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3 years ago

A 6.00 cm tall light bulb is placed a distance of 54.2 cm from a double convex lens

Physics

1 answer:

BigorU [14]3 years ago

7 0

Answer:

1 / f = 1 / i + 1 / o thin lens equation

1 / i = 1 / f - 1 / o = (o - f) / (o * f)

i = o * f / (o - f)

i = 54.2 * 12.7 / (54.2 - 12.7) = 16.6 cm image distance

Image is real and inverted and 16.6 / 54.2 * 6 = 1.94 cm tall

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(Fill the blank) the ________ g____ makes saliva that make your food wet and easy to swallow

blagie [28]

Answer:

What make saliva that make your food wet and easy to swallow?

enzyme amylase

Explanation:

The digestive functions of saliva include moistening food, and helping to create a food bolus, so it can be swallowed easily. Saliva contains the enzyme amylase that breaks some starches down into maltose and dextrin. Thus, digestion of food occurs within the mouth, even before food reaches the stomach.

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2 years ago

The heater element of a 120 V toaster is a 5.4 m length of nichrome wire, whose diameter is 0.48 mm. The resistivity of nichrome

Ksenya-84 [330]

Answer:

The power drawn by the toaster is closest to:

(A) 370 W

Explanation:

First we calculate the resistance of the nichrome wire (R).

$R=\frac{pL}{A} =\frac{pL}{\pi r^{2} } \\$

Where radious (r), resistance coefficient (p), and Length (L)

$r=\frac{0.48}{2} 10^{-3} m\\\\L=5.4 m\\p=1.3*10^{-3}\varOmega \\\\\\Replacing:\\\\R=\frac{1.3(10)^{-6} *5.4}{\pi* 0.24^{2} (10)^{-6}} =38.7940$

After replace the value in the ohm law power formula to obtain the power consumed:

$P=\frac{V^2}{R} =\frac{120^2}{38.7940} =371.191 Watts$

7 0

3 years ago

An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst

sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

$T_1 =$ 500°C

$T_2$ = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

$\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5$

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to $T_{avg}$ = 535.5K $\approx 550K$

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

$P_r = 0.63$

Number for the first strips is equal to

$R_e_x = \frac{u_o.L}{v}$

$R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4$

Calculating heat transfer coefficient from the first strip

$h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3$

$h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2$

The rate of convection heat transfer from the first strip is

$q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W$

The rate of convection heat transfer from the fifth trip is equal to

$q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2$

Calculating $h_o_-_4$

$h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2$

The rate of convection heat transfer from the tenth strip is

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2$

Calculating

$h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W$

The rate of convection heat transfer from 25th strip is equal to

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)$

Calculating $h_o_-_2_5$

$h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2$

Calculating $h_o_-_2_4$

$h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W$

6 0

3 years ago

All matter in the Universe consists of many substances called elements. true or faluse

Mrrafil [7]

The answer to this statement is true!

8 0

3 years ago

Read 2 more answers

A materials density is the same , no matter how large or small the sample is or what it’s shape is as long as it is solid unifor

Scorpion4ik [409]

Answer:

See the explanation below

Explanation:

Density is defined as the relationship between mass and volume, i.e. the following equation can be used:

density = m/v

where:

density [kg/m^3]

m = mass [kg]

v = volume [m^3]

If we change the volume of a body by reducing its size, its mass will also decrease proportionally with a density as seen in the equation.

m = density*v

To understand this concept more clearly, let's use the following example:

We know that the density of water is equal to 1000 [kg/m^3], that is, 1 cubic meter of water contains 1000 kilograms of water, using the equation.

1000 = m /1

m = 1000*1 = 1000 [kg]

Now if we have 500 kilograms of water, that would pass with the volume so that the density remains constant.

1000 = 500/v

v = 500/1000

v = 0.5 [m^3]

We can see that the volume of water has halved. Since the mass of water was reduced by half. That is, the relationship between mass and volume is proportional to the density of the material or substance.

8 0

3 years ago