Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon return, the people on earth will have advanced exactly 120 years into the future. According to special relativity, how fast must you travel

Answer:

 λ = 605.80 nm

Explanation:

These double-slit experiments the equation for constructive interference is

          d sin θ = m λ

where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.

In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm

Let's use trigonometry to find the angle

         tan θ = y / L

as the angles are very small

          tan θ = sin θ / cos θ = sin θ

we substitute

         sin θ = y / L

we substitute in the first equation

         d y / L = m λ          

          λ = d y / m L

let's calculate

           λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)

           λ = 6.05805 10⁻⁷ m

let's reduce to nm

          λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)

          λ = 605.80 nm

B. A chemical change

Explanation:

I'm guessing ?

Answer:

The direction of net magnetic force on the circular section of the loop is in the positive y-axis

The net magnetic force on the circular section of the loop is 3.74 N

Explanation:

The magnetic field strength [tex]B[/tex] = 1.7 T

the current [tex]I[/tex] = 0.7 A

The diameter of the loop = 2 m

the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2

==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m

The force on the semi-circular is given as

F = [tex]BIl[/tex] sin ∅

but the loop is perpendicular to the field, therefore

sin ∅ = sin 90° = 1

F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N

The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".

According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis

Answer:

E. None of these statements are true.

Explanation:

We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.

for A, the tension on the rope does not depend on if both teams pull are in equilibrium.

for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.

for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.

for D,  just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.

we go with option E.

Answer:

Therefore the characteristics to be found are:

* the focal length must be large and the focal length of the eyepiece must be small

* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light

* The system must be configured to the far sight tip,

Explanation:

The length of the telescope is

         L = f_ocular + f_objetive

the magnification of the telescope is

         m = - f_objective / f_ocular

These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:

* the focal length must be large and the focal length of the eyepiece must be small

* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light

* The system must be configured to the far sight tip,

Each electron winds up with kinetic energy of

(270 keV)

plus

(whatever KE it had when it started accelerating).

Answer:

A)     K / K₀ = 4   b)     v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

      Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

      [tex]Em_{f}[/tex] = K = ½ m v²

        Em₀ = Em_{f}

        ½ k 4 x₀² = K

        (½ K x₀²) = K₀

         K = 4 K₀

          K / K₀ = 4

B) the speed value

          ½ k 4 x₀² = ½ m v²

          v = 4 (k / m) x₀

if we call

           v₀ = k / m x₀

          v = 4 v₀

         v / v₀ = 4

Answer:

8.65x10^3m

Explanation:

See attached file

Answer:

The mas unit is the the 'Kilogram' written as 'kg'

Volume is 10 L

Explanation:

The complete question is

If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First, what are the mass units?

Second, what is the volume

mass units is the 'Kilogram', written as 'kg'

density = mass/volume = 100 kg/L

the mass  = 1000 kg

volume = mass/density = 1000/100 = 10 L

Answer:

The charge is  [tex]Q = 2.177 *10^{-9} \ C[/tex]

Explanation:

From the question we are told that

     The electric potential is  [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]

     The  mass of the drop is  [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]

      The  speed is  [tex]v = 18.8 \ m/s[/tex]

Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question  is mathematically represented as

       [tex]Q = \frac{m v^2 }{ 2 * V }[/tex]

Substituting values

      [tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]

       [tex]Q = 2.177 *10^{-9} \ C[/tex]

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:

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:

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:

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:

The current is  [tex]I = 8.9 *10^{-5} \ A[/tex]

Explanation:

From the question we are told that

     The  radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]

      The current density is  [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]

      The distance we are considering is  [tex]r = 0.5 R = 0.001585[/tex]

Generally current density is mathematically represented as

          [tex]J = \frac{I}{A }[/tex]

Where A is the cross-sectional area represented as

         [tex]A = \pi r^2[/tex]

=>      [tex]J = \frac{I}{\pi r^2 }[/tex]

=>    [tex]I = J * (\pi r^2 )[/tex]

Now the change in current per unit length is mathematically evaluated as

        [tex]dI = 2 J * \pi r dr[/tex]

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         [tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]

         [tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]

        [tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]

substituting values

        [tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]

        [tex]I = 8.9 *10^{-5} \ A[/tex]

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:

Explanation:

The equation for a traveling wave as expressed as y(x, t) = A cos(kx − [tex]\omega[/tex]t) where An is the amplitude f oscillation, [tex]\omega[/tex] is the angular velocity and x is the horizontal displacement and y is the vertical displacement.

From the formula; [tex]k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f[/tex] where;

[tex]\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency[/tex]

Before we can get the transverse speed, we need to get the frequency and the wavelength.

frequency = 1/period

Given period = 2/15 s

Frequency = [tex]\frac{1}{(2/15)}[/tex]

frequency = 1 * 15/2

frequency f = 15/2 Hertz

Given wavelength [tex]\lambda[/tex] = 2m

Transverse speed [tex]v = f \lambda[/tex]

[tex]v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s[/tex]

Hence, the transverse speed at that point is  15m/s

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:

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:

433

Explanation:

Answer:

Option (c)

Explanation:

In a Series circuit, as the components are connected end-to-end ,the current can flow through only one path from start to finish.

(C.) is the only correct statement in the list of choices.

In a series circuit, the current can flow through only one path from start to finish.

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:

[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:

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:

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

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].

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:

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:

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

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