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3 years ago

15

Explain how a charged amber rod can lift tiny bits of paper despite being under the gravitational force of the entire mass of th

e earth.

1 answer:

nika2105 [10]3 years ago

4 0

Explanation:

When a charged amber rod is brought near the tiny bits of paper. The bits of paper are neutral. There is an electrostatic force of attraction between the amber rod and tiny bits of paper.

The charged amber rod attracts or repel the bits of paper depending on the charge of amber rod. The electrostatic force is much stronger than the gravitational force.

Therefore, a charged amber rod can lift tiny bits of paper despite being under the gravitational force of the entire mass of the earth.

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You observe a star cluster with a main-sequence turn-off point at spectral type G2 (the same spectral type as the Sun). What is

Goryan [66]

Answer i dont even know im just putting this cus id ont care

Explanation:

3 0

2 years ago

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

7 0

3 years ago

You use 35 J of energy to move an object 5 m. What is the weight of the object

Vlad [161]

Explanation:

35 ÷ 5 = 7kg.

=======================

7 0

3 years ago

A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.

Fiesta28 [93]

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

$d=30,000 pc =9.26\cdot 10^{20} m$

The speed of the cosmic ray is

$v=0.98 c$

where

$c=3.0 \cdot 10^8 m/s$ is the speed of light. Substituting,

$v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s$

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

$T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s$

Converting into years,

$T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years$

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

$T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}$

And substituting v = 0.98c, we find:

$T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years$

3 0

3 years ago

A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?

anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}$

hence, the answer is 0.495 m/s

3 0

1 year ago