Answer:
Both the frequency f and velocity v will increase.
When the radius reduces, the circumference of the circular path becomes smaller which means that more number of revolutions can be made per unit time as long as the force is kept constant; this is an increase in frequency.
Explanation:
The centripetal force acting on a mass in circular motion is given by equation (1);
where m is the mass of the object and r is radius of the circle. From equation one we see that the centripetal force is directly proportional to the square of the velocity and inversely proportional to the radius of the circular path.
However, according to the problem, the force is constant while the radius and the velocity changes. Therefore we can write the following equation;
Also recall that m is constant so it cancels out from both sides of equation (2). Therefore from equation we can write the following;
By observing equation (2) carefully, the ratio 
will with the square root increase 
since 
is lesser than 
.
Hence by implication, the value of 
will be greater than .
As the radius changes from to , the velocity also changes from to .
According to Edge, the answers is
<h3>3100 V</h3>
and
<h3>200v</h3>
New experiment verify or find previous model
Answer:
Vrms = 291 m/s
Explanation:
The root mean square velocity or vrms is the square root of the average square velocity and is. vrms=√3RTM. Where M is equal to the molar mass of the molecule in kg/mol.
Temperature = 365 K
Root mean square velocity = ?
molar mass of oxygen = 16 g/mol.
But xygen gas (O2) is comprised of two oxygen atoms bonded together. Therefore:
molar mass of O2 = 2 x 16
molar mass of O2 = 32 g/mol
Convert this to kg/mol:
molar mass of O2 = 32 g/mol x 1 kg/1000 g
molar mass of O2 = 3.2 x 10-2 kg/mol
Molar mass of Oxygen = 3.2 x 10-2 kg/mol
Vrms = √[3(8.3145 (kg·m2/sec2)/K·mol)(365 K)/3.2 x 10-2 kg/mol]
Vrms = 291 m/s
<span>There is no special name for that. Physics is usually just concerned with "forces", and doesn't specify whether the force pushes or pulls. If you want to be more specific, you can just call it a "pulling force".
I hoped this was satisfying!:)</span>
