g When a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A. What is the resistance of the glowing thin-filament bulb

Answers

Answer 1

Answer:

R = 42.67 ohms

Explanation:

It is given that, a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A.

It means that when two batteries are connected in series, then the total voltage is 3.2 volts

Let R is the resistance of the glowing thin-filament bulb. So, using Ohm's law we get :

[tex]V=IR\\\\R=\dfrac{V}{I}[/tex]

So,

[tex]R=\dfrac{3.2}{0.075}\\\\R=42.67\ \Omega[/tex]

So, the resistance of the bulb is 42.67 ohms.


Related Questions

The Hermes spacecraft is traveling at 0.1c(1/10 the speed of light past Mars and shines a laser in front of the ship. You would see the light traveling at c (the speed of light )away from your ship. According to Einstein's special relativity how fast with a person on Mars observe the light to be traveling?

Answers

Answer:

So, according to Einstein's special relativity a person on Mars observe the light to be traveling at c = 3 x 10⁸ m/s.

Explanation:

The special theory of relativity has two main postulates:

1- VALIDITY OF PHYSICAL LAWS

The laws of physics such as Newton's Laws and Maxwell's Equations are valid in all inertial frame of references.

2- CONSTANCY OF SPEED OF LIGHT

The speed of light in vacuum is the same for all observers in uniform translational relative motion, and it is independent of the motion of the source or the observer. Thus, speed of light is a universal constant and its value is c = 3 x 10⁸ m/s.

So, according to Einstein's special relativity a person on Mars observe the light to be traveling at c = 3 x 10⁸ m/s.

An electron, moving toward the west, enters a uniform magnetic field. Because of this field the electron curves upward. The direction of the magnetic field is

Answers

Answer:

The magnetic field's direction is towards the north

Explanation:

The force on a positive charge in a uniform magnetic field is represented by the right hand rule. To determine the direction of the force, place your right hand with your palm up, and your thumb at 90° to the other fingers. If the fingers represent the magnetic field, and the thumb the direction of the positive charge, then the palm will push up in the direction of the force. But a negative charge like an electron pushes in exactly the opposite direction. So if you follow this rule, you will find that the magnetic field points towards the north.

The direction of the magnetic field is towards the North. This can be

determined using the right hand rule by Fleming.

The right hand rule states that to determine the direction of the magnetic

force, the right thumb should be pointed in in the direction of the velocity,

index finger in the direction of the magnetic field and middle finger in the

direction of magnetic force.

When this is applied, we will discover that the index finger will point towards

the north region.

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A ball is thrown at 23.2 m/s inside a boxcar moving along the tracks at 34.9 m/s. What is the speed of the ball relative to the ground if the ball is thrown forward

Answers

Answer:

The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s

Explanation:

Given;

speed of the ball thrown inside boxcar, [tex]V_B[/tex] = 23.2 m/s

speed of the boxcar moving along the tracks, [tex]V_T[/tex] = 34.9 m/s

Determine the speed of the ball relative to the ground if the ball is thrown forward.

If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.

[tex]V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s[/tex]

Therefore,  the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.

soaring birds and glider pilots can remain aloft for hours without expending power. Discuss why this is so.

Answers

Answer:

Since their wings and body develop the drag. When there is warm air then they expand their wings. Since,soaring birds and glider pilots have no engine, they always maintain their high speed to lift their weight in air for hours without expending power by convection

Explanation:

Suppose you wish to make a solenoid whose self-inductance is 1.8 mH. The inductor is to have a cross-sectional area of 1.6 x 10-3 m2 and a length of 0.066 m. How many turns of wire are needed

Answers

Answer:

The number of turns of the wire needed is 243 turns

Explanation:

Given;

self inductance of the solenoid, L = 1.8 mH

cross sectional area of the inductor, A = 1.6 x 10⁻³ m²

length of the inductor, l = 0.066 m

The self inductance of long solenoid is given by;

L = μ₀n²Al

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

n is number of turns per length

A is the area of the solenoid

l is length of the solenoid

[tex]n = \sqrt{\frac{L}{\mu_o Al} } \\\\n = \sqrt{\frac{1.8*10^{-3}}{(4\pi*10^{-7}) (1.6*10^{-3})(0.066)} } \\\\n = \sqrt{13562583.184} \\\\n = 3682.74 \ turns/m[/tex]

The number of turns is given by;

N = nL

N = (3682.74)(0.066)

N = 243 turns

Therefore, the number of turns of the wire needed is 243 turns

A beach ball filled with air is pushed about 1 m below the surface of a swimming pool and released from rest. Which of the following statements are valid, assuming the size of the ball remains the same?
a) The buoyant force on the ball decreases as the ball approaches the surface of the pool.
b) As the ball rises in the pool, the buoyant force on it increases.
c) The buoyant force on the ball equals its weight and remains constant as the ball rises.
d) The buoyant force on the ball while it is submerged is approximately equal to the weight of the volume of water that could fill the ball.
e) When the ball is released, the buoyant force exceeds the gravitational force, and the ball accelerates upward.

Answers

Answer:

e is correct

Explanation:

When a ball is pushed below the surface of a pool, it is submerged when the buoyant force is approximately equal to the water's weight of the volume that could fill the ball. When the ball is released, the buoyant force becomes greater than the gravitational force so that the ball accelerates upward.

What is buoyant force?

The buoyant force can be described as the upward force exerted on an object wholly or partially immersed in a fluid and is also called Upthrust. A body submerged partially or fully in a fluid due to the buoyant force appears to lose its weight.

The following factors affect buoyant force the density of the fluid, the volume of the fluid displaced, and the local acceleration due to gravity.

When an object immerses in water, the object experiences a force from the downward direction opposite to the gravitational pull, which causes a decrease in its weight. The difference in this pressure gives the upward force on the object, as buoyancy.

Therefore, options (d), (e) are correct.

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An apple falls from a tree and hits your head with a force of 9J. The apple weighs 0.22kg. How far did the apple fall?

Answers

Answer:

The apple fell at a distance of 4.17 m.

Explanation:

Work is defined as the force that is applied on a body to move it from one point to another. When a force is applied, an energy transfer occurs. Then it can be said that work is energy in motion.

When a net force is applied to the body or a system and this produces displacement, then that force is said to perform mechanical work.

In the International System of Units, work is measured in Joule. Joule is equivalent to Newton per meter.  

The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves.  

Work=Force*distance* cosine(angle)

On the other hand, Newton's second law says that the acceleration of a body is proportional to the resultant of forces on it acting and inversely proportional to its mass. This is represented by:

F=m*a

where F is Force [N], m is Mass [kg] and a Acceleration [m / s²]

In this case, the acceleration corresponds to the acceleration of gravity, whose value is 9.81 m / s². So you have:

Work= 9 JF=m*a=0.22 kg*9.81 m/s²= 2.1582 Ndistance= ?angle=0 → cosine(angle)= 1

Replacing:

9 J= 2.1582 N* distante* 1

Solving:

[tex]distance=\frac{9J}{2.1582 N*1}[/tex]

distance= 4.17 m

The apple fell at a distance of 4.17 m.

a small bar magnet is suspended horizontally by a string. When placed in a uniform horizontal magnetic field, it will

Answers

Answer:

It will neither translate in the opposite direction nor .rotate so as to be at right angles, it will also neither rotate so as to be vertical direction

Astronomers have recently observed stars orbiting at very high speeds around an unknown object near the center of our galaxy. For stars orbiting at distances of about 1014 m from the object, the orbital velocities are about 106 m/s. Assume the orbits are circular, and estimate the mass of the object, in units of the mass of the sun (MSun = 2x1030 kg). If the object was a tightly packed cluster of normal stars, it should be a very bright source of light. Since no visible light is detected coming from it, it is instead believed to be a supermassive black hole.

Answers

Answer:

The mass of the object is 745000 units of the sun

Explanation:

We know that the centripetal force with which the stars orbit the object is represented as

[tex]F_{c}[/tex] = [tex]\frac{mv^{2} }{r}[/tex]

and this centripetal force is also proportional to

[tex]F_{c}[/tex] = [tex]\frac{kMm}{r^{2} }[/tex]

where

m is the mass of the stars

M is the mass of the object

v is the velocity of the stars = 10^6 m/s

r is the distance between the stars and the object = 10^14 m

k is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

We can equate the two centripetal force equations to give

[tex]\frac{mv^{2} }{r}[/tex] = [tex]\frac{kMm}{r^{2} }[/tex]

which reduces to

[tex]v^{2}[/tex] = [tex]\frac{kM}{r}[/tex]

and then finally

M = [tex]\frac{rv^{2} }{k}[/tex]

substituting values, we have

M = [tex]\frac{10^{14}*(10^{6})^{2} }{6.67*10^{-11} }[/tex] = 1.49 x 10^36 kg

If the mass of the sun is 2 x 10^30 kg

then, the mass of the the object in units of the mass of the sun is

==> (1.49 x 10^36)/(2 x 10^30) = 745000 units of sun

The cost of buying shirts is partly constant and partly varies with the number of shirts bought. When the number of shirts is 5 the cost is #240, also, 10 shirts costs #400. find the cost when 300 shirts were bought ​

Answers

Answer:

The cost of the buying the shirts is #9680

Explanation:

let the cost of buying shirt = C

let the number of shirt bought = N

The following equation can be generated based on the statement above;

C = k + Nb

When the cost, C = #240, the number of shirt = 5

240 = k + 5b ------ equation (1)

where;

k and b are constants

When the cost, C = #400, the number of shirt = 10

400 = k + 10b ------ equation (2)

From equation (1), make k the subject of the formula;

k = 240 - 5b ---- equation (3)

Substitute in the value of k into equation (2)

400 = k + 10b

400 = (240 - 5b) + 10b

400 = 240 - 5b + 10b

400 - 240 = -5b + 10b

160 = 5b

b = 160 / 5

b = 32

From equation (3), calculate k

k = 240 - 5b

k = 240 -5(32)

k = 240 - 160

k = 80

When the number of shirts bought = 300, the cost of the buying the shirts =

C =  k + Nb

C = 80 +32N

Where;

N is the number of shirts

C = 80 + 32(300)

C = 80 + 9600

C = #9680

Therefore, the cost of the buying the shirts is #9680

In practice, a good insulator In practice, a good insulator A. slows heat flow. B. speeds negative heat flow. C. stops heat flow. D. all of the above

Answers

Answer:

The answer is  A. slows heat flow.

Explanation:

 An insulator is a material that impedes the movement of heat or electric current from flowing.

Theoretically good heat insulators stops the movement of heat, while practically this insulation can only be slowed down.

Hence from the options listed the correct answer practically is

 A. slows heat flow.

1/1
2. Fluid flows at 2.0 m/s through a pipe of diameter 3.0 cm. What is the
volume flow rate of the fluid in m^3/s *​

Answers

Answer:

9.42*10^-4 m^3/s

Explanation:

d=3/100 m

=0.03 m

A=3.14*0.03^2/6

=4.71*10^-4 m^2

Volume flow rate V = A * s

= 4.71*10^-4 * 2

= 9.42*10^-4

So, the volume flw rate of fluid is 9.42*10^-4 m^3/s

Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diameter and 6 meters long is carried on the back of a truck and is used to fuel tractors. The axis of the tank is horizontal. The opening on the tractor tank is 5 meters above the top of the tank in the truck. Find the work done in pumping the entire contents of the fuel tank into the tractor.

Answers

Answer:

work done in pumping the entire fuel is 1399761 J

Explanation:

weight per volume of the gasoline = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 6 m

The height of the tractor tank above the top of the tank = 5 m

The total volume of the fuel is gotten below

we know that the tank is cylindrical.

we assume that the fuel completely fills the tank.

therefore, the volume of a cylinder =  

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x  x 6 = 42.417 m^3

we then proceed to find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 42.417 = 279952.2 N

therefore,

the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 279952.2 x 5 = 1399761 J

A long straight wire carries a conventional current of 0.7 A. What is the approximate magnitude of the magnetic field at a location a perpendicular distance of 0.053 m from the wire due to the current in the wire

Answers

Answer:

2.64 x 10⁻⁶T

Explanation:

The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;

B = (μ₀ I) / (2π r)                ----------------(i)

B is magnetic field

I is current through the wire

r is the distance from the wire

μ₀  is the magnetic constant = 4π x 10⁻⁷Hm⁻¹

From the question;

I = 0.7A

r = 0.053m

Substitute these values into equation (i) as follows;

B =  (4π x 10⁻⁷ x 0.7) / (2π x 0.053)

B = 2.64 x 10⁻⁶T

Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T

A long solenoid (1500 turns/m) carries a current of 20 mA and has an inside diameter of 4.0 cm. A long wire carries a current of 2.0 A along the axis of the solenoid. What is the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire

Answers

Answer:

The magnitude of the magnetic field is 55μT

Explanation:

Given;

number of turns of the solenoid per length, n = N/L = 1500 turns/m

current in the solenoid, I = 20 mA = 20 x 10⁻³ A

diameter of the solenoid, d = 4 cm = 0.04 m

The magnetic field at a point that is inside the solenoid;

B₁ = μ₀nI

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³

B₁ = 3.77 x 10⁻⁵ T

Given;

current in the wire, I = 2 A

distance of magnetic field from the wire, r = 1 cm = 0.01 m

The magnetic field at 1.0 cm from the wire;

[tex]B_2 = \frac{\mu_0I}{2\pi r} \\\\B_2 = \frac{4\pi*10^{-7}*2}{2\pi *0.01}\\\\B_2 = 4 *10^{-5} \ T[/tex]

The magnitude of the magnetic field;

[tex]B = \sqrt{B_1^2 +B_2^2} \\\\B = \sqrt{(3.77*10^{-5})^2 + (4*10^{-5})^2} \\\\B = 5.5 *10^{-5} \ T\\\\B = 55 \mu T[/tex]

Therefore, the magnitude of the magnetic field is 55μT

The magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]

Given the following parameters from the question  

Number of turns of the solenoid per length, n = N/L = 1500 turns/m  current in the solenoid, I = 20 mA = 20 x 10⁻³ A  Diameter of the solenoid, d = 4 cm = 0.04 m

The magnetic field at a point that is inside the solenoid is expressed according to the formula;  

B₁ = μ₀nI  

Where;  

μ₀ is the permeability of free space = 4π x 10⁻⁷ m/A  

B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³  

B₁ = 3.77 x 10⁻⁵ T

Next is to get the magnetic field strength in the second wire.

Current in the wire, I = 2 A  Distance of magnetic field from the wire, r = 1 cm = 0.01 mThe magnetic field at 1.0 cm from the wire

Substitute into the formula:

[tex]B_2=\dfrac{\mu_0 I}{2 \pi r} \\B_2=\frac{4\pi \times 10^{-7}\times 2}{2 \times 3.14\times 0.01} \\B_2 =4.0 \times 10^{-5}T[/tex]

Get the resultant magnetic field:

[tex]B = \sqrt{(0.00003771)^2+(0.00004)^2} \\B =5.5 \times 10^{-7}T[/tex]

Therefore the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]

Learn more on the magnetic field here: https://brainly.com/question/15277459

What are the approximate dimensions of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 275 km above the Earth

Answers

Answer:

   s_400 = 16.5 m , s_700 = 29.4 m

Explanation:

The limit of the human eye's solution is determined by the diffraction limit that is given by the expression

                   θ = 1.22 λ / D

where you lick the wavelength and D the mediator of the circular aperture.

In our case, the dilated pupil has a diameter of approximately 8 mm = 8 10-3 m and the eye responds to a wavelength between 400 nm and 700 nm.

by introducing these values ​​into the formula

                 

λ = 400 nm      θ = 1.22 400 10⁻⁹ / 8 10⁻³ = 6 10⁻⁵ rad

λ = 700 nm     θ = 1.22 700 10⁻⁹ / 8 10⁻³-3 = 1.07 10⁻⁴ rad

Now we can use the definition radians

          θ= s / R

where s is the supported arc and R is the radius. Let's find the sarcos for each case

λ = 400 nm       s_400 = θ R

                         S_400 = 6 10⁻⁵ 275 10³

                         s_400 = 16.5 m

λ = 700 nm s_ 700 = 1.07 10⁻⁴ 275 10³

                          s_700 = 29.4 m

A sailor strikes the side of his ship just below the surface of the sea. He hears the echo of the wave reflected from the ocean floor directly below 2.5 ss later.
How deep is the ocean at this point? (Note: Use the bulk modulus method to determine the speed of sound in this fluid, rather than using a tabluated value.)
_____ m

Answers

Answer:

1248m

The time that wave moves from the wave source to the ocean floor is half the total travel time: t = 2.5/2 = 1.25s

The speed of sound in seawater is 1560 m/s

Therefore, s = vt = (1560 m/s)(1.25s) =1248 m = 1.2km

A segment of wire of total length 3.0 m carries a 15-A current and is formed into a semicircle. Determine the magnitude of the magnetic field at the center of the circle along which the wire is placed.

Answers

Answer:

4.9x10^-6T

Explanation:

See attached file

An electron initially at rest is accelerated over a distance of 0.210 m in 33.3 ns. Assuming its acceleration is constant, what voltage was used to accelerate it

Answers

Answer:

V = 451.47 volts

Explanation:

Given that,

Distance, d = 0.21 m

Initial speed, u = 0

Time, t = 33.3 ns

Let v is the final velocity. Using second equation of motion as :

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

a is acceleration, [tex]a=\dfrac{v-u}{t}[/tex] and u = 0

So,

[tex]d=\dfrac{1}{2}(v-u)t[/tex]

[tex]v=\dfrac{2d}{t}\\\\v=\dfrac{2\times 0.21}{33.3\times 10^{-9}}\\\\v=1.26\times 10^7\ m/s[/tex]

Now applying the conservation of energy i.e.

[tex]\dfrac{1}{2}mv^2=qV[/tex]

V is voltage

[tex]V=\dfrac{mv^2}{2q}\\\\V=\dfrac{9.1\times 10^{-31}\times (1.26\times 10^7)^2}{2\times 1.6\times 10^{-19}}\\\\V=451.47\ V[/tex]

So, the voltage is 451.47 V.

The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular velocity. If the torque is controlled so that the angle θ between the total acceleration of point A on the rim and the radial line to A is equal to 57° and remains constant, determine the angular velocity and the angular acceleration at time t = 0.22 s.

Answers

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

Which examples are simple machines?
Select all correct answers.
a hammer
an automobile
O a pulley
an inclined plane

Answers

A hammer and a pulley

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.

What explains why a prism separates white light into a light spectrum?
A. The white light, on encountering the prism, undergoes both reflection and refraction; some of the reflected rays re-enter the prism merging with refracted rays changing their frequencies.
B. The white light, on entering a prism, undergoes several internal reflections, forming different colors.
C. The different colors that make up a white light have different refractive indexes in glass.
D. The different colors that make up a white light are wavelengths that are invisible to the human eye until they pass through the prism.
E. The different rays of white light interfere in the prism, forming various colors.

Answers

Answer:

I think the answer probably be B

Answer :QUESTION①)

What explains why a prism separates white light into a light spectrum ?

C. The different colors that make up a white light have different refractive indexes in glass.

✔ Indeed, depending on the radiation (and therefore colors), which each have different wavelengths, the refraction index varies: the larger the wavelength (red) the less the reflection index is important and vice versa (purple).

✔ That's why purple is more deflected so is lower than red radiation.  

What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an angle of 62.5°?

Answers

Answer:

The distance is [tex]d = 1.747 *10^{-6} \ m[/tex]  

Explanation:

From the question we are told that  

       The order of maximum diffraction is  m =  2

         The wavelength is   [tex]\lambda = 775 nm = 775 * 10^{-9} \ m[/tex]

         The angle is  [tex]\theta = 62.5^o[/tex]

Generally the   condition for  constructive  interference for diffraction grating  is mathematically represented as

          [tex]dsin \theta = m * \lambda[/tex]

where  d is  the distance between the lines on a  diffraction grating

     So  

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

substituting values  

           [tex]d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}[/tex]

          [tex]d = 1.747 *10^{-6} \ m[/tex]

   

Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person in second with apparent frequency of 3400 Hertz what was the speed of cars

Answers

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-[tex]v_{s}[/tex])

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

A particle with mass m = 700 g is found to be moving with velocity v vector (-3.50i cap + 2.90j cap) m/s. From the definition of the scalar product, v^2 = v vector. v vector.
a. What is the particle's kinetic energy at this time? J If the particle's velocity changes to v vector = (6.00i cap - 5.00j cap) m/s,
b. What is the net work done on the particle? J

Answers

Answer:

Explanation:

v₁² = v₁ . v₁

= ( - 3.5 i + 2.9 j ).( - 3.5 i + 2.9 j )

= 12.25 + 8.41

= 20.66 m /s

a ) kinetic energy = 1/2 m v₁²

= 1/2 x .7 x 20.66

= 7.23 J

b )

changed velocity v₂ = v₂.v₂

= (6i - 5 j ) . (6i - 5 j )

= 36 + 25

= 61 m /s

kinetic energy = 1/2 m v₂²

= 1/2 x .7 x 61

= 21.35 J

Work done = change in energy

= 21.35 - 7.23

= 14.12 J .

If 50 km thick crust having an average density of 3.0 g/cm3 has a surface elevation of 2.5 km above sea level, what would you predict about the surface elevation for 50 km thick crust with an average density of 2.8 g/cm3

Answers

Answer:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

Explanation:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

a uniform rod of 30cm is pivoted at its center.a 40N weight is hung 5cm from left.from where 50N weight be hung to maintain equilibrium?

Answers

Answer:

The 50N weight be hung  at 23 cm to maintain equilibrium

Explanation:

Given;

length of the uniform rod = 30 cm

center of the uniform rod = 15 cm

weight of 40N is hung at 5 cm mark

weight of 50 N will be hung at ?

     0------5cm-----------------15cm-------------P---------30cm

              ↓             10cm      Δ       xcm    ↓

             40N                                             50N

Take moment about the pivot point and apply the principle of moment

50N (x cm) = 40N (10 cm)

x = (400) / 50

x = 8cm

P = x cm + 15 cm

P = 8 cm + 15 cm

P = 23 cm

Therefore, the 50N weight be hung  at 23 cm to maintain equilibrium

Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?

Answers

Answer:

 x = 0.006 m

Explanation:

The potential at one point is given by

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

remember that the potential is to scale, let's apply to our case

          V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)

in this case they indicate that the potential is zero

          0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + ​​3 10⁻⁶ / x)

         3 / x = + 2 / 10⁻² + ​​3 / 10⁻²

         3 / x = 500

          x = 3/500

          x = 0.006 m

A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a)How wide on the screen is the central bright fringe

Answers

Answer:

0.0127m

Explanation:

Using

Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m

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