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igor_vitrenko [27]
3 years ago
11

2. A ball tied to a pole by a rope swings in a circular path with a centripetal acceleration of 2.7 m/s. If the ball has a

Physics
1 answer:
Helga [31]3 years ago
3 0

Answer: The diameter of the circular path is 2.96m

Explanation: centripetal acceleration = tangential speed^2 / radius of the circular path.

Centripetal acceleration = 2.7m/s^2

Tangential speed = 2.0m/s

Radius = 2.0^2 / 2.7 = 4/2.7

= 1.48m

Diameter = radius*2

= 1.48*2 = 2.96m.

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What is the relationship between electric motors, generators, and transformers?
KengaRu [80]

Answer:

A generator turns rotary motion into electricity. It is basically the inverse of a motor. Generally a transformer changes one voltage into another based on the number of conductor windings on each side. There are two sets of windings called the “primary” and the “secondary”.

Explanation:

6 0
3 years ago
I had gotten this one wrong please help thank you
Nata [24]
Check the 1st 2nd 3rd and 4th boxes
4 0
3 years ago
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
2 years ago
a fly wheel of mass 12kg and radius 0.5m with wrapped around it.on the end of the rope is 2kg load.intially it is at rest.calcul
leva [86]

Answer:

19.6 N of torque. The 2kg load is being affected by acceleration due to gravity which is 9.8 m/s^s

Explanation:

2×9.8=19.6

6 0
2 years ago
Pls help I really need help w this one Tnks
lyudmila [28]
Sorry I do not know srry
6 0
3 years ago
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