Slow pain originating from an internal organ is often perceived as coming from an area of the body totally unrelated to the source. this is known as

There is a pattern of current flow. Based on the above scenario,  Option b is the correct answer as bulb B will be brighter than before (Image attached).

Based on the current flowing through bulb A  is said to remains the same in both scenario. It is the current through bulb B that changes when the switch  are said to be closed.

There is something that happens to the brightness of a bulb when the switch is said to be closed. Note that when the switch is closed, the light bulb often operates because of the current flows via the circuit. The bulb (B) tends to glows at its full brightness because it receives more volts.

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Answer:

The work is  371 lb ft

Explanation:

The work can be calculated with yhe scalar product between the force and the displacement:

[tex] W = F.d [/tex]

The scalar product is obtained multiplying component to comoponent and summing the terms:

[tex] W [/tex] = <47, 30> · <13, -8>  [lb ft]

[tex] W [/tex] = 47*13  + 30*(-8) [lb ft]

Finally:

W = 371 [lb ft]

The answer in this question is scale. The scale is used to determine the actual distance between two points in a planimetric map. This map only represents the horizontal position of features. This is also called by others as the line map.  This is usually consisted of man-made and natural features. These are from aerial shots by the use of modern technology. This map’s common features are street and water centerlines, sidewalks, culverts, utility lines, building footprints and vegetation. Sometimes, there are also other useful data attached in this map like titles of the properties, owner, assessed value, etc. It is because of these, this map becomes a storage of valuable data which can be easily and quickly accessed. 

The scale on a planimetric map is used to determine the actual distance between two points by converting measured map distances to real-world distances.

On a planimetric map, the scale is used to determine the actual distance between two points. By measuring the distance between two points on the map with a ruler and then using the map scale to find the real distance on the land surface, it is possible to convert map distances to real-world distances. For example, if you measure 10 centimeters between two points on a map with a scale of 1:10,000, you multiply the measured distance by the scale factor, resulting in an actual distance of 100,000 centimeters on the Earth's surface.

It is important to note that large-scale maps, like a standard topographic map, typically have less distance distortion, making it easier to measure straight line distances. However, when measuring distances along curved features, techniques such as using multiple straight-line segments or tools like an opisometer may be employed for greater accuracy. Map scales can be complex, as they can vary with latitude due to the Earth's curvature, particularly on maps that cover large areas such as equidistant projections which are true-to-scale only along meridians.

Answer:

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

Explanation:

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t²      (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²    

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the  equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m