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Mandarinka [93]
3 years ago
8

What is the experimental group(s)?

Physics
1 answer:
Vinil7 [7]3 years ago
5 0
<h3>I think it B The group(s) that gets the special treatment.</h3><h3 /><h3>I hope this is correct.</h3><h3 /><h3 /><h3 />
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true or false: a image created by an object located inside the focal point is virtual, enlarged, and upright
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true or false: a image created by an object located inside the focal point is virtual, enlarged, and upright

True :D

4 0
3 years ago
Pls help im begging you
Lostsunrise [7]

Answer:

I think it's TRUE because forces change an object's motion but dont quote me on it ok? Cause I'm not a 100 percent sure

7 0
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Cool facts about satellites ???
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2 years ago
A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3
4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

8 0
3 years ago
A force of 100N is applied to move an object a horizontal distance of 20m to the right. The work done by this force on the objec
horsena [70]
WORKDONE = FORCE * DISPLACEMENT
W=F*S
HERE, THE FORCE = 100N AND DISTANCE = 20M
WORKDONE = 100*20
WORKDONE=2000
ITS S.I UNIT IS JOULE OR J
SO, 2000J
5 0
3 years ago
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