The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.

E = 1/2mv²+mgh+Fd.................... Equation 1

Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.

But,

d = vt.................... Equation 2

Where v = velocity, t = time.

Substitute equation 2 into equation 1

E = 1/2mv²+mgh+F(vt)................... Equation 3

Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²

Substitute into equation 1

E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)

E = 260260+1622390+49608000

E = 51490650 J

Using,

E = qV................. Equation 4

Where q = charge of the battery, V = Voltage.

make q the subject of the equation

q = E/V............... Equation 5

Given: E = 51490650 J, V = 120 V

Substitute into equation 5

q = 51490650/120

q = 429088.75 C

q = 4.29×10⁵ C

7 0

3 years ago