Ask question

Ask question

All categories

Ira Lisetskai [31]

3 years ago

15

A soap bubble is essentially a thin film of water surrounded by air. The colors you see in soap bubbles are produced by interfer

ence. What visible wavelengths of light are strongly reflected from a 390-nm-thick soap bubble? What color would such a soap bubble appear to be?

1 answer:

drek231 [11]3 years ago

6 0

Answer:

Violet

Explanation:

Wavelength spectrum of visible light for the range of 390nm to 460nm is seen as violet. So the violet colour will be seen and reflected as what the eyes will see

You might be interested in

What is the momentum of a 10kg ball moving at a velocity of 4m/s east?

Katyanochek1 [597]

Answer:

40 kg.m/s

Explanation:

Momentum, p is defined as the product of mass and velocity of an object. Numerically, it is represented as, p=mv where m is mass of the object and v is the velocity in which the object moves, with keen observation on the direction before and after collision. Substituting 10 kg for m and 4 m/s for v then momentum, P=10*4=40 kg.m/s

6 0

3 years ago

How do you use vectors daily without realizing it was a vector?<br><br> I need help ASAP !!!

jonny [76]

Explanation:

Vectors are quantities that has both magnitude and direction. A vector is described in terms of quantity and the direction it is headed.

There are different forms of vectors that are used in our daily life.

When we describe the motion of a car and the direction it is heading, we are simply talking about its velocity which is a vector.
When we say James treks 2km from his school to the house everyday, we are simply describing his displacement.
When we apply a force to push the table from one corner of the room to another point, is simply a vector
When we change position, it is a vector application.

6 0

3 years ago

As the result of a thermal inversion the prevailing air temperature profile increases 1°C/100m above the ground level. To what m

astra-53 [7]

Answer:

909.1 m

Explanation:

Rate of temperature increase with 100 m elevation = 1°C

h = Maximum Height

Adiabatic lapse rate = -0.65°C/100 m

We have the relation

$\dfrac{1^{\circ}C}{100}=\dfrac{15^{\circ}C}{h}+\left(-\frac{0.65}{100}\right)\\\Rightarrow \dfrac{1^{\circ}C}{100}+\left(-\frac{0.65}{100}\right)=\dfrac{15^{\circ}C}{h}\\\Rightarrow h=\dfrac{15^{\circ}C}{1.65^{\circ}C}\times 100\\\Rightarrow h=909.1\ m$

The maximum height is 909.1 m

5 0

3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

A sprinter runs for 6.45 s at 7.44 m/s. How far does she get?

Gwar [14]

Answer:

D is the answer

Explanation:

6.45×7.44= 47.98800

Which if we round of we get 48m

8 0

3 years ago