Under what conditions does the magnitude of the average velocity equal the average speed?

Answer:

The molar mass of the unknown gas is 17.3 g/mol. The molar mass matches that of ammonia (NH₃) the most (17 g/mol)

Explanation:

Let the unknown gas be gas 1

Let N₂O₄ gas be gas 2

Rate of effusion ∝ [1/√(Molar Mass)]

R ∝ [1/√(M)]

R = k/√(M) (where k is the constant of proportionality)₁₂

R₁ = k/√(M₁)

k = R₁√(M₁)

R₂ = k/√(M₂)

k = R₂√(M₂)

k = k

R₁√(M₁) = R₂√(M₂)

(R₁/R₂) = [√(M₂)/√(M₁)]

(R₁/R₂) = √(M₂/M₁)

R₁ = 2.3 R₂

M₁ = Molar Mass of unknown gas

M₂ = Molar Mass of N₂O₄ = 92.01 g/mol

(2.3R₂/R₂) = √(92.01/M₁)

2.3 = √(92.01/M₁)

92.01/M₁ = 2.3²

M₁ = 92.01/5.29

M₁ = 17.3 g/mol

The molar mass matches that of ammonia the most (17 g/mol)

The unknown gas in the system has been ammonia.

The rate of diffusion of the two gases has been proportional to the molar mass of the gases.

The ratio of the rate of two gases can be given as:

[tex]\rm \dfrac{RateA}{RateB}\;=\;\sqrt{\dfrac{Molar\;mass\[A}{Molar\;mass\;B} }[/tex]

The two gases can be given as:

Gas A = Nitrogen tetraoxide = 2.3x

Gas B = x

[tex]\rm \dfrac{2.3x}{x}\;=\;\sqrt{\dfrac{92.011}{m} }[/tex]

Mass of the unknown gas = 17.39 grams.

The mass has been equivalent to the mass of the Ammonia. Thus, the unknown gas in the system has been ammonia.

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Answer:

25 m

Explanation:

The relationship between Velocity, distance and time is given as

S = v/t........................... Equation 1

Where S = average velocity of the car, d = distance covered by the car, t = time

Making d the subject of the equation,

d = vt.................... Equation 2

Given: v = 6.25 m/s, t = 4.00 s.

Substitute into equation 2,

d = 6.25(4)

d = 25 m.

Hence, the distance traveled by the car = 25 m

The distance traveled by the car will be "25 m".

The given values are:

Speed,

Time,

As we know the formula,

→ [tex]Distance = Speed\times Time[/tex]

or,

→ [tex]d = v\times t[/tex]

By substituting the values, we get

     [tex]= 6.25\times 4[/tex]

     [tex]=25 \ m[/tex]  

Thus the above answer is right.  

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Astronomers apply the Doppler effect because from there it is possible to obtain information about the change of light, which in turn affects the light spectrum and determines the movement of a body moving away or approaching us. The extent of the shift is directly proportional to the source's radial velocity relative to the observer.

The phenomenon that occurs to determine this process is linked to the wavelength. When the wave source moves towards you, the wavelength tends to decrease. This leads to a change in the color of the light moving towards the end of the spectrum, that is, towards the color blue. (It is really violet, but by convention the color blue was chosen as it is a more common color) When the source moves away from you and the wavelength lengthens, we call the color change a shift to red. Because the Doppler effect was first used with visible light in astronomy, the terms "blue shift" and "red shift" were well established.

Final answer:

Astronomers use the Doppler effect to calculate the velocities of stars and galaxies by observing changes in light wavelengths due to motion towards or away from the observer. It also helps in exoplanet detection and measuring a star's rotation speed by analyzing the broadened spectral lines.

Explanation:

Astronomers utilize the Doppler effect to determine the velocities of astronomical objects such as stars and galaxies. To calculate the radial velocity of an object, they require the speed of light, the original wavelength of the light emitted by the object, and the observed change in this wavelength due to the Doppler shift. This shift occurs because the object is moving relative to Earth—approaching objects cause a blue shift, where the wavelength shortens, while receding objects cause a red shift, where the wavelength lengthens.

The Doppler effect is also instrumental in exoplanet detection through stellar radial velocity measurements. When a planet orbits a star, it imparts a gravitational tug that causes the star to wobble slightly. This wobble changes the star's radial velocity, which can be detected as small shifts in the star's spectral lines, irrespective of the star's distance, as long as it can be observed with a high-resolution spectrograph.

Additionally, the Doppler effect helps measure the rotation speed of distant stars. By analyzing broadened spectral lines, which result from the spread of Doppler shifts due to the rotating star's edges moving towards and away from us, astronomers can infer how fast a star is spinning.

Answer:

Explanation:

All objects on Earth are subjected to a constant gravitational acceleration g = 9.8m/s2, wherever they are on the surface of Earth and whatever their speed is. So if an object is being thrown to its highest point and stopped moving at that instant, that means the velocity at that instant is 0, not the acceleration. The acceleration is still g = 9.8m/s2

Answer:

dimensions of the drag coefficient is [tex][M^0 L^0 T^0][/tex]

Drag coefficient is a dimensionless quantity

Explanation:

force is given by[tex]F=\frac{C_{D} \rho V^2 A}{2}[/tex]

we get expression for drag coefficient [tex]C_{D} =\frac{2F}{\rho V^2 A}[/tex]

By substituting the dimensions  of the F,V,A and density , we get

[tex]C_{D} =\frac{[F]}{[\rho ][V]^2[A]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L T^{-1}]^2[L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L^2 T^{-2}][L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[MLT^{-2}]}\\C_{D}=[M^0 L^0 T^0][/tex]

Drag coefficient is a dimensionless

The dimensions of the drag coefficient, CD, are kg/m.

The dimensions of the drag coefficient, CD, can be determined by examining the equation for force, F, of the wind blowing against a building. In this equation, the dimensions for force are mass x acceleration, which are kg x m/s^2. On the other side of the equation, the wind speed, V, has dimensions of m/s, the density, rho, has dimensions of kg/m^3, and the cross-sectional area, A, has dimensions of m^2. Therefore, in order for the equation to be balanced, the dimensions of the drag coefficient must be kg/m.

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Answer:

She must run 59 min to run 11.8 km.

Explanation:

Hi there!

First let's convert mi/h into km/min:

7.4 mi/h · (1.61 km /1 mi) · (1 h / 60 min) = 0.20 km/min (notice how the units mi and h cancel).

The runner runs at 0.20 km/ min, i.e., every minute she travels 0.20 km.

If 0.20 km are traveled in 1 min, then 11.8 km will be traveled in:

11.8 km / 0.20 km/min = 59 min

She must run 59 min to run 11.8 km.

Answer:

38.35 bar

Explanation:

We are given that

Temperature=T=36.6 degree Celsius=36.6+273=309.6 K

Given mass of glucose=9.18 g

Molar mass of glucose([tex]C_6H_{12}O_6=6(12)+12(1)+6(16)[/tex]=180 g

Mass of c=12 g,mass of hydrogen=1 g, mass of O=16 g

Volume of solution=34.2 mL

Molarity of solution=[tex]\frac{given\;mass}{molar\;mass\times volume}\times 1000[/tex]

Where volume (in mL)

Molarity of solution=[tex]\frac{9.18}{180\times 34.2}\times 1000=1.49 M[/tex]

We know that

Osmotic pressure=[tex]\pi=MRT[/tex]

Where M=Molarity of solution

R=Constant=0.08314 Lbar/mol k

T=Temperature in kelvin

Using the formula

[tex]\pi=1.49\times 0.08314\times 309.6=38.35 bar[/tex]

Hence, the osmotic pressure=38.35 bar

Answer:

[tex]5.4\times 10^{-10}C-m[/tex]

Explanation:

We are given that

Charge=[tex]q=0.30 nC=0.3\times 10^{-9} C[/tex]

[tex]1 nC=10^{-9}C[/tex]

Distance between two balls=l=0.90 m

We have to find the magnitude of dipole moment of the two balls.

We know that

Dipole moment=[tex]\mid p\mid=2lq[/tex]

Where q= Charge

l=Distance between two bodies

Using the formula

Magnitude of dipole moment=[tex]\mid P\mid=2\times 0.3\times 10^{-9}\times 0.9=5.4\times 10^{-10}C-m[/tex]

Hence, the magnitude of the dipole moment of the two balls=[tex]5.4\times 10^{-10}C-m[/tex]

Answer:

2.75 g/cm³

Explanation:

given,

Volume of unknown sample, V = 3.61 cm³

mass of the sample, m = 9.93 g

density = ?

We know,

[tex]density = \dfrac{mass}{volume}[/tex]

[tex]\rho= \dfrac{9.93}{3.61}[/tex]

[tex]\rho = 2.75\ g/cm^3[/tex]

Hence, density of the unknown sample is equal to 2.75 g/cm³

Answer:

1205.77 J/kg.K

Explanation:

Heat lost by alloy = heat gained by water + heat gained by the calorimeter

c₁m₁(t₂-t₃) = c₂m₂(t₃-t₁) + c₃m₃(t₃-t₁)................. Equation 1

Where c₁ = specific heat capacity of the alloy, m₁ = mass of the alloy, t₂ = initial temperature of the alloy, t₃ = equilibrium temperature, c₂ = specific heat capacity of water, m₂ = mass of water, t₁ = initial temperature of water and calorimter, c₃ = specific heat capacity of calorimter, m₃ = mass of calorimter.

Making c₁ the subject of the equation,

c₁ = c₂m₂(t₃-t₁) + c₃m₃(t₃-t₁)/m₁(t₂-t₃)........................ Equation 2

Given: c₂ = 4190 J/kgK, m₂ = 250 g = 0.25 kg, m₁ = 75 g = 0.075 kg, m₃ = 55 g = 0.055 kg, c₃ = 377 J/kg.K, t₁ = 18 °C, t₂ = 100 °C, t₃ = 24.4 °C.

Substitute into equation 2

c₁ = [0.25×4190×(24.4-18) + 0.055×377×(24.4-18)]/[0.075(100-24.4)]

c₁ = (6704+132.704)/5.67

c₁ = 6836.704/5.67

c₁ = 1205.77 J/kg.K

Thus the specific heat capacity of the alloy = 1205.77 J/kg.K

Final answer:

The student's question is about calculating the specific heat capacity of a metal alloy using the principles of calorimetry and the conservation of energy in a heat exchange process.

The student is asking about finding the specific heat capacity of a metal alloy using calorimetry. We know that when objects at different temperatures are combined, they will exchange heat energy until they reach thermal equilibrium. We can use the equation Q = mc ext{ extdegree}T (where Q is heat energy, m is mass, c is specific heat capacity, and  ext{ extdegree}T is the change in temperature) to find the specific heat capacity. In this scenario, the heat lost by the metal alloy will equal the heat gained by the copper calorimeter and the water contained within it. By setting these two equations equal to each other and solving for the specific heat capacity of the alloy, we can find that value.

Answer:

Total 3 holes are available for conduction of current at 300K.

Explanation:

In order to develop a semiconductor, two type of impurities can be added as given below:

Now for estimation of extra electrons in the impured structure is as

[tex]N_{electrons-free}=n_{pentavalent \, atoms}\\N_{electrons-free}=4\\[/tex]

Now for estimation of "holes"  in the impured structure is as

[tex]N_{holes}=n_{trivalent \, atoms}\\N_{holes}=7\\[/tex]

Now when the free electrons and "holes" are available in the structure ,the "holes" will be filled by the free electrons therefore

[tex]N_{holes-net}=N_{holes}-N_{electrons-free}\\N_{holes-net}=7-4\\N_{holes-net}=3[/tex]

So total 3 "holes" are available for conduction of current at 300K.

Answer:

The speed of the two carts after the collision is 10 m/s.

Explanation:

Hi there!

The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:

initial momentum = mA · vA1 + mB · vB1

final momentum = (mA + mB) · vAB2

Where:

mA = mass of cart A = 0.500 kg

vA1 = velocity of cart A before the collision = 100 m/s

mB = mass of cart B = 1.50 kg.

vB1 = velocity of cart B before the collision = - 20 m/s

vAB2 = velocity of the carts that move as a single object = unknown.

(notice that we have considered leftward as negative direction)

Since the momentum of system remains constant:

initial momentum = final momentum

mA · vA1 + mB · vB1 = (mA + mB) · vAB2

Solving for vAB2:

(mA · vA1 + mB · vB1) / (mA + mB) = vAB2

(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2

vAB2 = 10 m/s

The speed of the two carts after the collision is 10 m/s.

The post-collision speed of the two carts is 10 m/s moving in the positive x-direction.

In order to determine the post-collision speed of the two carts, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.

Before the collision, Cart A has a mass of 0.500 kg and a velocity of 100 m/s, while Cart B has a mass of 1.50 kg and a velocity of -20 m/s (negative because it is moving leftward). After the collision, the two carts stick together, so their masses add up to 2 kg.

Using the conservation of momentum equation: (Momentum before collision) = (Momentum after collision)

(0.500 kg × 100 m/s) + (1.50 kg × -20 m/s) = 2 kg × v

By solving this equation, we find that the post-collision speed of the two carts is 10 m/s moving in the positive x-direction.

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The minimum height that the sphere should start from to complete a loop-the-loop is 23.75 meters, as calculated through the conservation of energy and dynamics principles.

In this physics problem, the minimum height (h) that the solid sphere needs to start from to ensure it completes the loop-the-loop involves applying principles of conservation of energy and dynamics. Initially, the sphere has potential energy equal to mgh, and no kinetic energy as it starts from rest. As it descends, it gains kinetic energy and loses potential energy.

For the ball to successfully complete the loop, the force at the top must be equivalent to the weight of the sphere plus the force necessary to maintain circular motion. This can be written as: mg + mv²/r = 5mg. From here, we can derive the equation for v²: v² = 4gr.

Since the kinetic energy at the top of the loop is (1/2)mv² and the potential energy is 2mgr, by equating total energy at the top of the loop (potential plus kinetic) to the initial potential energy (mgh), we obtain: mgh = (1/2)m(4gr) + 2mgr.

From this equation, we can solve for h and find that h = 5r = 5*4.75m = 23.75m. This is the minimum height the sphere must start from to complete the loop.

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Answer:

Energy, 9 kWh or 32400 kJ

Explanation:

Given that,

The power of heater, P = 3 kW

It runs for 3 hours to raise the water temperature to the desired level. We need to find the amount of electric energy used. We know that the electrical power of an object is given by total energy delivered per unit time. It is given by :

[tex]P=\dfrac{E}{t}[/tex]

[tex]E=P\times t[/tex]

[tex]E=3\ kW\times 3\ h[/tex]

E = 9 kWh

Since, 1 kWh = 3600 kJ

E = 32400 kJ

So, the amount of electric energy used is 9 kWh or 32400 kJ. Hence, this is the required solution.

The amount of electric energy used by a 3-kW resistance heater running for 3 hours would be 9kWh, which is equivalent to 32,400 kJ.

To determine the amount of electric energy used, we use the formula E = Pt, where E represents energy, P is power, and t is time. Here, the power used is 3kW (or kilowatts) and time is 3 hours. So, E = 3kW * 3 hours = 9 kWh (kilowatt-hours).

Moving forward, 1 kWh is equal to 3600 kilojoules (kJ). Therefore, to convert the energy we obtained in kilowatt-hours to kilojoules, we multiply it by 3600. This amounts to: 9 kWh * 3600 kJ/kWh = 32,400 kJ.

Therefore, the amount of electric energy used is 9 kWh or 32,400 kJ.

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The x and y components of the electric field at point A are [tex]EAx = 3.11 * 10^4 N/C~ and~ EAy = 0 N/C.[/tex]

The net electric field at point A due to two point charges can be found by calculating the electric field contributed by each charge independently and then summing the components to find the net electric field as an ordered pair.

Finally, we sum the x-components and sum the y-components of the electric fields from both charges to get the net electric field at point A as an ordered pair (EAx, EAy).

For q1:

r1x = 16.0 m

r1y = 0.012 m (converting 12.0 mm to meters)

[tex]r1 = sqrt(r1x^2 + r1y^2)\\r_1 = sqrt(16.0^2 + 0.012^2) \\r_1 = 16.0001 m[/tex]

For q2:

[tex]r2x = -9.0 m\\r2y = 0.012 m (same as for q1)\\r2 = sqrt(r2x^2 + r2y^2) \\r2 = sqrt((-9.0)^2 + 0.012^2)\\ r2 = 9.0001 m[/tex]

Now, we can calculate the electric field contributions from each charge:

[tex]E1x = (8.99 * 10^9) * (8 * 10^-9) / (16.0001)^2 \\E1x = 1.94 * 10^4 N/C\\E1y = 0 E2y = (8.99 * 10^9) * (6 * 10^-9) / (9.0001)^2 \\E2y= 1.17 * 10^4 N/C\\E2y = 0 E1x[/tex]

Finally, we add the x-components of the electric fields vectorially:

[tex]EAx = E1x + E2x \\= 1.94 * 10^4 + 1.17 * 10^4 \\= 3.11 * 10^4 N/C[/tex]

The y-component of the electric field, EAy, is 0 since both charges are on the x-axis.

So, the x and y components of the electric field at point A are [tex]EAx = 3.11 * 10^4 N/C~ and~ EAy = 0 N/C.[/tex]

Answer:

A. 8.175 N

B. 49.05 N

Explanation:

A.

Acceleration due to gravity (moon) = 1/6 * (acceleration due to gravity (earth)

Acceleration due to gravity (earth) = 9.81 m/s2

Acceleration due to gravity (moon) = 9.81/6

= 1.635 m/s

Weight. Fm = acceleration due to gravity * mass

= 1.635 * 5

= 8.175 N

B. Acceleration due to gravity (earth) * mass = Fe

= 9.81 * 5

= 49.05 N

Answer:

T₂ = 111.57 °C

Explanation:

Given that

Initial pressure P₁ = 9.8 atm

T₁ = 32°C  = 273 + 32 =305  K

The final pressure   P₂ = 11.2 atm

Lets take the final temperature = T₂

We know that ,the ideal gas equation  

If the volume  of the gas is constant ,then we can say that

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

[tex]T_2=\dfrac{P_2}{P_1}\times T_1[/tex]

Now by putting the values in the above equation ,we get

[tex]T_2=\dfrac{11.2}{9.8}\times 305\ K[/tex]

[tex]T_2=348.57\ K[/tex]

T₂ = 384.57 - 273 °C

T₂ = 111.57 °C

Answer:

 h = v₀ g / a

Explanation:

We can solve this problem using the kinematic equations. As they indicate that the air does not influence the vertical movement, we can find the time it takes for the body to reach the floor

          y = [tex]v_{oy}[/tex] t - ½ g t²

The vertical start speed is zero

            t² = 2t / g

The horizontal document has an acceleration, with direction opposite to the speed therefore it is negative, the expression is

            x = v₀ₓ t - ½ a t²

Indicates that it reaches the same exit point x = 0

           v₀ₓ t = ½ a t2

           v₀ₓ = ½ a (2h / g)

           v₀ₓ = v₀

           h = v₀ g / a

Answer:

[tex]f_{2}=180Hz,f_{3}=270Hz,f_{4}=360Hz\\[/tex]

Explanation:

Given data

Frequency f=90 Hz

To find

First three overtones of bassoon

Solution

The fundamental frequency of bassoon is found by substituting n=1 in below equation

f=v/λ=nv/2L

[tex]f_{1}=v/2L[/tex]

The first overtone of bassoon is found by substituting n=2

So

[tex]f_{2}=2v/2L\\f_{2}=2(v/2L)\\as \\f_{1}=v/2L\\So\\f_{2}=2f_{1}\\f_{2}=2(90Hz)\\f_{2}=180Hz[/tex]

The second overtone of bassoon is found by substituting n=3

So

[tex]f_{3}=3v/2L\\f_{3}=3(v/2L)\\as \\f_{1}=v/2L\\So\\f_{3}=3f_{1}\\f_{3}=3(90Hz)\\f_{3}=270Hz[/tex]

The third overtone of bassoon is found by substituting n=4

So

[tex]f_{4}=4v/2L\\f_{4}=4(v/2L)\\as \\f_{1}=v/2L\\So\\f_{4}=4f_{1}\\f_{4}=4(90Hz)\\f_{4}=360Hz[/tex]

The first three overtones of a bassoon with a fundamental frequency of 90.0 Hz are 180.0 Hz, 270.0 Hz and 360.0 Hz. The calculation is based on the behaviour of the bassoon as a tube open at both ends where overtones occur at integer multiples of the fundamental frequency.

The question is asking for the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz. The bassoon is assumed to act like a tube that is open at both ends. For a tube open at both ends, the overtones, also called harmonics, occur at integer multiples of the fundamental frequency.

In this case, the fundamental frequency (first harmonic) is 90.0 Hz. The first overtone (which is the second harmonic) is then 2 * 90.0 Hz = 180.0 Hz. The second overtone (third harmonic) is 3 * 90.0 Hz = 270 Hz, and the third overtone (fourth harmonic) is 4 * 90.0 Hz = 360 Hz.

Thus, the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz, and is open at both ends, are 180.0 Hz, 270.0 Hz, and 360.0 Hz respectively.

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Answer:

Three advantages of Reflecting Telescope over refracting Telescope:

1. there is no chromatic aberration in the Reflecting Telescope being mirror as an objective, while Refracting Telescope can suffer chromatic aberration being using lenses.

2. image of Reflecting Telescope is brighter due to large, polished curved mirrors than Refracting Telescope

3. Reflecting Telescope is compact and portable in size than Refracting Telescope.

Final answer:

Reflecting telescopes have three advantages over refracting telescopes: they gather more light, do not suffer from chromatic aberration, and are easier and less expensive to manufacture.

Explanation:

There are three advantages of reflecting telescopes over refracting telescopes:

Explanation:

Formula to calculate resistivity is as follows.

            [tex]\rho = \frac{1}{qN \mu}[/tex]

                       = [tex]\frac{1}{(1.6 \times 10^{-19}) \times 2.72 \times 10^{17} \times 3900}[/tex] ohm/cm

                       = [tex]5.89 \times 10^{-3} ohm/cm[/tex]

As germanium is an intrinsic conductor. Hence, resistivity of Ge is as follows.

             [tex]\rho_{1} = \frac{1}{2qN_{o}\sqrt{\mu_{e}\mu_{r}}}[/tex]

                        = [tex]\frac{1}{2q(N_{A})^{\frac{1}{2}}\sqrt{\mu_{e}\mu_{r}}}[/tex]

                        =  [tex]\frac{1}{2 \times (1.6 \times 10^{-19}) \times (\sqrt{4.42 \times 10^{22}})\sqrt{(3900)(1900)}}[/tex] ohm/cm

                        = 0.546 [tex]ohm (\mu m)^{-1}[/tex]

Thus, we can conclude that resistivity of the Ge is 0.546 [tex]ohm (\mu m)^{-1}[/tex].

The final velocity of the mass is calculated using the principle of conservation of energy, which states that the potential energy in the spring is converted into kinetic energy and work done against friction as the mass moves down the incline. This total energy is then equal to the kinetic energy at the end of the path, from which the final velocity of the mass can be found.

To calculate the final velocity of the mass after it has been released from the spring, we need to realize that the energy in the system is conserved. The potential energy stored in the spring when it is compressed is converted into kinetic energy and work done against friction as the mass skids down the incline. The spring energy is given by (1/2)kx^2, where 'k' is the spring constant and 'x' is the amount of compression. In this case, it is (1/2)*2.2 N/cm*(28 cm)^2. We convert all values to standard SI units before calculation for accurate results.

The work done against the friction in the path of 0.8 m is friction force times the distance, which = μmgcosθ*d, where 'μ' is the friction coefficient, 'm' the mass, 'g' gravity, 'θ' the angle of incline, and 'd' the distance with friction. Here, it is 0.3*5 kg*9.8 m/s^2*cos(18 degrees)*0.8 m.

The total energy is thus the sum of spring energy and work done against friction. As this energy is equal to the kinetic energy at the end of the path (1/2)*m*v^2, we can solve this to find the final velocity 'v'

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The final velocity of a 5 kg mass sliding down a frictional incline is calculated using energy conservation and work-energy principles, resulting in a final velocity of approximately 2.25 m/s.

Calculating the Final Velocity:

To determine the final velocity of a 5 kg mass after it slides down a frictional incline, we need to consider both energy conservation and work-energy principles.

Step-by-Step Solution

Spring Potential Energy: The spring is compressed by 28 cm, and the spring constant k is 2.2 N/cm. First, convert these units to meters and N/m: k = 220 N/m and compression x = 0.28 m. The potential energy stored in the spring is given by:

PE_spring = 0.5 x k x x².

PE_spring = 0.5 x 220 x (0.28)² = 8.624 J.

Gravitational Potential Energy: As the mass slides down the incline of height h = 0.31 m, the gravitational potential energy is converted to kinetic energy. PE_gravity = m x g x h

PE_gravity = 5 kg x 9.8 m/s² x 0.31 m = 15.19 J.

Energy Loss Due to Friction: The mass encounters a frictional patch of length d = 0.8 m with a friction coefficient μ = 0.3. The work done by friction (which is energy lost) is calculated by:

W_friction = μ x m x g x cos(θ) x d.

Given θ = 18°, cos(18°) ≈ 0.9511.

W_friction = 0.3 x 5 kg x 9.8 m/s² x 0.9511 x 0.8 m = 11.16 J.

Total Mechanical Energy: The initial potential energy (spring + gravity) is:

8.624 J + 15.19 J = 23.814 J.

After accounting for the work done by friction, the remaining energy will be:

KE_final = 23.814 J - 11.16 J = 12.654 J.

Final Velocity: The kinetic energy at the bottom of the incline is converted into the final velocity.

Using the relation:

KE = 0.5 x m x vf².

12.654 J = 0.5 x 5 kg x vf²

vf² = (2 x 12.654 J) / 5 kg = 5.06 m²/s²

vf = √5.06 m²/s² ≈ 2.25 m/s.

Thus, the final velocity of the mass is approximately 2.25 m/s.

Answer:

Part A:

[tex]d_{min} = \frac{v_0^2}{2g\mu_s}[/tex]

Part B:

[tex]d_{min} = 6.37~m[/tex]

Explanation:

Part A:

We should determine the free-body diagram of the small box.

For the first box, the only force exerted to the box is the static friction force in the direction of the motion.

(The direction of the static friction is always confusing to the students. The wrong idea is that the static friction is in the opposite direction with the motion. However, if you look at the Newton's Second Law, it states that the net force acting on an object is equal to the mass times acceleration. And in this case acceleration of the total system is equal to that of the small box, since it sits on the larger box.)

We can use the equations of kinematics to find the minimum distance to stop without the small box slipping.

[tex]v^2 = v_0^2 + 2a(\Delta x)\\0 = v_0^2 + 2ad_{min}[/tex]

The acceleration can be found by Newton's Second Law:

[tex]F = ma\\mg\mu_s = m(-a)\\a = -g\mu_s[/tex]

The negative sign comes from the fact that in order for the boxes to stop they have to apply a negative acceleration.

Now, we can combine the two equations to find the distance x:

[tex]0 = v_0^2 + 2ad_{min} = v_0^2 + 2(-g\mu_s)d_{min}\\d_{min} = \frac{v_0^2}{2g\mu_s}[/tex]

Part B:

We can apply the above formula to the truck and file cabinet.

[tex]d_{min} = \frac{v_0^2}{2g\mu_s} = \frac{10^2}{2(9.8)(0.80)} = 6.37~m[/tex]

The expression for the shortest distance which the large box can stop without the small box slipping is [tex]d_{min} = \frac{v_0^2 }{2\mu_s g}[/tex]

The shortest distance in which the pickup can stop without the file cabinet sliding is 6.38 m.

The given parameters;

The expression for the shortest distance which the large box can stop without the small box slipping is calculated as follows;

Apply work-energy theorem;

[tex]\mu_s F \times d_{min} = \frac{1}{2} Mv_0^2\\\\\mu_s (Mg)d_{min} = \frac{1}{2} Mv_0^2\\\\\mu_s gd_{min} = \frac{v_0^2}{2} \\\\d_{min} = \frac{v_0^2}{2\mu_s g}\\\\[/tex]

At the given speed and coefficient of static friction, the shortest distance in which the pickup can stop without the file cabinet sliding is calculated as;

[tex]d_{min} = \frac{v^2}{2\mu_s g} = \frac{10^2 }{2\times 0.8 \times 9.8} = 6.38 \ m[/tex]

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Answer:

[tex]s=203.149\ m[/tex]

Explanation:

Given:

height of the initial projection point above the ground, [tex]y=35\ m[/tex]

Vertical component of the velocity:

[tex]u_y=u.\sin\theta[/tex]

[tex]u_y=41\times \sin38[/tex]

[tex]u_y=25.242\m.s^{-1}[/tex]

The time taken in course of going up:

(at top the final velocity will be zero)

[tex]v_y=u_y-g.t[/tex]

[tex]0=25.242-9.8\times t[/tex]

[tex]t=2.576\ s[/tex]

In course of going up the maximum height reached form the initial point:

(at top height the final velocity is zero. )

using eq. of motion,

[tex]v_y^2=u_y^2-2\times g.h[/tex]

where:

[tex]v_y=[/tex] final vertical velocity while going up.=0

[tex]h=[/tex] maximum height

[tex]0^2=25.242^2-2\times 9.8\times h[/tex]

[tex]h=32.5081\ m[/tex]

Now the total height to be descended:

[tex]h'=h+y[/tex]

[tex]h'=32.5081+35[/tex]

[tex]h'=67.5081\ m[/tex]

Now the time taken to fall the gross height in course of falling from the top:

[tex]h'=v_y.t'+\frac{1}{2} g.t'^2[/tex]

[tex]67.5081=0+4.9\times t'^2[/tex]

[tex]t'=3.7118\ s[/tex]

Now the total time the projectile spends in the air:

[tex]t_t=t+t'[/tex]

[tex]t_t=2.576+3.7118[/tex]

[tex]t_t=6.2878\ s[/tex]

Now the horizontal component of the initial velocity:

(it remains constant throughout the motion)

[tex]u_x=u.\cos\theta[/tex]

[tex]u_x=41\times \cos38[/tex]

[tex]u_x=32.3084\ m.s^{-1}[/tex]

Therefore the horizontal distance covered in the total time;

[tex]s=u_x\times t_t[/tex]

[tex]s=32.3084\times 6.2878[/tex]

[tex]s=203.149\ m[/tex]

Answer:

Explanation:

initial velocity, u = 41 m/s

angle, θ = 38 °

height, h = 35 m

Let the time is t.

Use second equation of motion in vertical direction

h = ut + 1/2 gt²

- 35 = 41 Sin 38 t - 0.5 x 9.8 x t²

4.9t² - 25.2 t - 35 = 0

[tex]t = \frac{25.2 \pm \sqrt{25.2^{2}+4\times 4.9\times 35}}{2\times 4.9}[/tex]

t = 6.3 second

Horizontal distance traveled in time t is

d = uCos 38 x t

d = 41 x Cos 38 x 6.3

d = 203.54 m

Answer:

455N/C

Explanation:

from  the question, the following data can be derived

Distance between plates=2cm=0.02m

length of plates=4cm=0.04m

initial speed of electron=8*10^6m/s

Note: the speed giving is the speed associated with the horizontal motion since it moves to cover the 4cm distance

we solve the equation component by component.

For the horizontal component, the time it takes to cover distance of 0.04m can be calculated as

[tex]time=\frac{distance }{velocity} \\t=\frac{0.04}{8*10^{6}}\\ t=5*10^{-9}secs[/tex]

this same time is used to cover the vertical distance which is midway between the plate,

Hence vertical distance covered is 0.02/2=0.01m

The acceleration in the vertical component can be calculated as

[tex]y=ut+1/2at^{2}\\u=0,\\y=0.01m\\a=\frac{2y}{t^{2}}\\ a=\frac{2*0.01}{5*10^{-9}}\\ a=8*10^{14}m/s^{2}[/tex]

since

F=qE

also F=ma

then

qE=ma

E=(ma)/q

m=mass of electron=9.1*10^-32kq

q=charge of electron=1.6*10^-19c

a=acceleration

if we substitute values  

[tex]E=\frac{9.1*10^{-32}*8*10^{14}}{1.6*10^{-19}} \\E=455N/C[/tex]

To find the magnitude of the electric field when an electron is projected into a uniform electric field between two parallel plates, consider the forces acting on the electron.

An electron is projected horizontally into the uniform electric field directed vertically downward between two parallel plates.

The plates are 2.00 cm apart, and the initial speed of the electron is 8.00 × 10^6 m/s.

To calculate the magnitude of the electric field, you would need to consider the forces acting on the electron as it moves between the plates.

Answer:

975.28 W.

Explanation:

Using,

R' = R(1+αΔt)....................... Equation 1

Where R' = Resistance at the final temperature, R = Resistance at the initial temperature, α = temperature coefficient of resistivity of Nichorome, Δt = Temperature rise.

Given: R = 9 Ω, α = 0.0004/°C, Δt = 1090-20 = 1070 °C

Substitute into equation 1

R' = 9(1+0.0004×1070)

R' = 9(1.428)

R' = 12.862  Ω.

Note: Operating wattage of the heater means the operating power of the heater

The power of the heater is given as,

P = V²/R'...................... Equation 2

Where P = Operating wattage of the heater, V = Voltage, R' = Operating resistance.

Given: V = 112 V, R' = 12.862 Ω

Substitute into equation 2

P = 112²/12.862

P = 975.28 W.

The operating wattage for this heater can be calculated using Ohm's Law, resulting in approximately 1405.33 Watts assuming constant resistance. However, in reality, resistance alters with temperature, reflecting the importance of considering temperature effects in physics.

The operating wattage for this electric heater, or the power (P), can be calculated using Ohm's Law where power equals voltage (V) times current (I), or P=IV. Because I = V/R, where R is resistance, the formula can also be written as P = V2/R. With the provided values, we have P = (112V)2 / 9Ω, which gives approximately 1405.33 Watts, assuming that the resistance remains constant over the temperature change.

However, in reality, the resistance changes with temperature according to the equation R = R0[1 + α(T - T0)] where R0 is the original resistance, α is the temperature coefficient of resistivity, T is the final temperature, and T0 is the initial temperature. Considering the provided values and the significant temperature increase, we would need to adjust the resistance for the increased temperature before calculating the power, underlining the importance of temperature effects in practical physics.

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Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

[tex]T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}[/tex]

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

[tex]T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}[/tex]

T = 7.83 X10⁻⁷ s

Final answer:

The force exerted on a proton in a magnetic field can be calculated using the given formula with specified values.

Explanation:

To find the time it takes for the proton to return to the horizontal plane, we need to consider the motion in both the horizontal and vertical directions separately.

Given:

[tex]- Initial velocity (\(v_0\)) of the proton = \(3.0 \times 10^4\) m/s\\- Launch angle (\(\theta\)) = 30°\\- Electric field (\(E\)) = 400 N/C\\- We'll assume the gravitational acceleration (\(g\)) as \(9.8 \, \text{m/s}^2\) downward.[/tex]

Vertical Motion:

In the vertical direction, the proton undergoes uniformly accelerated motion under gravity.

Using the formula for vertical motion:

[tex]\[ v = u + at \] \\where: \\- \( v \) is the final velocity (which is 0 when the proton returns to the horizontal plane),\\- \( u \) is the initial vertical velocity,\\- \( a \) is the acceleration due to gravity,\\- \( t \) is the time taken.\\[/tex]

We can resolve the initial velocity into vertical and horizontal components:

[tex]\[ u_y = v_0 \sin(\theta) \]\[ u_y = 3.0 \times 10^4 \times \sin(30^\circ) \]\[ u_y \approx 3.0 \times 10^4 \times 0.5 \]\[ u_y = 1.5 \times 10^4 \, \text{m/s} \][/tex]

Now, let's find the time it takes for the vertical velocity to become zero:

[tex]\[ 0 = u_y - gt \]\[ t = \frac{u_y}{g} \]\[ t = \frac{1.5 \times 10^4}{9.8} \]\[ t \approx 1530.61 \, \text{s} \][/tex]

Horizontal Motion:

In the horizontal direction, there is no acceleration or deceleration acting on the proton. So, the time taken for horizontal motion is the same as the time taken for vertical motion.

The time it takes for the proton to return to the horizontal plane is approximately [tex]\( 1530.61 \) seconds.[/tex]

Answer:

0.080 m

Explanation:

According to Hooke's law, a spring with stiffness k will stretch a distance of Δx when a force F is applied:

F = k Δx

If we say the weight of the object is W, then the stiffness of the original spring is:

W = k (0.160 m)

k = W / 0.160

When the spring is cut in half, the stiffness of each new spring is the same as the original.  This time, the weight of the object is evenly distributed between each spring, so the force on each is W/2.

F = k Δx

W/2 = (W/0.160) Δx

1/2 = Δx / 0.160

Δx = 0.080

Each spring stretches 0.080 meters.

Answer:

a. 2.645 * 10^7 m/s^2

b. 2.645 * 10^4 N

Explanation:

Parameters given:

Velocity of rod = 2010m/s

Length of rod = 15.3cm = 0.153m

Mass of object placed at the end of the rod = 1g = 0.001kg

a. Centripetal acceleration is given as:

a = (v*v)/r

Where v = velocity

r = radius of curvature.

The radius of curvature in this case is equal to the length of the rod, since the rod makes the circular path of the motion.

Hence, centripetal acceleration at the end of the rod:

a = (2010*2010)/(0.153)

a = 26432156.86 m/s^2 = 2.64 * 10^7 m/s

b. The force needed to hold the object at the end of the rod is equal to the centripetal force at the end of the rod. Centripetal force is given as:

F = ma = (m*v*v)/r

Where a = centripetal acceleration

F = 0.001 * 2.64 * 10^7

F = 2.64 * 10^4N

The Centripetal acceleration at the end of rod is [tex]\bold {2.64 x 10^7\ m/s}[/tex] and Centripetal force required to hold the object is [tex]\bold { 2.64x 10^4\ N}[/tex].

Given  here:

Velocity of rod = 2010 m/s

Length of rod = 15.3cm = 0.153 m

Mass of object placed at the end of the rod = 1g = 0.001 kg

(A). Centripetal acceleration can be calculated by  

[tex]\bold {a =\dfrac { v^2}{r}}[/tex]

Where

v = velocity

r = radius of curvature.

Put the values,

[tex]\bold {a = \dfrac {(2010^2)}{(0.153)}}\\\\\bold {a = 26432156.86\ m/s^2 } \\\\\bold {a = 2.64 x 10^7\ m/s}[/tex]

(B).  Centripetal force can be calculated by

[tex]\bold {F = ma = \dfrac {(m v^2)}{r}}[/tex]

Where,

a = centripetal acceleration

Thus the force

[tex]\bold {F = 0.001 \times 2.64x10^7}\\\\\bold {F = 2.64x 10^4\ N}[/tex]

Therefore, the Centripetal acceleration at the end of rod is [tex]\bold {2.64 x 10^7\ m/s}[/tex] and Centripetal force required to hold the object is [tex]\bold { 2.64x 10^4\ N}[/tex].

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Answer:

Δx=629.35 m

The pilot release the hay 629.35 m in front of the cattle so that the bales land at the point where the cattle are stranded.

Explanation:

Step 1:

Finding initial velocity components:

Initial velocity=v=75 m/s

α=55

[tex]v_{ox}=vcos\alpha\\v_{ox}=75cos55^o\\v_{ox}=43.018 m/s\\v_{oy}=vsin\alpha\\v_{oy}=75sin55^o\\v_{oy}=61.436 m/s[/tex]

Step 2:

[tex]y_o=150\ m[/tex]

Newton Second Equation:

[tex]y-y_o=v_{oy}t+\frac{1}{2}g t^2[/tex]

g=-9.8 m/s^2 (Downward direction)

[tex]v_{oy}=61.436\ m/s[/tex]

y=0 m

Above equation will become:

-150=(61.436)t-(4.90)t^2

Solving the above quadratic equation we will get:

t=-2.09 sec           ,        t=14.63 sec

t= 14.63 sec

Step 3:

Finding the distance:

Using Again Newton equation of motion in x-direction:

[tex]x-x_o=v_{ox}t+\frac{1}{2}a_{x} t^2[/tex]

Since velocity is constant in x- direction, [tex]a_x[/tex] will be zero.

Above equation will be:

[tex]\Delta x=v_{ox}t[/tex]

Δx=(43.018)(14.63)

Δx=629.35 m

The pilot release the hay 629.35 m in front of the cattle so that the bales land at the point where the cattle are stranded.

The pilot should release the hay at a height of 629.35 m.

Given information,

Initial velocity = 75 m/s

Velocity For x-component,

[tex]\bold {V_0x = Vcos \alpha}\\\\\bold {V_0x = 75 cos 55^o}\\\\\bold {V_0x = 43. 018m/s}[/tex]

Velocity for Y-component

[tex]\bold {V_0y = Vsin \alpha}\\\\\bold {V_0y = 75 sin 55^o}\\\\\bold {V_0y = 61. 43m/s}[/tex]

Using Newton's second equation for y-axis,

[tex]\bold {y-y_0 = V_0t + \dfrac {1}{2} gt^2}[/tex]

Where,

g - gravitational acceleration

put the values in the equation,

[tex]\bold {-150=(61.436)t-(4.90)t^2}[/tex]

Solving this quadratic equation, we get 2 values

t = 14.29 s

To find the distance, use Newton's second equation,

[tex]\bold {x-x_0 = V_0t + \dfrac {1}{2} gt^2}[/tex]

Since acceleration is zero because the velocity is constant in x-axis hence .

So,

[tex]\bold {x-x_0 = V_0_xt }[/tex]

[tex]\bold {x- x_0=(43.018)(14.63)}\\\\\bold {x - x_0=629.35 m}[/tex]

Therefore, the pilot should release the hay at 629.35 m.

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