Diane is training for a marathon and goes to the track (400 m) to run every morning.

Physics

1 answer:

stiv31 [10]3 years ago

3 0

Her displacement after any times round the track is zero.
The distance after 6 laps is 2400m

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After you enlarge a map, which one of the following scale remains correct?

9966 [12]

Answer:

None

Explanation:

An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:

Verbal scale use of words to represent scale information on the map. The distance or linear units are used for depicting this scale on the map. For example: 1 inch = 1 Kilo meter.
Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio. Example: 1: 10,000 or 1/10,000
In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given. Example: 1:500
Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map. Example: 1:25,000
A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.

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6 0

3 years ago

A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius

ahrayia [7]

Answer:

The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.

at r < R:

Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.

E = 0.

at R < r < 2R:

The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.

at 2R < r:

The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.

Explanation:

Gauss’ Law is straightforward when applied to spheres. The area of the sphere is $A = 4\pi r^2$ , and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.