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3 years ago

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A photon in a laboratory experiment has an energy of 5 eV. What is the frequency of this photon? (using the idea of the electron

volt)

1 answer:

andreyandreev [35.5K]3 years ago

7 0

Answer and working shown on photo

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Suppose you are walking in an airliner in flight. Use Newton’s third law to describe the effect of your walk on the motion on th

Novosadov [1.4K]

Answer:

The effect of your walking will create a force on the airline acting downwards, due to the weight. By Newton's thirds law the airline will exert an equal and opposite reaction force directed downwards.

Explanation:

The weight of all the passengers acts downwards on the floor of the Airplane.

The Airplane exerts an equal and opposite force on the passengers, with the help of its propulsion due to which the flight keeps on flying without falling down.

Hence by changing the frame of reference we can observe the force which is responsible for the reaction.

4 0

2 years ago

Twin space probes have a mass of 722 kg each. If the gravitational force between the two space probes is 8.61

IceJOKER [234]

Answer:200×10^5 meters

Explanation:

6 0

3 years ago

Read 2 more answers

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago

She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observ

Reika [66]

Answer:

The boats are 934.65 feet apart

Explanation:

Given:

The angles of depression to the two boats are 42 degrees and 29 degrees

Height of the observation deck i = 1,353 feet

To Find:

How far apart are the boats (y )= ?

Solution:

<em><u>Step 1 : Finding the value of x(Refer the figure attached)</u></em>

We can use the tangent ratio to find the x value

$tan(42^{\circ}) = \frac{1353}{x}$

$x = \frac{1353}{tan(42^{\circ}) }$

x = 590.47 feet

<em><u>Step 2 : Finding the value of z (Refer the figure attached)</u></em>

$tan(29^{\circ}) = \frac{1353}{z }$

$z = \frac{1353}{tan(29^{\circ})}$

z = 1525.12 feet

<em><u>Step 3 : Finding the value of y (Refer the figure attached</u></em>)

y = z -x

y = 1525.12 - 590.47

y = 934.65 feet

Thus the two boats are 934.65 feet apart

3 0

3 years ago

mario62 [17]

It's D. If it absorbed it would be turning to steam. I am taking honors chem in high school we are learning this.

7 0

3 years ago

Read 2 more answers