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2 years ago

why can scientists ignore the forces of attraction between particles in a gas under ordinary conditions?

Physics

2 answers:

krok68 [10]2 years ago

8 0

They are too weak to have an effect at great distances between gas particles.

Anna35 [415]2 years ago

5 0

<span>The particles in a gas are apart and moving fast, so the forces of attraction are too weak to have a noticeable effect.</span>

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7. A particle is forced to move in a straight line path. It returns to the starting point after 10 s. The total distance covered

lutik1710 [3]

d. All three statements are true.

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1 year ago

In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

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3 years ago

When Dr. Montero was observing the endoplasmic reticulum of a cell using an electron microscope, she noticed that it was covered

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The answer is Convoluted endoplasmic reticulum

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2 years ago

If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in

Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

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3 years ago

A stone tumbles into a mine shaft strikes bottom after falling for 4.2 seconds. How deep is the mine shaft

ziro4ka [17]

$d = u \times t \: + \frac{1}{2} \times a \times {t}^{2}$
Since initial velocity is zero hence , u = 0

=> d = 1/2 * a * t2

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on solving we get

d = 86.436 metres

Note ; Here Gravitational Acceleration is take as , g = 9.8 m/s2

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3 years ago