A converging lens is used to focus light from a small bulb onto a book. The lens has a focal length of 10.0 cm and is located 40.0 cm from the book. Determine the distance from the lens to the light bulb.

Answers

Answer 1

Answer:

[tex]u=13.3cm[/tex]

Explanation:

From the question we are told that:

Focal Length [tex]F=10.0cm[/tex]

Distance [tex]d=40cm[/tex]

Generally the equation for Focal length  is mathematically given by

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]

[tex]\frac{1}{10}=\frac{1}{u}+\frac{1}{40}[/tex]

[tex]\frac{1}{u}=\frac{3}{40}[/tex]

[tex]u=13.3cm[/tex]

Answer 2

Focal length is the distance from the center of the lens to principle foci.  The distance of the from the lens to the light bulb is 13.3 cm.

The distance can be determined by the formula,

[tex]\bold {\dfrac 1{f} = \dfrac 1{u} + \dfrac 1{v} }[/tex]

Where,

f - focal length = 10 cm

u - distance of object = ?

v = distance of image = 40 cm

Put the values in the equation,

[tex]\bold {\dfrac 1{10} = \dfrac 1{u} + \dfrac 1{40} }\\\\\bold {\dfrac 1{u} = \dfrac 3{40}}\\\\\bold {\dfrac 1{u} = 13.3 cm}[/tex]

Therefore, the distance of the from the lens to the light bulb is 13.3 cm.

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Related Questions

In a mass spectrometer chlorine ions of mass 35u and charge +5e are emitted from a source and accelerated through a potential difference of 250 kV. They then enter a region with a magnetic field that is perpendicular to their original direction of motion. The chlorine ions exit the spectrometer after being bent along a path with radius of curvature 3.5 m. What is the speed of the chlorine ions as they enter the magnetic field region?
(u = 1.66 × 10^(–27) kg, e = 1.6 × 10^(–19) C)
2.6 × 106 m/s
1.2 × 106 m/s
1.5 × 107 m/s

Answers

Answer:

v=26.23*105 m/s

or 2.6 × 106 m/s

Explanation:

Force generated by magnetic field will only provide centripetal acceleration thus the entering speed will be same as the exit speed

so,

.5mv2=eV potential differnce*charge= kinetic energy

.5*35*1.66*10-27*v2= 1.6*10-19*5*250000

v2=68.84*1011

v=26.23*105 m/s

or 2.6 × 106 m/s

An ideal double slit interference experiment is performed with light of wavelength 640 nm. A bright spot is observed at the center of the resulting pattern as expected. For the 2n dark spot away from the center, it is known that light passing through the more distant slit travels the closer slit.
a) 480 nm
b) 600 nm
c) 720 nm
d) 840 nm
e) 960 nm

Answers

Answer:

960 nm

Explanation:

Given that:

wavelength = 640 nm

For the second (2nd) dark spot;  the order of interference m = 1

Thus, the path length difference is expressed by the formula:

[tex]d sin \theta = (m + \dfrac{1}{2}) \lambda[/tex]

[tex]d sin \theta = (1 + \dfrac{1}{2}) 640[/tex]

[tex]d sin \theta = ( \dfrac{3}{2}) 640[/tex]

dsinθ = 960 nm

The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.18 Hz , and the acceleration of the top of the building can reach 1.9 % of the free-fall acceleration, enough to cause discomfort for occupants.

Required:
What is the total distance, side to side, that the top of the building moves during such an oscillation?

Answers

Answer:

The total distance, side to side, that the top of the building moves during such an oscillation is approximately 0.291 meters.

Explanation:

Let suppose that the building is experimenting a Simple Harmonic Motion due to the action of wind. First, we determine the angular frequency of the system ([tex]\omega[/tex]), in radians per second:

[tex]\omega = 2\pi\cdot f[/tex] (1)

Where [tex]f[/tex] is the frequency, in hertz.

If we know that [tex]f = 0.18\,hz[/tex], then the angular frequency of the system is:

[tex]\omega = 2\pi\cdot (0.18\,hz)[/tex]

[tex]\omega \approx 1.131\,\frac{rad}{s}[/tex]

The maximum acceleration experimented by the system is represented by the following formula, of which we estimate amplitude of the oscillation:

[tex]r\cdot g = \omega^{2}\cdot A[/tex] (2)

Where:

[tex]r[/tex] - Ratio of real acceleration to free-fall acceleration, no unit.

[tex]g[/tex] - Free-fall acceleration, in meters per square second.

[tex]A[/tex] - Amplitude, in meters.

If we know that [tex]\omega \approx 1.131\,\frac{rad}{s}[/tex], [tex]r = 0.019[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the amplitude of the oscillation is:

[tex]A = \frac{r\cdot g}{\omega^{2}}[/tex]

[tex]A = \frac{(0.019)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{\left(1.131\,\frac{rad}{s} \right)^{2}}[/tex]

[tex]A \approx 0.146\,m[/tex]

The total distance, side to side, is twice the amplitude, that is to say, a value of approximately 0.291 meters.

Current is a measure of…

Answers

Current is a measure of an Electric current. It’s more then likely C. Hope this helps Good luck !!

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle

Answers

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

a) The mass flow rate through the nozzle can be calculated with the following equation:

[tex] \dot{m_{i}} = \rho_{i} v_{i}A_{i} [/tex]

Where:

[tex]v_{i}[/tex]: is the initial velocity = 20 m/s

[tex]A_{i}[/tex]: is the inlet area of the nozzle = 60 cm²  

[tex]\rho_{i}[/tex]: is the density of entrance = 2.21 kg/m³

[tex] \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s [/tex]  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

[tex] \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} [/tex]

[tex] 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} [/tex]

[tex] A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} [/tex]

Therefore, the exit area of the nozzle is 23.6 cm².

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a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.

We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:

[tex]\dot m_{in} = \dot m_{out}[/tex] (1)

Where:

[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.

[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.

Given that air flows at constant rate, we expand (1) by dimensional analysis:

[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex] (2)

Where:

[tex]\rho_{in}, \rho_{out}[/tex] - Air density at inlet and outlet, in kilograms per cubic meter.

[tex]A_{in}, A_{out}[/tex] - Inlet and outlet area, in square meters.

[tex]v_{in}, v_{out}[/tex] - Inlet and outlet velocity, in meters per second.

a) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex] and [tex]v_{in} = 20\,\frac{m}{s}[/tex], then the mass flow rate through the nozzle is:

[tex]\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}[/tex]

[tex]\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)[/tex]

[tex]\dot m = 0.265\,\frac{kg}{s}[/tex]

The mass flow rate through the nozzle is 0.265 kilograms per second.

b) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex], [tex]v_{in} = 20\,\frac{m}{s}[/tex], [tex]\rho_{out} = 0.762\,\frac{kg}{m^{3}}[/tex] and [tex]v_{out} = 150\,\frac{m}{s}[/tex], then the exit area of the nozzle is:

[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex]

[tex]A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}[/tex]

[tex]A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}[/tex]

[tex]A_{out} = 2.320\times 10^{-3}\,m^{2}[/tex]

[tex]A_{out} = 23.202\,cm^{2}[/tex]

The exit area of the nozzle is 23.202 square centimeters.

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars

Answers

Answer:

the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

Explanation:

Given the data in the question;

Time period = 6 months = 1.577 × 10⁷ s

orbital speed v = 64000 m/s

since its a circular orbit,

v = 2πr / T

we solve for r

r = vT/ 2π

r = ( 64000 × 1.577 × 10⁷ ) / 2π

r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU

Now, from Kepler's law

T² = r³ / ( m₁ + m₂ )

T = 6 months = 0.5 years

we substitute

(0.5)² = (1.0737)³ / ( m₁ + m₂ )

0.25 = 1.2378 / ( m₁ + m₂ )

( m₁ + m₂ ) = 1.2378 / 0.25

( m₁ + m₂ ) = 4.9512

m₁ = m₂  = 4.9512 / 2 = 2.4756 solar mass

we know that solar mass = 1.989 × 10³⁰ kg

so

m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg

m₁ = m₂ = 4.92 × 10³⁰ kg

Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg

A 2.5 kg block slides along a frictionless surface at 1.5 m/s.A second block, sliding at a faster 4.1 m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/s. What was the mass of the second block?

Answers

Answer:

1.5kg

Explanation:

Given data

mass m1= 2.5kg

mass m2=??

velocity of mass one v1= 1.5m/s

velocity of mass two v2= 4.1m/s

common velocity after impact v= 2.5m/s

Let us apply the formula for the conservation of linear momentum for inelastic collision

The expression is given as

m1v1+ m2v2= v(m1+m2)

substitute

2.5*1.5+ m2*4.1= 2.5(2.5+m2)

3.75+4.1m2= 6.25+2.5m2

collect like terms

3.75-6.25= 2.5m2-4.1m2

-2.5= -1.6m2

divide both sides by -1.6

m2= -2.5/-1.6

m2= 1.5 kg

Hence the second mass is 1.5kg

A light bulb, attached (by itself) to an ideal 10 V battery, becomes hotter as time goes on. The bulb's filament is made of tungsten, a metal, which becomes more resistive as its temperature increases. Which statement below is true?

a. As time goes on and the bulb grows hotter, the voltage across the bulb increases so the bulb would grow brighter.
b. As time goes on and the bulb grows hotter, the current through the bulb increases so the bulb would grow brighter.
c. As time goes on and the bulb grows hotter, the voltage across the bulb decreases so the bulb would grow dimmer.
d. As time goes on and the bulb grows hotter, the current through the bulb decreases so the bulb would grow dimmer.

Answers

Answer:

the correct one is d

Explanation:

The filament of the bulb fulfills the law of ohm

          V = i R

indicate that the filament resistance increases with temperature, as the voltage remains constant if the resistance of the filament increases the current should decrease,

When examining the different statements, the correct one is d

A man standing in an elevator holds a spring scale with a load of 5 kg suspended from it. What would be the reading of the scale, if the elevator is accelerating downward with an acceleration 3.8 m/s?.

Answers

Answer:

3.1 kg

Explanation:

Applying,

R = m(g-a)..................... Equation 1

Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.

From the question,

Given: m = 5 kg, a = 3.8 m/s²

Constant: g = 9.8 m/s²

Substitute these values into equation 1

R = 5(9.8-3.8)

R = 5(6)

R = 30 N

Hence the spring scale is

m' = R/g

m' = 30/9.8

m' = 3.1 kg

Determine the density in kg \cm of solid whose Made is 1080 and whose dimension in cm are length=3 ,width=4,and height=3 ​

Answers

Answer:

d = 30kg/cm³

Explanation:

d = m/v

d = 1080kg/(3cm*4cm*3cm)

d = 30kg/cm³

calculate the force on an object with mass of 50kg and gravity of 10​

Answers

Answer: 500 N

Explanation:

The formula to find the force exerted by a mass, we may use F = mg, where g, the gravity, and a, the acceleration, can be interchangeable in the formula.

1) F = 50 x 10
2) F = 500 N

Hope this helps, brainliest would be appreciated :)

If a jet travels 350 m/s, how far will it travel each second?

Answers

Answer:

350

Explanation:

Since it travels 350 meters per second, the jet will travel 350 meters in one second.

A wheel is rotating freely at angular speed 530 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with 9 times the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels

Answers

Answer: [tex]53\ rev/min[/tex]

Explanation:

Given

angular speed of wheel is [tex]\omega_1 =530\ rev/min[/tex]

Another wheel of 9 times the rotational inertia is coupled with initial wheel

Suppose the initial wheel has moment of inertia as I

Coupled disc has [tex]9I[/tex] as rotational inertia

Conserving angular momentum,

[tex]\Rightarrow I\omega_1=(I+9I)\omega_2\\\\\Rightarrow \omega_2=\dfrac{I}{10I}\times 530\\\\\Rightarrow \omega_2=53\ rev/min[/tex]

List what sources of uncertainty go into calculating the wavelength of the laser (no explanation necessary here). (b) Accurately report the uncertainties for these quantities. (c) Explain which of these contributes the most to the final uncertainty on the laser wavelength

Answers

Answer:

thanks for da 5points hoi

Explanation: thanks dawg

There can be uncertainty in calculating the wavelength of a laser light  due to experimental errors

All measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.

What are uncertainity in measuring ?

Uncertainty  means the range of possible values within which the true value of the measurement lies.

What are errors?

The deviation  in the value of the measured quantity from the actual quantity or true value is called an error

(a) For the calculation of wavelength of laser light , the sources which can lead to uncertainty are

1. least count of measuring instruments like spectrometer or interferometer

2. Parallax error in the measurement

3. Error in identifying the order of fringes

4.. unable to identify the accurate  reading of Vernier or circular scales present in the measuring instruments.

5. Propagating errors

What is least count?

The least count of a measuring instrument is the smallest and accurate value in the measured quantity that can be measured by instrument.

What is propagating error?

When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is that  all the errors add up. which increases the uncertainty

b. The uncertainty in measurement due to  least count depends on the instrument used for measurement f wavelength. A  Michelson's

interferometer has the least count of .0001mm. whereas spectrometer has a least count of 0.5⁰. Hence uncertainty in the measurement by Michelson's interferometer is very less as compared to any other instrument.

C. The maximum uncertainty arises due to the least count , as all other errors can be minimized by taking an average value of many observations but the least count of an instrument do not change so uncertainty within the least count arises.

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A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects

Answers

Answer:

P = -1 D

Explanation:

For this exercise we must use the equation of the constructor

       / f = 1 / p + 1 / q

where f is the focal length, p and q is the distance to the object and the image, respectively

The far view point is at p =∞  and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image  

        [tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]

         f = 1 m

         P = 1/f

          P = -1 D

A parallel-plate capacitor is constructed of two horizontal 16.8-cm-diameter circular plates. A 1.8 g plastic bead, with a charge of -4.4 nC is suspended between the two plates by the force of the electric field between them.
a. Which plate, the upper or the lower, is positively charged?
b. What is the charge on the positive plate?

Answers

Answer:

Explanation:

Given that:

diameter of the circular plates = 16.8 cm

mass of the plastic bead = 1.8 g

charge q = -4.4 nC

From above, the area of the circular plates is:

[tex]Area = \pi r^2[/tex]

[tex]Area = \pi (\dfrac{d}{2})^2[/tex]

[tex]Area = \pi (\dfrac{16.8}{2*100} m)^2 \[/tex]

Area = 0.022 m²

Thus, as the plastic beads glide between the two plates of the capacitor, there exists a  weight acting downwards while the weight is balanced by the force of the plates acting upwards.

Hence, this can be achieved only when the upper plate is positively charged.

b)

Recall that

Force (F) = qE

where;

F = mg

mg = qE

[tex]E = \dfrac{mg}{q}[/tex]

[tex]E = \dfrac{1.8*10^{-3}*9.8}{4.4*10^{-9}}[/tex]

E = 4.0 × 10⁶ N/C

From the electric field;

[tex]E = \dfrac{\Big(\dfrac{Q}{A}\Big)}{e_o}[/tex]

[tex]4.0*10^{6} = \dfrac{\Big(\dfrac{Q}{0.022}\Big)}{8.85*10^{-12}}[/tex]

[tex]4.0*10^{6}*8.85*10^{-12} = {\Big(\dfrac{Q}{0.022}\Big)}{}[/tex]

[tex]Q= 4.0*10^{6}*8.85*10^{-12}*0.022[/tex]

Q = 7.788 × 10⁻⁷ C

Q = 779 nC

A cylindrical swimming pool has a radius 2m and depth 1.3m .it is completely filled with salt water of specific gravity 1.03.The atmospheric preassure is 1.013 x 10^5 Pa.

a.calculate the density of salt water.

Answers

Answer:

the density of the salt water is 1030 kg/m³

Explanation:

Given;

radius of the cylindrical pool, r = 2 m

depth of the pool, h = 1.3 m

specific gravity of the salt water, γ = 1.03

The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa

Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³

The density of the salt water is calculated as;

[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]

Therefore, the density of the salt water is 1030 kg/m³

đổi đơn vị
42 ft2/hr to cm2/s

Answers

Answer:

X = 10.8387 cm²/s

Explanation:

In this exercise, you're required to convert a value from one unit to another.

Converting 42 ft²/hr to cm²/s;

Conversion:

1 ft² = 929.03 cm²

42 ft² = X cm²

Cross-multiplying, we have;

X = 42 * 929.03

X = 39019.26 cm²

Next, we would divide by time in seconds.

1 hour = 3600 seconds

X = 39019.26/3600

X = 10.8387 cm²/s

A cylindrical wire made of an unknown alloy hangs from a support in the ceiling. You measure the relaxed length of the wire to be 16 m long; and the radius of the wire to be 3.5 m. When hang a 5 kg mass from the wire, you measure that it stretches a distance of 4 x 10 m The average bond length between atoms is 2.3 x 10^0 m for th alloy.

Required:
What is the stiffness of a typical interatomic bond in the alloy

Answers

Answer: hello  some of your values are wrongly written hence I will resolve your question using the right values

answer:

stiffness =  1.09 * 10^-6 N/m

Explanation:

Given data:

Length ( l ) = 16 m

radius of wire ( r ) = 3.5 m

mass ( m ) = 5kg

Distance stretched (  Δl ) = 4 * 10^-3 m ( right value )

average bond length ( between atoms ) = 2.3 * 10^-10 m ( right value)

first step : calculate the area

area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2

        γ          = MgL / A Δl

                    = [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]

                    = 784.8 / 0.165 = 4756.36 N/m^2

hence : stiffness =   γ  * bond length

                           =  4756.36 * 2.3 * 10^-10  = 1.09 * 10^-6 N/m

Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:

Answers

The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,

[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]

I assume the path itself is a line segment, which can be parameterized by

[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]

with 0 ≤ t ≤ 1. Then the work performed by F along C is

[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]

In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves) ____. This is in accordance with the right-hand-rule.

Answers

This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.

We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:

F = L*(IxB)

if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:

I = i*(1, 0, 0)

And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.

To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.

Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.

For example, if the magnetic field is in the positive z-axis, we will point upwards.

Now the palm of our hand tells us in which direction the force is applied.

This is the right-hand rule.

For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3, is stirred until its temperature is 500 K. Assuming the ideal gas model, for the air, and ignoring kinetic and potential energy, determine

Answers

Answer:

The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".

Explanation:

As we know,

PV = nRT

[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]

then,

⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]

⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]

        [tex]=\frac{10}{3} \ Bar[/tex]

Thus the above is the correct answer.

Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An electron at rest at point 1 is accelerated by the electric field to point 2.

Required:
Write an equation for the change of electric potential energy ΔU of the electron in terms of the symbols given.

Answers

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

a sperical ballon with a diameter of 6 m filled with helium at 20 degree centigrade and 200kpa determine mole number and the mass of helium

Answers

Answer:

A. 9280.78 moles.

B. 37123.12 g.

Explanation:

We'll begin by calculating the volume of the spherical balloon. This can be obtained as follow:

Diameter (d) = 6 m

Radius (r) = d/2 = 6/2 = 3 m

Pi (π) =3.14

Volume (V) =?

V = 4/3πr³

V = 4/3 × 3.14 × 3³

V = 4/3 × 3.14 × 27

V = 113.04 m³

Next, we shall convert 20°C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 20°C

T(K) = 20°C + 273

T(K) = 293 K

Next, we shall convert 200 KPa to Pa. This can be obtained as follow:

1 KPa = 1000 Pa

Therefore,

200 KPa = 200 KPa × 1000 Pa / 1 KPa

200 KPa = 2×10⁵ Pa

A. Determination of the number of mole of He in the spherical balloon.

Volume (V) = 113.04 m³

Temperature (T) = 293 K

Pressure (P) = 2×10⁵ Pa

Gas constant (R) = 8.314 m³Pa/Kmol

Number of mole (n) =?

PV = nRT

2×10⁵ × 113.04 = n × 8.314 × 293

22608000 = n × 2436.002

Divide both side by 2436.002

n = 22608000 / 2436.002

n = 9280.78 moles

B. Determination of the mass of He.

Mole of He (n) = 9280.78 moles

Molar mass of He = 4 g/mol

Mass of He =?

Mass = mole × molar mass

Mass of He = 9280.78 × 4

Mass of He = 37123.12 g

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks are released at the same time. The block on the left hits the table at exactly the same instant as the block on the right first comes to an instantaneous rest. What is the spring constant?

Answers

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kgThe distance of the block from the table = 30 cmLength of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

[tex]\mathsf{S = ut + \dfrac{1}{2}gt^2}[/tex]

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 mThen, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

inputting the values into the equation above, we have;

[tex]\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}[/tex]

[tex]\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}[/tex]

[tex]\mathsf{0.3 =4.9*(t^2)}[/tex]

By dividing both sides by 4.9, we have:

[tex]\mathsf{t^2 = \dfrac{0.3}{4.9}}[/tex]

[tex]\mathsf{t^2 = 0.0612}[/tex]

[tex]\mathsf{t = \sqrt{0.0612}}[/tex]

[tex]\mathbf{t =0.247 \ seconds}[/tex]

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

[tex]\mathbf{T = 2 \pi \sqrt{\dfrac{m}{k}}}[/tex]

where the relation between time (t) and period (T) is:

[tex]\mathsf{t = \dfrac{T}{2}}[/tex]

T = 2t

T = 2(0.247)

T = 0.494 seconds

[tex]\mathbf{T = 2 \pi \sqrt{\dfrac{m}{k}}}[/tex]

By making the spring constant (k) the subject of the formula:

[tex]\mathbf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}[/tex]

[tex]\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}[/tex]

[tex]\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}[/tex]

[tex]\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}[/tex]

[tex]\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}[/tex]

[tex]\mathbf{ k =8.25 \ N/m}[/tex]

Therefore, we can conclude that the spring constant between the two 51 g blocks held at a distance 30 cm above a table as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

https://brainly.com/question/17315536?referrer=searchResults

Why are hydraulic brakes used?​

Answers

Answer:

Hydraulic brake systems are used as the main braking system on almost all passenger vehicles and light trucks. Hydraulic brakes use brake fluid to transmit force when the brakes are applied.

Explanation:

A 50-turn coil has a diameter of . The coil is placed in a spatially uniform magnetic field of magnitude so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf, , induced in the coil if the magnetic field is reduced to zero unfiformly

Answers

Answer:

EMF = 51.01 Volt

Explanation:

A 50-turn coil has a diameter of 15 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.500~\text{T}0.500 T so that the plane of the coil makes an angle of 30^\circ30 ​∘ ​​ with the magnetic field. Find the magnitude of the emf induced in the coil if the magnetic field is reduced to zero uniformly in 0.100~\text{s}0.100 s

We have,

Number of turn in the coil, N = 50

The diameter of the coil, d = 15 cm

Radius, r = 7.5 cm = 0.075 m

Initial magnetic field, [tex]B_i=0.5\ T[/tex]

The plane of the coil makes an angle of 30° with the magnetic field.

The magnetic field reduced to zero in 0.1 seconds

We need to find the emf induced in the coil. We know that, emf is equal to the rate of change of magnetic flux. So,

[tex]\epsilon=\dfrac{BNA\cos\theta}{t}\\\\\epsilon=\dfrac{0.5\times 50\times \pi \times 0.075\cos(30)}{0.1}\\\\\epsilon=51.01\ V[/tex]

So, the induced emf in the coil is 51.01 V.

Two harmonic sound waves reach an overseveer simulatenouslt. the obsever hears the sound intensity rise and fall with a time of 0.2 between the maximmu intensity and the successing minimum intensity. What is the difference in frequency of the two sound waves?

Answers

Answer:

[tex]dF=2.5Hz[/tex]

Explanation:

From the question we are told that:

Time [tex]T=0.2sec[/tex]

Generally the Period is given as

 [tex]T= 2 * 0.2 = 0.4[/tex]

Therefore difference in frequency dF

 [tex]dF=\frac{1}{T}[/tex]

 [tex]dF=\frac{1}{0.4}[/tex]

 [tex]dF=2.5Hz[/tex]

An energy efficient light bulb uses 15 W of power for an equivalent light output of a 60 W incandescent light bulb. How much energy is saved each month by using the energy efficient light bulb instead of the incandescent light bulb for 4 hours a day? Assume that there are 30 days in one month
A. 7.2 kW⋅hr
B. 21.6 kW⋅hr
C. 1.8 kW⋅hr
D. 5.4 kW⋅hr
E. 1.35 kW⋅hr

Answers

Answer: (d)

Explanation:

Given

15 W is equivalent to 60 W light that is, it save 45 W

So, for 4 hours it is, [tex]4\times 45=180\ W.hr[/tex]

For 30 days, it becomes

[tex]\Rightarrow 180\times 30=5400\ W.hr\\\Rightarrow 5.4\ kWh[/tex]

Thus, [tex]5.4\ kWh[/tex] is saved in 30 days

option (d) is correct.

Astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c relative to an observer on Earth. When Jill left Earth, the spaceship was equipped with all kinds of scientific instruments, including a meter stick. Now that Jill is underway, how long does she measure the meter stick to be?
A) 0.280 m
B) 1.00 m
C) 0.960 m
D) 1.28 m
E) 1.04 m

Answers

(B) 1.00 m

Explanation:

Since the meter stick is traveling with Jill, it will have the same speed as she does so relative to Jill, the meter stick is stationary so its length remains 1.00 m as measured by her.

When astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c and measure the meter stick to be 1 meter. Hence, option B is correct.

What is length contraction?

Length contraction is defined as the phenomenon of the moving object being shorter than its appropriate length, measured in the object's rest frame.

When the object travels with the speed of light, the length of the object gets more contracted than its original length, relative to the observer. It is also known as the Lorentz-Fitgerald contraction.

Length contraction, L = L₀√(1-v²/c²), where L is the original length, L₀ is the contracted length. c² is known as the velocity of light. v² is the velocity of the speed of the object.

From the given,

speed of the spaceship = 0.280c      (c is the speed of the light)

Length contraction, L = L₀ √(1-v²/c²)

The stick also travels in the spaceship. Hence, the length of the meter stick does not change. It remains at its original length of one meter.  Thus, the ideal solution is option B.

To learn more about length contraction:

https://brainly.com/question/10272679

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