What type of star has a low temperture but a high luminosity?

Frying an egg conduction convention radiation

Allushta [10]

Answer:

frying an egg is conduction because the pan is touching the stove and the pan is touching the eggs

Explanation:

4 0

3 years ago

Which is present when iodine changes from brown to blue or purple?

vladimir1956 [14]

If iodine is added to a starch solution, they react with each other and the iodine darkens to an almost pitch black.

however, if iodine is added to a solution containing no starch, it will show up only as an extremely pale brown. almost colorless and hardly visible.

when following the changes in some inorganic oxidation reduction reactions, iodine may be used as an indicator to follow the changes of iodide ion and iodine element. soluble starch solution is added. only iodine element in the presence of iodide ion will give the characteristic blue black color. neither iodine element alone nor iodide ions alone will give the color result.

hope this answer really helps your question :)

5 0

3 years ago

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

3 years ago

A stone is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 20 m/s. Th

Hitman42 [59]

Answer:

Explanation:

Given

inclination $\theta =30^{\circ}$

initial speed $u=20\ m/s$

Point of release is 45 m above the ground

Considering stone to be a projectile, so time taken by projectile for its zero vertical displacement is

$t_1=\frac{2u\sin \theta }{g}$

$t_1=\frac{2\times 20\times \sin 30}{10}$

$t_1=2\ s$

Now after completing zero vertical displacement , stone needs to travel another 45 m in downward direction with initial speed u=20\sin 30

$h=u_yt+\frac{1}{2}a_yt^2$

where, $h=height$

$u_y=vertical velocity$

$a_y=vertical acceleration$

$t_0=time$

$45=20\sin 30+\frac{1}{2}(9.8)(t_0)^2$

$t_0^2=\frac{70}{9.8}$

$t_0=2.64\ s$

thus total time time required is $t=t_0+t_1=2.64+2=4.64\ s$

vertical velocity just before hitting

$v_y=\sqrt{u_y^2+2\times a_y\times s}$

$v_y=\sqrt{10^2+2\times 10\times 45}$

$v_y=\sqrt{1000}=31.622\ m/s$

Horizontal velocity $v_x=u\cos 30=17.32\ m/s$

Net velocity Just before hitting $=\sqrt{v_x^2+v_y^2}$

$=\sqrt{(17.32)^2+(31.62)^2}$

$=\sqrt{1299.82}=36.05\ m/s$

7 0

3 years ago

PLEASE HELP ASAP!!!!!!!!!!

Marianna [84]

Hey,

I think the answer's are 1,3

Hope this helpss

~Girlygir101~

7 0

3 years ago