What needed to be present in marine mud to form fossils fuels?

What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm?

ivann1987 [24]

Answer:

$2.46\cdot 10^5 J$

Explanation:

The energy of a single photon is given by:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength

For the photon in this problem,

$\lambda=486 nm=4.86\cdot 10^{-7}m$

So, its energy is

$E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J$

One mole of photons contains a number of photons equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

So, the total energy of one mole of photons is

$E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J$

7 0

3 years ago

A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co

lidiya [134]

The position vector can be transcribed as:

A = 6 i + y j

i points in the x-direction and j points in the y-direction.

The magnitude of the vector is its dot product with itself:

|A|2 = A·A

102 = (6 i + y j)•(6 i+ y j) Note that i•j = 0, and i•i = j•j = 1

100 = 36 + y2

64 = y2

get the square root of 64 = 8

The vertical component of the vector is 8 cm.

3 0

3 years ago

Consider a situation where you are playing air hockey with a friend. The table shoots small streams of air upward to keep the pu

artcher [175]

When we hit the puck from tap the puck will move forward.

This is due to the impulse provided by us at the time of hit. Due to this impulse the puck will move forward and start moving in some direction.

As soon as puck move forward the force on it is zero as the weight of the puck is counterbalanced by the air stream force and there is no other force on it so puck will continue its motion till it will hit at some other point.

So here the motion of the puck will be uniform motion till it will collide with some other points.

So here the correct option will be given as

moves with a constant speed until hitting the other end.

5 0

3 years ago

A 4.10 g bullet moving at 837 m/s strikes a 820 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

atroni [7]

Answer:

(a) 1.85 m/s

(b) 4.1 m/s

Explanation:

Data

bullet mass, Mb = 4.10 g

initial bullet velocity, Vbi = 837 m/s

wooden block mass, Mw = 820 g

initial wooden block velocity, Vwi = 0 m/s

final bullet velocity, Vbf = 467 m/s

(a) From the conservation of momentum:

Mb*Vbi + Mw*Vwi = Mb*Vbf + Mw*Vwf

Mb*(Vbi - Vbf)/Mw = Vwf

4.1*(837 - 467)/820 = Vwf

Vwf = 1.85 m/s

(b) The speed of the center of mass speed is calculated as follows:

V = Mb/(Mb + Mw) * Vbi

V = 4.1/(4.1 + 820) * 837

V = 4.1 m/s

6 0

3 years ago

Elect the correct answer. Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of laun

allsm [11]

Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction $= u\cos\theta$

and the speed in the vertical direction is $= u\sin\theta$ upward.

For A:

The speed in the horizontal direction $= u\cos75^{\circ}$

and the speed in the vertical direction is $= u\sin75^{\circ}$ upward.

For B:

The speed in the horizontal direction $= u\cos15^{\circ}$

and the speed in the vertical direction is $= u\sin15^{\circ}$ upward.

Let $t_A$ and $t_B$ are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A $= u\cos75^{\circ}\times t_A\cdots(i)$

The range for the projectile B = $u\cos15^{\circ}\times t_B\cdots(ii)$

At the highest point, the vertical velocity is 0.

Bu using the equation of motion $v^2=u^2 +2a s.$

Here, the final velocity v=0, the initial velocity $u = u \sin \theta$ , h= vertical distance up to the highest point, and $a= -g$ (as per sign convention).