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Pani-rosa [81]
3 years ago
11

In scientific notation, 474,000 is written:

Physics
1 answer:
ioda3 years ago
4 0

Answer:

The answer is 4.74 × 10⁵

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3. When two liquids are mixed and a solid
sashaice [31]

Answer:

D

Explanation:

6 0
2 years ago
You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of el
nordsb [41]

Answer:

q = 2.066* 10⁻¹³ C.

n = 1,291,250 electrons.

Explanation:

1)

  • If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:

       F_{g} = F_{c} (1)

  • where Fg is the gravitational attraction, that can be written as follows        according Newton's Universal Law of Gravitation:

       F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)

  • Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming  we can treat both spheres as point charges), as follows:

       F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)

  • since m₁ = m₂ = 0.0024 kg, and  r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:

       G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)

  • Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:

       Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)

  • Since both charges are the same, the charge on each sphere is just the square root of (5):
  • Q = 2.066* 10⁻¹³ C.

2)

  • Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
  • Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:
  • n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)
4 0
3 years ago
A sound wave is a
Stolb23 [73]
A sound wave is a longitudinal wave
6 0
3 years ago
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t
Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = 854\times 10^{3}\ Hz

Power, P = 50 kW = 50\times 10^{3}\ W

I_{p} = 1\ photon/s/m^{2}

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s

Area of the sphere, A = 4\pi r^{2}

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is 1\ photon/s/m^{2}

I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}

1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}

r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m

Now,

We know that:

1 ly = 9.4607\times 10^{15}\ m

Thus

r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly

5 0
3 years ago
Can there be situation when velocity is constant but speed is not
klasskru [66]

Answer:

no I don’t think there can be so my answer is No.

Okay then yes sorry that I must have gotten it wrong before.

Explanation:

4 0
2 years ago
Read 2 more answers
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