Calculate the electric force an electron exerts upon a proton inside a He atom if they are d=2.7⋅10^-10m apart.
Hint: Fe=k⋅qp⋅qp/d^2 where k=9⋅10^9.

A) 6.8E-8N;
B) -212.7E-9N;
C) -6.31E-9N;
D) -57.6E-10N;

Answer:

[tex]P_{1}-P_{2}=5400lb/ft^{2}[/tex]

Explanation:

We shall use newtons second law to evaluate the pressure difference

For the system the forces that act on it as shown in the figure

Thus by Newton's second law

[tex]F_{1}-F_{2}=mass\times acceleration\\\\P_{1}\times Area-P_{2}\times Area=mass\times acceleration\\\\\because Force=Pressure\times Area\\\\\therefore P_{1}-P_{2}=\frac{mass\times acceleration}{Area}[/tex]

Mass of the gasoline can be calculated from it's density[tex]45lb/ft^{3}[/tex]

[tex]Mass=Density\times Volume\\\\Mass= 45lb/ft^{3}\times \pi \frac{d^{2}}{4}L\\\\Mass=45lb/ft^{3}\times\frac{\pi 8^{2}}{4}\times 24\\Mass=54286.72lbs[/tex]

Using the calculated values we get

[tex]P_{1}-P_{2}=\frac{54286.72\times 5}{\frac{\pi 8^{2}}{4}}[/tex]

[tex]P_{1}-P_{2}=5400lb/ft^{2}[/tex]

(a) 107.2 Ω

The voltage in the circuit is written in the form

[tex]V=V_0 sin(\omega t)[/tex]

where in this case

V0 = 35.0 V is the maximum voltage

[tex]\omega=100 rad/s[/tex]

is the angular frequency

Given the inductance: L = 165 mH = 16.5 H, the reactance of the inductance is:

[tex]X_L = \omega L=(100)(0.165)=16.5 \Omega[/tex]

Given the capacitance: [tex]C=99\mu F=99\cdot 10^{-6} F[/tex], the reactance of the capacitor is:

[tex]X_C = \frac{1}{\omega C}=\frac{1}{(100)(99\cdot 10^{-6})}=101.0 \Omega[/tex]

And given the resistance in the circuit, R = 66.0 Ω, we can now find the impedance of the circuit:

[tex]Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{66^2+(16.5-101.0)^2}=107.2\Omega[/tex]

(b) 0.33 A

The maximum current can be calculated by using Ohm's Law for RLC circuit. In fact, we know the maximum voltage:

V0 = 35.0 V

The equivalent of Ohm's law for an RLC circuit is

[tex]I=\frac{V}{Z}[/tex]

where

Z=107.2 Ω

is the impedance. Substituting these values into the formula, we find:

[tex]I=\frac{35.0}{107.2}=0.33 A[/tex]

(c) 100 rad/s

The equation for the current is

[tex]I=I_0 sin(\omega t-\Phi)[/tex]

where

I0 = 0.33 A is the maximum current, which we have calculated previously

[tex]\omega[/tex] is the angular frequency

[tex]\Phi[/tex] is the phase shift

In an RLC circuit, the voltage and the current have the same frequency. Therefore, we can say that

[tex]\omega=100 rad/s[/tex]

also for the current.

(d) -0.907 rad

Here we want to calculate the phase shift [tex]\Phi[/tex], which represents the phase shift of the current with respect to the voltage. This can be calculated by using the equation:

[tex]\Phi = tan^{-1}(\frac{X_L-X_C}{R})[/tex]

Substituting the values that we found in part a), we get

[tex]\Phi = tan^{-1}(\frac{16.5-101.0}{66})=-52^{\circ}[/tex]

And the sign is negative, since the capacitive reactance is larger than the inductive reactance in this case. Converting in radians,

[tex]\theta=-52 \cdot \frac{2\pi}{360}=-0.907 rad[/tex]

So the complete equation of the current would be

[tex]I=0.33sin(100t+0.907)[/tex]

The impedance of the circuit is 107.2 Ω.

The maximum current is 0.33 A.

The numerical value for ω in the equation i = Imax sin (ωt − ϕ). rad/s is 100 rad/s.

The numerical value for ϕ in the equation i = Imax sin (ωt − ϕ) is -0.907 rad.

1. To find the voltage, we already know that V0=35V

And the voltage in the circuit is V= V0sin(wt)

Hence, the angular frequency, w= 100 rad/s

Given the inductance is 16.5Ω and the capacitance is 101Ω.

And the resistance is 66Ω

Hence, the impedance of the circuit is [tex]\sqrt{66^{2} + (16.5-101)^2} =107.2[/tex]Ω

2. To find the maximum current, since Z= 107.2Ω,

Recall, I = V/Z

Put the values into the formula

I = 0.33A.

3. To find the numerical value for ω in the equation i = Imax sin (ωt − ϕ)
We recall that I0 = 0.33A

In an RLC circuit, the voltage and the current have the same frequency. Therefore, we can say that

w= 100 rad/s.

4. To find the numerical value for ϕ in the equation i = Imax sin (ωt − ϕ)

Because the sign is in the negative and since the capacitive reactance is larger than the inductive reactance in this case.

Converting in radians,

θ = -52. 2π/360

= -0.907rad.

Hence, the complete equation would be:

I = 0.33sin(100t + 0.907)

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Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

Where, F is the force on ram, A be the area of ram, f be the force on plunger and a be the area of plunger.

Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

f = 67.5 Kgf

Answer:

v/U=0.79

Explanation:

Given u=[tex]u=\frac{Uy}{\partial }=\frac{Uy}{cx^{1/2}}[/tex]

Now for the given flow to be possible it should satisfy continuity equation

[tex]\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0[/tex]

Applying values in this equation we have

[tex]\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0\\\\\frac{\partial u}{\partial x}=\frac{\partial (\frac{Uy}{cx^{1/2}})}{\partial x}\\\\\frac{\partial u}{\partial x}=\frac{-1}{2}\frac{Uy}{cx^{3/2}}\\\\[/tex]

Thus we have

[tex]\frac{\partial v}{\partial y}=\frac{1}{2}\frac{Uy}{cx^{3/2}}\\\\\therefore \int \partial v=\int \frac{1}{2}\frac{Uy}{cx^{3/2}}\partial x\\\\v=\frac{1}{4}\frac{Uy^{2}}{cx^{3/2}}\\\\v=\frac{1}{4}\frac{uy}{x}[/tex] Hence proved [tex]\because u=\frac{Uy}{cx^{1/2}}[/tex]

For maximum value of v/U put y =[tex]\partial[/tex]

[tex]v=\frac{1}{4}\frac{Uy^{2}}{cx^{3/2}}[/tex]

[tex]v=\frac{1}{4}\frac{Uy^{2}}{cx^{3/2}}\\\\\frac{v}{U}=\frac{\partial ^{2}}{4cx^{3/2}}\\\\[/tex]

Thus solving we get using the given values

v/U=0.79

The y-component of velocity in a boundary layer flow is derived using the principle of continuity for an incompressible fluid. Upon evaluation, the ratio v /U has a maximum of 0.25 at the specified location (x = 0.5 m and δ = 5 mm).

The y-component v of the velocity can be solved using the continuity equation, d/dx (u A) + d/dy (v A) = 0, where A is the cross-sectional area of the pipe. This equation arises from the principle of conservation of mass for an incompressible fluid. In this case, our A is the width of the plate (into the page) times the y-distance from the plate or y*b. When the equation derived for velocity, u = Uy/δ, and the equation for the boundary layer thickness, δ = cx1/2, are plugged into the continuity equation and simplified, we derive the expression v = uy/4x.  Substituting for δ, we get v = Uy/(4δ*sqrt(x)), and at x = 0.5 m and δ = 5 mm, the maximum value of v /U is 1/4 or 0.25.

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Answer:

[tex]K.E =0.081eV[/tex]

[tex]\vec{\Omega}=(0.54\hat{i}+0.83\hat{j}+0.039\hat{k}[/tex]

Explanation:

Given:

Velocity vector [tex]\vec{v}=(2132\hat{i}+3300\hat{j}+154\hat{k})m/s[/tex]

the mass of neutron, m = 1.67 × 10⁻²⁷ kg

Now,

the kinetic energy (K.E) is given as:

[tex]K.E =\frac{1}{2}mv^2[/tex]

[tex]\vec{v}^2 = \vec{v}.\vec{v}[/tex]

or

[tex]\vec{v}^2 = (2132\hat{i}+3300\hat{j}+154\hat{k}).(2132\hat{i}+3300\hat{j}+154\hat{k})[/tex]

or

[tex]\vec{v}^2 =15.45\times 10^6 m^2/s^2[/tex]

substituting the values in the K.E equation

[tex]K.E =\frac{1}{2}1.67\times 10^{-27}kg\times 15.45\times 10^6 m^2/s^2[/tex]

or

[tex]K.E =1.2989\times 10^{-20}J[/tex]

also

1J = 6.242 × 10¹⁸ eV

thus,

[tex]K.E =1.2989\times 10^{-20}\times 6.242\times 10^{18}[/tex]

[tex]K.E =0.081eV[/tex]

Now, the direction vector [tex]\vec{\Omega}[/tex]

[tex]\vec{\Omega}=\frac{\vec{v}}{\left | \vec{v} \right |}[/tex]

or

[tex]\vec{\Omega}=\frac{(2132\hat{i}+3300\hat{j}+154\hat{k})}{\left |\sqrt{((2132\hat{i})^2+(3300\hat{j})^2+(154\hat{k}))^2} \right |}[/tex]

or

[tex]\vec{\Omega}=\frac{(2132\hat{i}+3300\hat{j}+154\hat{k})}{3930.65}[/tex]

or

[tex]\vec{\Omega}=0.54\hat{i}+0.83\hat{j}+0.039\hat{k}[/tex]

Answer:

V=4v

Explanation:

To perform this operation, we will refer to Ohm's Law. This Law relates the terms current, voltage and resistance.

The current intensity that passes through a circuit is directly proportional to the voltage or voltage of the circuit and inversely proportional to the resistance it presents.

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

Where

I=Current

V= Voltage

R=Resistance.

So, in this case:

R=200 ohm

I= 2mA  = 0.002 A

V= searched variable

[tex]V=r*I[/tex]

[tex]V=(200ohm)(0.002A)=4V[/tex]

Answer:

2.14×10⁸ m/s

Explanation:

L=Required Length=0.7 m

L₀=Initial length=1 m

c=speed of light=3×10⁸ m/s

[tex]From\ Lorentz\ Contraction\ relation\\L=L_0\sqrt{1-\frac {v^2}{c^2}}\\\Rightarrow 0.7=1\sqrt {1-\frac {v^2}{c^2}}\\\Rightarrow 0.49=1-\frac {v^2}{c^2}\\\Rightarrow 0.49-1=-\frac {v^2}{c^2}\\\Rightarrow -0.51=-\frac {v^2}{c^2}\\\Rightarrow 0.51=\frac {v^2}{c^2}\\\Rightarrow v^2=0.51\times c^2\\\Rightarrow v=\sqrt{0.51} \times c\\\Rightarrow v=0.71\times 3\times 10^8\\\Rightarrow v=2.14\times 10^8\ m/s\\\therefore velocity\ of\ stick\ should\ be\ 2.14\times 10^8\ m/s[/tex]

Answer:

(a) 250 kg/m^3

(b) 2 m^3

(c) 1000 kg

Explanation:

(a) Let V be the total volume of the boat.

Volume immersed = 25 % of total volume = 25 V / 100   = V / 4

Let the density of boat is d.

Density of water = 1000 kg/m^3

Use the principle of flotation

Buoyant force on the boat = weight of the boat

Volume immersed x density of water x g = Volume x density of boat x g

V / 4 x 1000 x g = V x d x g

d = 250 kg/m^3

(b)

mass = 500 kg

density = 250 kg/m^3

V = mass / density = m / d = 500 / 250 = 2 m^3

(c) Let the mass of cargo is m.

Volume immeresed = 75 % of total volume = 75 x 2 / 100 = 1.5 m^3

Buoyant force = weight of boat + weight of cargo

1.5 x 1000 x g = 500 g + m g

1500 = 500 + m

m = 1000 kg

Explanation:

We know

the frequency of the tube with one end open and the other end closed follows the given relations as

[tex]f_{1}[/tex] : [tex]f_{2}[/tex] : [tex]f_{3}[/tex] : [tex]f_{4}[/tex] = 1 : 3 : 5 : 7

∴ the 4th allowed wave is [tex]f_{4}[/tex] = 7 [tex]f_{1}[/tex]

                                                                   = [tex]\frac{7v}{4l}[/tex]

We know  [tex]f_{4}[/tex] = 1975 Hz and v = 343 m/s ( as given in question )

∴[tex]l = \frac{7\times v}{4\times f_{4}}[/tex]

 [tex]l = \frac{7\times 343}{4\times 1975}[/tex]

           = 0.303 m

We know that v = [tex]f_{4}[/tex] x [tex]λ_{4}[/tex]

[tex]\lambda _{4}= \frac{v}{f_{4}}[/tex]

[tex]\lambda _{4}= \frac{343}{1975}[/tex]

                            = 0.17 m

Now when the warmer air is flowing, the speed gets doubled and the mean temperature increases. And as a result the wavelength increases but the amplitude and the frequency remains the same.

So we can write

v ∝ λ

or  [tex]\frac{v_{1}}{v_{2}}= \frac{\lambda _{1}}{\lambda _{2}}[/tex]

Therefore, the wavelength becomes doubled = 0.17 x 2

                                                                             = 0.34 m

Now the new length of the air column becomes doubled

∴ [tex]l^{'}[/tex] = 0.3 x 2

                        = 0.6 m

∴ New speed, [tex]v^{'}[/tex] = 2 x 343

                                               = 686 m/s

∴ New frequency is [tex]f^{'}=\frac{v^{'}}{4\times l^{'}}[/tex]

                                 [tex]f^{'}=\frac{686}{4\times 0.6}[/tex]

                                               = 283 Hz

∴ The new frequency remains the same.

Now we know

[tex]v_{s}[/tex] = 12 m/s, [tex]v_{o}[/tex] = 4 m/s, [tex]f_{o}[/tex] = 1975 Hz

Therefore, apparent frequency is [tex]f^{'}=f^{o}\left ( \frac{v+v_{s}}{v+v_{o}} \right )[/tex]

[tex]f^{'}=1975\left ( \frac{343+12}{343+4} \right )[/tex]

            = 2020.5 Hz

Answer:

The drift velocity of electrons in a copper wire is [tex]1.756\times10^{-4}\ m/s[/tex]

Explanation:

Given that,

Density [tex]\rho=8.93\ g/cm^3[/tex]

Mass [tex]M=63.5\ g/mol[/tex]

Radius = 0.625 mm

Current = 3 A

We need to calculate the drift velocity

Using formula of drift velocity

[tex]v_{d}=\dfrac{J}{ne}[/tex]

Where, n = number of electron

j = current density

We need to calculate the current density

Using formula of current density

[tex] J=\dfrac{I}{\pi r^2}[/tex]

[tex] J=\dfrac{3}{3.14\times(0.625\times10^{-3})^2}[/tex]

[tex] J=2.45\times10^{6}\ A/m^2[/tex]

Now, we calculate the number of electron

Using formula of number of electron

[tex]n=\dfrac{\rho}{M}N_{A}[/tex]

[tex]n=\dfrac{8.93\times10^{6}}{63.5}\times6.2\times10^{23}[/tex]

[tex]n=8.719\times10^{28}\ electron/m^3[/tex]

Now put the value of n and current density into the formula of drift velocity

[tex]v_{d}=\dfrac{2.45\times10^{6}}{8.719\times10^{28}\times1.6\times10^{-19}}[/tex]

[tex]v_{d}=1.756\times10^{-4}\ m/s[/tex]

Hence, The drift velocity of electrons in a copper wire is [tex]1.756\times10^{-4}\ m/s[/tex]

Answer:

The International Space Station move at 7.22 km/s.

Explanation:

Orbital speed of satellite is given by  [tex]v=\sqrt{\frac{GM}{r}}[/tex], where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.

r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m

G = 6.673 x 10⁻¹¹ Nm²/kg²

M = 5.98 x 10²⁴ kg

Substituting

              [tex]v=\sqrt{\frac{6.673\times 10^{-11}\times 5.98\times 10^{24}}{7.75\times 10^6}}=7223.86m/s=7.22km/s[/tex]

  The International Space Station move at 7.22 km/s.      

Answer:

Q = 116.8 J

Explanation:

Here given that the temperature of 1 L hydrogen is increased by 90 degree C at constant pressure condition.

So here we will have

[tex]Q = n C_p \Delta T[/tex]

here we know that

n = number of moles

[tex]n = \frac{1}{22.4}[/tex]

[tex]n = 0.0446[/tex]

for ideal diatomic gas molar specific heat capacity at constant pressure is given as

[tex]C_p = \frac{7}{2}R[/tex]

now we have

[tex]Q = (0.0446)(\frac{7}{2}R)(90)[/tex]

[tex]Q = 116.8 J[/tex]

Answer:

a)

51764.7 ohm

b)

28.5 A

Explanation:

a)

t = time taken to charge to 90.0% of its final value = 16.2 s

T = time constant

C = capacitance = 136 x 10⁻⁶ F

R = resistance of the resistor

Q₀ = Maximum charge stored

Q = Charge after time "t" = 0.90 Q₀

Using the equation

[tex]Q = Q_{o}(1 - e^{\frac{-t}{T}})[/tex]

[tex]0.90 Q_{o} = Q_{o}(1 - e^{\frac{-16.2}{T}})[/tex]

[tex]0.90 = (1 - e^{\frac{-16.2}{T}})[/tex]

T = 7.04 s

Time constant is given as

T = RC

7.04 = R (136 x 10⁻⁶)

R = 51764.7 ohm

b)

V = Potential difference of voltage source = 265 volts

q = Amount of charge discharged = 0.90 Q₀ = 0.90 (CV) = (0.90) (136 x 10⁻⁶) (265) = 0.032436 C

t = time taken to discharge = 1.14 x 10⁻³ s

Current is given as

[tex]i = \frac{q}{t}[/tex]

i = 28.5 A

For the RC charging circuit of a camera flash unit has the value of resistance and current as,

RC circuit is the circuit in which the voltage or current passes through the resistor and capacitor consist by the circuit.

The formula for the RC circuit can be given as,

[tex]Q=Q_o\left(1-e^{\dfrac{-1}{RC}}\right )[/tex]

Here, [tex](Q_o)[/tex] is the initial charge and (R) is the resistance.

The  capacitor charges to 90.0% of its final value. Thus, the charge for the first case can be given as,

[tex]Q=0.90 Q_o[/tex]

The capacitance of 136 μF. Put the values in the above formula as,

[tex]0.90Q_o=Q_o\left(1-e^{\dfrac{-1}{R\times136\times10^{-6}}}\right )\\R=51765\rm ohm[/tex]

Thus, the value of the resistance R is 51765 ohms.

The voltage source of the RC circuit is 265 V  and the capacitor discharges 90.0% of its full charge in 1.14 ms. Thus, the amount of charge discharged is,

[tex]Q=0.9\times136\times10^{-6}\times265\\Q=3.2436\times10^{-2}\rm C[/tex]

The value of current is the ratio of discharge per second time (1.14 ms.). Therefore,

[tex]I=\dfrac{3.2436\times10^{-2}}{1.14\times10^{-3}}\\I=28.5\rm A[/tex]

Thus, for the RC charging circuit of a camera flash unit has the value of resistance and current as,

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in increasing order of kinetic energy, the joggers are: A, D, C, B

To rank the joggers in order of increasing kinetic energy, we'll use the formula for kinetic energy:

[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]

Where:

- ( m ) is the mass of the jogger

- ( v ) is the speed of the jogger

Let's calculate the kinetic energy for each jogger:

1. Jogger A:

[tex]\[ KE_A = \frac{1}{2} m v^2 \][/tex]

2.Jogger B:

[tex]\[ KE_B = \frac{1}{2} \left(\frac{m}{3}\right) (3v)^2 = \frac{1}{2} \left(\frac{m}{3}\right) 9v^2 = \frac{9}{2} \left(\frac{1}{3} m v^2\right) = \frac{9}{2} KE_A \][/tex]

3. Jogger C:

[tex]\[ KE_C = \frac{1}{2} (4m) \left(\frac{v}{3}\right)^2 = \frac{1}{2} (4m) \left(\frac{1}{9}\right) v^2 = \frac{2}{9} (4m v^2) = \frac{8}{9} KE_A \][/tex]

4. Jogger D:

[tex]\[ KE_D = \frac{1}{2} (3m) \left(\frac{v}{2}\right)^2 = \frac{1}{2} (3m) \left(\frac{1}{4}\right) v^2 = \frac{3}{8} (m v^2) = \frac{3}{4} KE_A \][/tex]

Now, let's rank the joggers based on their kinetic energies:

- Jogger A: [tex]\( KE_A \)[/tex]

- Jogger D: [tex]\( KE_D = \frac{3}{4} KE_A \)[/tex]

- Jogger C: [tex]\( KE_C = \frac{8}{9} KE_A \)[/tex]

- Jogger B: [tex]\( KE_B = \frac{9}{2} KE_A \)[/tex]

So, in increasing order of kinetic energy, the joggers are: A, D, C, B

Answer:

2.15 m

Explanation:

u = 2.2 m/s, v = 0, uk = 0.115

a = uk x g = 0.115 x 9.8 = 1.127 m/s^2

Let s be the distance traveled before stopping.

v^2 = u^2 - 2 a s

0 = 2.2^2 - 2 x 1.127 x s

s = 2.15 m

Answer:

COP = 4.25

Explanation:

It is given that,

Cooling temperature, [tex]T_c=-26^{\circ}C=273-26=247\ K[/tex]

Heating temperature, [tex]T_h=50^{\circ}C=323\ K[/tex]

We need to find the coefficient of performance. It is given by :

[tex]COP=\dfrac{T_h}{T_h-T_c}[/tex]

[tex]COP=\dfrac{323}{323-247}[/tex]

COP = 4.25

So, the best coefficient of performance of a refrigerator is 4.45 Hence, this is the required solution.

Answer:

0.48

Explanation:

m = mass of disc shaped grindstone = 50 kg

d = diameter of the grindstone = 0.52 m

r = radius of the grindstone = (0.5) d = (0.5) (0.52) = 0.26 m

w₀ = initial angular speed = 850 rev/min = 850 (0.10472) rad/s = 88.97 rad/s

t = time taken to stop = 7.5 sec

w = final angular speed  0 rad/s

Using the equation

w = w₀ + α t

0 = 88.97 + α (7.5)

α = - 11.86 rad/s²

F = normal force on the disc by the ax = 160 N

μ = Coefficient of friction

f = frictional force

frictional force is given as

f = μ F

f = 160 μ

Moment of inertia of grindstone is given as

I = (0.5) m r² = (0.5) (50) (0.26)² = 1.69 kgm²

Torque equation is given as

r f = I |α|

(0.26) (160 μ) = (1.69) (11.86)

μ = 0.48

Answer:

a)

14 rad/s

b)

11.2 m/s

Explanation:

a)

d = diameter of tire = 0.80 m

r = radius of tire = (0.5) d = (0.5) (0.80) = 0.40 m

v = speed of bicycle = 5.6 m/s

w = angular speed of the tire

Speed of cycle is given as

v = r w

5.6 = (0.40) w

w = 14 rad/s

b)

v' = speed of blue dot

Speed blue of dot is given as

v' = v + rw

v' = 5.6 + (0.40) (14)

v' = 11.2 m/s

The angular speed of the bicycle tire is 14 rad/s. The speed of the blue dot when it is 0.80 m above the road is the same as the bicycle's speed which is 5.6 m/s.

This is a problem which involves the calculation  of the angular speed and linear speed of an object in circular motion, here the circular motion being the rotation of a bicycle tire.

Given the linear speed of the bicycle and the radius of the tire, we would use the equation that links linear speed v and angular speed a, expressed as v = ra. We can find angular speed by rearranging this formula as a = v/r. Hence, we can calculate the angular speed of the tire as 5.6 m/s divided by the radius of the tire, which is half of the diameter, so 0.4m, which equals 14 rad/s.

Next, when a point on the tire (the blue dot) is 0.8 m above the ground it is at the top of tire's circular path, so its speed is equivalent to the linear speed of the bicycle, 5.6 m/s. This is because at this point the dot is not in contact with the ground hence it isn't stationary relative to the ground unlike the point at the bottom of the tire.

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D = 497.4x10⁻⁶m. The diameter of a mile of 24-gauge copper wire with resistance of 0.14 kΩ and resistivity of copper 1.7×10−8Ω⋅m is 497.4x10⁻⁶m.

In order to solve this problem we have to use the equation that relates resistance and resistivity:

R = ρL/A

Where ρ is the resistivity of the matter, the length of the wire, and A the area of ​​the cross section of the wire.

If a mile of 24-gauge copper wire has a resistance of 0.14 kΩ and the resistivity of copper is 1.7×10⁻⁸ Ω⋅m. Determine the diameter of the wire.

First, we have to clear A from the equation R = ρL/A:

A = ρL/R

Substituting the values

A = [(1.7×10⁻⁸Ω⋅m)(1.6x10³m)]/(0.14x10³Ω)

A = 1.9x10⁻⁷m²

The area of a circle is given by A = πr² = π(D/2)² = πD²/4, to calculate the diameter D we have to clear D from the equation:

D = √4A/π

Substituting the value of A:

D = √4(1.9x10⁻⁷m²)/π

D = 497.4x10⁻⁶m

The diameter of the wire is approximately 4.5 × 10^-6 meters.

To find the diameter of the wire, we can use the formula for resistance:

R = (ρ * L) / (A),

where R is the resistance, ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire. Rearranging the formula to solve for the area:

A = (ρ * L) / R.

Given that the mile of 24-gauge copper wire has a resistance of 0.14 kΩ (or 0.14 * 10^3 Ω), the resistivity of copper is 1.7 × 10^-8 Ω ⋅ m, and 1 mile is equal to 1.6 km, we can substitute the values in the formula:

A = (1.7 × 10^-8 Ω ⋅ m * (1.6 * 10^3 m)) / (0.14 * 10^3 Ω).

Calculating the area, we find:

A ≈ 1.94 × 10^-11 m^2.

Next, we use the formula for the area of a circle (A = π * r^2) to find the radius (r) of the wire:

r = √(A / π).

Substituting the values, we have:

r ≈ √(1.94 × 10^-11 m^2 / π).

Calculating the radius, we find:

r ≈ 2.25 × 10^-6 m.

Finally, we can double the radius to find the diameter of the wire:

D = 2r ≈ 2 * 2.25 × 10^-6 m = 4.5 × 10^-6 m.

Therefore, the diameter of the wire is approximately 4.5 × 10^-6 meters.

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Answer:

Work done, W = 39.2 J

Explanation:

It is given that,

Mass of the block, m = 2 kg

The block is lifted vertically 2 m by the man i.e the distance covered by the block is, h = 2 m. The man is doing work against the gravity. It is given by :

[tex]W=mgh[/tex]

Where

g is acceleration due to gravity

[tex]W=2\ kg\times 9.8\ m/s^2\times 2\ m[/tex]

W = 39.2 J

So, the work done by the man is 39.2 J. Hence, this is the required solution.

Explanation:

It is given that,

A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.

In 2 seconds, distance covered by the mass is 12 cm.

In 1 seconds, distance covered by the mass is 6 cm

So, in 16 seconds, distance covered by the mass is 96 cm

So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.

Answer:

12 × 10^12 Nm^2/C

Explanation:

According to the Gauss's theorem

The total electric flux by the enclosed charge is given by

Charge enclosed / €0

So flux by each surface

= Q enclosed / 6 × €0

= 641 / ( 6 × 8.854 × 10^-12)

= 12 × 10^12 Nm^2/C

Answer:

(a) 4323.67 N

(b) 421.13 degree

Explanation:

A = 2580 N 36 degree South of east

B = 2270 N due South

A = 2580 ( Cos 36 i - Sin 36 j) = 2087.26 i - 1516.49 j

B = 2270 (-j) = - 2270 j

A + B = 2087.26 i - 1516.49 j - 2270 j

A + B = 2087.26 i - 3786.49 j

(a) Magnitude of A + B = [tex]\sqrt{2087.26^{2}+(-3786.49)^{2}}[/tex]

Magnitude of A + B = 4323.67 N

(b) Direction of A + B

Tan theta = - 3786.49 / 2087.26

theta = - 61.13 degree

Angle from ast = 360 - 61.13 = 421.13 degree.

To determine the magnitude and direction of the resultant force acting on a tree stump when two forces are applied, one must resolve the forces into components, sum the components, then apply the Pythagorean theorem for magnitude and inverse tangent for direction.

We need to resolve each force into its horizontal (east-west) and vertical (north-south) components. For Force A, we can use trigonometry to find the components:

Force B is already pointing due south, so its components are:

The net force components are:

With these components, the magnitude of the resultant force can be calculated using the Pythagorean theorem:

Fnet = √(Fnet,x² + Fnet,y²)

To find the direction of the resultant force relative to due east, we can use the inverse tangent function (tan-1):

θ = tan-1(Fnet,y / Fnet,x)

This angle should be positive, as we are measuring it with respect to due east.

Answer:

(a) 15.4 second

(b) 769.23 m

Explanation:

u = 0, a = 6.5 m/s^2, v = 100 m/s

(a) Let t be the time taken

Use First equation of motion

v = u + a t

100 = 0 + 6.5 x t

t = 15.4 second

(b) Let height covered is h.

Use third equation of motion

v^2 = u^2 + 2 a h

100^2 = 0 + 2 x 6.5 x h

10000 = 13 x h

h = 769.23 m

Answer:

[tex]P = 2.94 \times 10^5 Pa[/tex]

Explanation:

Normal force due to four tires is counter balancing the weight of the car

So here we will have

[tex]4F_n = mg[/tex]

[tex]F_n = \frac{mg}{4}[/tex]

[tex]F_n = \frac{1.20 \times 10^3 \times 9.81}{4}[/tex]

[tex]F_n = 2943 N[/tex]

now we know that pressure in each tire is given by

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

Here we know that

[tex]A = 1.00 \times 10^2 cm^2 = 1.00 \times 10^{-2} m^2[/tex]

[tex]P = \frac{2943}{1.00 \times 10^{-2}}[/tex]

[tex]P = 2.94 \times 10^5 Pa[/tex]

Answer:

P = 294300Pa or 42.67psi by conversion.

Explanation:

Since Four tyres were inflated, we have that area of the four tyres are

4×1×10²cm²

Pressure is given as:

P = f/a but f = mg

P = m×g/a

Therefore,

P = 1.20x10³kg × 9.81m/s² / (4 ×1x10² cm²)

P = 1.20x10³kg×9.81m/s² / (0.04m²)

P = 294300Pa or 42.67psi by conversion.

Answer:

- 0.23

0.56

Explanation:

[tex]\underset{v_{o}}{\rightarrow}[/tex] = initial velocity of the particle = [tex](5.24 Cos35.2) \hat{i} + (5.24 Sin35.2) \hat{j}[/tex] = [tex](4.28) \hat{i} + (3.02) \hat{j}[/tex]

[tex]\underset{v_{f}}{\rightarrow}[/tex] = final velocity of the particle = [tex](6.25 Cos57.5) \hat{i} + (6.25 Sin57.5) \hat{j}[/tex] = [tex](3.36) \hat{i} + (5.27) \hat{j}[/tex]

t = time interval = 4.00 sec

[tex]\underset{a}{\rightarrow}[/tex] = average acceleration = ?

Using the kinematics equation

[tex]\underset{v_{f}}{\rightarrow}[/tex] = [tex]\underset{v_{o}}{\rightarrow}[/tex] + [tex]\underset{a}{\rightarrow}[/tex] t

[tex](3.36) \hat{i} + (5.27) \hat{j}[/tex] = [tex](4.28) \hat{i} + (3.02) \hat{j}[/tex] + [tex]\underset{a}{\rightarrow}[/tex] (4)

[tex](- 0.92) \hat{i} + (2.25) \hat{j}[/tex] = 4 [tex]\underset{a}{\rightarrow}[/tex]

[tex]\underset{a}{\rightarrow}[/tex] = [tex](- 0.23) \hat{i} + (0.56) \hat{j}[/tex]

hence

Average acceleration along x-direction = - 0.23

Average acceleration along y-direction = 0.56

Answer:

Heat generated in the rod is 8.33 watts.

Explanation:

It is given that,

Length of rod, l = 2 m

Area of cross section, [tex]A=2\ mm\times 2\ mm=4\ mm^2=4\times 10^{-6}\ m^2[/tex]

Resistivity of rod, [tex]\rho=6\times 10^{-8}\ \Omega-m[/tex]

Potential difference, V = 0.5 V

The value of resistance is given by :

[tex]R=\rho\dfrac{l}{A}[/tex]

[tex]R=6\times 10^{-8}\ \Omega-m\times \dfrac{2\ m}{4\times 10^{-6}\ m^2}[/tex]

R = 0.03 ohms

Let H is the rate at which heat is generated in the rod . It is given by :

[tex]\dfrac{H}{t}=I^2R[/tex]

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

[tex]\dfrac{H}{t}=\dfrac{V^2}{R}[/tex]

[tex]\dfrac{H}{t}=\dfrac{(0.5)^2}{0.03}[/tex]

[tex]\dfrac{H}{t}=8.33\ watts[/tex]

So, the at which heat is generated in the rod is 8.33 watts. Hence, this is the required solution.

Answer:

670400 J

Explanation:

m = mass of water = 2 kg

T₀ = initial temperature of water = 20 ºC

T = initial temperature of water = 100 ºC

c = heat capacity of water = 4190 J /kg ºC

Q = energy needed to raise the temperature of water

Energy needed to raise the temperature of water is given as

Q = m c (T - T₀)

Inserting the values

Q = (2) (4190) (100 - 20)

Q = 670400 J

Answer:

Explanation:

Force, F = - mg j

r = - 7x i + y j

Torque is defined as the product f force and the perpendicular distance.

It is also defined as the cross product of force vector and the displacement vector.

[tex]\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}[/tex]

[tex]\overrightarrow{\tau }=(- 7 x i + yj)\times (-mgj)[/tex]

[tex]\overrightarrow{\tau  }= 7 m g x k

Here, we observe that the torque is independent of y coordinate.

Answer:

1.1 sec

Explanation:

m = mass of the box = 8 kg

k = spring constant of the spring = 69 N/m

v = initial speed of the box = 1.5 m/s

t = time period of oscillation of box in contact with the spring

Time period is given as

[tex]t = \pi \sqrt{\frac{m}{k}}[/tex]

Inserting the values

[tex]t = (3.14) \sqrt{\frac{8}{69}}[/tex]

t = 1.1 sec