The orbital motion of Earth around the Sun leads to an observable parallax effect on the nearest stars. For each star listed, calculate the distance in parsecs before converting that distance to astronomical units. A. Sirius (0.38") B. Alpha Centauri A (0.75") C. Procyon (0.28") D. Wolf 359 (0.42") E. Epsilon Eridani (0.31") D(pc) = 1/parallax(arcsecs), D(a.u.) = D(pc) * 206265 (arcsecs per radian)

Answer:

Explanation:

The e.m.f induced in the coil depend on the following :

(a) No. of turns in the coil

(b) Cross-sectional Area of the coil

(c) Magnitude of Magnetic field

(d) Angular velocity of the coil

Answer:

Explanation:

The formula for calculating the voltage across an inductor is expressed as

[tex]V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)[/tex]

Given parameters

current amplitude I = 1.50mA = 1.5*10⁻³A

inductance L = 0.450mH = 0.450*10⁻³H

Voltage across the inductor [tex]V_l[/tex] = 13.0V

Required

frequency f

Substituting the given parametres into the formula, we have;

[tex]V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz[/tex]

Hence, the frequency required is 3.067MHz

Answer:

[tex]\theta=10.20^{\circ}[/tex]  

[tex]\Delta l=0.10 ft[/tex]    

Explanation:

First of all, we analyze the system of blocks before starting to move.

[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]  

[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]

[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]  

[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]  

[tex]tan(\theta)=0.18[/tex]  

[tex]\theta=arctan(0.18)[/tex]  

[tex]\theta=10.20^{\circ}[/tex]  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]

Where:

[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]

[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]

[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]

[tex]\Delta l=0.10 ft[/tex]    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

(a) The inclined angle for which both blocks begin to slide is 10.3⁰.

(b) The compression of the spring is 0.22 ft.

The given parameters;

The normal force on block A and B:

[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]

The frictional force on block A and B:

[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]

The net force on the blocks when they starts sliding;

[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]

The change in the energy of the blocks is the work done in compressing the spring;

[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]

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Answer:

Explanation:

Given data:

power rating of the heater P= 1010 W

voltage of the heater V= 115 volts

current taken by the heater I= ?

We can apply the power formula to solve for the current in the heater

i.e P= IV

Making I the current subject of formula we have

I= P/V

Substituting our given data into the expression for I we have

I=1010/115= 8.78 A

Answer:

1.94cm³/s

Explanation:

1L = 1000cm³

Ihr = 3600s

So

Using

Average flow rate

Fr= 1L/100Km x 700Km/1hr x 1hr/3600s x 1000cm³/ 1L

= 1.94cm³/s

B. A chemical change

Explanation:

I'm guessing ?

Explanation:

Using Equations of Motion :

[tex]s = ut + \frac{1}{2} g {t}^{2} [/tex]

Height = 0 * 4 + 4.9 * 16

Height = 78.4 m

Answer:

50 cm long

When 35cm long wrench is compared to 50cm long wrench, we find that the 50cm long wrench produces more turning effect of force because it has longer distance between fulcrum and line of action of force. At conclusion, the more the turning effect of force the more it is easy to unscrew bolts.

Answer:

8.65x10^3m

Explanation:

See attached file

Each electron winds up with kinetic energy of

(270 keV)

plus

(whatever KE it had when it started accelerating).

Answer:

0.35 m³/s

Explanation:

When the pipe's depth is 0.4 m, the area of the circular segment is:

A = ½ R² (θ − sin θ)

The depth of the water is:

h = R (1 − cos(θ/2))

Solving for θ:

0.4 = 0.5 (1 − cos(θ/2))

0.8 = 1 − cos(θ/2)

cos(θ/2) = 0.2

θ/2 = acos(0.2)

θ = 2 acos(0.2)

θ ≈ 2.74 rad

The area is therefore:

A = ½ (0.5 m)² (2.74 − sin 2.74)

A = 0.338 m²

The cross-sectional area when the pipe is full is:

A = π (0.5 m)²

A = 0.785 m²

The flow velocity is constant:

v = v

Q / A = Q / A

(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)

Q = 0.35 m³/s

Answer:

Explanation:

To convert from °C to °F , the formula is

( F-32 ) / 9 = C / 5

F is reading fahrenheit scale and C is in centigrade scale .

F = 205 , C = ?

(205 - 32) / 9 = C / 5

C = 96°C approx .

Let us calculate the heat required .

Total heat required = heat required to heat up the ice at - 20 °C  to 0°C  + heat required to melt the ice + heat required to heat up the water at  0°C to

96°C.

=  5 x 2.04 x (20-0) +  5 x 336 + 5 x ( 96-0 ) x 4.2  kJ .

= 204 + 1680 + 2016

= 3900 kJ .

Answer: Astro 1 will have a 10x larger field of view than the Hubble.

Explanation: The hubble will also be extremely light weight that way it can go further into space and the mission will be able to last a longer amount of time.

Answer:

The potential will be Va/b

Explanation:

So Let sphere A charged Q to potential V.

so, V= KQ/a. ....(1

Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.

therefore, potential will be ,

V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]

Answer:

v/c = 0.76

Explanation:

Formula for Length contraction is given by;

L = L_o(√(1 - (v²/c²))

Where;

L is the length of the object at a moving speed v

L_o is the length of the object at rest

v is the speed of the object

c is speed of light

Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.

Thus;

0.65L_o = L_o[√(1 - (v²/c²))]

Dividing both sides by L_o gives;

0.65 = √(1 - (v²/c²))

Squaring both sides, we have;

0.65² = (1 - (v²/c²))

v²/c² = 1 - 0.65²

v²/c² = 0.5775

Taking square root of both sides gives;

v/c = 0.76

Answer:

y ’= y / 2

thus when the slit width is doubled the pattern width is halved

Explanation:

The diffraction of a slit is given by the expressions

          a sin θ = m λ

where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.

          sin θ = m λ / a

If this equation

          a ’= 2 a

we substitute

          2 a sin θ'= m λ

          sin θ'= (m λ / a)  1/2

          sin θ ’= sin θ / 2

We can use trigonometry to find the width

         tan θ = y / L

as the angle is small

         tan θ = sin θ / cos θ = sin θ

         sin θ = y / L  

we substitute

        y ’/ L = y/L   1/2

        y ’= y / 2

thus when the slit width is doubled the pattern width is halved

Answer:

The wavelength is  [tex]\lambda = 419 \ nm[/tex]

Explanation:

From the question we are told that

   The  distance of separation is   [tex]d = 3.34 *10^{-5} \ m[/tex]

   The  distance of the screen is  [tex]D = 3.30 \ m[/tex]

      The  order of the fringe is  n =  7

     The distance of separation of  fringes is y =  29.0 cm = 0.29 m

Generally the wavelength of the light is mathematically represented as

          [tex]\lambda = \frac{y * d }{ n * D}[/tex]

substituting values

         [tex]\lambda = \frac{0.29 * 3.34*10^{-5} }{ 7 * 3.30}[/tex]

        [tex]\lambda = 4.19*10^{-7}\ m[/tex]

        [tex]\lambda = 419 \ nm[/tex]

Answer:

a)  A = 1.99 μm , b) λ = 0.4134 m , c)  v = 57.2 m / s , d)   s = - 1,946 nm ,

e)      v_max = 1,739 mm / s

Explanation:

A sound wave has the general expression

           s = s₀ sin (kx - wt)

where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is

           s = 1.99 sin (15.2 x - 869 t)

a) the amplitude of the wave is

        A = s₀

        A = 1.99 μm

b) wave spectrum is

      k = 2π /λ

in the equation k = 15.2 m⁻¹

      λ = 2π / k

      λ = 2π / 15.2

     λ = 0.4134 m

c) the speed of the wave is given by the relation

       v = λ f

angular velocity and frequency are related

       w = 2π f

        f = w / 2π

        f = 869 / 2π

        f = 138.3 Hz

        v = 0.4134 138.3

         v = 57.2 m / s

d) To find the instantaneous velocity, we substitute the given distance and time into the equation

       s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

       s = 1.99 sin (0.77368 - 2.55486)

remember that trigonometry functions must be in radians

       s = 1.99 (-0.98895)

       s = - 1,946 nm

The negative sign indicates that it shifts to the left

e) the speed of the oscillating part is

           v = ds / dt)

           v = - s₀(-w) cos (kx -wt)

the maximum speed occurs when the cosines is 1

           v_maximo = s₀w

           v_maximum = 1.99 869

           v_maximo = 1739.31 μm / s

let's reduce to mm / s

          v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

          v_max = 1,739 mm / s

a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s

When A sound wave has the general expression is:

Then, s = s₀ sin (kx - wt)

Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is

s is = 1.99 sin (15.2 x - 869 t)

a) When the amplitude of the wave is

A is = s₀

Thus, A = 1.99 μm

b) When the wave spectrum is

k is = 2π /λ

Now, in the equation k = 15.2 m⁻¹

Then, λ = 2π / k

After that, λ = 2π / 15.2

Thus, λ = 0.4134 m

c) When the speed of the wave is given by the relation is:

Then, v = λ f

Now, the angular velocity and frequency are related is:

w is = 2π f

Then, f = w / 2π

After that, f = 869 / 2π

Now, f = 138.3 Hz

Then, v = 0.4134 138.3

Thus, v = 57.2 m / s

d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation

Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

After that, s = 1.99 sin (0.77368 - 2.55486)

Then remember that trigonometry functions must be in radians

After that, s = 1.99 (-0.98895)

Thus, s = - 1,946 nm

When The negative sign indicates that it shifts to the left

e) When the speed of the oscillating part is

Then, v = ds / dt)

Now, v = - s₀(-w) cos (kx -wt)

When the maximum speed occurs when the cosines is 1

Then, v_maximo = s₀w

After that, v_maximum = 1.99 869

v_maximo = 1739.31 μm / s

Now, let's reduce to mm / s

Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

Therefore, v_max = 1,739 mm / s

Finf more informmation about Wavelength here:

https://brainly.com/question/6352445

Answer:

Around the center of the mirror

Explanation:

Change in Velocity = final velocity - initial velocity

Change in velocity = 30km/h - (- 45km/h )

= 75 km/h due west

Answer:

The  angle is  [tex]\theta = 36.24 ^o[/tex]

Explanation:

From the question we are told that

    The  mass is  [tex]m = 0.6 \ kg[/tex]

     The radius is  [tex]r = 1.1 \ m[/tex]

     The speed is  [tex]v = 3.57 \ m /s[/tex]

According to  the law of energy conservation

  The  potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e

      [tex]m * g * h = \frac{1}{2} * m * v^2[/tex]

 =>    [tex]h = \frac{1}{2 g } * v^2[/tex]

Here h is the vertical distance traveled by the mass  which is also mathematically represented as

      [tex]h = r * sin (\theta )[/tex]

So

     [tex]\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2][/tex]

substituting values

     [tex]\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2][/tex]

     [tex]\theta = 36.24 ^o[/tex]

Answer:

The magnitude and direction are

7.638×10-4m

80.01°

Explanation:

We know that the coefficient of linear expansion for copper = 16.6×10^-6 m/m-C

ΔT = 40.2 - 18.0 = 28.5 C°

The expansion of horizontal pipe length can be calculated as

= (0.28)(16.6×10^-6)(28.5) = 13247×10^-8

= 0.0001325 m

The expansion of vertical pipe length = (1.28)(16.6×10^-6)(28.5) = 60557×10^-8 = 0.000752229 m

horizontal displacement = 0.1325 mm

= 1.356×10^-4m

vertical displacement = 0.75223mm

=7.5223×10-4m

size of total displacement can be calculated as

√(x²+y²)

Where x and y are vertical and horizontal displacement respectively

= √(0.1325)²+(0.75223)² =

= 0.7638 mm

= 7.638×10-4m

Angle below horizontal = arctan Θ

= 0.75223/0.1325

=5.6772

= arctan (5.6772)

= 80.01°

Therefore, the the magnitude and direction of the displacement of the pipe elbow when the water flow is turned at (7.638×10-4m) 0.7638 mm and 80.01°

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

Answer:

1. 6.99x 10^-6V/m

2. 18m

Explanation:

See attached file

Answer:

[tex]I=2.71\times 10^{-5}\ A[/tex]

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.  

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

[tex]C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}[/tex], r is radius

Let I is the displacement current. It is given by :

[tex]I=C\dfrac{dV}{dt}[/tex]

Here, [tex]\dfrac{dV}{dt}[/tex] is rate of increasing potential difference

So

[tex]I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A[/tex]

So, the value of displacement current is [tex]2.71\times 10^{-5}\ A[/tex].

Answer:

she need to use 20 cm thick

Explanation:

given data

wants the attic insulated = R-30

purchased = 10 cm thick

solution

as per given we can say that

10 cm is for the R 15

but she want for R 30

so  

R 30 thickness = [tex]\frac{30}{15} \times 10[/tex]  

R 30 thickness = 20 cm

so she need to use 20 cm thick

Explanation:

EE = ½ kx²

EE = ½ (800 N/m) (6 m)²

EE = 14,400 J

Answer:

14,400 J

Explanation:

Its the answer

Answer:

d)    α = 1693.5 rad / s² , a = 392.7 m / s² ,   a_total = α √(R² +1) ,

e)   tan θ = a / α

Explanation:

This is an exercise in linear and angular kinematics.

We initialize reduction of all the magnitudes to the SI system

   w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s

   w = 6000 rev / mi = 628.32 rad / s

   θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad

d) ask for centripetal, tangential and total acceleration.

Let's start by looking for centripetal acceleration, let's use the formula

          w² = w₀² + 2 α θ

          α = (w²- w₀²) / 2θ

we calculate

           α = (628.32²2 - 314.16²) / 2 75.398

           α = 1693.5 rad / s²

the quantity is linear and angular are related

the linear or tangential acceleration is

            a =    α  R

where R is the radius of the drum

            a = 1693.5 R

Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m

           a = 1693.5 0.20

           a = 392.7 m / s²

the total acceleration is

           a_total = √(a² + α²)

           a_total = √ (α² R² + α²)

           a_total = α √(R² +1)

e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant

Tangential acceleration is tangency to radius and its value varies proportionally radius

the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry

            tan θ = a / α

the angular velocity increases linearly when with centripetal acceleration

Answer:

196 nm

Explanation:

Given that

Value of wavelength, = 490 nm

Time spent in air, t(a) = 17.5 ns

Thickness of glass, th = 0.8 m

Time spent in glass, t(g) = 21.5 ns

Speed of light in a vacuum, c = 3*10^8 m/s

To start with, we find the difference between the two time spent

Time spent on glass - Time spent in air

21.5 - 17.5 = 4 ns

0.8/(c/n) - 0.8/c = 4 ns

Note, light travels with c/n speed in media that has index of refraction

(n - 1) * 0.8/c = 4 ns

n - 1 = (4 ns * c) / 0.8

n - 1 = (4*10^-9 * 3*10^8) / 0.8

n - 1 = 1.2/0.8

n - 1 = 1.5

n = 1.5 + 1

n = 2.5

As a result, the wavelength of light in a medium with index of refraction would then be

490 / 2.5 = 196 nm

Therefore, our answer is 196 nm

Answer:

433

Explanation: