If the volume of a gaseous system is increased by a factor of 3 and the temperature is raised by a factor of 6, then the pressur
1 answer:
You might be interested in
1) the weight of an object at Earth's surface is given by 
, where m is the mass of the object and 
is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is
2) On Mars, the value of the gravitational acceleration is different:
. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: 
3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
<span>6) On Earth, the gravity acceleration is </span><span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
</span>
<span>
</span>
2) On Mars, the value of the gravitational acceleration is different:
3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
<span>6) On Earth, the gravity acceleration is </span>
</span>
</span>
Answer:
Number of electrons,
Explanation:
It is given that,
Resistance, R = 4 ohms
Current, I = 3 A
Time, t = 5 min = 300 s
We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.
i.e. q = n e ...............(1)
And current,
e is the charge of an electron
So, the number of electrons pass through the resistor is . Hence, this is the required solution.