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Elodia [21]
3 years ago
12

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

observes that a particular spring-mass system has a frequency of oscillation of 10 Hz. the spring constant of the spring is 250 N/m. what is the mass?​
Physics
1 answer:
pochemuha3 years ago
6 0

Answer:

m = 63 grams

Explanation:

ω = 10 cycles/s(2π radians/cycle) = 20π rad/s

ω = √(k/m)

m = k/ω² = 250/(20π)² = 0.06332... kg

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Answer:

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Explanation:

For the first blank, the answer is decreases. For the second blank, the answer is increases. And finally for the third blank, the answer is decreases.

When a liquid is cooled, the kinetic energy of the particles decreases. The force of attraction between the particles increases, the space between the particles decreases, and the matter changes its state to solid.

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Batman chased the joker on his batbike for 20 minutes, traveling at the speed of 30 kilometers per minute. How far did Batman go
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123 kilometer

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Statistically, consumers tell more people about negative experiences in business encounters than about positive experiences. A)
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A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis
bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

B=\frac{\mu_o I_r}{2\pi r}     (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

I_r=JA_r

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by  A_r=\pi(r^2-R_1^2)

The current density is given by the total area and the total current:

J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}  (2)

You replace the last result of equation (2) into the equation (1):

B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})

Finally. you replace the values of all parameters:

B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0
3 years ago
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