The kinetic theory states that the higher the temperature, the faster the
1 answer:
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Answer:
Time, t = 80 seconds
Explanation:
Given that,
The frequency of the oscillating mass, f = 1.25 Hz
Number of oscillations, n = 100
We need to find the time in which it makes 100 oscillations. We know that the frequency of an object is number of oscillations per unit time. It is given by :
t = 80 seconds
So, it will make 100 oscillations in 80 seconds. Hence, this is the required solution.
Answer:
Radius of the loop is 0.18 m or 18 cm
Explanation:
Given :
Current flowing through the wire, I = 45 A
Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T
Number of turns in circular wire, N = 1
Consider R be the radius of the circular wire.
The magnetic field at the center of the current carrying circular wire is determine by the relation:
Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.
Substitute the suitable values in the above equation.
R = 0.18 m
Answer:
The correct answer is
Explanation:
The formula for the electron drift speed is given as follows,
where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.
Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is . Converting this number to m³ using very elementary unit conversion we get
. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,
if we convert the area from mm³ to m³ we get .So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,
which is our final answer.