A 2 meter long bar has a longitudinal/axial wave speed of 5,000 m/s. If the bar is fixed at each end, what is the second natural frequency an for this system in rad/s? a. 12,450 b. 14,850 e 15,710 d. 16,800 e. 18,780 f. None of the above

Answer:

Tangential Speed equals 5.57m/s

Explanation:

In the figure shown for equilibrium along y- axis we have

[tex]\sum F_{y}=0[/tex]

Resolving Forces along y axis we have

[tex]Tcos(\theta )=mg............(i)[/tex]

Similarly along x axis

[tex]\sum F_{x}=ma_{x}[/tex]

[tex]Tsin(\theta )=m[tex]\frac{v^{2} }{r}[/tex]............(ii)[/tex]

Dividing ii by i we have

[tex]tan(\theta )=\frac{v^{2}}{rg}[/tex]

In the figure below we have [tex]r=lsin(\theta )[/tex]

Thus solving for v we have

[tex]v=\sqrt{lgsin(\theta) tan(\theta )}[/tex]

Applying values we get

v=5.576m/s

Answer:

The pressure at constant velocity is 1.1 psi.

(1). is correct option.

Explanation:

Given that,

Diameter = 8 inch = 20.32 cm = 0.2032 m

Flow rate =2000 g/m

We need to calculate the velocity

[tex]v = \dfrac{Flow\ rate}{Area}[/tex]

[tex]2000\ g/m=\dfrac{0.00378\times2000}{60}[/tex]

[tex]2000\ g/m=0.1262 m^3/s[/tex]

[tex]v=\dfrac{0.1262}{\pi (0.1016)^2}[/tex]

[tex]v=3.89\ m/s[/tex]

We need to calculate the pressure

Using formula of pressure

[tex]P = \dfrac{1}{2}\rho v^2[/tex]

[tex]P=\dfrac{1}{2}\times1000\times(3.89)^2[/tex]

[tex]P=7566\ Pa[/tex]

[tex]P=1.09\ psi=1.1\ psi[/tex]

Hence, The pressure at constant velocity is 1.1 psi.

Answer:

should be the answer B

Explanation:

Answer:

47.14 Km/h

Explanation:

distance = 1650 km

time = 35 hours

The average speed is defined as the ratio of total distance to the total time taken.

Average speed = total distance / total time

Average speed = 1650 / 35 = 47.14 Km/h

A magnetic field of 0.80 T is there, then the magnitude of the magnetic force on the electron will be equal to - 6.4 x 10⁻¹⁴ k.

Charged material exerts a force when it is introduced to an electromagnetic field because of the chemical property of electric charge. You can have a negative or a positive electric charge. Unlike charge is created one another while like charges repel one another. Neutral refers to an item that carries no net charge.

q = - 1.6 x 10⁻¹⁹ c

v(x) = 5 x 10⁵ m/s,

v(y) = 3 x 10⁵ m/s,

B = 0.8 T along Y-axis

The velocity vector is given by

v = 5 x 10⁵ i + 3 x 10⁵ j

B = 0.8 j

Now calculate the force,

F = q (v x B)

F = -1.6 x 10⁻¹⁹ {(5 x 10⁵ i + 3 x 10⁵ j) x (0.8 j)}

F = - 1.6 x 10⁻¹⁹ (5 x 0.8 x 10^5 k)

F = - 6.4 x 10⁻¹⁴ k

The force's amplitude and direction are along the negative Z axis Is 6.4 x 10⁻¹⁴ N.

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The magnitude of the magnetic force on the electron, given its charge, velocity components, and a magnetic field of 0.80 T in the positive y direction, is 8.0 x 10^(-14) N.

The magnetic force on a charged particle moving in a magnetic field can be calculated using the following formula:

F = q * v * B * sin(θ)

Where:

- F is the magnetic force.

- q is the charge of the particle.

- v is the velocity of the particle.

- B is the magnetic field strength.

- θ is the angle between the velocity vector and the magnetic field vector.

In this case, the charge of the electron, q, is -1.6 x 10^(-19) C, the velocity components are given as vx = 5.0 x 10^5 m/s and vy = 3.0 x 10^5 m/s, and the magnetic field strength is 0.80 T in the positive y direction.

The angle θ between the velocity and the magnetic field is 90 degrees (π/2 radians) because the electron is moving in the xy plane, and the magnetic field is in the positive y direction.

Now, let's calculate the magnitude of the magnetic force:

F = (-1.6 x 10^(-19) C) * (5.0 x 10^5 m/s) * (0.80 T) * sin(π/2)

F = (-8.0 x 10^(-14) N) * 1

F = -8.0 x 10^(-14) N

So, the magnitude of the magnetic force on the electron is 8.0 x 10^(-14) N.

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Explanation:

Mass of ball A, [tex]m_A=7\ kg[/tex]

Mass of ball B, [tex]m_B=3\ kg[/tex]

Initial velocity of ball A, [tex]u_A=12\ m/s[/tex]

Initial velocity of ball B, [tex]u_B=-1\ m/s[/tex]

We need to find the final velocity of each ball. For a perfectly elastic collision, the coefficient of restitution is equal to 1. It is given by :

[tex]e=\dfrac{v_B-v_A}{u_A-u_B}[/tex]

[tex]v_A\ and\ v_B[/tex] are final velocities of ball A and B

[tex]1=\dfrac{v_B-v_A}{13}[/tex]

[tex]v_B-v_A=13[/tex]...........(1)

Using the conservation of linear momentum as :

[tex]m_Au_A+m_Bu_B=m_Av_A+m_Bu_B[/tex]

[tex]7(12)+3(-1)=7v_A+3u_B[/tex]

[tex]7v_A+3u_B=81[/tex]..............(2)

On solving equation (1) and (2) using calculator we get :

[tex]v_A=4.2\ m/s[/tex]

[tex]v_B=17.2\ m/s[/tex]

So, the final velocities of ball A and B are 4.2 m/s and 17.2 m/s. Hence, this is the required solution.

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

         Initial momentum = Final momentum

         4.4v = 12.364 i + 15.092 j

         v =2.81 i + 3.43 j

Magnitude

        [tex]v=\sqrt{2.81^2+3.43^2}=4.43m/s[/tex]

Direction

       [tex]\theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0[/tex]

       50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Answer:

The pressure is 2.167 psi.

Explanation:

Given that,

Diameter = 1.5 feet

Height = 10 feet

We need to calculate the psi at 5 feet

Using formula of pressure at a depth in a fluid

Suppose the fluid is water.

Then, the pressure is

[tex]P=\rho g h[/tex]

Where, P = pressure

[tex]\rho[/tex] = density

h = height

Put the value into the formula

[tex]P=1000\times9.8\times1.524[/tex]

[tex]P=14935.2\ N/m^2[/tex]

Pressure in psi is

[tex]P=2.166167621\ psi[/tex]

[tex]P=2.167\ psi[/tex]

Hence, The pressure is 2.167 psi.  

Answer:

[tex]T_{2}[/tex] =[tex]\frac{25T_{1}}{V_{1}}[/tex]

where

[tex]V_{1}[/tex] is initial volume in liters

[tex]T_{1}[/tex] is initial temperature in kelvins

Explanation:

Let the initial volume be [tex]V_{1}[/tex] and the initial temperature be  [tex]P_{1}[/tex]

Now by ideal gas law

[tex]\frac{P_{1}V_{1}}{T_{1}}=nR..............(i)[/tex]

Similarly let

[tex]V_{2}[/tex] be final volume

[tex]T_{2}[/tex] be the final volume

thus by ideal gas law we again have

[tex]\frac{P_{2}V_{2}}{T_{2}}=nR..............(ii)[/tex]

Equating i and ii we get

[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex]

For system at constant pressure the above expression reduces to

[tex]\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}[/tex]

Solving for [tex]T_{2}[/tex] we get

[tex]T_{2}[/tex] =[tex]\frac{25T_{1}}{V_{1}}[/tex]

where

[tex]V_{1}[/tex] is initial volume in liters

[tex]T_{1}[/tex] is initial temperature in kelvins

Answer:

j

Explanation:

x = 4 t^2 - 2 t - 4.5

Position at t = 3 s

x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m

Velocity at t = 3 s

v = dx / dt = 8 t - 2

v ( t = 3 s) = 8 x 3 - 2 = 22 m/s

Acceleration at t = 3 s

a = dv / dt = 8

a ( t = 3 s ) = 8 m/s^2

When is the velocity = 0

v = 0

8 t - 2 = 0

t = 0.25 second

When is the position = 0

x = 0

4 t^2 - 2 t - 4.5 = 0

[tex]t = \frac{2 \pm \sqrt{4 + 72}}{8}[/tex]

t = 1.4 second

Answer:

a) Initial angular velocity of the turbine = 102.66 rad/s

b) Time elapsed while the turbine is speeding up = 157 s

Explanation:

a) Considering angular motion of turbine:-

Initial angular velocity, u =  ?

Acceleration , a = 0.155 rad/s²

Final angular velocity, v  = 127 rad/s  

Angular displacement, s = 2π x 2870 = 18032.74 rad

We have equation of motion v² = u² + 2as

Substituting

   v² = u² + 2as

    127² = u² + 2 x 0.155 x 18032.74

    u = 102.66 rad/s

Initial angular velocity of the turbine = 102.66 rad/s

b) We have equation of motion v = u + at

Initial angular velocity, u =  102.66 rad/s

Acceleration , a = 0.155 rad/s²

Final angular velocity, v  = 127 rad/s  

Substituting

  v = u + at

  127  = 102.66 + 0.155 x t = 157 s

Time elapsed while the turbine is speeding up = 157 s

Answer:

31.3 m/s

Explanation:

The relation between the coefficient of friction and the velocity, radius of curve path is given by

μ = v^2 / r g

v^2 = μ r g

v^2 = 0.4 x 250 x 9.8 = 980

v = 31.3 m/s

Answer:

3.92 cm

Explanation:

k = spring constant of the spring scale = 1500 N/m

m = mass of the brick = 3 kg

x = stretch caused in the spring

h = height dropped by the brick = x

Using conservation of energy

Spring potential energy gained by the spring scale = Potential energy lost

(0.5) k x² = mgh

(0.5) k x² = mgx

(0.5) k x = mg

(0.5) (1500) x = (3) (9.8)

x = 0.0392 m

x = 3.92 cm

Answer:

Frequency, f = 0.274 Hz

Explanation:

It is given that,

Mass of the girl, m = 40 kg

Length of the rope, L = 3.3 m

We need to find the frequency of her swinging. It is an example of simple harmonic motion whose time period is given by :

[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]

[tex]T=2\pi\sqrt{\dfrac{3.3\ m}{9.8\ m/s^2}}[/tex]

T = 3.64 seconds

The frequency, [tex]f=\dfrac{1}{T}[/tex]

[tex]f=\dfrac{1}{3.64\ s}[/tex]

f = 0.274 Hz

So, the frequency of her swinging is 0.274 Hz. Hence, this is the required solution.

Answer:

The speed of her at the bottom is 24.035 m/s.

Explanation:

Given that,

Speed of cyclist = 12 m/s

Height = 30 m

Distance d = 450 m

Mass of cyclist and bicycle =70 kg

Drag force = 12 N

We need to calculate the speed at the bottom

Using conservation of energy

K.E+P.E=drag force+K.E+P.E

Potential energy is zero at the bottom.

K.E+P.E=drag force+K.E

[tex]\dfrac{1}{2}mv^2+mgh=Fx+\dfrac{1}{2}mv^{2}[/tex]

[tex]\dfrac{1}{2}\times70\times12^2+70\times9.8\times30=12\times450+\dfrac{1}{2}\times70\timesv^2[/tex]

[tex]25620=5400+35v^2[/tex]

[tex]35v^2=25620-5400[/tex]

[tex]35v^2=20220[/tex]

[tex]v^2=\dfrac{20220}{35}[/tex]

[tex]v=\sqrt{\dfrac{20220}{35}}[/tex]

[tex]v=24.035\ m/s[/tex]

Hence, The speed of her at the bottom is 24.035 m/s.

Answer:

Speed at the bottom of the slope is 24.03 m /sec

Explanation:

We have given speed of the cyclist v = 12 m/sec

She starts down a 450 m long that is 30 m high

Si distance = 450 m and height h = 30 m

Combined mass of cyclist and bicycle m = 70 kg

Drag force = 12 N

We have to find the speed at the bottom

According to conservation theory

KE +PE = work done by drag force + KE + PE

As we know that at the bottom there will be no potential energy

So [tex]\frac{1}{2}\times 70\times 12^2+70\times 9.8\times 30=12\times 450+0+\frac{1}{2}\times 70\times v^2[/tex]

[tex]\frac{1}{2}\times 70\times v^2=20220[/tex]

[tex]v=24.03m/sec[/tex]

Answer:

1184 kJ/kg

Explanation:

Given:

water pressure P= 28 bar

internal energy U= 988 kJ/kg

specific volume of water v= 0.121×10^-2 m^3/kg

Now from steam table at 28 bar pressure we can write

[tex]U= U_{f}= 987.6 kJ/Kg[/tex]

[tex]v_{f}=v=1.210\times 10^{-3}m^{^{3}}/kg[/tex]

therefore at saturated liquid we have specific enthalpy at 55 bar pressure.

that the specific enthalpy h =  h at 50 bar +(55-50)/(60-50)*( h at 50 bar - h at 60 bar)

[tex]h= 1154.5 + \frac{5}{10}\times(1213-1154)[/tex]

h= 1184 kJ/kg

Answer:

8.6 J

Explanation:

Work done by the block = change in energy in the spring

W = ½ kx²

W = ½ (254.0 N/m) (0.26 m)²

W = 8.6 J

Final answer:

To find the maximum height, we need to use the initial velocity and the angle of the baseball's trajectory. By breaking down the initial velocity into its horizontal and vertical components, we can then use the equation h = v_y^2 / (2g) to find the maximum height.

Explanation:

To find the maximum height, we first need to break down the initial velocity into its horizontal and vertical components. The initial velocity of the baseball is given as 100 ft/s at an angle of 45° with respect to the ground.

The horizontal component of velocity can be found using the equation: vx = v * cos(θ), where v is the initial velocity and θ is the angle.

The vertical component of velocity can be found using the equation: vy = v * sin(θ).

Once we have the vertical component of velocity, we can use the equation h = vy2 / (2g) to find the maximum height.
Substituting the given values, we have:
h = (100 * sin(45°))2 / (2 * 32)

Calculating this will give us the maximum height of the baseball above its initial position.

Answer:

The ball covered the distance in 2 second is 20 m.

(b) is correct option.

Explanation:

Given that,

Time = 2 sec

Using equation of motion

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

Where, s = height

u = initial velocity

g = acceleration due to gravity

t = time

Put the value in the equation

[tex]s = 0+\dfrac{1}{2}\times9.8\times2^2[/tex]

[tex]s = 19.6\ approx\ 20\ m[/tex]

Hence, The ball covered the distance in 2 second is 20 m.

Answer:

Part a)

55.3 cm

Part b)

41.5 cm

Explanation:

Pipe A is open at both ends so the fundamental frequency of this pipe is given as

[tex]f_o = \frac{V}{2L}[/tex]

here we know that

V = 343 m/s

[tex]f_o = 310 Hz[/tex]

now we have

[tex]310 = \frac{343}{2L}[/tex]

[tex]L = 55.3 cm[/tex]

Now we also know that second harmonic of pipe A and third harmonic of pipe B has same frequency

so we will have

[tex]\frac{2V}{2L_a} = \frac{3V}{4L_b}[/tex]

[tex]L_b = \frac{3}{4}L_a[/tex]

[tex]L_b = \frac{3}{4}(55.3) = 41.5 cm[/tex]

Answer:

vf = (m1*v1i)/(m1 + m2)

Explanation:

take * as multiplication

from the conservation of linear momentum, we know that:

sum of pi = sum of pf,   where sum of pi = m1*v1i + m2*v2i

                                                              pf = vf(m1 + m2) since the masses stick                        

                                                                                           together.

v2i = o, since mass m2 is initially stationary.

m1*v1i + m2*(0) = vf(m1 + m2)

m1*v1i = vf(m1 + m2)

therefore

vf = m1*v1i/(m1 + m2)

Answer:

[tex]q_1 = 6.22 \times 10^{-11} C[/tex]

[tex]q_2 = -6.22 \times 10^{-11} [/tex]

Explanation:

Force on the charge Q = -1.85 mC is along the line joining the two charges

so here we can say that net force on it is given by

[tex]F = ma[/tex]

[tex]F = (0.00550)(314)[/tex]

[tex]F = 1.727 N[/tex]

Now this is the force due to two charges which are in same magnitude but opposite sign

so this force is given as

[tex]F = 2\frac{kq_1q}{r^2} cos\theta[/tex]

here we know that

[tex]F = 2\frac{(9\times 10^9)(1.85 \times 10^{-3})(q)}{0.03^2}cos\theta[/tex]

here we know that

[tex]cos\theta = \frac{2.25}{3} = 0.75[/tex]

now we have

[tex]1.727 = 2\frac{(9\times 10^9)(1.85 \times 10^{-3})(q)}{0.03^2}(0.75)[/tex]

[tex]1.727 = 2.775 \times 10^{10} q[/tex]

[tex]q = 6.22 \times 10^{-11} C[/tex]

Answer:

a)

3.25 x 10⁵ m/s

b)

8.7 x 10²⁰ N

Explanation:

(a)

w = angular speed of the star = 1.8 x 10⁻¹⁵ rad/s

r = radius of the orbit = 1.9 x 10⁴ ly =  1.9 x 10⁴ (9.5 x 10¹⁵) m = 18.05 x 10¹⁹ m

tangential speed of the star is given as

v = r w

v = (18.05 x 10¹⁹) (1.8 x 10⁻¹⁵)

v = 32.5 x 10⁴ m/s

v = 3.25 x 10⁵ m/s

b)

m = mass of the star = 1.48 x 10³⁰ kg

Net force on the star to keep it moving is given as

F = m r w²

F = (1.48 x 10³⁰) (18.05 x 10¹⁹) (1.8 x 10⁻¹⁵)²

F = 8.7 x 10²⁰ N

Answer:

a) k  = 191.67 N\m

b) V = 22 m/s

c) t = 3.83s

d) 17.36m

e) 45.89 m

Explanation:

given:

F = 23 N

x = 12 cm = 0.12 m

mass of projectile, m = 5.7g = 5.7×10⁻³ kg

a) For a spring

F = kx

where,

F = Applied force

k = spring constant

x = change in in spring length

thus, we have

23 = k×0.12

or

k = 23/0.12

⇒ k  = 191.67 N\m

b) From the conservation of energy between the start and the point of interest we have

[tex]\frac{1}{2}\times kx^{2}=\frac{1}{2}mV^{2}[/tex]

where,

V = velocity of the projectile at 1.37 m above the floor

[tex]\frac{1}{2}\times 191.67\times 0.12^{2}=\frac{1}{2}5.7\times 10^{-3}V^{2}[/tex]

V = 22 m/s

c) time in air (t)

applying the Newtons's equaton of motion

[tex]h = Vsin\Theta\times  t +\frac{1}{2}a_{y}t^{2}[/tex]

substituting the values in the above equation we get

[tex]-1.37 = 22sin57^{\circ} \times  t +\frac{1}{2}(-9.8)t^{2}[/tex]

or

[tex]4.9t^{2}-18.45t-1.37 = 0[/tex]

solving the qudratic equation for 't', we get

t = 3.83s

d) For maximum height ([tex]H_{max}[/tex])

we have from the equations of projectile motion

[tex]H_{max} =\frac{V^{2}sin^{2}\theta }{2g}[/tex]

substituting the values in the above equation we get

[tex]H_{max} =\frac{22^{2}sin^{2}57^{\circ} }{2\times 9.8}[/tex]

or

[tex]H_{max} =17.36 m[/tex]

the height with respect to the ground surface will be = 17.36m + 1.37 m 18.73m

e)For horizontal distance traveled (R) we have the formula

[tex]R = Vcos\theta\times t[/tex]

substituting the values in the above equation we get

[tex]R =22\times cos57^{\circ} \times 3.83[/tex]

or

[tex]R =45.89 m[/tex]

To calculate the potential difference VA - VB, first determine the displacement from point A to B. This results in a vector, which you dot product with the electric field vector to find the work done. This work done is the potential difference.

To compute the potential difference VA - VB, we first need to determine the displacement from point A to point B. This is determined by subtracting the coordinates of point A from point B, so (5 - 2)i + (7 - 3)j = 3i + 4j (meters). The next step involves calculating the work done by the electric field in moving a unit positive charge from A to B, which is obtained by taking the dot product of the displacement and the electric field vectors. Hence, the work done is (3i + 4j) . (4i + 3j) = 24 Joules/Coulomb. Now, since potential difference is defined as the work done per unit charge in moving a positive charge from one point to another, the potential difference VA - VB in this case would just be equal to this work done. So VA - VB = 24 Volts.

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Answer:

4.2 m/s²

Explanation:

m = combined mass of car and driver = 869 kg

a = maximum acceleration reached by the car with driver in it = 5.6 m/s²

Force provided by the engine is given as

[tex]F_{eng}[/tex] = ma

[tex]F_{eng}[/tex] = (869) (5.6)

[tex]F_{eng}[/tex] = 4866.4 N

M = Combined mass of car, driver and his three friends = 869 + 300 = 1169 kg

a' = maximum acceleration reached with friends in the car = ?

Force provided by the engine remains same and is then given as

[tex]F_{eng}[/tex] = M a'

4866.4 = (1169) a'

a' = 4.2 m/s²

Answer:

Electric force, [tex]F=2.88\times 10^{-14}\ N[/tex]

Explanation:

It is given that,

Magnitude of electric field, [tex]E=6\times 10^4\ N/C[/tex]

Charge, [tex]q=4.8\times 10^{-19}\ C[/tex]

The electric field is directed parallel to the positive y axis. We need to find the  force on the charge particle. It is given by :

[tex]F=q\times E[/tex]

[tex]F=4.8\times 10^{-19}\ C\times 6\times 10^4\ N/C[/tex]

[tex]F=2.88\times 10^{-14}\ N[/tex]

So, the electric force on the charge is [tex]2.88\times 10^{-14}\ N[/tex]. Hence, this is the required solution.

Answer:

[tex]E' = \frac{E}{8}[/tex]

Explanation:

As we know that that electric field inside the solid non conducting sphere is given as

[tex]\int E.dA = \frac{q_{en}}{\epsilon_0}[/tex]

[tex]\int E.dA = \frac{\frac{Q}{R^3}r_1^3}{\epsilon_0}[/tex]

[tex]E(4\pi r_1^2) = \frac{Qr_1^3}{R^3 \epsilon_0}[/tex]

so electric field is given as

[tex]E = \frac{Qr_1}{4\pi \epsilon_0 R^3}[/tex]

now if another sphere has same charge but twice of radius then the electric field at same position is given as

[tex]E' = \frac{Qr_1}{4\pi \epsilon_0 (2R)^3}[/tex]

so here we have

[tex]E' = \frac{E}{8}[/tex]

Final answer:

The electric field at distance r1 from the center of a uniformly charged sphere with charge Q and radius 2R would be one-eighth of the field when the charge is within a sphere with radius R.

Explanation:

To find the electric field at a distance r1 from the center of a uniformly charged sphere, we can use Gauss's law. Inside a uniformly charged sphere, the electric field is proportional to the distance from the center because only the charge enclosed by a Gaussian surface contributes to the field at a point. The field inside the sphere can be described using the formula E = (Qr) / (4πε₀R³), where Q is the total charge, r is the distance from the center, R is the radius of the sphere, and ε0 is the permittivity of free space.

When the same total charge Q is distributed uniformly throughout a sphere of radius 2R, for a point at a distance r1 inside the sphere, which is still less than 2R, the electric field magnitude would be calculated with the new radius. Since the enclosed charge within the distance r1 would be the same and the volume of the larger sphere is greater, the electric field at r1 would be one-eighth of its original value, due to the volume of the sphere increasing by eight times when the radius is doubled (since the volume is proportional to the cube of the radius). Therefore, the new electric field magnitude at distance r1 would be E/8.

Answer:

3. Is 180◦ out of phase with the original wave at the end.

Explanation:

Here when wave is reflected by the rigid boundary then due to the rigidly bounded particles at the end or boundary they have tendency not to move and remains fixed at their position.

Due to this fixed position we can say when wave reach at that end the particles will not move and they apply equal and opposite force at the particles of string

Due to this the reflected wave is transferred back into the string in opposite phase with respect to the initial wave

so here correct answer will be

3. Is 180◦ out of phase with the original wave at the end.

Answer:

a) P = 44850 N

b) [tex]\delta l =0.254\ mm[/tex]

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

[tex]\sigma=\frac{\textup{Load}}{\textup{Area}}[/tex]

on substituting the values, we get

[tex]345\times10^6=\frac{\textup{Load}}{0.00013}[/tex]

or

Load, P = 44850 N

Hence the maximum load that can be applied is 44850 N = 44.85 KN

b)The deformation ([tex]\delta l[/tex]) due to an axial load is given as:

[tex]\delta l =\frac{PL}{AE}[/tex]

on substituting the values, we get

[tex]\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}[/tex]

or

[tex]\delta l =0.254\ mm[/tex]

The maximum load that may be applied to the specimen without causing plastic deformation is 44850 N. The maximum length to which the specimen can be stretched without causing plastic deformation is 76.254 mm.

To find the maximum load that may be applied to the specimen without causing plastic deformation, we need to calculate the stress.

Stress = Force / Area

Where the force can be calculated by multiplying the stress with the cross-sectional area of the specimen.

Hence, maximum load = Stress × Area = 345 MPa × 130 mm² = 44850 N

To find the maximum length to which the specimen can be stretched without causing plastic deformation, we need to calculate the strain.

Strain = Change in length / Original length

Maximum length = Original length + (Strain × Original length) = 76 mm + (345 MPa / 103 GPa × 76 mm) = 76 mm + 0.254 mm = 76.254 mm

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