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Liono4ka [1.6K]

2 years ago

6

Answers n = 2 energy level to the n = 4 energy level has a __________ wavelength than the photon absorbed by an electron moving

from the n = 2 to the n = 3 energy level.

1 answer:

ludmilkaskok [199]2 years ago

3 0

The answer is The Bohr Atom. I hope it helped

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An automobile traveling on a straight, level road has an initial speed v when the brakes are applied. In coming to rest with a c

Molodets [167]

Answer:

4x

Explanation:

Use $v^{2} = u^{2} +2as$ to do the question.

For first instance,

$0 = v^{2} +2ax$ -------------------( 1 )

for second instance,

$0 = (2v)^{2} +2as$ -----------------( 2 )

So by (1) and (2),

s = 4x

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2 years ago

What is physics mean and branch

Jobisdone [24]

Physics is a branch of science that deals with the properties of matter and energy and relationship between them it also tries to explain the material world of the natural phenomenon of the universe the scope of physics is very wide and vast

5 0

2 years ago

Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr

Sedbober [7]

Answer:

(A) ratio of electric force to weight will be $23.469\times 10^{-10}$

(b) Electric field will be $E=4.26\times 10^{10}N/C$

Explanation:

We have given mass of bee = 100 mg = $m=100\times 10^{-3}=0.1kg$

Charge on bee $q=23pC=23\times 10^{-12}C$

Electric field E = 100 N/C

Weight of the bee $W=mg=0.1\times 9.8=0.98N$

Electric force on the bee $F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N$

So the ratio of electric force on the bee and weight is $=\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}$

(B) To hold the bee in air electric force must be equal to weight of bee

So $mg=qE$

$0.1\times 9.8=23\times 10^{-12}E$

$E=4.26\times 10^{10}N/C$

4 0

2 years ago

What is the area of a circle of a radius 4.1×10^4 cm

dmitriy555 [2]

Answer:

128805.2988 cm squared or 128000 to 3 significant figures.

Explanation:

Area of a circle : pi times radius squared or $r^{2} \pi$

pi times 4.1×10^4 cm = 41000 pi or 128805.2988

4 0

2 years ago

A marble rolling at a speed of 2.71 m/s falls of the end of a table that is 1.25 m high. How far from the base of the table does

katrin [286]

Answer:

The marble lands at a distance of 1.36 m from the base of the table.

Explanation:

The motion of the marble falling off the table is a projectile with initial velocity in the horizontal direction only.

The motion can be solved in two directions, the horizontal and vertical direction.

Along the vertical direction, the initial velocity is 0 m/s as it has only horizontal component initially. Also acceleration in the vertical direction is acceleration due to gravity.

Let us use equation of motion in vertical direction.

$y-y_0=v_{0y}t+\frac{1}{2}at^2$

Where,

$v_{0y}\rightarrow \textrm{vertical component of the initial velocity}.\\y\rightarrow \textrm{final position of the marble}\\y_0\rightarrow \textrm{initial vertical position}\\a_y\rightarrow \textrm{acceleration in the vertical direction}\\t\rightarrow \textrm{time taken to reach bottom}$

Now, plug in 0 for $y$ , $1.25$ for $y_0$ , 0 for $v_{0y}$ , -9.8 for $a_y$ . This gives,

$0-1.25=0+\frac{1}{2}(-9.8)t^2\\-1.25=-4.9t^2\\t^2=\frac{-1.25}{-4.9}\\t^2=0.255\\t=\sqrt{0.255}=0.501\ s$

Therefore, time to reach bottom is 0.501 s.

Now, consider the horizontal motion. There is no acceleration in the horizontal direction. So, distance is given as the product of horizontal velocity and time taken.

Horizontal distance covered by the marble is given as:

$x=v_{0x}\times t=2.71\times 0.501=1.36 \textrm{ m}$

Therefore, the marble lands at a distance of 1.36 m from the base of the table.

6 0

2 years ago