Ask question

Ask question

All categories

ludmilkaskok [199]

2 years ago

7

Why would the bulb not light?

1 answer:

padilas [110]2 years ago

8 0

are you a dmbss? the bulb and wire must be connected to both end

You might be interested in

Một cần trục có trọng lượng Q = 50 kN cẩu vật nặng có trọng lượng P = 10 kN

lesantik [10]

5

3373727227717177

Explanation:

yun hehew

8 0

2 years ago

Read 2 more answers

What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?

iren2701 [21]

Answer:

4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

$V(r) = \frac{4}{3} \ \pi \ r^3$

If we got an uncertainty $\Delta r$ the formula for the uncertainty of V is:

$\Delta V(r) = \sqrt{ (\frac{dV}{dr} \Delta r)^2 }$

We can calculate this uncertainty, first we obtain the derivative:

$\frac{dV}{dr} = 3 * \frac{4}{3} \ \pi \ r^2$

$\frac{dV}{dr} = 4 \ \pi \ r^2$

And using it in the formula:

$\Delta V(r) = \sqrt{ (4 \ \pi \ r^2\Delta r)^2 }$

$\Delta V(r) = \sqrt{ 4^2 \ \pi^2 \ r^4 \Delta r^2 }$

$\Delta V(r) = 4 \ \pi \ r^2 \Delta r$

The relative uncertainty is:

$\frac{\Delta V(r)}{V(r)}$

$\frac{ 4 \ \pi \ r^2 \Delta r }{ \frac{4}{3} \ \pi \ r^3}$

$\frac{ 3 \Delta r }{ r}$

Using the values for the problem:

$\frac{ 3 * 0.09 m }{ 5.66 m} = 0.0477$

This is, a percent uncertainty of 4.77 %

4 0

2 years ago

A bird flies from the South Pole to the North Pole. Part of the journey is 1000 miles that takes 2 weeks. What is the bird’s vel

viktelen [127]

1000 miles = 1610km = 1.61x10^6m
2 weeks = 14 days = 14x24x1440

V=d/t = 1.61x10^6/14x24x1440
= 3.33m/s

6 0

3 years ago

Read 2 more answers

A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.

ANTONII [103]

<u>Answer:</u> The number of photons are $3.7\times 10^8$

<u>Explanation:</u>

We are given:

Wavelength of microwave = $1.22\times 10^8nm=0.122m$ (Conversion factor: $1m=10^9nm$ )

To calculate the energy of one photon, we use Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.625\times 10^{-34}J.s$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = 0.122 m

Putting values in above equation, we get:

$E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J$

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

$E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J$

To calculate the mass of water, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g$

To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

$q=mc\Delta T$

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=100^oC-23^oC=77^oC$

Putting values in above equation, we get:

$q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J$

This energy is the amount of energy for 'n' number of photons.

To calculate the number of photons, we divide the total energy by energy of one photon, we get:

$n=\frac{q}{E}$

q = 53127.72 J

E = $1.44\times 10^{-24}J$

Putting values in above equation, we get:

$n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}$

Hence, the number of photons are $3.7\times 10^8$

4 0

3 years ago

Why are the coldest places on earth found near the poles

Nata [24]

Polar regions do not receive direct sunlight during the winter months due to the tilt in the Earth's<span> axis. Hence, polar regions can get very cold. Antarctica is the </span>coldest place on Earth. <span>The </span>coldest places on Earth<span> tend to be located </span>near the poles<span>. Hope this answers the question.</span>

6 0

2 years ago