laser beam is incident on two slits with a separation of 0.195 mm, and a screen is placed 5.10 m from the slits. If the bright interference fringes on the screen are separated by 1.61 cm, what is the wavelength of the laser light?

Answer:

   x = 2,864 m ,       Ra = 32.1 m                       

Explanation:

Let's solve this problem in parts, let's start by finding the intensity of the sound in each observer

observer A β = 64 db

             β = 10 log Iₐ / I₀

where I₀ = 1 10⁻¹² W / m²

              Iₐ = I₀ 10 (β/ 10)

let's calculate

              Iₐ = 1 10⁻¹² (64/10)

              Iₐ = 2.51 10⁻⁶ W / m²

Observer B β = 85 db

             I_b = 1 10-12 10 (85/10)

             I_b = 3.16 10⁻⁴ W / m²

now we use that the emitted power that is constant is the intensity over the area of ​​the sphere where the sound is distributed

              P = I A

therefore for the two observers

              P = Ia Aa = Ib Ab

the area of ​​a sphere is

               A = 4π R²

we substitute

               Ia 4pi Ra2 = Ib 4pi Rb2

               Ia Ra2 = Ib Rb2

Let us call the distance from the observer be to the haughty R = ax, so the distance from the observer A to the haughty is R = 35 ax; we substitute

             Ia (35 -x) 2 = Ib x2

we develop and solve

           35-x = Ra (Ib / Ia) x

           35 = [Ra (Ib / Ia) +1] x

           x (11.22 +1) = 35

           x = 35 / 12.22

            x = 2,864 m

This is the distance of observer B

The distance from observer A

            Ra = 35 - x

            Ra = 35 - 2,864

            Ra = 32.1 m

Answer:

(a) θ = 45° = 0.78 rad

(b) θ = 32.27° = 0.56 rad

(c) θ = 57.27° = 1 rad

Explanation:

When a vector is resolved into its rectangular components, the formula for the direction angle of the vector with positive x-axis is given as:

tan θ = Ay/Ax

θ = tan⁻¹(Ay/Ax)

(a)

Ax = 12 m

Ay = 12 m

θ = tan⁻¹(12 m/ 12 m)

θ = tan⁻¹(1)

θ = 45° = 0.78 rad

(b)

Ax = 19 m

Ay = 12 m

θ = tan⁻¹(12 m/19 m)

θ = tan⁻¹(0.6315)

θ = 32.27° = 0.56 rad

(c)

Ax = 12 m

Ay = 19 m

θ = tan⁻¹(19 m/12 m)

θ = tan⁻¹(1.58333)

θ = 57.27° = 1 rad

Answer:

V₀ₓ = 9.2 m/s

Nearest answer:

D) 8.9 m/s

Explanation:

First we find the time taken by the pumpkin to hit the car. For that purpose we apply 2nd equation of motion to the pumpkin:

h = V₀y t + (1/2)gt²

where,

h = height of building = 10.4 m

V₀y = vertical component of initial speed = 0 m/s

t = time = ?

g = 9.8 m/s²

Therefore,

10.4 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (10.4 m)(2)/(9.8 m/s²)

t = √[2.122 s²]

t = 1.45 s

Now, we analyze horizontal motion for horizontal component of initial velocity. We assume air friction to be zero so that the horizontal motion is uniform. Therefore,

s = V₀ₓ t

where,

s = horizontal distance between building and car = 13.4 m

V₀ₓ = Horizontal Component of Initial Velocity = ?

Therefore,

13.4 m = V₀ₓ(1.45 s)

V₀ₓ = 13.4 m/1.45 s

V₀ₓ = 9.2 m/s

Answer:

-63.6m/s

Explanation:

Given parameters:

Initial velocity  = 5m/s

Time of flight  = 7s

Unknown:

Velocity of the eraser after 7s = ?

Solution:

To solve this problem, we have to use the right motion equation which is given below;

          v  = u - gt

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity  = 9.8m/s²

t is the time taken;

  Now insert the parameters and solve for v;

     v  = 5  - (9.8 x 7)  

    v   = -63.6m/s

Answer:

The average power is  [tex]P = 195 .42 \ W[/tex]

Explanation:

From the question we are told that

   The  energy of the appliance is  [tex]E = 4.69 \ kWh = 4.69 *10^{3} \ \ Wh[/tex]

    The time considered is   [tex]t = 1 \ day = 24 \ hours[/tex]

Generally the average power consumption is mathematically represented as

      [tex]P = \frac{E}{t}[/tex]

=>   [tex]P = \frac{4.69 *10^{3}}{24 }[/tex]

=>   [tex]P = 195 .42 \ W[/tex]

Answer:

Q = 425 kJ

Explanation:

Given that,

Mass, m = 25 kg

The clown doll head that heats up from 25°C to 35°C

The specific heat is 1700 J/kg°C

We need to find the internal energy of it. The heat required to raise the temperature is given by the formula as follows :

[tex]Q=mc\Delta T\\\\Q=25\times 1700\times (35-25)\\\\Q=425000\ J\\\\Q=425\ kJ[/tex]

So, 425 kJ of thermal energy is severed.

Answer:

a

[tex]\lambda_{long} = 288.5 \ nm[/tex]

b

The velocity is  [tex]v = 3.7 *0^{5} \ m/s[/tex]

Explanation:

From the question we are told that

   The work function of Zinc is  [tex]W = 4.3 eV[/tex]

Generally the work function can be mathematically represented as

     [tex]E_o = \frac{hc}{\lambda_{long}}[/tex]

=>   [tex]\lambda_{long} = \frac{hc}{E_o}[/tex]

Here  h is the Planck constant with the value  [tex]h = 4.1357 * 10^{-15} eV s[/tex]

  and c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

So

     [tex]\lambda_{long} = \frac{4.1357 * 10^{-15} * 3.0 *10^{8}}{4.3}[/tex]

=>  [tex]\lambda_{long} = 2.885 *10^{-7} \ m[/tex]

=>  [tex]\lambda_{long} = 288.5 \ nm[/tex]

Generally the kinetic energy of the emitted electron is mathematically represented as

      [tex]K = E -E_o[/tex]

Here  E is the energy of the photon that strikes the surface

So

    [tex]E- E_o = \frac{1}{2} m * v^2[/tex]

Here m is the mass of electron with value  [tex]m = 9.11*10^{-31 } \ kg[/tex]

Generally  [tex]1 ev = 1.60 *10^{-19} \ J[/tex]

=>   [tex]v = \sqrt{ \frac{2 (E - E_o ) }{ m } }[/tex]

=>    [tex]v = \sqrt{ \frac{2 (4.7 - 4.3 )* 1.60 *10^{-19} }{ 9.11 *10^{-31} } }[/tex]

=>    [tex]v = 3.7 *0^{5} \ m/s[/tex]

Answer:

A. The current in the wire increases.

Explanation:

Increasing the speed at which a magnet moves in and out of a wire coil increases the current in the wire.

This phenomenon shows the inter-relationship between electricity and magnetic fields.

Answer:

It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.

Explanation:

Answer:

It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.

Answer:

Yuh

Explanation:

Answer:

m = 4.9 10⁸ kg

Explanation:

The expression for the density is

           ρ = m / V

           m = ρ V

the volume of the atmosphere is the volume of the sphere of the outer layer of the atmosphere minus the volume of the plant

          V = V_atmosphere - V_planet

           V = 4/3 π R_atmosphere³ - 4/3 π R_venus³

           V = 4/3 π (R_atmosphere³ - R_venus³

)

the radius of the planet is R_venus = 6.06 10⁶ m.

The radius of the outermost layer of the atmosphere

          R_atmosphere = 50 10³ + R_ venus = 50 10³ + 6.06 10⁶

           R_atmosphere = 6.11 10⁶ m

let's find the volume

           V = 4/3 pi [(6,11 10⁶)³ - (6,06 10⁶)³]

            V = 23,265 10⁶ m³

let's calculate the mass

          m = 21  23,265 10⁶

          m = 4.89   10⁸ kg

with two significant figurars is

          m = 4.9 10⁸ kg

Answer:

P_abs = 105120.2 N/m²

Explanation:

We are given;

Specific gravity of oil; ρ_oil = 0.6 g/cm³ = 600 kg/m³

Depth of water; h_w = 21 cm = 0.21 m

Depth of oil; h_o = 35 cm = 0.35 m

From tables specific gravity of water is; ρ_w = 1000 kg/m³

Thus, to get the absolute pressure at the bottom of the container, we will use the formula;

P_abs = (ρ_w × g × h_w) + (ρ_oil × g × h_oil) + P_a

Where P_a is atmospheric pressure with a standard value of 1.01 × 10^(5) N/m²

g is gravitational acceleration = 9.81 m/s²

Thus;

P_abs = (1000 × 9.81 × 0.21) + (600 × 9.81 × 0.35) + (1.01 × 10^(5))

P_abs = 105120.2 N/m²

Answer:

a motor is used to covert electrical energy to mechanical energy

Answer:

Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion.

Explanation:

please mark me brainlist

Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion. they still use mercury even though it is the poorest conductor of heat.

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

[tex]\tau = I\alpha[/tex]

Where [tex]\tau[/tex] is the torque

[tex]I[/tex] is the moment of inertia

[tex]\alpha[/tex] is the angular acceleration

But, the angular acceleration is given by

[tex]\alpha = \frac{\omega}{t}[/tex]

Where [tex]\omega[/tex] is the angular speed

and [tex]t[/tex] is time

Then, we can write that

[tex]\tau = \frac{I\omega}{t}[/tex]

Hence,

[tex]\omega = \frac{\tau t}{I}[/tex]

Now, to determine the angular speed, we first determine the Torque [tex]\tau[/tex] and the moment of inertia [tex]I[/tex].

Here, The torque is given by,

[tex]\tau = rF[/tex]

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ [tex]\tau = 3.00 \times 195[/tex]

[tex]\tau = 585[/tex] Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

[tex]I = \frac{1}{2}MR^{2}[/tex]

Where M is the mass and

R is the radius

∴[tex]I = \frac{1}{2} \times 325 \times (3.00)^{2}[/tex]

[tex]I = 1462.5[/tex] kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

[tex]\omega = \frac{\tau t}{I}[/tex]

[tex]\omega = \frac{585 \times 2.05}{1462.5}[/tex]

[tex]\omega = 0.82[/tex] rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

Answer:

The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N

Explanation:

Forces on block 4.3 kg are:

63N to the right and R21 (contact force from the 6.3 kg block) to the left

Net force on 4.3 kg block is: 63 N - R21

Forces on the 6.3 kg block are:

R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.

So net force on the 6.3 kg block is: R12 - 11 N

According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").

Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:

a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2

solve for R by cross multiplication

6.3 (63 - R) = 4.3 (R - 11)

396.9 - 6.3 R = 4.3 R - 47.3

369.9 + 47.3 = 10.6 R

444.2 = 10.6 R

R = 444.2 / 10.6

R = 41.90 N

Answer:

775 N  due North.

Explanation:

If the crate is pulled South with 350 N force, and the net force on the crate results into 425 N due North, then the other force (F) acting must be larger than the 350 N, and pointing North:

F - 350 N = 425 N

F = 425 N + 350 N = 775 N  due North.

Answer:

643N.m

Explanation:

From this question we have:

Mass flow = 4kg/s

Velocity V = 400m/s

Rotation N = 1500rev/min

We get the relative velocity at exit to be:

V2 = V - r2w

400-0.5x [(2*π*1500)/60]

= 400-78.5

= 321.5m/s

Then we have to calculate the frictional torque My

Mt = Mr2 x V2

= 4x0.5x321.5

= 643Nm

From the calculations above, we get the frictional torque M on the tube to be 643Nm.

Answer:

a bowling ball because it has the most mass.

Answer:

1.7333333m/s²

Explanation:

Tension of the line = the weight + force from pulling up the fish

30N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6a

∴ a = 1.7333333m/s²

You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is  1.7333333 m/s².

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.

Tension of the line = the weight + force from pulling up the fish

30 N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6 a

a = 1.7333333 m/s²

You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is  1.7333333 m/s².

To learn more about acceleration refer to the link:

brainly.com/question/12550364

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Answer:

its something that hold the air for forceing liy by the exgen

Explanation:

Answer:c

Explanation:its the answer because its the answer

Answer:

delivery truck

Explanation:

because i got it right

Answer:

You have an apple pen. :)

Answer:

I have a pen

I have a apple

apple pen

Explanation:

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

[tex] E_{i} = E_{f} [/tex]

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]                              

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s [/tex]

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]    

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]      

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s [/tex]      

3) When the object is at the equilibrium position, the speed of the object is:

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]    

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]      

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s [/tex]

I hope it helps you!                                                                                        

(1) the speed of the object when compression of the spring is 8 cm is 0.449 m/s

(2) the speed of the object when compression of the spring is 5 cm is 0.648 m/s

(3) the speed of the object when the spring is at equilibrium is 0.748 m/s

Given that the mass of the object, m = 5 kg

spring constant, k = 280 N/m

compression of the spring , x = 10 cm = 0.1m

(i) the spring compression is at d = 8cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.08)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.2016[/tex]

v = 0.449 m/s

(ii) the spring compression is at d = 5cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.05)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.42[/tex]

v = 0.648 m/s

(iii) the spring is at equilibrium so compression is at d = 0cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.56[/tex]

v = 0.748 m/s

Learn more about conservation of energy:

https://brainly.com/question/14245799?referrer=searchResults