It took a student 30 minutes to drive from his home to campus on
Monday, and it took him 20 minutes on Tuesday driving the same
route. If on Monday he drove 36 mi/hr on average, what was his
average speed on Tuesday?
O 12 mi/hr
O 18 mi/hr
O 48 mi/hr
O 54 mi/hr
O 72 mi/hr​

Answers

Answer 1

Answer:

48 i believe

Explanation:


Related Questions

A cube has one corner at the origin and the opposite corner at the point (L,L,L)(L,L,L). The sides of the cube are parallel to the coordinate planes. The electric field in and around the cube is given by

Answers

Answer:

Net charge = E• b • L^3.

Explanation:

NB: here, the symbol representation of the flux is "p" = electric Field • Area(dot Product).

So, we will take a look at the flux through -x face, through x face and through -y face, through y face and through - z face and through z face.

(1). Starting from -z and z faces which are the back and front faces of the cube:

Thus, We have that the flux,p = 0 for -z and z.

(2). Recall that we are given that E = =(a+bx)i^+cj^.

Thus, p_-y = (a + bx)i + cj (-j) (L^2)

Where y = 0

p_-y = -cL^2.

Obviously for p_j, we will have cL^2 and y = L

(3). For p_-x = =(a + bx)i + cj (-i) (L^2).

p_-x = -aL^2

Where x = 0.

When x = L and p_x = (a + bL)L^2.

This, adding all together gives Net charge = E • b • L^3.

differences between
hor
rse
and horse​

Answers

The proper difference between hor Rse and horse is both shows the same thing that is the horse
the difference is that they are spelled differently

Receiver maxima problem. When the receiver moves through one cycle, how many maxima of the standing wave pattern does the receiver pass through

Answers

The number of maxima of the standing wave pattern is two.

Maxima problem:

At the time when the receiver moves via one cycle so here two maximas should be considered. At the time when the two waves interfere by traveling in the opposite direction through the same medium so the standing wave pattern is formed.

learn more about the waves here: https://brainly.com/question/3004869?referrer=searchResults

A vertically polarized light wave of intensity 1000 mW/m2 is coming toward you, out of the screen. After passing through this polarizing filter, the wave's intensity is

Answers

Answer:

The intensity is [tex]I = 500 mW/m^2[/tex]

Explanation:

From the question we are told that

    The  intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]

Generally the intensity of the light emerging from the polarizer is  mathematically represented as

          [tex]I = \frac{I_o}{2}[/tex]

substituting values

         [tex]I = \frac{1000 *10^{-3}}{2}[/tex]

         [tex]I = 500 *10^{-3} W/m^2[/tex]

         [tex]I = 500 mW/m^2[/tex]

A centrifuge's angular velocity is initially at 159.0 radians/second to test the stability
of a high speed drill component. It then increases its angular velocity to 999.0
radians/second. If this is achieved in 4,100.0 radians what is the angular acceleration
of the centrifuge?

Answers

Answer:

118.6 rad/s²

Explanation:

Δθ = 4100.0 rad

ω₀ = 159.0 rad/s

ω = 999.0 rad/s

Find: α

ω² = ω₀² + 2αΔθ

(999.0 rad/s)² = (159.0 rad/s)² + 2α (4100.0 rad)

α = 118.6 rad/s²

1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?

Answers

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = [tex]\sqrt{3^{2} +2^{2} }[/tex] = [tex]\sqrt{13}[/tex] = 3.61cm = 0.036m

r₂ = [tex]\sqrt{4^{2} + 3^{2} }[/tex] = [tex]\sqrt{25}[/tex] = 5cm = 0.05m

electric potential V = [tex]\frac{kq}{r}[/tex]

change in potential ΔV = [tex]V_{1} - V_{2}[/tex]

ΔV = [tex]\frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} }[/tex] , where [tex]q_{1} = q_{2}=[/tex]2.00μC

ΔV = [tex]2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })[/tex]

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × [tex](\frac{1}{0.036} - \frac{1}{0.05} )[/tex]

ΔV= 2.789×10⁵

[tex]\frac{1}{2}mv^{2}[/tex] = ΔV × q₃

[tex]\frac{1}{2}[/tex] ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC  = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

Read more about point charges at;https://brainly.com/question/13914561

I wish to use a step up transformer to turn an initial RMS AC voltage of 100 V into a final RMS AC voltage of 200 V. What is the ratio of the number of turns in the primary to the secondary

Answers

Answer:

1:2

Explanation:

It is given that,

Initial RMS AC voltage is 100 V and final RMS AC voltage is 200 V.

We need to find the ratio of the number of turns in the primary to the secondary  for step up transformer.

For a transformer, [tex]\dfrac{V_1}{V_2}=\dfrac{N_1}{N_2}[/tex]

So,

[tex]\dfrac{N_1}{N_2}=\dfrac{100}{200}\\\\\dfrac{N_1}{N_2}=\dfrac{1}{2}[/tex]

So, the ratio of the number of turns in the primary to the secondary is 1:2.

An RC circuit is connected across an ideal DC voltage source through an open switch. The switch is closed at time t = 0 s. Which of the following statements regarding the circuit are correct?
a) The capacitor charges to its maximum value in one time constant and the current is zero at that time.
b) The potential difference across the resistor and the potential difference across the capacitor are always equal.
c) The potential difference across the resistor is always greater than the potential difference across the capacitor.
d) The potential difference across the capacitor is always greater than the potential difference across the resistor
e) Once the capacitor is essentially fully charged, There is no appreciable current in the circuit.

Answers

Answer:

e)

Explanation:

In an RC series circuit, at any time, the sum of the voltages through the resistor and the capacitor must be constant and equal to the voltage of the DC voltage source, in order to be compliant with KVL.

At= 0, as the voltage through the capacitor can't change instantaneously, all the voltage appears through the resistor, which means that a current flows, that begins to charge the capacitor, up to a point that the voltage through the capacitor is exactly equal to the DC voltage, so no current flows in the circuit anymore, and the charge in the capacitor reaches to its maximum value.

21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases

Answers

Answer:

B.distance between lines increases

Answer:

A. Lines become narrower

Explanation:

I  got it right on my quiz!

I hope this helps!! :))

An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.25 m/s2. Determine the orbital period of the satellite.

Answers

Answer:

118 minutes( 2 hours approximately )

Explanation:

Here, we are interested in calculating the orbital period of the satellite

Please check attachment for complete solution

Answer:

T = 7101 s = 118.35 mins = 1.9725 hrs

Explanation:

To solve the question, we apply the formula for gravitational acceleration

a = GM/r², where

a = acceleration due to gravity

G = gravitational constant

M = mass of the earth

r = distance between the satellite and center of the earth

Now, if we make r, subject of formula, we have

r = √(GM/a)

Recall also, that

a = v²/r, making v subject of formula

v = √ar

If we substitute the equation of r into it, we have

v =√a * √r

v =√a * √[√(GM/a)]

v = (GM/a)^¼

Again, remember that period,

T = 2πr/v, we already have v and r, allow have to do is substitute them in

T = 2π * √(GM/a) * [1 / (GM/a)^¼]

T = 2π * (GM/a³)^¼

T = 2 * 3.142 * [(6.67*10^-11 * 5.97*10^24) / (6.25³)]^¼

T = 6.284 * [(3.982*10^14) / 244.140]^¼

T = 6.284 * (1.63*10^12)^¼

T = 6.284 * 1130

T = 7101 s

T = 118.35 mins

T = 1.9725 hrs

A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the switch is open and the capacitor is uncharged. What is the voltage across the resistor and the capacitor at the moment the switch is closed

Answers

Answer:

The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.

Explanation:

This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery

A wheel on a car is rolling without slipping along level ground. The speed of the car is 36 m/s. The wheel has an outer diameter of 50 cm. The speed of the top of the wheel is

Answers

Answer:

The speed of the top of the wheel is twice the speed of the car.

That is: 72  m/s

Explanation:

To find the speed of the top of the wheel, we need to combine to velocities: the tangential velocity of the rotating wheel due to rotational motion [tex](v_t=\omega\,R=\omega\,(0.25\,m)\,)[/tex] - with [tex]\omega[/tex] being the wheel's angular velocity,

plus the velocity due to the translation of the center of mass (v = 36 m/s).

The wheel's angular velocity (in radians per second) can be obtained using the tangential velocity for the pure rotational motion and it equals:[tex]\omega=\frac{v_t}{r} =\frac{36}{0.25} \,s^{-1}[/tex]

Then the addition of these two velocities equals:

[tex]\omega\,R+v=\frac{36}{0.25} (0.25)\,\,\frac{m}{s} +36\,\,\frac{m}{s} =72\,\,\frac{m}{s}[/tex]

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

What is the major cause of the muffled noise from a radio station?

Answers

Answer:

The major cause is "lack of high frequencies in a sound wave".

Explanation:

Muffling derives from either the absence of such a radio signal of a higher or specific frequency. This very same phenomenon has been observed whenever you overhear conversations through some kind of wall and perhaps door. The approach is equalization. This method helps them to raise those frequencies although these overprotective wavelengths decrease.

So that the above would be the correct solution.

The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine

Answers

Answer:

0.306mm

Explanation:

The radius of the conductor is 3mm, or 0.003m

The area of the conductor is:

A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2

The current density is:

J = 130/2.8*10^-5 = 4.64*10^6 A/m

According to the listed reference:

Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s

The distance traveled is:

x = v*t = 0.34 * .90 = 0.306 mm

g The current in a series circuit is 15.0 A. When an additional 8.00-% resistor is inserted in series, the current drops to 12.0 A. What is the resistance in the original circuit

Answers

Answer:

Explanation:

Let the original resistance be R and voltage be V

Applying ohm's law

V / R = 15

V = 15 R

In second case

V / (R+8 ) = 12

V = 12 R + 96

15 R = 12 R + 96

3R = 96

R = 32 ohm .

A boat floating in fresh water displaces 16,000 N of water. How many newtons of salt water would it displace if it floats in salt water of specific gravity 1.10

Answers

Answer:

It will displace the same weight of fresh water i.e.16000N. The point is the body 'floats'- which is the underlying assumption here, and by Archimedes Principle, for this body or vessel or whatever it may be, to float it should displace an equal weight of water

Explanation:

The right-hand rule, which is a convention for identifying the direction of the force on a current or a moving charged positively charged particle, has several correct versions. Which one of the descriptions below is the right-hand rule for the magnetic force exerted on a current or a moving charged particle recommended in this textbook?
A. Thumb of the right hand points in the direction of current or the velocity of the charged particle, the fingers in the direction of B, and the force (F) is directed perpendicular to the right hand palm.
B. Keeping your right hand flat, point your thumb in the direction of the current or the velocity of the charged particle, the remaining four fingers perpendicular to the thumb in the direction of magnetic field. The magnetic force, as the result of the magnetic field on the current, is the direction your palm is facing.
C. Using your right hand, point your thumb in the direction of the current or the velocity of the charged particle, your fingers in the direction of magnetic field, and your palm points in the direction of the cross-product.
D. Using your right-hand, point your index finger in the direction of the current or the velocity of the charged particle. Point your middle finger in the direction of the magnetic field. Your thumb now points in the direction of the magnetic force.
E. Using the right hand, the direction of the thumb is the direction of the force, the direction of the index finger indicates the direction of the magnetic field, and the direction of the middle finger is the direction of the electric current. Submit

Answers

Answer:

B. Keeping your right hand flat, point your thumb in the direction of the current or the velocity of the charged particle, the remaining four fingers perpendicular to the thumb in the direction of magnetic field. The magnetic force, as the result of the magnetic field on the current, is the direction your palm is facing.

Explanation:

This is the Fleming's right hand rule, which was stated to explain the relationship or induction ability of the magnetic field, current or velocity of charged particles and magnetic force. These three variables are held mutually perpendicularly to one another.

The most suitable description of the right-hand rule is option B which clarifies the perpendicular mutual relationship of the thumb in the direction of the current or the velocity of the charged particle, the remaining four fingers perpendicular to the thumb in the direction of magnetic field. The magnetic force, as the result of the magnetic field on the current, is the direction your palm is facing.

"A parcel moving in a horizontal direction with speed v0 = 13 m/s breaks into two fragments of weights 1.4 N and 1.9 N, respectively. The speed of the larger piece remains horizontal immediately after the separation and increases to v1.9 = 29 m/s. Find the necessary speed and direction of the smaller piece immediately after the separation. (Assume the initial direction of the parcel is positive. Indicate the direction with the sign of your answer.)"

Answers

Answer:

the smaller particle moves with speed of 8.706 m/s in the opposite direction to the bigger particle.

Explanation:

Speed of the original particle = 13 m/s

We designate particles as A and B

The final weights of the component particles are

Particle A = 1.4 N

particle B = 1.9 N

The speed of the larger piece (particle B) = 29 m/s

We know that weight is the product of a body's mass and acceleration due to gravity g which is equal to 9.81 m/s^2, therefore, masses of the particles are

particle A = 1.4/9.81 = 0.143 kg

Particle B = 1.9/9.81 = 0.194 kg

The momentum of a body is the product of its mass and its velocity i.e

P = mv

This means that the mass of the particle before splitting is  

0.143 kg + 0.194 kg = 0.337 kg

Momentum of the initial whole particle = mv

==> 0.337 x 13 = 4.381 kg-m/s

The bigger particle B remains horizontal, and has a momentum of

mv = 0.194 x 29 = 5.626 kg-m/s

According to the conservation of momentum, the total initial momentum of a system must be equal tot the total final momentum of the system.

Initial total momentum of the system = 4.381 kg-m/s (momentum of original particle before splitting)

Final total momentum of the system = Total momentum of the particles after splitting = 5.626 kg-m/s + ( 0.143 kg x [tex]V_{B}[/tex])

where  [tex]V_{B}[/tex]  is the velocity of smaller particle A

final total momentum of the system = 5.626 + 0.143[tex]V_{B}[/tex]

Equating the two momenta of the system, we'll have

4.381 = 5.626 + 0.143[tex]V_{B}[/tex]

4.381 - 5.626 = 0.143[tex]V_{B}[/tex]

-1.245 = 0.143[tex]V_{B}[/tex]

[tex]V_{B}[/tex]  = -1.245/0.143 = -8.706 m/s

The negative sign indicates that the smaller particle moves in the opposite direction to the bigger particle

11. A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is fingered at a length of only 70% of its original length

Answers

Answer:

The frequency is  [tex]f_n = 257.1 \ Hz[/tex]

 

Explanation:

From the question we are told that

    The third harmonic frequency of the tight guitar string is  [tex]f_3 = 540 \ Hz[/tex]

     

Let the original length be  L  

   Then the length at which it is fingered is  0.7 L

Generally the fundamental  is mathematically represented as

         [tex]f = \frac{v_s}{ 2L}[/tex]

Now when it finger at 70% it original length is

      [tex]f_n = \frac{v}{2 * (0.7 L)}[/tex]

      [tex]f_n = \frac{v}{1.4 L}[/tex]

Here v  the velocity of sound

  So  

         [tex]\frac{f_n}{f} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }[/tex]

Also the fundamental frequency for the original length can also be represented as

       [tex]f = \frac{f_3}{3}[/tex]

substituting values

          [tex]f = \frac{540}{3}[/tex]

          [tex]f = 180 \ Hz[/tex]

So

       [tex]\frac{f_n}{180} = \frac{\frac{v}{1.4L} }{\frac{v}{2L} }[/tex]

=>  [tex]f_n =\frac{180}{0.7}[/tex]

=>   [tex]f_n = 257.1 \ Hz[/tex]

 

     

The fundamental frequency, if it is fingered at a length of only 70% of its original length, will be 257.1  Hz.

What is the frequency?

Frequency is defined as the number of repetitions of a wave occurring waves in 1 second.

f is the frequency of tight guitar string = 540 Hz

Let's call the original length L.

The amount of time it is fingered is then 0.7 L.

In general, the fundamental frequency is expressed mathematically as;

[tex]\rm f = \frac{v_0}{2L} \\\\[/tex]

For the given conditions;

[tex]\rm f_n=\frac{v}{2 \times 0.7L} \\\\ \rm f_n=\frac{v}{1.4L}[/tex]

The ratio of the frequency is;

[tex]\rm \frac{f_n}{f} =\frac{\frac{v}{1.4L} }{\frac{V}{2L} }[/tex]

Also, the fundamental frequency for the original length can also be represented as;

[tex]\rm f= \frac{f'}{3} \\\\ f=\frac{540}{3} \\\\ \rm f=180\ Hz[/tex]

On putting the given data;

[tex]\rm \frac{f_n}{180} =\frac{\frac{v}{1.4L} }{\frac{V}{2L} }\\\\ \rm f_n=\frac{180}{0.7}\\\\\ \rm f_n=257.1\ Hz[/tex]

Hence the fundamental frequency, if it is fingered at a length of only 70% of its original length, will be 257.1  Hz.

To learn more about the frequency reference the link;

https://brainly.com/question/14926605

The cart now moves toward the right with an acceleration toward the right of 2.50 m/s2. What does spring scale Fz read? Show your calculations, and explain.

Answers

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The spring scale [tex]F_2[/tex] reads  [tex]F_2 = 2.4225 \ N[/tex]

Explanation:

From the question we are told that

      The first force is  [tex]F_1 = 10.5 \ N[/tex]

      The acceleration by which the cart moves to the right is  [tex]a = 2.50 \ m/s^2[/tex]

      The mass of the cart is  m  = 3.231  kg

       

Generally the net force on the cart is  

       [tex]F_{net} = F_1 - F_2[/tex]

This net force is mathematically represented as

      [tex]F_{net} = m * a[/tex]

So  

        [tex]m* a = 10 - F_2[/tex]

        [tex]F_2 = 10.5 - 2.5 (3.231)[/tex]

        [tex]F_2 = 2.4225 \ N[/tex]

 

change in entropy of universe during 900g of ice at 0 degree celcus to water at 10 degree celcius at room temp=30 degree celcius

Answers

Answer:

4519.60 J/K

Explanation:

Change in entropy is expressed as ΔS = ΔQ/T where;

ΔQ is the total heat change during conversion of ice to water.

T is the room temperature

First we need to calculate the total change in heat using the conversion formulae;

ΔQ = mL + mcΔθ (total heat energy absorbed during phase change)

m is the mass of ice/water = 900g = 0.9kg

L is the latent heat of fusion of ice = 3.33 x 10⁵J/kg

c is the specific heat capacity of water = 4200J/kgK

Δθ is the change in temperature of water = 10°C - 0C = 10°C = 283K

Substituting the given values into ΔQ;

ΔQ = 0.9(333000)+0.9(4200)(283)

ΔQ = 299700 + 1069740

ΔQ = 1,369,440 Joules

Since Change in entropy ΔS = ΔQ/T

ΔS =  1,369,440/30+273

ΔS = 1,369,440/303

ΔS = 4519.60 J/K

Hence, the change in entropy of the universe is 4519.60 J/K

A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40 MΩ. After a time of 4.00 s, the voltmeter reads 3.0 V.
A) What are the capacitance?
B) What is the time constant of the circuit?

Answers

Answer:

a. 0.849 micro farad

b. 2.89 s

Explanation:

a) V=V0 e^-t/RC

3=12*e^-4/3.4*10^6*C

3/12=e^-4/3.4*10^6*C

-1.3863 =-4/3.4*10^6*C

C=8.49*10^-7 F

=0.849 micro farad

B) time constant= R*C

=3.4*10^6*8.49*10^-7

=2.89 S

a. The capacitance is 0.849 micro farad

b. The  time constant of the circuit is 2.89 s

Calculation of capacitance & time constant:

a)

We know that

V=V0 e^-t/RC

3=12*e^-4/3.4*10^6*C

3/12=e^-4/3.4*10^6*C

-1.3863 =-4/3.4*10^6*C

C=8.49*10^-7 F

=0.849 micro farad

B)

Now

time constant= R*C

=3.4*10^6*8.49*10^-7

=2.89 S

Learn more about capacitor here: https://brainly.com/question/24708010

Find the distance to a Sun-like star (L=3.8x1026 watts) whose apparent brightness at Earth is 1.0 x10-10 watt/m2.

Answers

Answer:

5.49 x 10^17 m  is the distance between the sun-like star to the earth

Explanation:

Radiation intensity on Earth = 1.0 x 10^-10 W/m^2

Power of radiation of the star = 3.8 x 10^26 W

Recall that the intensity of radiation is given as

[tex]I[/tex] = [tex]\frac{P}{A}[/tex]

where

[tex]I[/tex] = intensity of radiation

P = power of radiation

A is the area through which the radiation spreads out in all three dimensional direction.

A = [tex]\frac{P}{I}[/tex] = [tex]\frac{3.8*10^{26} }{1.0*10^{-10} }[/tex] = 3.8 x 10^36 m^2

This area is spread out in the form of a sphere of area

A = [tex]4\pi r^{2}[/tex] = 4 x 3.142 x [tex]r^{2}[/tex]

3.8 x 10^36 = 12.568[tex]r^{2}[/tex]

[tex]r^{2}[/tex] =  (3.8 x 10^36)/12.568 = 3.02 x 10^35

r = [tex]\sqrt{3.02*10^{35} }[/tex] = 5.49 x 10^17 m   this is the distance of the star to the Earth

Three m^3 of air in a rigid, insulated container fitted with a paddle wheel is initially at 295 K, 200 kPa. The air receives 1546 kJ of work from the paddle wheel. Assuming the ideal gas model, determine for the air the mass, in kg, final temperature, in K, and the amount of entropy produced, in KJ/K

Answers

Answer:

1. 7.08Kg

2. 311K

3. 0.268KJ/K

Explanation:

See attached file

15. The blank
of a sine wave is the time it takes to complete one cycle of the wave.
O A. maximum amplitude
O B. minimum amplitude
O C. average value
O D. wavelength

Answers

That time is the "period" of the wave.

(It's not one of the choices.)

The blank of a sine wave is the time it takes to complete one cycle of the wavelength, the correct answer is D.

What is Wavelength?

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

It is the total length of the wave for which it completes one cycle.

The wavelength is inversely proportional to the frequency of the wave as from the following relation.

C = νλ

where c is the speed of light

ν is the frequency of the wave

λ is the wavelength of the wave

The time taken by the sine wave to complete one cycle of the wavelength is called blank the correct answer is D.

Learn more about wavelength from here

brainly.com/question/7143261

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A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium:__________.
1. the electric fields at the surfaces of the two spheres are equal.
2. the amount of charge on each sphere is q/2.
3. both spheres are at the same potential. the potentials are in the ratio V2/V1 = q2/q1.
4. the potentials are in the ratio V2/V1 = r2/r1 .

Answers

Answer:

Option 3 = both spheres are at the same potential.

Explanation:

So, let us complete or fill the missing gap in the question above;

" A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium BOTH SPHERES ARE AT THE SAME POTENTIAL"

The reason both spheres are at the same potential after the charges on the spheres are in equilibrium is given below:

=> So, if we take a look at the Question again, the kind of connection described in the question above (that is a charged sphere, say X is connected another charged sphere, say Y by a conducting wire) will eventually cause the movement of charges(which initially are not of the same potential) from X to Y and from Y to X and this will continue until both spheres are at the same potential.

The Goliath six flags Magic Mountain roller coaster ride starts at 71.6 m (235 feet) above the ground. Assuming the coaster starts from rest and ignoring any friction, what is the speed of the coaster when it reaches the ground level

Answers

Answer:

The  velocity is [tex]v = 37 .46 \ m/s[/tex]

Explanation:

From the question we are told that

    The start distance above the ground is  [tex]h = 71.6 \ m[/tex]

Generally according to the  law of energy conservation we have that

     [tex]PE_{top} = KE_{bottom }[/tex]

Where [tex]PE_{top}[/tex] is potential energy at the top which is mathematically represented as

      [tex]PE_{top} = m * g * h[/tex]

And  [tex]KE_{bottom }[/tex] is the kinetic energy at the bottom which is mathematically represented as

     [tex]KE_{bottom } = \frac{1}{2} * m * v^2[/tex]

Therefore  

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

=>    [tex]v = \sqrt{2 * g * h }[/tex]

substituting value

     [tex]v = \sqrt{2 * 9.8 * 71.6 }[/tex]

     [tex]v = 37 .46 \ m/s[/tex]

A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)

Answers

Answer:

[tex]3.1\times 10^{5}m/s[/tex]

Explanation:

The computation of the speed does the proton gain is shown below:

The potential difference is the difference that reflects the work done as per the unit charged

So, the work done should be

= Potential difference × Charge

Given that

Charge on a proton is

= 1.6 × 10^-19 C

Potential difference = 500 V

[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]

[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]

Simply we applied the above formulas

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature of 1950 K. (kb is Boltzmann's constant, 1.38x10-23 J/K).

Answers

Answer:

The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m

The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m

Explanation:

Thermal kinetic energy of electron or proton = KE

∴ KE = 3kbT/2

given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K

so we substitute

KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2

kE = 4.0365 × 10⁻²⁰ (  is the kinetic energy for both electron and proton at temperature T )

Now we know that

mass of electron M'e = 9.109 ×  10⁻³¹

mass of proton M'p = 1.6726 ×  10⁻²⁷

We also know that

KE = p₂ / 2m

from the equation, p = √ (2mKE)

{ p is momentum, m is mass }

de Broglie wavelength = β

so β = h / p = h / √ (2mKE)

h = Planck's constant = 6.626 ×  10⁻³⁴

βe =  h / √ (2m'e × KE)

βe = 6.626 ×  10⁻³⁴ / √ (2 × 9.109 ×  10⁻³¹ × 4.0365 × 10⁻²⁰ )

βe = 6.626 ×  10⁻³⁴ / √  7.3536957 × 10⁻⁵⁰

βe = 6.626 × 10⁻³⁴  / 2.71176984642871 × 10⁻²⁵

βe = 2.443422 × 10⁻⁹ m

βp =  h / √ (2m'p ×KE)

βp = 6.626 ×  10⁻³⁴ / √ (2 × 1.6726 ×  10⁻²⁷ × 4.0365 × 10⁻²⁰ )

βp = 6.626 ×  10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶

βp =  6.626 ×  10⁻³⁴ / 1.16201978468527 ×  10⁻²³

βp = 5.702140 × 10⁻¹¹ m

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