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Oksana_A [137]
3 years ago
8

Displacement time squared graph

Physics
1 answer:
Scilla [17]3 years ago
6 0
If distance versus time<span> is a straight line then Free-Fall is at a constant velocity and the slope of the </span>graph<span> measures that velocity. If distance versus </span>time-squared<span> is a straight line then Free-Fall is at a constant acceleration and the slope of the </span>graph<span>measures one-half of that acceleration.</span>
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what volume of alcohol will have the same mass as 4.2m^3 of petrol? (density of alcohol 0.4kg/m^3 and petrol is 0.3kg/m^3)​
Alexxx [7]

Answer:

3.15m³

Explanation:

To solve this problem, let us first find the mass of the petrol from the given dimension.

       Mass  = density x volume

Volume of petrol  = 4.2m³

Density of petrol  = 0.3kgm⁻³  

       Mass of petrol  = 4.2 x 0.3  = 1.26kg

So;

      We can now find the volume of the alcohol

 Volume of alcohol = \frac{mass}{density}  

Mass of alcohol  = 1.26kg

Density of alcohol  = 0.4kgm⁻³  

  Volume of alcohol  = \frac{1.26}{0.4}   = 3.15m³

7 0
3 years ago
Which is the result of mitosis and cytokinesis?
Leno4ka [110]
During cytokinesis, the cytoplasm of the cell is divided in half, and the cell membrane grows to enclose each cell, forming two separate cells as a result. The end result of mitosis and cytokinesis is two genetically identical cells where only one cell existed before.
3 0
3 years ago
What effect does a tripling of the net force have upon the acceleration of the object?
dsp73
The formula of net Force is:F = mawhere m is the mass of the objecta is the acceleration of the object
thus, if we triple the net force applied to the object:
3F = maa = 3F / m
The acceleration is also tripled since the force is directly proportional to the acceleration.


4 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.
Citrus2011 [14]

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}

Here, R_{\infty} is the "general" Rydberg constant, m_e is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, M=1.67*10^{-27}kg:

R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}

Now, we calculate the wavelength for hydrogen:

\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm

For deuterium, we have M=2(1.67*10^{-27}kg):

R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm

5 0
3 years ago
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