Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air? b. What is the E-field amplitude? c. What is the intensity of this wave? What pressure does such wave exert on a totally-absorbing surface?

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

[tex]B = \frac{\mu_oI}{2\pi R}[/tex]

Where [tex]\mu_o[/tex] is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

[tex]B_1 = \frac{\mu_oI_1}{2\pi R}[/tex]

[tex]B_2 = \frac{\mu_oI_2}{2\pi R}[/tex]

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  [tex]\frac{\mu_oI_1}{2\pi R}[/tex] - [tex]\frac{\mu_oI_2}{2\pi R}[/tex]

or

4.0×10⁻⁶T =  [tex]\frac{\mu_o}{2\pi R}\times (I_1-I_2 )[/tex]

also

[tex]\frac{I_1}{I_2} = \frac{3}{1}[/tex]

⇒[tex]I_1 = 3\times I_2[/tex]

substituting the values in the above equation we get

4.0×10⁻⁶T =  [tex]\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)[/tex]

⇒[tex]I_2 = 1A[/tex]

also

[tex]I_1 = 3\times I_2[/tex]

⇒[tex]I_1 = 3\times 1A[/tex]

⇒[tex]I_1 = 3A[/tex]

Hence, the larger of the two currents is 3A

Answer:

1.77 x 10^-8 C

Explanation:

Let the surface charge density of each of the plate is σ.

A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2

d = 2 mm

E = 2.5 x 10^6 N/C

ε0 = 8.85 × 10-12 C2/N ∙ m2

Electric filed between the plates (two oppositively charged)

E = σ / ε0

σ = ε0 x E

σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2

The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

                                                                 = ± 11.0625 x 10^-6 C/m^2  

Charge on each plate = Surface charge density on each plate x area of each plate

Charge on each plate = ± 11.0625 x 10^-6  x 16 x 10^-4 = ± 1.77 x 10^-8 C

Answer:

[tex]F_{net} = 220.8 N[/tex]

Explanation:

It is pulled by three forces as given below

1. Jack pulls directly ahead of the donkey with a force of 61.3 N,

2. Jill pulls with 83.9 N in a direction 45° to the left, and

3. Jane pulls in a direction 45° to the right with 137 N.

Now net force directly in forward direction given as

[tex]F_x = 61.3 N + 83.9 cos45 + 137cos45[/tex]

[tex]F_x = 217.5 N[/tex]

Now similarly in perpendicular to this we have

[tex]F_y = 137 sin45 - 83 sin45 [/tex]

[tex]F_y = 38.2 N[/tex]

Now net force is given by them

[tex]F_{net} = \sqrt{F_x^2 + F_y^2}[/tex]

[tex]F_{net} = \sqrt{217.5^2 + 38.2^2}[/tex]

[tex]F_{net} = 220.8 N[/tex]

Answer:

Part a)

[tex]\frac{v_A}{v_B} = -\frac{m_B}{m_A}[/tex]

Part b)

[tex]\frac{K_A}{K_B} = \frac{m_B}{m_A}[/tex]

Explanation:

Part a)

As we know that initially the two blocks are connected by a spring and initially stretched by some amount

Since the two blocks are at rest initially so its initial momentum is zero

since there is no external force on this system so final momentum is also zero

[tex]m_Av_{1i} + m_Bv_{2i} = m_Av_A + m_Bv_B[/tex]

now for initial position the speed is zero

[tex]0 = m_Av_A + m_Bv_B[/tex]

now we have

[tex]\frac{v_A}{v_B} = -\frac{m_B}{m_A}[/tex]

Part b)

now for ratio of kinetic energy we know that the relation between kinetic energy and momentum is given as

[tex]K = \frac{P^2}{2m}[/tex]

now for the ratio of energy we have

[tex]\frac{K_A}{K_B} = \frac{P^2/2m_A}{P^2/2m_B}[/tex]

since we know that momentum of two blocks are equal in magnitude so we have

now we have

[tex]\frac{K_A}{K_B} = \frac{m_B}{m_A}[/tex]

Final answer:

To calculate the ratios of velocities and kinetic energies for two blocks connected by a massless spring released from rest, use conservation of momentum and the kinetic energy formula. The velocity ratio is mB/mA and the kinetic energy ratio is (mB/mA)².

Explanation:

The question is about two blocks connected by a massless spring on a frictionless surface. When they are released from rest after stretching the spring, we want to find the ratio of their velocities and kinetic energies during their motion.

Ratio of Velocities

The ratio of velocities is obtained through conservation of momentum. Since no external forces are involved and the spring does not have mass, the momentum of the system is conserved. Assuming block A and block B move in opposite directions after being released, their velocities will be in different directions but they will have the same magnitude of momentum.

Momentum conservation dictates that mAvA = mBvB, where vA and vB are their speeds respectively. Thus, the ratio of their velocities vA/vB = mB/mA.

Ratio of Kinetic Energies

The kinetic energy of each block is given by KE = (1/2)[tex]mv^2[/tex]. The ratio of their kinetic energies KEA/KEB is therefore [tex](m_Av_A^2)/(m_Bv_B2^)[/tex]. Inserting the ratios found in the velocities section, we find that the ratio of the kinetic energies is [tex](mB/mA)^2[/tex].

Answer:

The final gauge pressure will be 1.74 atm

Explanation:

Assume air tire as an ideal gas, therefore, it is considered:

The expression to describe the ideal gas process is:

[tex]P.v=Rg.T[/tex]

Where v is the specific volume or the inverse of the density p:

[tex]P/p=Rg.T[/tex] (2)

Then the P and T are the absolute pressure and temperature respectively. Rg represents the particular gas constant for air, Rg is equal to 287 J/kg-K. Take into account that P and T must be expressed in Pascal and Kelvin respectively.

By reorganizing  the expression (2) as below is doing:

[tex]P/T=Rg.p[/tex]  

Can be noticed that the product Rg*p is constant therefore relation P/T will also be constant. For two different states 1 and 2 of the ideal gas, it follows:

[tex]P_{1}/T_{1}=P_{2}/T_{2}[/tex] (3)

Here, states 1 and 2 will represent the state before and after arriving in Alaska.

Note that the temperatures given are in °C so it must be converted:

[tex]T(K)=273+T(°C)[/tex]  

[tex]T_{1} (K)=273+35[/tex]  

[tex]T_{1}=308 K[/tex]  

[tex]T_{2} (K)=273-42[/tex]

[tex]T_{2}=231 K[/tex]

Note also that the pressure given is the gauge pressure therefore it must be expressed as absolute pressure:

[tex]Pa (Pa)=Patm(Pa)+Pg(Pa)[/tex]  

Where Patm is the atmosphere pressure and is equal to 101325 Pa then for Pg of 2.7 10^5 Pa:

[tex]Pa_{1} (Pa)= 101325 Pa+ 270000 Pa[/tex]  

[tex]Pa_{1} (Pa)= 371325 Pa[/tex]  

Solving equation (3) for pressure at state 2:

[tex]P_{2}=P_{1}*T_{2}/T_{1}[/tex]

For the temperatures and pressure calculated values:

[tex]P_{2}= 371325 Pa * \frac231K}/{308K}[/tex]

[tex]P_{2}= 278493.75 Pa [/tex]

As is required the gauge pressure and not the absolute pressure:

[tex]Pg (Pa)=Pa(Pa)-Patm(Pa)[/tex]  

[tex]Pg_{2} (Pa)= 278493.75 -101325(Pa)[/tex]  

[tex]Pg_{2} (Pa)= 177168 Pa[/tex]  

Finally re expressing the pressure in atm units:

[tex]P (atm)=P (Pa)*\frac{1 atm}{101325 Pa}[/tex]  

[tex]P_{2} (atm)=177168*\frac{1 atm}{101325 Pa}[/tex]  

[tex]P_{2} (atm)=1.74 atm[/tex]

Final answer:

Using Gay-Lussac's law, the gauge pressure of the car tires drops from 2.50 imes [tex]10^5 N/m^2[/tex] at 35.0°C to approximately 1.85 atm at -42.0°C, after converting the final pressure to atmospheres.

Explanation:

The question pertains to the pressure change of a car tire due to a temperature drop, which is a direct application of the Ideal Gas Law. To determine the new gauge pressure in the tires after the temperature drop from 35.0°C to -42.0°C, we can use the combined gas law which relates the pressure, volume, and temperature of a gas.

Since the volume of the tires is constant (the tire has not gained or lost any air), the combined gas law simplifies to Gay-Lussac's law:

P1/T1 = P2/T2

Here, P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature. We must convert the temperatures to kelvin, which is the absolute temperature scale used in gas law calculations:

T1(K) = 35.0 + 273.15 = 308.15 K
T2(K) = -42.0 + 273.15 = 231.15 K

We can now rearrange the equation to solve for P2:

P2 = P1 * (T2/T1)

Substituting the given values:

P2 = 2.50 × 105 N/m² * (231.15 K / 308.15 K)
P2 = 1.8725 × 105 N/m²

However, the question asks for the pressure in atmospheres. We convert using the conversion factor 1 atm = 1.013 × 105 N/m²:

P2(atm) = 1.8725 × 105 N/m² / 1.013 × 105 N/m²
P2(atm) \  1.85 atm

Therefore, the gauge pressure of the car tires when the temperature has dropped to -42.0°C is approximately 1.85 atm.

Answer:

1.718 N , attractive

Explanation:

r = 0.66 m, n = 5.7 x 10^13

q1 = 5.7 x 10^13 x 1.6 x 10^-19 = 9.12 x 10^-6 C

q2 = - 5.7 x 10^13 x 1.6 x 10^-19 = - 9.12 x 10^-6 C

F = K q1 q2 / r^2

F = 9 x 10^9 x  9.12 x 10^-6 x 9.12 x 10^-6 / (0.66)^2

F = 1.718 N

As both the charges are opposite in nature, so the force between them is attractive.

Answer:2.172 GPa

0.370

0.139 mm

Explanation:

Load[P]=22 KN

thickness[t]=15 mm

width[w]=45mm

length[L]=200 mm

Longitudnal strain [tex]\varepsilon _{l0}=[tex]\frac{3}{200}[/tex]=0.015

modulus of elasticity[E]=[tex]\frac{PL}{\Delta L\ A}[/tex]

E=[tex]\frac{22\times 10^{3}\times 0.2}{15\times 45\times \10{-9}\times 3}[/tex]

E=2.172 GPa

[b]poisson's ratio [tex]\mu [/tex]

[tex]\mu =\frac{Lateral strain}{longitudnal strain}[/tex]

[tex]\mu =\frac{200\times 0.25}{3\times 45}[/tex]

[tex]\mu =0.370[/tex]

[c]Change in bar thickness

As volume remains constant

[tex]15\times 45\times 200=203\times 44.75\times t'[/tex]

t'=14.86mm

change in thickness =0.139 mm[compression]

The modulus of elasticity for the polymer bar is calculated as 1466666.67 Pa, and Poisson's ratio is determined to be 0.37. The change in the bar thickness is found to be 0.75 mm.

To find the modulus of elasticity (E) for the polymer bar, we use the formula for stress (σ) and strain (ε), where stress is the force per unit area (σ = F/A) and strain is the deformation per unit length (ε = ΔL/L0).

Given an axial load (F) of 22 kN, a thickness (t) of 15 mm, a width (w) of 45 mm, and an original length (L0) of 200 mm with the deformation (ΔL) being 3.0 mm, E can be calculated using the rearranged form: E = σ/ε = (F/A) / (ΔL/L0). Converting all measurements to meters for consistency, we get:

E = (22000 N / (0.015 m * 0.045 m)) / (0.003 m / 200 mm) = 1466666.67 Pa

To calculate Poisson's ratio (ν), we use the lateral strain (εt) over the axial strain (ε), where εt = -Δw/w and ε = ΔL/L0.

ν = -εt/ε = (0.00025 m / 0.045 m) / (0.003 m / 0.200 m) = 0.37

For the change in the bar thickness, since we do not have lateral deformation in the thickness direction and assuming a constant volume, we can find the change in thickness by using the relation: Δt = -(w * Δw) / t. Plugging in the values, we get:

Δt = -(0.045 m * 0.00025 m) / 0.015 m = 0.00075 m or 0.75 mm

Using the kinematic equations of motion, we can calculate that a projectile launched at an initial speed of 1360 m/s from the moon's surface will reach two-fifths its initial speed at an altitude of roughly 680 kilometers.

A projectile launched vertically from the moon's surface decelerates due to the moon's gravity at a rate of 1.6 m/s². Given the initial speed of the projectile is 1360 m/s, we want to find out the altitude at which the speed is two-fifths of this initial velocity. To find out, we use the kinematic equation v² = u² + 2as.

Substituting the values into the equation, we get (2/5 * 1360)² = (1360)² - 2 * 1.6 * s. Solving this equation gives us the distance s, where s comes out to be approximately 680000 meters or 680 km.

This means that the projectile's speed is two-fifths its initial value at an altitude of approximately 680 km from the moon's surface.

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To find the altitude at which the projectile's speed is two-fifths its initial value, the loss of kinetic energy must equal the gain in potential energy. Solving for the altitude (h), we find the answer.

This problem involves physics concepts of projectile motion and gravitational acceleration. Given that the projectile is launched with an initial velocity of 1360 m/s, and we want to determine the altitude at which it has slowed to two-fifths of that speed, we will be considering the effects of the Moon's gravity on the deceleration of the projectile.

Using the formula for kinetic energy K.E = 1/2 m v^2, when the projectile's velocity slows to two-fifths of its original speed, the kinetic energy will be four-fifths less since kinetic energy is proportional to the square of the speed. This loss of kinetic energy must equal the gain in potential energy (since energy is conserved), which is given by the formula P.E = mgh (where m is mass, g is gravitational acceleration, and h is height or altitude).

Solving P.E = K.E for h, we get the altitude at which the projectile's speed is two-fifths its initial value.

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Answer:

13 sec

Explanation:

Hello

by definition the acceleration is a vector derived magnitude that indicates the speed variation per unit of time

[tex]a=\frac{V_{f}-V_{i}}{t_{f}-t_{i}}\\a(t_{f}-t_{i})={V_{f}-V_{i}  }}\\t_{f}-t_{i}=\frac{V_{f}-V_{i}}{a} \\\\t_{f}=\frac{V_{f}-V_{i}}{a}+t_{i} \\\\[/tex]

Let

Vi=4.0 m/s

a=1.0 m/s2

Vf=17.0 m/s

ti= (0)sec

t2= unknown

[tex]t_{f}=\frac{V_{f}-V_{i} }{a}+t_{i}\\t_{f}=\frac{17-4 }{1}+0\\t_{f}=\ 13\ sec\\[/tex]

Answer: 13 sec

I hope it helps

Answer:

 D . 1.47 x 10⁻² Ω-m

Explanation:

L = length of the cylindrical resistor = 2 m

d = diameter = 0.1 m

A = Area of cross-section of the resistor = (0.25) [tex]\pi[/tex] d² = (0.25) (3.14) (0.1)² = 0.785 x 10⁻² m²

V = battery Voltage = 12 volts

[tex]i [/tex] = current flowing through the resistor = 3.2 A

R = resistance of the resistor

Resistance of the resistor is given as

[tex]R = \frac{V}{i}[/tex]

[tex]R = \frac{12}{3.2}[/tex]

R = 3.75 Ω

[tex]\rho[/tex] = resistivity

Resistance is also given as

[tex]R = \frac{ \rho L}{A}[/tex]

[tex]3.75 = \frac{ \rho (2)}{(0.785\times 10^{-2})}[/tex]

[tex]\rho[/tex] =  1.47 x 10⁻² Ω-m

Answer:

11.515 Joule

Explanation:

Volume of aluminium = V = 4.89×10⁻³ m³

Coefficient of volume expansion for aluminum = α = 69×10⁻⁶ /°C

Initial temperature = 19.1°C

Final temperature = 357°C

Pressure of air = 1.01×10⁵ Pa

Change in temperature = ΔT= 357-19.1 = 337.9 °C

Change in volume

ΔV = αVΔT

⇒ΔV = 69×10⁻⁶×4.89×10⁻³×337.9

⇒ΔV = 114010.839×10⁻⁹ m³

Work done

W = PΔV

⇒W = 1.01×10⁵×114010.839×10⁻⁹

⇒W = 11.515 J

∴ Work is done by the expanding aluminum is 11.515 Joule

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

[tex]F = qvB[/tex]

here we have

[tex]B = \frac{F}{qv}[/tex]

here we know that

[tex]F = 4.8 \times 10^{-13} N[/tex]

[tex]q = 1.6 \times 10^{-19} C[/tex]

[tex]v = 4 \times 10^6 m/s[/tex]

now from above equation we have

[tex]B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}[/tex]

[tex]B = 0.75 T[/tex]

Answer:

[tex]K = 0.076 W/m K[/tex]

Explanation:

Heat required to melt the complete ice is given as

[tex]Q = mL[/tex]

here we have

m = 5.0 kg

[tex]L = 3.35 \times 10^5 J/kg[/tex]

now we have

[tex]Q = (5 kg)(3.35 \times 10^5)[/tex]

[tex]Q = 1.67 \times 10^6 J[/tex]

now the power required to melt ice in 8.4 hours is

[tex]P = \frac{Q}{t} = \frac{1.67\times 10^6}{8.4 \times 3600 s}[/tex]

[tex]P = 55.4 Watt[/tex]

now by formula of conduction we know

[tex]P = \frac{KA(\Delta T)}{x}[/tex]

now we have

[tex]55.4 = \frac{K(0.73 m^2)(25 - 5)}{0.02}[/tex]

[tex]K = 0.076 W/m K[/tex]

The heat needed to evaporate 31.5 g of boiling water is approximately 7,150 J.

To calculate the heat needed to completely evaporate water, we use the equation Q = m × ΔHvap, where Q is the heat, m is the mass, and ΔHvap is the heat of vaporization.

The heat of vaporization for water is approximately 40.7 kJ/mol. To find the heat for 31.5 g of water, we need to convert the mass to moles by dividing by the molar mass of water (18.015 g/mol).

Using the equation, Q = (31.5 g ÷ 18.015 g/mol) × (40.7 kJ/mol × 1000 J/kJ), the answer is approximately 7,150 J.

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Final answer:

To evaporate 31.5 g of boiling water at 100°C, 70,875 joules of heat are needed. This is determined by multiplying the mass by the heat of vaporization for water, which is 2,250 J/g.

Explanation:

To calculate the amount of heat required to completely evaporate 31.5 g of boiling water at 100°C, the heat of vaporization of water is needed. The heat of vaporization of water is 2,250 J/g. To find the total heat we will multiply the mass of the water by the heat of vaporization:

Heat required = mass × heat of vaporization

Heat required = 31.5 g × 2,250 J/g

Heat required = 70,875 J

Therefore, 70,875 joules of heat would be needed to completely evaporate 31.5 g of boiling water at 100°C.

Answer:

63.6 %

Explanation:

TH = 1100 K , Tc = 400 k

Efficiency is given by

η = 1 - Tc / TH

η = 1 - 400 / 1100

η = 1 - 0.36

η = 0.636

η = 63.6 %

Answer:

Coefficient of static friction between the tires and the road is 0.92.

Explanation:

It is given that,

Radius of the curve, r = 120 m

Speed, v = 119 km/h = 33.05 m/s

We need to find the coefficient of static friction between the tires and the road. In a curved road the safe velocity is given by :

[tex]v=\sqrt{\mu rg}[/tex]

[tex]\mu[/tex] is the coefficient of static friction

g is acceleration due to gravity

[tex]\mu=\dfrac{v^2}{rg}[/tex]

[tex]\mu=\dfrac{(33.05\ m/s)^2}{120\ m\times 9.8\ m/s^2}[/tex]

[tex]\mu=0.92[/tex]

So, the coefficient of static friction between the tires and the road is 0.92. Hence, this is the required solution.

Answer:

The change in the area of the circle is [tex]8.88\times10^{-7}\ m^2[/tex]

Explanation:

Given that,

Radius = 0.250 cm

Temperature = 20.0°C

Final temperature =800.0°C

Coefficient of linear thermal expansion for lead[tex]\alpha =29.0\times10^{-6}\ /\°C [/tex]

We calculate the change in temperature,

[tex]\Delta T=800.0-20.0=780^{\circ}[/tex]

Now, We calculate the area of the disc

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

Put the value into the formula

[tex]A=3.14\times(2.5\times10^{-3})^2[/tex]

[tex]A =1.9625\times10^{-5}\ m^2[/tex]

We need to calculate the areal expansion

[tex]\Delta A=2\alpha\times A\times\Delta T[/tex]

[tex]\Delta A=2\times29.0\times10^{-6}\times1.9625\times10^{-5}\times780[/tex]

[tex]\Delta A=8.88\times10^{-7}\ m^2[/tex]

Hence, The change in the area of the circle is [tex]8.88\times10^{-7}\ m^2[/tex]

Final answer:

The change in area of the lead circular disc can be calculated using the formulas for linear thermal expansion. The change in radius is calculated using the coefficient of linear expansion and the change in temperature. The change in area is then calculated using the change in radius and the original radius.

Explanation:

To determine the change in area of the circular disc made of lead, we need to calculate the change in its radius using the coefficient of linear thermal expansion. The formula for linear thermal expansion is given by AL = a * L * AT, where AL is the change in length, a is the coefficient of linear expansion, L is the original length, and AT is the change in temperature.

In this case, we are interested in the change in radius, so we can use the formula AR = a * R * AT, where AR is the change in radius and R is the original radius.

Substituting the given values, we have:

AR = (29.0 × 10^-6/°C) * (0.250 cm) * (800.0 °C - 20.0 °C)

AR = 0.00145 cm

The change in area can be calculated using the formula AE = 2π * R * AR. Substituting the values, we have:

AE = 2π * (0.250 cm) * (0.00145 cm)

AE = 2.31 x 10^-3 cm²

Answer:

W = 2.158 eV

fo = 5.23 x 10^14 Hz

Explanation:

f = 7.9 x 10^14 Hz, Vo = 1.1 V

Let W be the work function.

Use the Einstein equation

Energy = W + eVo

hf = W + eVo

where, h is the Plank's constant and e be the electronic charge.

W = hf - eVo

W = (6.6 x 10^-34 x 7.9 x 10^14) - (1.6 x 10^-19 x 1.1)

W = 5.214 x 10^-19 - 1.76 x 10^-19 = 3.454 x 10^-19 J

W = (3.454 x 10^-19) / (1.6 x 10^-19) = 2.158 eV

Let fo be the cut off frequency

W = h fo

fo = W / h = (3.454 x 10^-19) / (6.6 x 0^-34) = 5.23 x 10^14 Hz

Answer:

(a) 1.468 km/s

(b) 129.5 m

Explanation:

Use the formula for the wavelength is given by

[tex]\lambda = \frac{h}{mv}[/tex]

mass of electron, m = 9.1 x 10^-31 kg

(a) λ = 4.94 x 10^-7 m

[tex]\lambda = \frac{h}{mv}[/tex]

[tex]v = \frac{6.6\times 10^{-34}}{9.1\times 10^{-31}\times 4.94\times 10^{-7}}[/tex]

v = 1.468 x 10^3 m/s = 1.468 km/s

(b) v = 5.60 x 10^6 m/s

[tex]\lambda = \frac{h}{mv}[/tex]

[tex]\lambda  = \frac{6.6\times 10^{-34}}{9.1\times 10^{-31}\times 5.60\times 10^{6}}[/tex]

λ = 129.5 m

Answer:

The internal energy is 73.20 J.

Explanation:

Given that,

Weight of helium = 86 mg

Temperature = 0°C

We need to calculate the internal energy

Using formula of internal energy

[tex]U =nc_{v}T[/tex]

Where, [tex]c_{v}[/tex] = specific heat at constant volume

He is mono atomic.

So, The value of [tex]c_{v}= \dfrac{3}{2}R[/tex]

now, 1 mole of Helium = 4 g helium

n =number of mole of the 86 mg of helium

[tex]n = \dfrac{86\times10^{-3}}{4}[/tex]

[tex]n =2.15\times10^{-2}\ mole[/tex]

T = 0°C=273 K

Put the value into the formula

[tex]U = 2.15\times10^{-2}\times\dfrac{3}{2}\times8.314\times273[/tex]

[tex]U = 73.20\ J[/tex]

Hence, The internal energy is 73.20 J.

Answer:

explained

Explanation:

Yes, the heating of filament is what causes the light production (photon emission), and this heating is caused because of current in the light bulb

(H= i^2*R*t i=current, H= heat, t= time and R= resistance).But using constant current source is not a good idea because in constant current source resistance is very low that can cause short circuit and ultimately fusing it. Whereas in constant voltage source current adjusts itself and prevents fusing because of high resistance in the circuit.

Final answer:

Using a constant current source for light bulbs is not ideal because the filament's resistance changes as it heats up, which would lead to inefficient operation and potential damage. A constant voltage source is preferred as it adapts better to the changing resistance, ensuring a stable operation.

Explanation:

The heating of the filament in light bulbs, which leads to light production, is inherently linked to the current flowing through the filament. Using a constant current source instead of a constant voltage source is not advisable due to the changing resistance of the filament. As the filament heats up, its resistance increases significantly, which, according to Ohm's law (V=IR), would necessitate a higher voltage to maintain the constant current. Initially, when the light bulb is turned on and the filament is at a lower temperature, the required voltage to maintain a constant current would be lower. As the temperature and resistance rise, so would the required voltage to maintain that current. This varying voltage could lead to inefficient operation and potentially damage the bulb or reduce its lifespan. The constant voltage approach is preferred as it naturally adapts to the changing resistance of the filament, providing a more stable and predictable operation of the bulb.

Answer:

[tex]\rho = 2940 kg/m^3[/tex]

Explanation:

When pendulum is in air then the tension in the string is given as

[tex]T = mg[/tex]

[tex]T = 38.7 N[/tex]

now when it is submerged in water then tension is decreased due to buoyancy force of water

so we will have

[tex]T_{water} = mg - F_b[/tex]

[tex]32 = 38.7 - F_b[/tex]

[tex]F_b = 6.7 N[/tex]

When rock is immersed in unknown liquid then tension is given by

[tex]T = mg - F_b'[/tex]

[tex]19 = 38.7 - F_b'[/tex]

[tex]F_b' = 19.7 N[/tex]

from buoyancy force of water we can say

[tex]F_b = 1000(V)(9.8) = 6.7[/tex]

[tex]V = 6.84 \times 10^{-4} m^3[/tex]

now for other liquid we will have

[tex]19.7 = \rho(6.84 \times 10^{-4})(9.8)[/tex]

[tex]\rho = 2940 kg/m^3[/tex]

To find the density of the unknown liquid, we can use Archimedes' principle and the difference in tension when the rock is immersed in water and in the unknown liquid. The formula for buoyant force can be used to calculate the density of the unknown liquid.

To find the density of the unknown liquid, we can use Archimedes' principle. Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

In this case, the difference in tension when the rock is in air and when it is totally immersed in water represents the buoyant force acting on the rock. We can use the formula:

Buoyant force = weight of the fluid displaced

We know that the tension in the string when the rock is in water is 32.0 N and when it is in the unknown liquid is 19.0 N. Subtracting the tension in water from the tension in the unknown liquid, we get:

Tension in unknown liquid - Tension in water = Buoyant force

19.0 N - 32.0 N = -13.0 N

The negative sign indicates that the buoyant force is acting upward on the rock. Since the buoyant force is equal to the weight of the fluid displaced, we can equate the buoyant force to the weight of the unknown liquid displaced by the rock:

-13.0 N = -(density of unknown liquid) x (gravitational acceleration) x (volume of rock)

To find the density of the unknown liquid, we can rearrange the equation as follows:

Density of unknown liquid = (-13.0 N) / (9.8 m/s^2) x (volume of rock)

Plugging in the values, the density of the unknown liquid can be calculated.

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The angular speed of the truck's wheels is approximately 3422.35 rad/min.

The angular speed [tex](\( \omega \))[/tex] is related to the linear speed [tex](\( v \))[/tex] and the radius [tex](\( r \))[/tex] of the wheels by the formula [tex]\( v = r \cdot \omega \)[/tex].

Firstly, we need to find the radius of the wheels, which is half of the diameter. So, [tex]\( r = \frac{34}{2} = 17 \) inches[/tex].

Now, convert the speed from miles per hour to inches per minute. There are 5280 feet in a mile, 12 inches in a foot, and 60 minutes in an hour.

[tex]\[ v = 55 \, \text{mi/h} \times \frac{5280 \, \text{ft}}{1 \, \text{mi}} \times \frac{12 \, \text{in}}{1 \, \text{ft}} \times \frac{1 \, \text{h}}{60 \, \text{min}} \][/tex]

[tex]\[ v \approx 58080 \, \text{in/min} \][/tex]

Now, use the formula [tex]\( v = r \cdot \omega \)[/tex] to find [tex]\( \omega \)[/tex]:

[tex]\[ \omega = \frac{v}{r} \][/tex]

[tex]\[ \omega = \frac{58080 \, \text{in/min}}{17 \, \text{in}} \][/tex]

[tex]\[ \omega \approx 3422.35 \, \text{rad/min} \][/tex]

Therefore, the angular speed of the truck's wheels is approximately 3422.35 rad/min.

Answer:

Lifting force, F = 21240 N

Explanation:

It is given that,

Mass of the helicopter, m = 1800 kg

It rises with an upward acceleration of 2 m/s². We need to find the lifting force  supplied by its rotating blades. It is given by :

F = mg + ma

Where

mg is its weight

and "ma" is an additional acceleration when it is moving upwards.

So, [tex]F=1800\ kg(9.8\ m/s^2+2\ m/s^2)[/tex]

F = 21240 N

So, the lifting force supplied by its rotating blades is 21240 N. Hence, this is the required solution.

The lifting force supplied by the helicopter's blades is calculated using Newton's second law. It's the sum of the force needed to counteract the weight of the helicopter and the force needed to provide the upward acceleration. Therefore, the lifting force is 21240 N.

To calculate the lifting force, we need to use the principle of Newton's second law applied vertically, which says that Force equals mass times acceleration (F = ma). For a helicopter to rise, the upward force supplied by the rotating blades (Lift) must be greater than the downward force of gravity (Weight).

The weight of the helicopter is given by its mass times the gravitational acceleration, which is approximately 9.8 m/s^2. So, the weight of the helicopter is 1800 kg * 9.8 m/s^2 = 17640 N.

The total force needed for the helicopter to rise with an acceleration of 2 m/s^2 upwards is given by its mass times this acceleration, so that's 1800 kg * 2 m/s^2 = 3600 N.

Therefore, the lifting force supplied by the blades is the sum of the force needed to counteract the weight and the force needed to provide the upward acceleration. That is, 17640 N + 3600 N = 21240 N.

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Answer:

[tex]v = 1.11 \times 10^3 m/s[/tex]

Explanation:

By energy conservation law we will have

[tex]U_i + KE_i = U_f + KE_f[/tex]

as we know that as the projectile is rising up then due to gravitational attraction of earth it will slow down

At the highest position the speed of the projectile will become zero

So here we will have

[tex]-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h} + 0[/tex]

[tex]-\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(100)}{6.37 \times 10^6} + \frac{1}{2}(100)v^2 = -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{-11})(100)}{(6.37 \times 10^{6} + 65000)}[/tex]

[tex] - 6.25 \times 10^7 + 0.5 v^2 = -6.19 \times 10^7[/tex]

[tex]v = 1.11 \times 10^3 m/s[/tex]

Answer:

[tex]f_{min} = 574.3 Hz[/tex]

[tex]f_{max} = 604.5 Hz[/tex]

Explanation:

As per Doppler's effect of sound we know that when source and observer moves relative to each other then the frequency of sound observed by the observer is different from real frequency of sound.

As the source is moving here in this case so the frequency is given as

[tex]f = f_o\frac{v}{v\pm v_s}[/tex]

part a)

for lowest frequency we will have

[tex]f_{min} = 589(\frac{343}{343 + R\omega})[/tex]

here we know that

R = 54.6 cm

[tex]\omega = 16.1 rad/s[/tex]

now we have

[tex]f_{min} = 589(\frac{343}{343 + 0.546(16.1)})[/tex]

[tex]f_{min} = 574.3 Hz[/tex]

part b)

for maximum frequency we will have

[tex]f_{max} = 589(\frac{343}{343 - R\omega})[/tex]

here we know that

R = 54.6 cm

[tex]\omega = 16.1 rad/s[/tex]

now we have

[tex]f_{max} = 589(\frac{343}{343 - 0.546(16.1)})[/tex]

[tex]f_{max} = 604.5 Hz[/tex]

Answer:

[tex]INTNESITY = 0.88W/m^{2}[/tex]

Explanation:

given data:

power P =  40W

Distance of light bulb from screen is  R  = 1.9 m

The light intensity considered as  the light energy falling on the surface per unit area per unit time

Intensity of light can be determined by using following formula

[tex]intesnity = \frac{P} {4\pi R^{2}}[/tex]

[tex]intesnity = \frac{40} {4\pi *1.9^{2}}[/tex]

[tex]INTNESITY = 0.88W/m^{2}[/tex]

Answer:

32 J

Explanation:

Power is V²/R = 20²/1500 = 4/15 . . . watts

In 120 seconds, the energy is ...

(4/15 J/s)×(120 s) = 32 J

In a 2-minute period 32 joules are dissipated as heat.

We calculate the power dissipation rate of the 1.5-kΩ resistor under a constant potential difference of 20- V, which is 0.27 Watts or 0.27 Joules per second. Multiplying this rate by the time interval of two minutes (or 120 seconds), we find the total energy dissipated as heat, which is 32.4 Joules.

To understand the amount of energy dissipated during the given time interval, we first need to determine the power of the 1.5-kΩ resistor. This can be done using the equation P = V²/R, where V is the voltage, and R is the resistance. For this case, we have P = (20 V) ² / (1.5 kΩ) = 0.27 Watts.

Since the power dissipation rate is 0.27 Watts, which is equivalent to 0.27 Joules per second, we can now find out the energy dissipated during the two-minute interval using the equation E = Pt. Here, t represents the time in seconds, so t = 2 minutes * 60 seconds/minute = 120 seconds. Therefore, substituting the known values, we get E = (0.27 J/s)(120 s) = 32.4 Joules.

So, this is the total energy dissipated as heat by the 1.5- kΩ resistor under a constant potential difference of 20- V during the two-minute interval.

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Answer:

For no friction condition there is no range of speed only on possible speed is 41.1 km/h

while for the case of 0.300 friction coefficient the range of speed is from 6.5 m/s to 15.75 m/s

Explanation:

When there is no friction on the turn of road then the centripetal force is due to the component of Normal force only

So we will have

[tex]Ncos\theta = mg[/tex]

[tex]Nsin\theta = \frac{mv^2}{R}[/tex]

so we have

[tex]tan\theta = \frac{v^2}{Rg}[/tex]

[tex]\theta = tan^{-1}\frac{v^2}{Rg}[/tex]

v = 41.1 km/h = 11.42 m/s

[tex]\theta = tan^{-1}\frac{11.42^2}{28(9.8)}[/tex]

[tex]\theta = 25.42^o[/tex]

so here in this case there is no possibility of range of speed and only one safe speed is possible to take turn

Now in next case if the coefficient of static friction is 0.300

then in this case we have

[tex]v_{max} = \sqrt{(\frac{\mu + tan\theta}{1 - \mu tan\theta})Rg}[/tex]

[tex]v_{max} = \sqrt{(\frac{0.3 + 0.475}{1 - (0.3)(0.475)})(28 \times 9.8)}[/tex]

[tex]v_{max} = 15.75 m/s[/tex]

Similarly for minimum speed we have

[tex]v_{min} = \sqrt{(\frac{\mu - tan\theta}{1 + \mu tan\theta})Rg}[/tex]

[tex]v_{min} = \sqrt{(\frac{-0.3 + 0.475}{1 + (0.3)(0.475)})(28 \times 9.8)}[/tex]

[tex]v_{min} = 6.5 m/s[/tex]

So the range of the speed is from 6.5 m/s to 15.75 m/s

Power dissipated by a resistor = (current)² x (resistance).

0.25 W = (0.02 A)² x (resistance)

This resistor = (0.25 W) / (0.02 A)²

This resistor = 625 ohms

If we want it to dissipate 0.5 W, then

0.5 W = (current)² x (625 ohms)

Current = √(0.5/625)

Current = √(0.0008)

Current = 28.28 mA

============================

A slightly easier way:

Since the power is = I²R, it grows in proportion to (current)² .

We want to double the power dissipated, so we only need to increase the current by the factor of √2 .

(20 mA) x (√2) = 28.28 mA

Final answer:

Using the power dissipation formula, we calculate the resistance as 625 ohms. To dissipate 0.50 W, a current of sqrt(0.50 W / 625 ohms) is required, which equals 28 mA.

Explanation:

To determine how much current would be needed for a resistor to dissipate 0.50 W, we'll use the formula for electric power dissipation in a resistor which is P = I^2 R, where P is the power in watts, I is the current in amperes, and R is the resistance in ohms. In the given example, a resistor dissipates 0.25 W when a 20 mA (or 0.02 A) current passes through it. With P = I^2 R, we can derive R as R = P / I^2 which gives a resistance value through this initial condition.

The first step is to calculate the resistance of the resistor using the initial power dissipation value:

R = 0.25 W / (0.02 A)^2 = 0.25 / 0.0004 = 625 ohms.

Now, to find the current for 0.50 W power dissipation, we use

I = sqrt(P / R) = sqrt(0.50 W / 625 ohms) = sqrt(0.0008) = 0.0283 A or 28.3 mA.

Therefore, 28 mA of current is needed for the resistor to dissipate 0.50 W.