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sweet-ann [11.9K]

3 years ago

13

A person is standing on a completely frictionless surface. Can they walk on this surface? Explain! What could they do to move ac

ross the surface? Explain!

1 answer:

Digiron [165]3 years ago

4 0

Answer:

Explanation:

We cannot walk on a surface which has no friction.

To move across the surface, take a stone and throw it in the opposite direction of motion so that you get a reaction in the direction of motion and then you move across the surface.

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Installed on your roof is a solar panel that has a surface area of 1.00 m2 . How many photons from the Sun reach the panel? Assu

kirill115 [55]

Answer:

$3.95241\times 10^{21}\ Photons/second$

Explanation:

h = Planck's Constant = $6.626\times 10^{-34}\ m^2\kgs$

A = Area = 1 m²

Distance between Earth and Sun = $1.5\times 10^{11}\ m$

$\lambda$ = Wavelength = 580 nm

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

Power output of Sun

$P=3.83\times 10^{26}\ W$

Intensity

$I=\frac{P}{4\pi d^2}\times A\\\Rightarrow I=\frac{3.83\times 10^{26}}{4\pi (1.5\times 10^{11})^2}\times 1\\\Rightarrow I=1354.5854\ W$

Energy of photons

$E=\frac{hc}{\lambda}\\\Rightarrow E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{580\times 10^{-9}}\\\Rightarrow E=3.42724\times 10^{-19}\ J$

Number of photons

$N=\frac{I}{E}\\\Rightarrow N=\frac{1354.5854}{3.42724\times 10^{-19}}\\\Rightarrow N=3.95241\times 10^{21}\ Photons/second$

The number of photons are $3.95241\times 10^{21}\ Photons/second$

8 0

3 years ago

Which describes an image that a concave mirror can make?

Zigmanuir [339]

An image that a concave mirror can make is that it <span>can form a real image that is projected out in front of the mirror. </span>A concave mirror<span> is a mirror that is curved inward in the middle. Using the mirror equation, you can deal with concave mirrors. It's like you're looking into a cave when try to look in a concave mirror.</span>

6 0

3 years ago

If a photon of Light leaves the sun than reachers the earth 494 second later. What is the distance from the earth to sun

denis-greek [22]

The distance from the Earth to the Sun is 92.96 million mi.

4 0

4 years ago

Read 2 more answers

g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide

vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring $\left ( m_1\right )=0.5 kg$

Compression in the spring $\left ( x\right )=20 cm$

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block $\left ( m_1\right )$

$\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2$

$v=x\sqrt{\frac{k}{m_1}}$

now conserving momentum

$m_1v=\left ( m_1+m_2\right )v_0$

$v_0=\frac{m_1}{m_1+m_2}v$

where $v_0$ is the final velocity

3 0

4 years ago

Barnard's star is a near neighbor of the Sun whose properties we know quite well. It is a type M4V with absolute magnitude 13.22

vitfil [10]

Answer:

The star is at a distance of 100 parsecs.

Explanation:

The distance can be determined by means of the distance modulus:

$M - m = 5log(d) - 5$ (1)

Where M is the absolute magnitude, m is the apparent magnitude and d is the distance in units of parsec.

Therefore, d can be isolated from equation 1

$log(d) = (M - m + 5)/5$

Then, Applying logarithmic properties it is gotten:

$d = 10^{(M - m + 5)/5}$ (2)

The absolute magnitude is the intrinsic brightness of a star, while the apparent magnitude is the apparent brightness that a star will appear to have as is seen from the Earth.

Since both have the same spectral type is absolute magnitude will be the same.

Finally, equation 2 can be used:

$d = 10^{(13.22 - 8.22+ 5)/5}$

$d = 100 pc$

Hence, the star is at a distance of 100 parsecs.

Key term:

Parsec: Parallax of arc seconds

8 0

3 years ago