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3 years ago

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physics Test 2 In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits

to a screen is 2.5 m, find the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits.

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Answer:0.015 m

Explanation:

Given

distance between two slits(d)=0.050 mm

Distance between slit and screen=2.5 m

Wavelength $(\lambda )=600 nm$

and distance between First order and second order $=\frac{D\lambda }{2d}$

$=\frac{2.5\times 600\times 10^{-9}}{2\times 5\times 10^{-5}}$

$=15\times 10^{-3}$

=0.015 m

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Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.

GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately $\boxed{\sf{2.9 \times 10^{17} \: N}}$

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

$\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}$

With the following condition :

F = gravitational force (N)
G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_1}$ = mass of the first object (kg)
$\sf{m_2}$ = mass of the second object (kg)
r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_X}$ = mass of the planet X = $\sf{1.55 \times 10^{22}}$ kg.
$\sf{m_Y}$ = mass of the planet Y = $\sf{3.95 \times 10^{28}}$ kg.
r = distance between two objects = $\sf{3.75 \times 10^{11}}$ m.

What was asked :

F = gravitational force = ... N

Step by step :

$\sf{F = G \times \frac{m_X \times m_Y}{r^2}}$

$\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}$

$\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}$

$\sf{F \approx 2.9 \times 10^{39 - 22}}$

$\sf{F \approx 2.9 \times 10^{17} \: N}$

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

$\boxed{\sf{2.9 \times 10^{17} \: N}}$

<h3>See More</h3>

Gravity is a thing has depends on ... brainly.com/question/26485200

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2 years ago

A car is travelling with a velocity of 10 m/s and has a mass of 550 kg. The<br> car has<br> energy.

Elenna [48]

Answer: The car has a kinetic energy (because it's in motion) of: $27500J$

Explanation:

$Formula: E=\frac{1}{2}mv^2$

$E=\frac{1}{2}mv^2$

$E=\frac{1}{2}(550kg)(10m/s)^2$

$E=\frac{1}{2}(550kg)(100m^2/s^2)$

$E=\frac{1}{2}(55000J)\\E=27500J$

3 0

3 years ago

The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a

Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

with m the mass , a the acceleration and $\sum\overrightarrow{F}$ the net force forces that is:

$(F-f)$ (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

$(F-f)=ma$

solving for a:

$a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}$

Now let's use the Galileo’s kinematic equation

$Vf^{2}=Vo^{2}+2a\varDelta x$ (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and $\varDelta x$ the time took to achieve that velocity, solving (3) for $\varDelta x$ :

$\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}$

$t=68.94 m$

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3 years ago

An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into th

Black_prince [1.1K]

Answer:

a) R = ρ₀ L /π(r_b² - R_a²) , b) ρ₀ = V / I π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of the cylindrical shell

A = π r_b² - π r_a²

A = π (r_b² - r_a²)

we substitute

R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

V = i R

we subsist the expression of resistance

V = I ρ₀ L /π (r_b² - R_a²)

ρ₀ = V / I π (r_b² - R_a²) / L

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3 years ago

In the above lightwave the property labeled a determines which characteristic of visible light

kap26 [50]

The light must be either very dim or else non-existent.
We can't see the light wave or the label.

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3 years ago