Charges that do not transfer

A physics instructor wants to project a spectrum of visible-light colors from 400 nm to 700 nm as part of a classroom demonstrat

Likurg_2 [28]

Answer:

Dr = 263 10⁻⁶ m

Explanation:

The diffraction pattern for constructive interference is described by

a sin θ = m λ

in this it indicates that the order of diffraction is m = 1

Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits

a = 2 lines 1/600

a = 2/600

a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm

let's use trigonometry

tan θ = y / L

as the measured angles are small

tan θ = sin θ / cos θ sin θ

sin θ = y / L

we substitute

a y/L = λ

y = λ L / a

for λ = 400 10-9 m

I = 400 10⁻⁹ 2.9 / 3.33 10⁻³

i = 346.89 10⁻⁶ m

f

or λ = 700 nm

y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³

y_f = 609.609 10⁻⁶ m

the separation of this spectrum

Δr = v_f - i

Dr = (609.609 - 346) 10 ⁻⁶

Dr = 263 10⁻⁶ m

4 0

3 years ago

The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second

just olya [345]

Answer:

$g=274\ m/s^2$

Explanation:

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

The radius of the Sun, $r=6.96\times 10^8\ m$

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

$g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2$

So, the value of acceleration due to gravity on the Sun is $274\ m/s^2$ .

8 0

3 years ago

How to create or build a homemade golf club. is this possible? If so, what suggestion did you find? Do you have an original idea

goldenfox [79]

Answer:

All you need is golfing equipment make a small hole in the ground and have like something to cover the sun off of you and boom you got a golfing spot and a golf club no skills needed. Oh and you need a small flag for your gold hole but then your good and you can have fun

Explanation:

8 0

2 years ago

An object thrown vertically upward from the surface of a celestial body at a velocity of 36 m/s reaches a height of sequalsminu

Anarel [89]

Answer:

v = -1.8t+36

20 seconds

360 m

40 seconds

36 m/s

The object speed will increase when it is coming down from its highest height.

Explanation:

$s=-0.9t^2+36t$

Differentiating with respect to time we get

$\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36$

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

$0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s$

b) The object will reach the highest point after 20 seconds

$s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m$

c) Highest point the object will reach is 360 m

$s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0$

$\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s$

d) Time taken to strike the ground would be 20+20 = 40 seconds

$[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s$

Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of $s=ut+\frac{1}{2}at^2$

e) The velocity with which the object strikes the ground will be 36 m/s

f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.

4 0

3 years ago

Using the same cost and time estimates, consider the time-effectiveness of each engineer.

sammy [17]

Answer:

Camilla

Explanation:

I got it right on edge. :)

4 0

3 years ago