Monochromatic light with wavelength 633 nn passes through a narrow slit and a patternappears on a screen 6.0 m away. The distance on the screen between the centers of thefirst minima on either side of the screen is 32 mm. How wide (in mm) is the slit

Answer:

Explanation:

b) light

Answer:

"In water, the speed of light is v01.33=2.26 x 108 m/s."

Explanation:

Answer:

The maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.

Explanation:

Given;

radius of the circular loop, r = 2.0 m

maximum induced emf, E = 5.0 V

The emf induced in a magnetic field is given as;

[tex]emf = \frac{d\phi}{dt} \\\\\phi = AB\\\\emf = A\frac{dB}{dt} \\\\\frac{dB}{dt} = \frac{emf}{A} \\\\where;\\A \ is \ the \ area \ circular \ l00p = \pi r^2 = \pi (2)^2 = 4\pi \ m^2\\\\\frac{dB}{dt} = \frac{5}{4\pi} \\\\\frac{dB}{dt} = 0.398 \ T/s[/tex]

Therefore, the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.

Answer:

If we use the equation for the transformation of velocities for moving frames:

v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'

v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)

or v' = 1.2 c / (1 + .36) = .88 c

v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u  (-.6 c) and we measure the speed of v as seen in the frame moving to the left

Answer:

1. F = 3400 N = 3.4 KN

2. [tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]

3. v = 14.9 m/s

Explanation:

1.

First, we will calculate the acceleration of Pam by using the third equation of motion:

[tex]2as = v_f^2-v_i^2[/tex]

where,

a = acceleration = ?

s = distance = 27.3 m

vf = final speed = 62 m/s

vi = initial speed = 0 m/s

Therefore,

[tex]2a(27.3\ m) = (62\ m/s)^2-(0\ m/s)^2\\\\a = 70.4\ m/s^2[/tex]

Now, we will calculate the force by using Newton's Second Law of Motion:

F = ma

F = (48.3 kg)(70.4 m/s²)

F = 3400 N = 3.4 KN

2.

Final kinetic energy is given as:

[tex]K.E_f = \frac{1}{2}mv_f^2\\\\K.E_f = \frac{1}{2} (48.3\ kg)(62\ m/s)^2[/tex]

[tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]

3.

According to the law of conservation of energy:

[tex]Potential\ Energy\ at\ top = Kinetic\ Energy\ at\ bottom\\mgh = \frac{1}{2}mv_2 \\\\v = \sqrt{2gh}[/tex]

where,

v = speed at bottom = ?

g = acceleration due to gravity = 9.81 m/s²

h = height at top = 11.3 m

Therefore,

[tex]v = \sqrt{(2)(9.81\ m/s^2)(11.3\ m)}[/tex]

v = 14.9 m/s

So they give us this

V=IR

V= 1.8

I=0.4

R=?

So we insert the thing that we know.

1.8=0.4*R

We need to leave our unknown value alone. So if our value of 0.4 is multiplying the unknown value it passes to the other side dividing.

So we have this.

Lastly we solve.

R=4.5ohms

The formula to find R is V=IR

V/I=R

So the resistance will be the Voltage divided by the Current

Answer:

static friction=0.126

The sum of the resistance = 2 ohms x 80 lights = 160 ohms.

Current = Total voltage / total resistance:

Current = 120V / 160 ohms

Current = 0.75 Amps

Answer:

[tex]F=1.159[/tex]

Explanation:

From the question we are told that:

Mass of pulley [tex]M=1kg[/tex]

Radius [tex]r=12cm[/tex]

Mass of block A [tex]M_a=2.1kg[/tex]

Mass of block B [tex]m_b=4.1kg[/tex]

Spring constant[tex]\mu= 358 J/m2[/tex]

Generally the equation for Torque is mathematically given by

Since [tex]\sumF=ma[/tex]

At mass A

 [tex]T_2-f_3=2.1a[/tex]

At mass B

 [tex]4.8-T_1=4.1a[/tex]

At  Pulley

 [tex]R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}[/tex]

 [tex]R(T_1-T_2)=0.55a[/tex]

Therefore the equation for total force F

At mass A+At mass B+At  Pulley

 [tex](T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a[/tex]

 [tex](T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a[/tex]

 [tex]-f_3+4.1=6.75a[/tex]

 [tex]-f_3=6.75a+4.8[/tex]

Since From above equation

[tex]M_{eff}=6.7kg[/tex]

Therefore

[tex]T=2\pi \sqrt{{\frac{M_{eff}}{k}}[/tex]

[tex]T=2\pi \sqrt{{\frac{6.75}{\mu}}[/tex]

[tex]T=0.862s[/tex]

Generally the equation for frequency is mathematically given by

[tex]F=\frac{1}{T} \\F=\frac{1}{0.862}[/tex]

[tex]F=1.159[/tex]

Answer:

The right solution is "0.511".

Explanation:

Given:

Initial moment of inertia,

= 1630 kg.m²

Radius,

= 5 m

Angular speed,

= 1.6 rad/s

Now,

The moment of inertia after stepping on will be:

= [tex]1630+2\times (69.5\times (5)^2)[/tex]

= [tex]1630+2\times (69.5\times 25)[/tex]

= [tex]5105 \ Kg.m^2[/tex]

hence,

As per the question, the angular speed is conserved, then

⇒ [tex]1630\times 1.6=5105\times \omega'[/tex]

            [tex]2608=5105\times \omega'[/tex]

                [tex]\omega'=\frac{2608}{5105}[/tex]

                     [tex]=0.511[/tex]

Answer:

the magnitude of the velocity of one particle relative to the other is 0.9988c

Explanation:

Given the data in the question;

Velocities of the two particles = 0.9520c

Using Lorentz transformation

Let relative velocity be W, so

v[tex]_r[/tex] = ( u + v ) / ( 1 + ( uv / c²) )

since each particle travels with the same speed,

u = v

so

v[tex]_r[/tex] = ( u + u ) / ( 1 + ( u×u / c²) )  

v[tex]_r[/tex] = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )  

we substitute

v[tex]_r[/tex] = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )  

v[tex]_r[/tex] = 1.904c / ( 1 + 0.906304 )

v[tex]_r[/tex] = 1.904c / 1.906304

v[tex]_r[/tex] = 0.9988c

Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c

Answer:

Height of the door = 2m = 2000 cm

1 in = 2.54 cm

So 1 cm = 1/2.54 in

2000 cm = 200000/ 254

=

787.401574803

So no.c is correct

The door is  78.7 inch tall. Hence, option (A) is correct.

Any arbitrarily selected and widely used reference standard for length measurement is referred to as a unit of length. The metric system, which is adopted by every nation on earth, is the most widely utilized in modern times.

The American customary units are also in use in the United States. In the UK and several other nations, British Imperial units are still used sometimes. There are SI units and non-SI units in the metric system.

Given that: the height of the door is = 2 meter

= 2*100 centimeter

= 200 centimeter.

There are 2.54 centimeter in 1 inch.

Hence,  the height of the door is = 2 meter  = 200 centimeter

= (200/2.54) inch

= 78.7 inch.

The door is 78.7 inch tall.

Learn more about length here:

https://brainly.com/question/17139363

#SPJ2

Answer:

That is equal to R1 + R2. If three or more unequal (or equal) resistors are connected in series then the equivalent resistance is: R1 + R2 + R3 +…, etc. One important point to remember about resistors in series networks to check that your maths is correct.

Answer:

♕ [tex]\large{ \red{ \tt{Step - By - Step \: Explanation}}}[/tex]

☃ [tex] \underline{ \underline{ \blue{ \large{ \tt{G \: I \: V \: E\: N}}}}} : [/tex]

♨ [tex] \underline {\underline{ \orange{ \large{ \tt{T \: O \: \: F \: I \: N\: D}}}} }: [/tex]

☄ [tex]\underline{ \underline{ \large{ \pink{ \tt{S\: O \: L \: U \: T\: I \: O \: N}}}}}: [/tex]

✧ [tex] \red{ \boxed{ \large{ \purple{ \sf{Wave \: velocity(v) = Frequency(f) \times Wavelength(λ)}}}}}[/tex]

~Plug the known values and then multiply!

↦ [tex] \large{ \tt{7 \times 42}}[/tex]

↦ [tex] \boxed{ \boxed{ \large{ \bold{ \tt{294 \: m {s}^ {- 1} }}}}}[/tex]

☥ [tex] \large{ \boxed{ \boxed{ \large{ \tt{Our \: Final \: Answer : \underline{ \large{ \tt{294 \: m {s}^{ - 1}}}}}}}}} [/tex]

---------------------------------------------------------------

❁ [tex] \underline{ \large{ \red{ \tt{D\: E\: T \: A \: I \: L\: E \: D \: \: I\: N \: F \: O}}}} : [/tex]

# KILL : Excuses

KISS : Opportunities

MARRY : Goals

♪ Hope I helped! ♡

☂ Have a wonderful day / night ! ツ

✎ [tex] \underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}}[/tex] ✔

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Answer:

= 1200m/s or 1.2 x [tex]10^{3}[/tex] m/s

Explanation:

Answer:

a. hydroelectric power plant

Answer:

d = 1.19 x 10⁻⁴ m = 0.119 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

[tex]Y = \frac{\lambda L}{d}[/tex]

where,

Y = fringe spacing = 32 mm = 0.032 m

L = slit to screen distance = 6 m

λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m

d = slit width = ?

Therefore,

[tex]0.032\ m = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{d}\\\\d = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{0.032\ m}[/tex]

d = 1.19 x 10⁻⁴ m = 0.119 mm

Answer:

5*10^2

Explanation:

A p e x

Answer:

speed = 0.8c

Explanation:

Given :

Distance from earth to the distant planet = 10 ly

Time taken by the astronauts for the entire journey = 25 years

The time taken to reach the planet is [tex]$t_1=\frac{25}{2}$[/tex]

                                                                  = 12.5 years

Therefore, speed of the starship can be calculated by :

[tex]$\text{Speed} = \frac{\text{distance}}{\text{time}}$[/tex]

[tex]$v=\frac{10 \times c \times 3.15 \times 10^7}{12.5 \times 3.15 \times 10^7}$[/tex]

  [tex]$=0.8c$[/tex]

Therefore the speed of the starship is 0.8c

Answer: how the government and teachers can address or help students experiencing stress and anxiety.

Explanation:

Answer:

377 nm

Explanation:

Number of lines per meter is, [tex]N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}[/tex]

Grating element is, [tex]d=\frac{1}{N}[/tex]

[tex]=1.8868 \times 10^{-6} \mathrm{~m}[tex]

Order is, n=5

Condition for maximum intensity is, [tex]d \sin \theta=n \lambda[/tex]

 [tex]\lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\ &=0.377 \times 10^{-6} \mathrm{~m} \\ &=377 \mathrm{~nm}[/tex]

Answer:

Explanation:

Rate of volume flow at two points will be same at two points .

A₁ V₁ = A₂V₂

A₁ and A₂ are area of cross section at two points and V₁ and V₂ are velocities .

A₁ = 18.2 x 3.55 = 64.61 m²

V₁ = 2 .55 x 10⁻² m/s

A₂ = 16.3 x d = 16.3 d  m²

d is depth at second point .

V₂ = 11.6 x 10⁻² m/s

64.61 m² x 2 .55 x 10⁻² m/s = 16.3 d  m² x 11.6 x 10⁻² m/s

d = .87 m

so canal is .87 m deep.

Answer:

[tex]175\ \text{N/C}[/tex]

Explanation:

[tex]E_1[/tex] = Initial electric field = 3.5 N/C

[tex]E_2[/tex] = Final electric field

[tex]r_1[/tex] = Initial distance = 10 m

[tex]r_2[/tex] = Final distance = 20 cm

Electric field is given by

[tex]E=\sqrt{\dfrac{2P}{\pi r^2c\varepsilon_0}}[/tex]

So,

[tex]E\propto \dfrac{1}{r}[/tex]

[tex]\dfrac{E_2}{E_1}=\dfrac{r_1}{r_2}\\\Rightarrow E_2=E_1\dfrac{r_1}{r_2}\\\Rightarrow E_2=3.5\dfrac{10}{0.2}\\\Rightarrow E_2=175\ \text{N/C}[/tex]

The electric field amplitude at the required point is [tex]175\ \text{N/C}[/tex].

Complete question is;

A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The

centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car. How does the Force of the small car FS compare to the force of the large car FL as they round the curve.

four times

twice

half

equal to

Answer:

Half

Explanation:

Formula for centripetal force is given as;

F = mv²/R

Where;

v is velocity

R is radius

Now, centripetal acceleration is given by;

a = v²/R

Since they both travel with the same velocity V and radius remains the same, we can say that;

F = ma

For the small car;

FS = ma

For the big car;

FL = 2ma

This means the force of the small car is half of that of the Large car

Thus;

FS = ½FL

Answer:

Cold

Explanation:

Im pretty sure im sorry if I am wrong

Rt = 3 ohms

Explanation:

Let R1 = 4-ohm resistor

R2 = 12-ohm resistor

For 2 resistors connected in parallel, the total resistance Rt is given by

1/Rt = 1/R1 + 1/R2

or

Rt = R1R2/(R1 + R2)

= (4 ohms)(12 ohms)/(4 ohms + 12 ohms)

= 48 ohms^2/16 ohms

= 3 ohms