What is the energy of a photon emitted with a wavelength of 518 nm?

During summer break, Jamie's family went camping in the mountains. There were no roads to their camp, 8 miles away, and it took

liraira [26]

The Answer is B

hope this helps

5 0

3 years ago

Read 2 more answers

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240

Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

$V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC$

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

3 0

3 years ago

Which sentence describes Newton's first law?

Marrrta [24]

$\Large{\red{\bf{\blue{\dag} Answer:-}}}$

Newton's First Law of Motion :

An object stays in rest or in uniform motion unless and until compelled by an external unbalanced force . This is also known as " Law of Inertia " .

Here this statement is best represented by the statement : Unbalanced forces cause an object to move. [ option c ]

More to know :-

Newton's 3 laws of motion :-

Newton's First Law of Motion :-

It states that an object stays in rest or in uniform motion unless and until compelled by an external unbalanced force . This is also known as " Law of Inertia " .

Ex - (i) We tend to fall backwards while standing in the bus , when it starts suddenly.

(ii) We tend to fall forward while standing in the bus , when it stops suddenly.

Newton's Second Law of Motion :-

It states that the rate of change of momentum is directly proportional to the applied force in the direction of the force . That is ∆p $\propto$ Force .

Ex - (i) When we push car & a truck with same force the car will have Greater acceleration due to less mass . (Imaginary case ).

(ii) We feel easy to push a empty cart than a full cart .

Newton's Third Law of Motion :-

It states that every action has equal and opposite reaction .

Ex - (i) We are able to walk on ground due to Newton's third law of motion .

(ii) When we push wall it also applies equal and opposite force .

8 0

3 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area $= \sigma = 1.99*10^{-7} C/m^2$

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates is $E = \frac{\sigma}{\epsilon}$

force acting on charge is F = q E

Work done by charge q id $\Delta X =\frac{ q\sigma \Delta x}{\epsilon}$

this work done is converted into kinectic enerrgy

$\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}$

solving for v

$v = \sqrt{\frac{2q\Delta x}{\epsilon m}$

$\epsilon = 8.85*10^{-12} Nm2/C2$

$v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}$

v = 1.15*10^{7} m/s

8 0

3 years ago

Charge is flowing through a conductor at the rate of 420 C/min. If 742 J. of electrical energy are converted to heat in 30 s., w

Y_Kistochka [10]

Answer:

3.53 V

Explanation:

Electric charge: The is the rate of flow of electric charge along a conductor.

The S.I unit of electric charge is C.

Mathematically it is expressed as,

Q = It ............................ Equation 1

Where Q = electric charge, I = current, t = time.

I = Q/t.......................... Equation 2

From the question, charge flows through the conductor at the rate of 420 C/mim

Which means in 1 min, 420 C of charge flows through the conductor.

Hence,

Q = 420 C, t = 1 min = 60 seconds

Substitute into equation 2

I = 420/60

I =7 A

Also

P = VI......................... Equation 3

Where P = power, V = potential drop, I = current.

V = P/I................... Equation 4

Note: Power = Energy/time

From the question, P = 742/30 = 24.733 W. and I = 7 A.

Substitute these values into equation 4

V = 24.733/7

V = 3.53 V

Hence the potential drop across the conductor = 3.53 V

4 0

3 years ago