All categories

3 years ago

I have a problem in this questions?

Physics

1 answer:

photoshop1234 [79]3 years ago

7 0

Answer:

8.46E+1

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = 39 C

Charge 2 (q₂) = –53 C

Force (F) of attraction = 26×10⁸ N

Electrical constant K) = 9×10⁹ Nm²/C²

Distance apart (r) =?

The distance between the two charges can be obtained as follow:

F = Kq₁q₂ / r²

26×10⁸ = 9×10⁹ × 39 × 53 / r²

26×10⁸ = 1.8603×10¹³ / r²

Cross multiply

26×10⁸ × r² = 1.8603×10¹³

Divide both side by 26×10⁸

r² = 1.8603×10¹³ / 26×10⁸

r² = 7155

Take the square root of both side

r = √7155

r = 84.6 m

r = 8.46E+1 m

You might be interested in

Which of these are a covalent compound?

nikdorinn [45]

C is a non-metal and so is O. So the answer is CO

8 0

3 years ago

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon-population sy

Genrish500 [490]

The question is incomplete. The complete question is :

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon population system move? Assume the population is 7 billion, the average human has a mass of 65 kg, and that the population is evenly distributed over both the Earth and the Moon. The mass of the Earth is 5.97×1024 kg and that of the Moon is 7.34×1022 kg. The radius of the Moon’s orbit is about 3.84×105 m.

Solution :

Given :

Mass of earth, $M_e = 5.97 \times 10^{24} \ kg$

Mass of moon, $M_m = 7.34 \times 10^{22} \ kg$

Mass of each human, $m_p =65 \ kg$

Therefore mass of total population, $M_p = 65 \times 7 \times 10^{9} \ kg$

$M_p = 4.55 \times 10^{11} \ kg$

Let the earth is at the origin of the coordinate system. Then,

Since $M_e>> M_p$

$M_m>> M_p$

Hence if we shift all the population on the moon there will be negligible change in the mass of the moon and earth. Hence there will not be any significant shift on the centre of mass. i.e.

$X_{cm} = \frac{5.97 \times 10^{24}+ 7.34 \times 10^{22} \times 3.84 \times 10^5}{5.97 \times 10^{24}+ 7.34 \times 10^{22}}$

$= 4.68 \times 10^6 \ m$

$4.68 \times 10^3 \ km$ from the earth.

3 0

3 years ago

imagine that you have a sample of an unknown element and that you know only the elements mass number without directly measuring

Ostrovityanka [42]

If the element's atomic mass is accurate enough you should be able to tell from that , however if it isn't you would most likely want the number of neutrons or electrons contained by the atom.

4 0

3 years ago

A bucket is filled partly with water such that its combined mass is 2.5 kg. It is tied to a rope and whirled in a circle with a

Ivenika [448]

Answer:

1. Simply τ = m x g x r = 54kg x 9.8m/s² x 0.050m = 26 N·m

2. The bucket creates a torque

τ = 75kg x 9.8m/s² x 0.075m = 55 N·m,

so we must create the same torque with the handle.

55 N·m = F x 0.25m

F = 220 N

Explanation:

Hope this is helpful

7 0

3 years ago

A magnetic pickup is used to detect the shaft speed of an experimental high speed compressor. a gear with 12 teeth is used to ex

worty [1.4K]

The answer would be 39 seconds

5 0

4 years ago

Read 2 more answers