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3 years ago

4. Which activity would be BEST for prolonged training within your target heart rate zone?

1 answer:

6 0

I am sorry I would answer this questions but I don’t have the options of the heart rates so I can give you the best answer

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Convert the following to relative uncertainties <br>a) 2.70 ± 0.05cm<br>b) 12.02 ± 0.08cm

DENIUS [597]

data which is expressed in form of following way

$a = a_o + \Delta a$

here in above expression

$a_o$ = true value

$\Delta a$ = uncertainty in the value

now the relative uncertainty is given as

$\frac{\Delta a}{a_o}$

now by above formula we can say

a) 2.70 ± 0.05cm

here

True value = 2.70

uncertainty = 0.05

Relative uncertainty = $\frac{0.05}{2.70}$ = 0.0185

b) 12.02 ± 0.08cm

here

True value = 12.02

uncertainty = 0.08

Relative uncertainty = $\frac{0.08}{12.02}$ = 0.00665

4 0

3 years ago

In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If t

Vitek1552 [10]

Answer:

So the ratio will be $\frac{T_L}{T_H}=-0.171$

Explanation:

We have given heat engine absorbs 450 joule from high temperature reservoir

So $Q=450j$

As the heat engine expels 290 j

So work done W = 290 J

We know that efficiency $\eta =\frac{W}{Q}=\frac{290}{450}=0.6444$

It is given that efficiency of the engine only 55 % of Carnot engine

So efficiency of Carnot engine $=\frac{0.6444}{0.55}=1.171$

Efficiency of Carnot engine is $\eta =1-\frac{T_L}{T_H}$

$1.171 =1-\frac{T_L}{T_H}$

$\frac{T_L}{T_H}=-0.171$

3 0

3 years ago

Read 2 more answers

How is the phenomenon of reflection used in making a kaleidoscope

dusya [7]

Answer:

The kaleidoscope

Explanation:

gives a number of images formed by reflection from the mirrors inclined to one another. Designers and artists use kaleidoscope to get ideas for new patterns to design wallpapers, jewellery and fabrics.

8 0

3 years ago

An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

Zigmanuir [339]

Answer:

(A) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )$

$\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )$

$\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

7 0

3 years ago

Nearly all liquids and solids are nearly opaque to radiation at nearly all wavelengths. why is that

anastassius [24]

As radiation enters a solid or liquid, the light waves are absorbed and reflected by the atoms. the resulting reflection from an atom can be in any direction and is random. you can understand opacity in terms of the scattering of radiation. The more opaque, the more scattering.

(The truth is far more complicated than this, check out Richard Feynmans lectures on "Quantum Electro Dynamics" to learn the reality of it. )

3 0

3 years ago