π=iMRT
Where, π is Osmotic pressure,
i=1 for non-electrolytes,
M is molar concentration of dissolved species (units of mol/L)
R is the ideal gas constant = 0.08206 L atm mol⁻¹K⁻¹,
T is the temperature in Kelvin(K),
Here, to calculate M convert into standard units mg tog, ml to L, c to Kelvin
M= (
*10⁻³ )/ 0.175 =(5.987 *10⁻⁵)mol / 0.175L = 34.21*10⁻⁵ mol/L
π=iMRT=(1)*(34.21*10⁻⁵)*(0.08206)*(298.15)=837×10⁻⁵= 8.37×10⁻³ atm
=6.36 torr
(1 atm=760 torr, 1 Kelvin =273.15 °C, 1L=1000ml, 1g=1000mg)
"Free fall" is the motion of an object when gravity is the ONLY force
acting on it.
In true 'free fall' the speed of an object increases at a constant rate
for the total duration of the fall. The rate of increase, on or near the
Earth's surface, is 9.8 meters per second for each second of fall.
True free fall is almost impossible to observe in everyday life, because
whenever we see anything falling, it's almost always falling through air,
so gravity is NOT the only force acting on it. The friction due to the
motion through air works against the gravitational force. In many cases,
the result is that the object's speed eventually stops increasing and
becomes constant, at a speed often described with the faux technical,
high-fallutin' sounding phrase "terminal velocity". It must be understood
that 'terminal velocity' is NOT a property of gravity or of free fall, but is
only a result of falling through some surrounding stuff that interferes with
the process of true 'free fall'.
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➷ It would be 'when they have like charges.'
'Like charges' means the same charge. For example, two positive charged objects have like charges.
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If an element has a charge of +1, there is 1 more proton than electrons.
A proton has a charge of +1
A neutron has a charge of 0
A electron has a charge of -1.
For there to be a charge of 0, there would be the same amount of charges for both proton and neutron. To get a charge of 1, you will need 1 more proton.
hope this helps
Answer:
Explanation:
The relation between orbital period T and orbital radius R is as follows .
T² ∝ R³
T ∝ R¹°⁵
So time period of orbit is proportional to radius of orbit . Higher the height , larger the orbital period . As the orbital period is larger than required , the altitude of satellite must have been larger than required .
As mass of satellite is not involved in the formula of orbital period , this is independent of mass of the satellite .
Hence the option C is correct .
