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svlad2 [7]
3 years ago
13

List the metric base units for the following:

Physics
2 answers:
Sedbober [7]3 years ago
7 0
The base unit for length is meters, which can be compared to yards in the American system.

Generally length and distance mean the same thing, but seeing as they are listed separately here I will assume it means distance as in travel. The metric system used kilometers for distance, which can be compared to American miles.

The mass of objects is measured in grams, which is sort of like ounces in the American system.

Finally, temperature is measured in Celsius, which is Fahrenheit with the American system.  
77julia77 [94]3 years ago
4 0
Meter
meter
gram
Celsius
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An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos
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Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v}    \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm

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What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?
MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

I = \frac{P}{A}

I = \frac{P}{4\pi r^2}

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

I = \frac{100}{4\pi (2.5)^2}

I = 1.2738W/m^2

The relation between intensity I and E_{max}

I = \frac{E_max^2}{2\mu_0 c}

Here,

\mu_0 = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

E_{max}^2 = 2I\mu_0 c

E_{max}=\sqrt{2I\mu_0 c }

E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)

E_{max} = 30.982 V/m

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

B_{max} = \frac{E_{max}}{c}

B_{max} = \frac{30.982 V /m}{3*10^8}

B_{max} = 1.03275 *10{-7} T

Therefore the maximum value of the magnetic field is B_{max} = 1.03275 *10{-7} T

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