A woman was told in 2020 that she had exactly 15 years to live. If she travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:

Answers

Answer 1

Answer:

2035

Explanation:

The doctor does not travel with the woman, and therefore, he won't experience any relativistic effect on his time. The doctor will judge time by the time here on earth. Technically, the last new year's day the doctor, who is here on earth, would expect the woman to celebrate will be in 2020 + 15 years = 2035


Related Questions

A city of punjab has 15 percemt chance of wet weather on any given day. What is probability that it will take a week for it three wet weather on 3 sepaprate days? Also find it standard deviation

Answers

Answer:

The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Explanation:

We are given that A city of Punjab has 15 percent chance of wet weather on any given day.

So, Probability of wet weather = 0.15

Probability of not being a wet weather = 1-0.15 =0.85

We are supposed to find probability that it will take a week for it three wet weather on 3 separate days

Total number of days in a week = 7

We will use binomial over here

n = 7

p =probability of failure = 0.15

q = probability of success=0.85

r=3

Formula :[tex]P(r=3)=^nC_r p^r q ^{n-r}[/tex]

[tex]P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166[/tex]

Standard deviation =[tex]\sqrt{n \times p \times q}[/tex]

Standard deviation =[tex]\sqrt{7 \times 0.15 \times 0.85}[/tex]

Standard deviation =0.9447

Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

A cube has a mass of 100 grams and its density is determined to be 1 g/cm3. The volume of the cube must be _____. 0.1 cm3 1 cm3 10 cm3 100 cm3

Answers

Answer:   The volume of the block will be [tex]100cm^3[/tex]

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{volume}[/tex]

Given : Mass of cube = 100 grams

Density of cube = [tex]1g/cm^3[/tex]

Putting in the values we get:

[tex]Volume=\frac{mass}{density}[/tex]

[tex]Volume=\frac{100g}{1g/cm^3}=100cm^3[/tex]

 Thus volume of the block will be [tex]100cm^3[/tex]

A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizontal. If the initial speed of the stream is 40 m/s the height that the stream of water will strike the building is

Answers

Answer:

We can think the water stream as a solid object that is fired.

The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)

The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).

The initial speed of the stream is 40m/s.

First, using the fact that:

x = R*cos(θ)

y = R*sin(θ)

in this case R = 40m/s and θ = 30°

We can use the above relation to find the components of the velocity:

Vx = 40m/s*cos(30°) = 34.64m/s

Vy = 20m/s.

First step:

We want to find the time needed to the stream to hit the buildin.

The horizontal speed is 34.64m/s and the distance to the wall is 50m

So we want that:

34.64m/s*t = 50m

t = 50m/(34.64m/s) = 1.44 seconds.

Now we need to calculate the height of the stream at t = 1.44s

Second step:

The only force acting on the water is the gravitational one, so the acceleration of the stream is:

a(t) = -g.

g = -9.8m/s^2

For the speed, we integrate over time and we get:

v(t) = -g*t + v0

where v0 is the initial speed: v0 = 20m/s.

The velocity equation is:

v(t) = -g*t + 20m/s.

For the position, we integrate again over time:

p(t) = -(1/2)*g*t^2 + 20m/s*t + p0

p0 is the initial height of the stream, this data is not known.

Now, the height at the time t = 1.44s is

p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po

             = 16.57m + p0

So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.

A pulley 326 mm in diameter and rotating initially at 4.00 revolutions per second receives a constant angular acceleration of 2.25 radians per second squared by a drive belt. What is the linear velocity of the belt after 5.00 seconds

Answers

Answer:

The linear velocity, v = 5.93 m/s

Explanation:

To find the linear velocity after 5 seconds, we find its angular velocity after 5 seconds using

ω' = ω + αt where ω = initial angular speed = 4.00 rev/s = 4.00 × 2π rad/s = 25.13 rad/s, ω' =  = final angular speed, α = angular acceleration = 2.25 rad/s² and t = time = 5.00 s

ω' = ω + αt

= 25.13 rad/s + 2.25 rad/s² × 5.00 s

= 25.13 rad/s + 11.25 rad/s

= 36.38 rad/s

The linear velocity v is gotten from v = rω' where r = radius of pulley = 326 mm/2 = 163 mm = 0.163 m

v = rω'

= 0.163 m × 36.38 rad/s

= 5.93 m/s

So, the linear velocity  v = 5.93 m/s

physics approach to study macromoelcues at nanoscales
in detail plx

Answers

Answer:

Abstracto

Los ácidos nucleicos y las proteínas comprenden una red de biomacromoléculas que almacenan y transmiten información que sustenta la vida de la célula. El estudio de estos mecanismos es un campo llamado biología molecular. El desarrollo de esta ciencia siempre ha ido acompañado de avances técnicos que permiten romper barreras metodológicas para probar hipótesis novedosas. Entre los métodos disponibles para los biólogos moleculares, destacan cinco: electroforesis, secuenciación, clonación, transferencia y reacción en cadena de la polimerasa. Su impacto llega a la genética, la medicina y la biotecnología. Aquí, se revisan la relevancia histórica, los fundamentos técnicos y las tendencias actuales de estos cinco métodos esenciales. La revisión pretende ser útil tanto para estudiantes como para científicos profesionales que buscan adquirir conocimientos avanzados sobre el valor de estos métodos para investigar los mecanismos moleculares que sostienen la vida.

What is the reason for the increase and decrease size of the moon and write down in a paragraph.

Answers

Answer:

The reason for the increase or decrease of the moon is due to the angular perception of the moon.

Explanation:

Also called lunar illusion, this phenomenon is due to the position in which the moon is, it can be at the zenith or on the horizon, both distances are different from each other with respect to the position of the person.

The zenith is the highest part of the sky and the horizon the lowest.

When there are landmarks such as trees, buildings or mountains on the horizon, the illusion of closeness is given and the illusion of distance is misinterpreted.

But when looking up at the sky as there is no reference point there will be a failure in the perception of size.

Convert 7,348 grams to kilograms

Answers

0.00735 kilograms is your answer. I’m sorry If I didn’t explain it right :(,

You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.2 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.8 cm .

Required:
What is the magnitude of the charge (in nC) on each bead?

Answers

Answer:

The magnitude of the charge is 54.9 nC.

Explanation:

The charge on each bead can be found using Coulomb's law:

[tex] F_{e} = \frac{k*q_{1}q_{2}}{r^{2}} [/tex]

Where:

q₁ and q₂ are the charges, q₁ = q₂  

r: is the distance of spring stretching = 4.8x10⁻² m

[tex]F_{e}[/tex]: is the electrostatic force

[tex] F_{e} = \frac{k*q^{2}}{r^{2}} \rightarrow q = \sqrt{\frac{F_{e}}{k}}*r [/tex]    

Now, we need to find [tex]F_{e}[/tex]. To do that we have that Fe is equal to the spring force ([tex]F_{k}[/tex]):

[tex] F_{e} = F_{k} = -kx [/tex]

Where:

k is the spring constant

x is the distance of the spring = 4.8 - 4.0 = 0.8 cm

The spring constant can be found by equaling the sping force and the weight force:

[tex] F_{k} = -W [/tex]

[tex] -k*x = -m*g [/tex]

where x is 5.2 - 4.0 = 1.2 cm, m = 1.8 g and g = 9.81 m/s²

[tex] k = \frac{mg}{x} = \frac{1.8 \cdot 10^{-3} kg*9.81 m/s^{2}}{1.2 \cdot 10^{-2} m} = 1.47 N/m [/tex]      

Now, we can find the electrostatic force:

[tex] F_{e} = F_{k} = -kx = -1.47 N/m*0.8 \cdot 10^{-2} m = -0.0118 N [/tex]

And with the magnitude of the electrostatic force we can find the charge:

[tex]q = \sqrt{\frac{F_{e}}{k}}*r = \sqrt{\frac{0.0118 N}{9 \cdot 10^{9} Nm^{2}/C^{2}}}*4.8 \cdot 10^{-2} m = 54.9 \cdot 10^{-9} C = 54.9 nC[/tex]

Therefore, the magnitude of the charge is 54.9 nC.

I hope it helps you!  

The magnitude of the charge (in nC) on each bead is equal to 55.21 nC.

Given the following data:

Original length = 4.0 cm to m = 0.04 mMass = 1.8 grams to kg = 0.0018New length = 5.2 cm to m = 0.052.Final length = 4.8 cm to m = 0.048 m.

Extension, e = [tex]0.052 - 0.048[/tex] = 0.012 m

Scientific data:

Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]Coulomb's constant = [tex]8.99 \times 10^9\; Nm^2/C^2[/tex]

To calculate the magnitude of the charge (in nC) on each bead, we would apply Coulomb's law:

First of all, we would determine the spring constant of this lightweight spring by using this formula:

[tex]W = mg = Ke \\\\K=\frac{mg}{e} \\\\K=\frac{0.0018 \times 9.8}{0.012} \\\\K=\frac{0.01764}{0.012}[/tex]

Spring constant, K = 1.47 N/m.

For the electrostatic force:

[tex]F = ke\\\\F = 1.47 \times 0.08[/tex]

F = 0.01176 Newton.

Coulomb's law of electrostatic force.

Mathematically, the charge in an electric field is given by this formula:

[tex]q = \sqrt{\frac{F}{k} } \times r[/tex]

Substituting the given parameters into the formula, we have;

[tex]q = \sqrt{\frac{0.01176 }{8.99 \times 10^9} } \times 0.048\\\\q=\sqrt{1.3228 \times 10^{-12}} \times 0.048\\\\q=1.1502 \times 10^{-6} \times 0.048\\\\q= 5.521 \times 10^{-8}\;C[/tex]

Note: 1 nC = [tex]1 \times 10^{-9}\;C[/tex]

Charge, q = 55.21 nC.

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In a double-slit experiment the distance between slits is 5.0 mm and the slits are 1.4 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 450 nm, and the other due to light of wavelength 590 nm. What is the separation in meters on the screen between the m = 5 bright fringes of the two interference patterns?

Answers

Answer:

 Δy = 1 10⁻⁴ m

Explanation:

In double-slit experiments the constructive interference pattern is described by the equation

           d sin θ = m λ

In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle

         tan θ = y / L

as the angles are small,

         tan θ = sin θ / cos θ = sin θ

substituting

         sin θ = y / L

         d y / L = m λ

         y = m λ / d L

let's apply this formula for each wavelength

λ = 450 nm = 450 10⁻⁹ m

m = 5

d = 5.0 mm = 5.0 10⁻³ m

      y₁ = 5 450 10⁻⁹ / (5 10⁻³  1.4)

      y₁ = 3.21 10⁻⁴ m

we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m

      y₂ = 5 590 10⁻⁹ / (5 10⁻³  1.4)

      y₂=  4.21 10⁻⁴ m

the separation of these two lines is

        Δy = y₂ - y₁

        Δy = (4.21 - 3.21) 10⁻⁴ m

        Δy = 1 10⁻⁴ m

Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of static friction are μA = 0.16 and μB = 0.23. Determine the incline angle θ for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 2.1 lb/ft .

Answers

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;

mass of block A, = 11 lbmass of block B, = 5 lbcoefficient of static friction for A, = 0.16coefficient of static friction for B, = 0.23 spring constant, k = 2.1 lb/ft

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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Current is the rate at which charge is flowing.

a. True
b. Fals

Answers

Answer:

A. True

Explanation:

In which type of indicating valve is the valve stem housed in a hollow metal post that contains a movable plate with a small glass window

Answers

Answer:

Post indicator valve

Explanation:

Post Indicator Valves are commonly used to control the water flow of sprinkler systems used in public and private buildings, warehouses, and factories for fire suppression. PIVs control water flow from the public system into the building's fire suppression system.

A typical ten-pound car wheel has a moment of inertia of about 0.35kg *m2. The wheel rotates about the axle at a constant angular speed making 70.0 full revolutions in a time interval of 4.00 seconds. What is the rotational kinetic energy K of the rotating wheel? Express answer in Joules

Answers

Answer:

The  rotational kinetic energy is  [tex]K = 2116.3 \ J[/tex]

Explanation:

From the question we are told that

    The moment of inertia  is  [tex]I = 0.35 \ kg \cdot m^2[/tex]

    The number of revolution is N  =  70 revolution

     The  time taken is  t  =  4.0  s

   

Generally the angular velocity is mathematically represented as

      [tex]w = \frac{2 \pi N }{t }[/tex]

substituting values

      [tex]w = \frac{2* 3.142 * 70 }{4 }[/tex]

       [tex]w = 109.97 \ rad/s[/tex]

     

The rotational kinetic energy K i mathematically represented as

       [tex]K = \frac{1}{ 2} * I * w^2[/tex]

substituting values

       [tex]K = \frac{1}{ 2} * 0.35 * (109.97)^2[/tex]

       [tex]K = 2116.3 \ J[/tex]

The intensity level 10 m from a point sound source is 85 dB. What is the intensity level 50 m away from the same source

Answers

Answer:

425dB

Explanation:

Given the intensity level 10 m from a point sound source is 85 dB, then;

L1 = 10m, I1= 85dB ...1

The intensity level 50 m away from the same source cal be calculated using the equivalent expression;

when L2 = 50m, I2 = ? ... 2

Solving equation 1 nad 2;

10m = 85db

50m = x

Cross multiplying;

50 * 85 = 10 * x

10x = 50*85

10x = 4250

Divide both sides by 10

10x/10 = 4250/10

x = 425 dB

Hence, the intensity level 50 m away from the same source is 425dB

Two protons moving with same speed in same direction repel each other but what about two protons moving with different speed in the same direction?

Answers

Answer:In the case of two proton beams the protons repel one another because they have the same sign of electrical charge. There is also an attractive magnetic force between the protons, but in the proton frame of reference this force must be zero! Clearly then the attractive magnetic force that reduces the net force between protons in the two beams as seen in our frame of reference is relativistic. In particular the apparent magnetic forces or fields are relativistic modifications of the electrical forces or fields. As such modifications, they cannot be stronger than the electrical forces and fields that produce them. This follows from the fact that switching frames of reference can reduce forces, but it can’t turn what is attractive in one frame into a repulsive force in another frame.

In the case of wires the net charges in two wires are zero everywhere along the wires. That makes the net electrical forces between the wires very nearly zero. Yet the relativistic magnetic forces and fields will be of the same sort as in the case of two beams of charges of a single sign. This is true even in the frame of reference of what we think as the moving charges, that is, the electrons. In the frame of reference moving at the drift velocity of these current-carrying electrons, it is the protons or positively charged ions that are moving in the other direction. Consequently in any frame of reference for current-carrying wires in parallel, the net electrical force will be essentially zero, and there will be a net attractive magnetic force

Explanation:                                                                              

Explanation:

Particles with similar charges (both positive or both negative) will always repel each other, regardless of their speed or direction.

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.


Required:

What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge?

Answers

Complete Question

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

Required:

What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?

Answer:

The velocity is [tex]v = 80.82 \ m/s[/tex]

Explanation:

From  the question we are told that

   The magnitude of charge one is  [tex]q_1 = 3.25 nC = 3.25 *10^{-9} \ C[/tex]

   The  magnitude of charge two  [tex]q_2 = 2.00 \ nC = 2.00 *10^{-9} \ C[/tex]

   The distance of separation is   [tex]d = 58.0 \ cm = 0.58 \ m[/tex]

 

Generally the electric potential of the electron at the midway point is mathematically represented as

         [tex]V = \frac{ q_1 }{\frac{d}{2} } + \frac{ q_2}{\frac{d}{2} }[/tex]

substituting values

         [tex]V = \frac{ 3.25 *10^{-9} }{\frac{ 0.58}{2} } + \frac{ 2 *10^{-9} }{\frac{ 0.58}{2} }[/tex]

          [tex]V = 1.8103 *10^{-8} \ V[/tex]

Now when the electron is 10 cm   =  0.10 m  from charge 1 , it is  (0.58 - 0.10 =  0.48 m ) m from charge two

Now the electric potential  at that point is mathematically represented as

       [tex]V_1 = \frac{q_1}{ 0.10} + \frac{q_2}{ 0.48}[/tex]

 substituting values

      [tex]V_1 = \frac{3.25 *10^{-9}}{ 0.10} + \frac{2.0*10^{-9}}{ 0.48}[/tex]

      [tex]V_1 = 3.67*10^{-8} \ V[/tex]

Now the law of energy conservation ,

   The  kinetic energy of the electron  =  potential energy of the electron

i.e     [tex]\frac{1}{2} * m * v^2 = [V_1 - V]* q[/tex]

where q is the magnitude of the charge on the electron with value

      [tex]q = 1.60 *10^{-19} \ C[/tex]

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

        [tex]\frac{1}{2} * 9.11 *10^{-19} * v^2 = [ (3.67 - 1.8103) *10^{-8}]* 1.60 *10^{-19}[/tex]

         [tex]v = \sqrt{6532.4}[/tex]

        [tex]v = 80.82 \ m/s[/tex]

 

Which scientist proposed a mathematical solution for the wave nature of light?

Answers

Answer:

Explanation:

Christian Huygens

Light Is a Wave!

Then, in 1678, Dutch physicist Christian Huygens (1629 to 1695) established the wave theory of light and announced the Huygens' principle.

Two 75 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. Part A How much power is dissipated by each bulb

Answers

Answer:

300 W

Explanation:

power of each bulb P = 75 W

voltage in the circuit = 120 V

we know that electrical power P = IV    ....1

and V = IR

we can also say that I = V/R

substituting for I in  equation 1, we have

P = [tex]V^{2}/R[/tex]    ....2

The total total power in the circuit = 75 x 2 = 150 W

from equation 2, we have

150 = [tex]120^{2} /R[/tex]

R = [tex]120^{2}/150[/tex] = 96 Ω    this is the resistance of the whole circuit.

This resistance is due to the two light bulbs, for each light bulb since they are arranged in series

R = 96/2 = 48 Ω

From P =  [tex]V^{2}/R[/tex]  

for each light bulb, power is

P = [tex]120^{2} /48[/tex] = 300 W

The intensity at a certain distance from a bright light source is 7.20 W/m2 .
A. Find the radiation pressures (in pascals) on a totally absorbing surface and a totally reflecting surface.
B. Find the radiation pressures (in atmospheres) on a totally absorbing surface and a totally reflecting surface.

Answers

Answer:

A) P_rad.abs = 2.4 × 10^(-8) Pa and P_rad.ref = 4.8 × 10^(-8) Pa

B) P_rad.abs = 2.369 × 10^(-13) atm and P_rad.ref = 4.738 × 10^(-13) atm

Explanation:

A) The formula for radiation pressure for absorbed light is given as;

P_rad = I/c

Where I is the intensity = 7.20 W/m² and c is the speed of light = 3 × 10^(8) m/s

Thus;

P_rad = 7.2/(3 × 10^(8))

P_rad.abs = 2.4 × 10^(-8) Pa

Now formula for radiation pressure for reflected light is given as;

P_rad = 2I/c

Thus;

P_rad = (2 × 7.2)/(3 × 10^(8))

P_rad.ref = 4.8 × 10^(-8) Pa

B) Now, 1.013 × 10^(5) Pa = 1 atm

Thus, for the absorbed surface, we have;

P_rad.abs = (2.4 × 10^(-8))/(1.013 × 10^(5))

P_rad.abs = 2.369 × 10^(-13) atm

For the reflecting surface, we have;

P_rad_ref = (4.8 × 10^(-8))/(1.013 × 10^(5))

P_rad.ref = 4.738 × 10^(-13) atm

3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000 rpm, completando un total de 12 revoluciones.

d. Determina la aceleración tangencial, centrípeta y la total en m.s-2 cuando el tambor a alcanzado los 60000 rpm

e. Explica lo que ocurre con la magnitud y dirección de los vectores aceleración tangencial, aceleración centrípeta, aceleración total, aceleración angular, velocidad angular cuando la lavadora ha girado desde 3000 rpm hasta 6000 rpm.

Answers

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

it is easier to drag a stone than to kick it?why.​

Answers

Answer:

you are going to expend energy to give a lot of velocity (and momentum) to your foot in order to transfer it the stone air drag this time the kicking speed is for superior to walking speed.

 

                                                  Thank You

A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. A parallel plate capacitor sets up an electric field E which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by 3 mm and the value of the magnetic field is 0.3 T, what voltage between the plates will allow particles of speed 5 x 105 m/s to pass straight through without deflection? A. 70 V B. 140 V C. 450 V D. 1,400 V E. 2,800 V

Answers

Answer:

C. 450v

Explanation:

Using

Voltage= B*distance of separation*velocity

3mm x 0.3T x 5E5m/s

= 450v

Specific heat is a measurement of the amount of heat energy input required for one gram of a substance to increase its temperature by one degree Celsius. Solid lithium has a specific heat of 3.5 J/g·°C. This means that one gram of lithium requires 3.5 J of heat to increase 1°C. Plot the temperature of 1g of lithium after 3.5, 7, and 10.5 J of thermal energy are added.

Answers

Answer:

ΔT = 1ºC , 2ºCand 3ºC

Explanation:

In this exercise they indicate the specific heat of lithium

let's calculate the temperature increase as a function of the heat introduced

          Q = m [tex]c_{e}[/tex] ΔT

          ΔT = Q / m c_{e}

calculate

 for Q = 3.5 J

         ΔT = 3.5 / (1 3.5)

         ΔT = 1ºC

For Q = 7.0 J

         ΔT = 7 / (1 3.5)

         ΔT = 2ºC

for Q = 10.5 J

         ΔD = 10.5 / (1 3.5)

         ΔT = 3ºC

we see that this is a straight line, see attached

An organ pipe of length 3.0 m has one end closed. The longest and next-longest possible wavelengths for standing waves inside the pipe are

Answers

Answer:

The longest wavelength for closed at one end and open at the other is

y / 4      where y is the wavelength - that is node - antinode

The next possible wavelength is 3 y / 4 -    node - antinode - node -antinode

y / 4 = 3 m     y = 12 meters    the longest wavelength

3 y / 4 = 3 m      y = 4 meters   1 / 3 times as long

the treasure map gives the following directions to the buried treasure ​

Answers

Answer:

North

South

East

West

Explanation:

please mark me as brainliest

A bullet is fired from a rifle pointed 45 degrees above horizontal. The bullet leaves the muzzle traveling 1400 m/s. How many seconds does it take the bullet to reach the high point of its trajectory?

Answers

The bullet's vertical velocity at time [tex]t[/tex] is

[tex]v=1400\dfrac{\rm m}{\rm s}-gt[/tex]

where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.

At its highest point, the bullet's vertical velocity is 0, which happens

[tex]0=1400\dfrac{\rm m}{\rm s}-gt\implies t=\dfrac{1400\frac{\rm m}{\rm s}}g\approx\boxed{142.857\,\mathrm s}[/tex]

(or about 140 s, if you're keeping track of significant figures) after being fired.

An interference pattern is produced by light with a wavelength 590 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.580 mm .


Required:

a. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?

b. What would be the angular position of the second-order, two-slit, interference maxima in this case?

Answers

Answer:

a. 0.058°

b.  0.117°

Explanation:

a. The angular position of the first-order is:

[tex] d*sin(\theta) = m\lambda [/tex]

[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{1* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.058 ^{\circ} [/tex]

Hence, the angular position of the first-order, two-slit, interference maxima is 0.058°.

b. The angular position of the second-order is:

[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{2* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.12 ^{\circ} [/tex]

Therefore, the angular position of the second-order, two-slit, interference maxima is 0.117°.

I hope it helps you!

With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor would the following change?
A. Kinetic energy when passing through the equilibrium position.
B. Speed when passing through the equilibrium position.

Answers

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

An electric heater draws 13 amperes of current when connected to 120 volts. If the price of electricity is $0.10/kWh, what would be the approximate cost of running the heater for 8 hours?
(A) $0.19
(B) $0.29
(C) $0.75
(D) $1.25
(E) $1.55

Answers

Answer:

C $0.75 my friend I wish it is right answer

what happened when aniline is treated with benzene diazonium chloride​

Answers

Answer:

p-aminoazobenzene is formed

Explanation:

The reaction of benzene diazonium chloride and aniline takes place in a basic medium and leads to the formation of an azo compound which is also a dye. The terminal diazonium nitrogen of the benzene diazonium ion is coupled to the aniline at the para-position. The product of the reaction, p-aminoazobenzene is a yellow dye.

Benzene diazonium chloride is prepared by diazotization of aniline in the presence of hydrochloric acid. The full reaction of aniline and benzene diazonium chloride​ is shown in the image attached to this answer.

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