It is desiredto FDM 30voice channels (each with a bandwidth of 4.5KHz) along with a guard band of 0.8KHz. Ignoring a guard band before the first channel and the one after the last channel, what is the total bandwidth, in KHz,required for such a system.

Answer:

The radius of the sphere is 3.6 m.

Explanation:

Given that,

Potential of first sphere = 450 V

Radial distance = 7.2 m

If the potential of sphere =150 V

We need to calculate the radius

Using formula for potential

For 450 V

[tex]V=\dfrac{kQ}{r}[/tex]

[tex]450=\dfrac{kQ}{r}[/tex]....(I)

For 150 V

[tex]150=\dfrac{kQ}{r+7.2}[/tex]....(II)

Divided equation (I) by equation (II)

[tex]\dfrac{450}{150}=\dfrac{\dfrac{kQ}{r}}{\dfrac{kQ}{r+7.2}}[/tex]

[tex]3=\dfrac{(r+7.2)}{r}[/tex]

[tex]3r=r+7.2[/tex]

[tex]r=\dfrac{7.2}{2}[/tex]

[tex]r=3.6\ m[/tex]

Hence, The radius of the sphere is 3.6 m.

The radius of the sphere whose surface has a potential difference of  450 V is 3.6 m.

We know that the potential difference can be written as,

[tex]V = k\dfrac{Q}{R}[/tex]

We know that at  R= R, Potential difference= 450 V,

[tex]450 = k\dfrac{Q}{R}[/tex]

Also, at R = (R+7.2), Potential difference = 150 V,

[tex]150 = k\dfrac{Q}{(R+7.2)}[/tex]

Taking the ratio of the two,

[tex]\dfrac{450}{150} = \dfrac{kQ}{R} \times \dfrac{(R+7.2)}{kQ}\\\\\dfrac{450}{150} = \dfrac{(R+7.2)}{R}\\\\R = 3.6\ m[/tex]

Hence, the radius of the sphere whose surface has a potential difference of  450 V is 3.6 m.

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Answer:

Contraction of Chicago concrete walkway = 8.64 x 10⁻³ m

Explanation:

Thermal expansion or compression is given by ΔL = LαΔT

Here Length of Chicago concrete walkway, L = 18 m

         Change in temperature, ΔT = (-16 - 24) = -40 °C

         Coefficient of thermal expansion for concrete, α = 12 x 10⁻⁶ °C⁻¹

Substituting

    ΔL = LαΔT = 18 x 12 x 10⁻⁶ x (-40) = -8.64 x 10⁻³ m

Contraction of Chicago concrete walkway = 8.64 x 10⁻³ m

Answer:

(a): The bird speed after swallowing the insect is V= 4.83 m/s

(b): The impulse on the bird is I= 0.3 kg m/s

(c): The force between the bird and the insect is F= 20 N

Explanation:

ma= 0.3 kg

va= 6 m/s

mb= 0.01kg

vb= 30 m/s

(ma*va - mb*vb) / (ma+mb) = V

V= 4.83 m/s (a)

I= mb * vb

I= 0.3 kg m/s  (b)

F*t= I

F= I/t

F= 20 N (c)

This physics problem uses the principle of conservation of momentum to calculate the bird's speed after swallowing the insect, the impulse experienced by the bird, and the force between the bird and the insect.

This is a physics problem relating to the conservation of momentum. Let's start by defining some facts, where m bird = 0.3 kg and v bird = 6.0 m/s are the mass and speed of the bird before the incident and m insect = 0.01 kg and v insect = 30 m/s are the mass and speed of the insect.

a. To find the bird's speed immediately after swallowing the insect, we need to apply the conservation of momentum principle: initial total momentum = final total momentum, which can be written as m bird * v bird + m insect * v insect = (m bird + m insect) * v final.

b. The impulse on the bird equals the change in momentum of the bird, thus equals to the final momentum of the bird - initial momentum of the bird.

c. The force between the bird and the insect is obtained from the definition of impulse: Force * time = impulse, or Force = Impulse/time.

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Answer:

E =  ρ ( R1²) / 2 ∈o R

Explanation:

Given data

two cylinders are parallel

distance = d

radial distance = R

d < (R2−R1)

to find out

Express answer in terms of the variables ρE, R1, R2, R3, d, R, and constants

solution

we have two parallel cylinders

so area is 2 [tex]\pi[/tex] R × l

and we apply here gauss law that is

EA = Q(enclosed) / ∈o   ......1

so first we find  Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( [tex]\pi[/tex] R1² × l )

so put all value in equation 1

we get

EA = Q(enclosed) / ∈o

E(2 [tex]\pi[/tex] R × l)  = ρ ( [tex]\pi[/tex] R1² × l ) / ∈o

so

E =  ρ ( R1²) / 2 ∈o R

Final answer:

The resulting electric field in the specified region can be calculated using Gauss' Law. The equation for the electric field in that region is [tex]E = 2\pi R_1^2ho_E[/tex].

Explanation:

The resulting electric field in the region R>R3 is:

[tex]E = 2\pi R_1^2ho_E[/tex]

where R_1 is the radius of the inner cylinder, and ρ_E is the charge density. This expression is obtained by applying Gauss' Law for the region where R1 < r < R2.

Given:

mass of satellite, m = 1710 kg

altitude, h = [tex]2.6\times 10^{6} m[/tex]

G =  [tex]6.67\times 10^{-11} [/tex]

we know

mass of earth, [tex]M_{E}[/tex] =  [tex]5.972\times 10^{24} kg[/tex]

Here, according to question we will consider

[tex]2M_{E}[/tex] =  [tex]11.944\times 10^{24} kg[/tex]

radius of earth,  [tex]R_{E}[/tex] =  [tex]6.371\times 10^{6} m[/tex]

Formulae Used and replacing [tex]M_{E}[/tex] by  [tex]2M_{E}[/tex] :

1). [tex]v = \sqrt{\frac{2GM_{E}}{R_{E} + h}}[/tex]

2). [tex]T = \sqrt{\frac{4\pi ^{2}(R_{E} + h)^{3}}{2GM_{E}}}[/tex]

3). [tex]KE = \frac{1}{2}mv^{2}[/tex]

4). [tex]Total Energy, E = -\frac{2GM_E\times m}{2(R_{E} + h)}[/tex]

where,

v = orbital velocity of satellite

T = time period

KE = kinetic energy

Solution:

Now, Using Formula (1), for orbital velocity:

 [tex]v = \sqrt{\frac{6.67 \times 10^{-11} \times 11.944 \times 10^{24}}{6.371 \times 10^{6} + 2.6 \times 10^{6}}[/tex]

v =  [tex]9.423 \times 10^{3}[/tex]  m/s

Using Formula (2) for time period:

[tex]T = \sqrt{\frac{4\pi ^{2}(6.371\times 10^{6} + 2\times 10^{6})^{3}}{6.67\times 10^{-11}\times 9.44\times 10^{24}}}[/tex]

[tex]T = 6.728\times 10^{3} s[/tex]

Now, Using Formula(3) for kinetic energy:

[tex]KE = \frac{1}{2}(9.44\times 10^{24})(9.42\times 10^{3})^{2}[/tex]

[tex]KE = \frac{1}{2}(1710)(9.42\times 10^{3})^{2} = 7.586\times 10^{10} J[/tex]

Now, Using Formula(4) for Total energy:

[tex]E = -\frac{6.67\times 10^{-11}\times 9.44\times 10^{24}\times 1710}{2( 6.371\times 10^{6} + 2.6\times 10^{6})}[/tex]

[tex]E = - 7.59\times 10^{10} J[/tex]

The amplitude of the oscillating mass is [tex]\(A = 4.05 \, \text{cm}\).[/tex]

To calculate the amplitude  A of the oscillating mass, we can use the equations of motion for simple harmonic motion (SHM). In SHM, the displacement [tex]\( y(t) \)[/tex] of the mass at time [tex]\( t \)[/tex] is given by:

[tex]\[ y(t) = A \sin(\omega t + \phi) \][/tex]

where:

- [tex]\( A \)[/tex]) is the amplitude,

- [tex]\( \omega \)[/tex] is the angular frequency, and

- [tex]\( \phi \)[/tex] is the phase angle.

Given that at [tex]\( t = t_0 = 0 \)[/tex], the mass is at [tex]\( y_0 = 4.05 \)[/tex] cm and moving upward at velocity [tex]\( v_0 = +4.12 \)[/tex] m/s, we can find the amplitude A

At [tex]\( t = 0 \),[/tex] we have:

[tex]\[ y(0) = A \sin(\phi) = y_0 = 4.05 \, \text{cm} \][/tex]

And also:

[tex]\[ v(0) = \omega A \cos(\phi) = v_0 = +4.12 \, \text{m/s} \][/tex]

To find [tex]\( A \),[/tex]A we'll use the fact that at [tex]\( t = 0 \)[/tex], the mass is at its maximum displacement, which means the velocity is zero at [tex]\( t = 0 \).[/tex] This gives us:

[tex]\[ \omega A \cos(\phi) = 0 \]\[ \cos(\phi) = 0 \][/tex]

Since [tex]\( \cos(\phi) = 0 \) when \( \phi = \frac{\pi}{2} \) or \( \phi = \frac{3\pi}{2} \), we'll consider \( \phi = \frac{\pi}{2} \).[/tex]

Now, from the equation [tex]\( A \sin(\phi) = y_0 \),[/tex] we can find [tex]\( A \):[/tex]

[tex]\[ A \sin\left(\frac{\pi}{2}\right) = 4.05 \]\[ A = 4.05 \, \text{cm} \][/tex]

So, the amplitude[tex]\( A \)[/tex] of the oscillating mass is [tex]\( 4.05 \, \text{cm} \).[/tex]

Answer:

0.5 Ns

Explanation:

When a large force acting on a body for a very small time it is called impulsive force.

Impulse = force × small time

Impulse = 25 × 0.02 = 0.5 Ns

It is a vector quantity

Answer:

[tex]1.88\cdot 10^{-5} T[/tex], inside the plane

Explanation:

We need to calculate the magnitude and direction of the magnetic field produced by each wire first, using the formula

[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]

where

[tex]\mu_0[/tex] is the vacuum permeability

I is the current

r is the distance from the wire

For the top wire,

I = 4.00 A

r = d/2 = 0.105 m (since we are evaluating the field half-way between the two wires)

so

[tex]B_1 = \frac{(4\pi\cdot 10^{-7})(4.00)}{2\pi(0.105)}=7.6\cdot 10^{-6}T[/tex]

And using the right-hand rule (thumb in the same direction as the current (to the right), other fingers wrapped around the thumb indicating the direction of the magnetic field lines), we find that the direction of the field lines at point P is inside the plane

For the bottom wire,

I = 5.90 A

r = 0.105 m

so

[tex]B_2 = \frac{(4\pi\cdot 10^{-7})(5.90)}{2\pi(0.105)}=1.12\cdot 10^{-5}T[/tex]

And using the right-hand rule (thumb in the same direction as the current (to the left), other fingers wrapped around the thumb indicating the direction of the magnetic field lines), we find that the direction of the field lines at point P is also inside the plane

So both field add together at point P, and the magnitude of the resultant field is:

[tex]B=B_1+B_2 = 7.6\cdot 10^{-6} T+1.12\cdot 10^{-5}T=1.88\cdot 10^{-5} T[/tex]

And the direction is inside the plane.

The magnitude and direction of the net magnetic field generated by the two wires will be [tex]1.55\times 10^{-8}[/tex].

What is magnetic field strength?

The number of magnetic flux lines on a unit area passing perpendicular to the given line direction is known as induced magnetic field strength .it is denoted by B.

[tex]B = \frac{u_00I}{2\pi r}[/tex]

(I) is the current

r is the distance from the probe

B is the induced magnetic field

r denotes the distance between the wire and the object.

[tex]u_0[/tex] is the permeability to vacuum

For magnetic field in the top wire

Given

0.105 m = r = d/2

[tex]B_1= \frac{4\pi\times10^{-7}\times4.00}{2\pi \times 0.105}[/tex]

[tex]\rm{B_1=7.6\times 10^{-6}}[/tex] T

As the current to the left other fingers wrapped around the thumb showing the direction of the magnetic field lines.

Given data for bottom wire

I= 5.90 A

r = 0.105 m r = 0.105 m r = 0.105 m

[tex]B_2= \frac{4\pi\times10^{-7}\times5.90}{2\pi \times 0.105}[/tex]

[tex]\rm{B_2=1.12 \times 10^-5[/tex] T

As the current to the left other fingers wrapped around the thumb showing the direction of the magnetic field lines. The field lines direction at point P is also inside the plane.

The megnitude of the resultant magnetic field be the sum of both the field

[tex]\rm{B=B_1+B_2}[/tex]

[tex]\rm{B=7.6\times 10^{-6}+ 1.12\times 10^{-5}}[/tex]

[tex]B = 1.88\times 10^{-5}[/tex] T

Hence The megnitude of the resultant magnetic field be the sum of both the fields  [tex]1.88\times 10^{-5}[/tex]T.

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Answer:

The average total energy density of this electromagnetic wave is [tex]72.5\ \mu\ J/m^3[/tex].

Explanation:

Given that,

Magnetic field [tex]B = 13.5\mu T[/tex]

We need to calculate the average total energy density

Using formula of energy density

[tex]Energy\ density =\dfrac{S}{c}[/tex]....(I)

Where, S = intensity

c = speed of light

We know that,

The intensity is given by

[tex]S = \dfrac{B^2c}{2\mu_{0}}[/tex]

Put the value of S in equation (I)

[tex]Energy\ density =\dfrac{\dfrac{B^2c}{2\mu_{0}}}{c}[/tex]

[tex]Energy\ density = \dfrac{(13.5\times10^{-6})^2}{2\times4\pi\times10^{-7}}[/tex]

[tex]Energy\ density = 0.0000725\ J/m^3[/tex]

[tex]Energy\ density = 72.5\times10^{-6}\ J/m^3[/tex]

[tex]Energy\ density = 72.5\ \mu\ J/m^3[/tex]

Hence, The average total energy density of this electromagnetic wave is [tex]72.5\ \mu\ J/m^3[/tex].

Answer:

A positive charge experiences a force in the direction of an electric field.

Explanation:

Electric field is defined as the electric force acting per unit positive charge. Mathematically, it is given by :

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

We know that like charges repel each other while unlike charges attract each other. The direction of electric field is in the direction of electric force. For a positive charge the field lines are outwards and for a negative charge the electric field lines are inwards.

So, the correct option is (b) "A positive charge experiences a force in the direction of an electric field".

Answer:

[tex]\rho = \rho_0 r[/tex]

Explanation:

As we know by Gauss law

[tex]\int E. dA = \frac{q}{\epsilon}[/tex]

here we know that

[tex]E = \frac{\rho_0 r^2}{4\epsilon}[/tex]

so here we have

[tex](\frac{\rho_0 r^2}{4\epsilon})(4\pi r^2) = \frac{(\int\rho dV)}{\epsilon}[/tex]

now we have

[tex]\frac{\pi \rho_0 r^4}{\epsilon} = \frac{(\int\rho dV)}{\epsilon}[/tex]

[tex]\pi \rho_0 r^4 = (\int\rho dV)[/tex]

now differentiate both sides by volume

[tex]\frac{d(\pi \rho_0 r^4)}{dV} = \rho [/tex]

[tex]\frac{\pi \rho_0 4r^3 dr}{4\pi r^2 dr} = \rho[/tex]

[tex]\rho = \rho_0 r[/tex]

Answer:· Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed on a screen that is 2.74 m from the grating. In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.16 m. What is the difference between these wavelengths? . I know to apply the equation din(theta) = m*wavelength but I'm not sure how to find all the missing variables or to get the difference in wavelengths

Answer:

1.176m

Explanation:

Local ambient pressure(P1) = 100 kPa

Absolute pressure(P2)=260kPa

Net pressure=absolute pressure-local ambient absolute pressure

Net pressure=P1(absolute pressure)-P2(local ambient absolute pressure)

Net pressure=260-100=160kPa

Pressure= ρgh

160kPa=13600*10*h

h=[tex]\frac{160000}{136000}[/tex]

h=1.176m

To calculate the diffusion flux at 1000˚C for the same concentration gradient, use the Arrhenius equation.

J ≈ 2.4 × 10-12 kg/m2-s

To calculate the diffusion flux at 1000˚C (1273 K) for the same concentration gradient, we can use the Arrhenius equation:

J = J0 * exp(-Q/RT)

Where J is the diffusion flux, J0 is the pre-exponential factor, Q is the activation energy for diffusion, R is the gas constant, and T is the absolute temperature.

Given that the diffusion flux at 1220˚C (1493 K) is 7.8 × 10-8 kg/m2-s and the activation energy for diffusion is 145,000 J/mol, we can calculate the diffusion flux at 1000˚C as:

J = (7.8 × 10-8) * exp(-145000/(8.314*1273))

J ≈ 2.4 × 10-12 kg/m2-s

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To calculate the steady-state diffusion flux at a different temperature, use the Arrhenius equation to find the diffusion coefficients at the two temperatures, and find the ratio based on the fact that the diffusion flux is proportional to the diffusion coefficient when the concentration gradient is constant. Input the known values into the equation to solve for the unknown diffusion flux.

The steady-state diffusion through a metal plate can be calculated using the Arrhenius equation, which relates the diffusion coefficient (D) to temperature. The equation is D = D0e^-(Q/RT), where D0 is the pre-exponential factor, Q is the activation energy for diffusion, R is the gas constant and T is the temperature in K.

Given that the diffusion flux (J) is defined as J = -D×(dc/dx), where dc/dx is the concentration gradient. We can find that when the concentration gradient remains the same, the ratio of the two diffusion fluxes at different temperatures can be represented as J1/J2 = D1/D2.

Substitute the Arrhenius equation into the ratio, we get J1/J2 = e^(Q/R)×(1/T1-1/T2). Then you can use the given values, namely Q = 145,000 J/mol, R = 8.314 J/(mol×K), and temperatures T1 = 1493K , T2 = 1273K, as well as the known J1, to calculate J2.

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Answer:

The input current is 0.05 A.

Explanation:

Given that,

Output voltage = 7.50 V

Output current = 1.6 A

Input voltage = 240 v

We need to calculate the input current

The efficiency is 100% so.

Input power = output power

[tex]\dfrac{V_{i}}{V_{o}}=\dfrac{I_{o}}{I_{i}}[/tex]

[tex]\dfrac{240}{7.50}=\dfrac{1.6}{I_{i}}[/tex]

[tex]I_{i}=\dfrac{1.6\times7.50}{240}[/tex]

[tex]I_{i}=0.05\ A[/tex]

Hence, The input current is 0.05 A.

The input current Ip for the plug-in transformer for a laptop computer in amps is 0.05 A.

The primary winding of the transformer converts the electric power into the magnetic power, and the secondary winding of the transformer converts it back into the required electric power.

The ratio of the input voltage to the output voltage is equal to the ratio of output current to the input current of the transformer (for 100 percent efficiency). Therefore

[tex]\dfrac{V_i}{V_o}=\dfrac{I_o}{I_i}[/tex]

The input and output voltage of the transformer is 240 V and 7.50 V respectively and the output current is 1.6 amp.

As, the efficiency of the transformer is 100 percent. Thus, put the values in the above formula as,

[tex]\dfrac{240}{7.50}=\dfrac{1.6}{I_i}\\I_i=0.05\rm A[/tex]

Thus, the input current Ip for the plug-in transformer for a laptop computer in amps is 0.05 A.

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The specific heat capacity of the alloy can be calculated using the formula q = m * c * ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. We can use the principle of conservation of heat to calculate the specific heat capacity of the alloy.

The specific heat capacity of the alloy can be calculated using the formula: q = m * c * ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the heat transferred to the water can be calculated by q_water = m_water * c_water * ΔT_water. Using the principle of conservation of heat, q_water = -q_alloy. We can rearrange the equation to solve for c_alloy: c_alloy = -q_water / (m_alloy * ΔT_alloy)

Plugging in the values, we have: c_alloy = -q_water / (m_alloy * (ΔT_final - ΔT_initial)). This gives us the specific heat capacity of the alloy.

In this case, the final temperature of the system is 31.10°C, which means that ΔT_alloy = 31.10°C - 93.00°C = -61.90°C. Plugging in the values, we get: c_alloy = -(-50.0 g * 4.184 J/g °C * (31.10°C - 22.00°C)) / (21.6 g * -61.90°C). After calculating, you will find the specific heat capacity of the alloy.

The specific heat capacity of the alloy can be calculated using the formula: q = m × c × ΔT.

The specific heat capacity of the alloy can be calculated using the formula:

q = m × c × ΔT

where q is the heat transferred, m is the mass of the alloy, c is the specific heat capacity of the alloy, and ΔT is the change in temperature. In this case, the heat transferred to the alloy is equal to the heat transferred from the water:

m1 × c1 × ΔT1 = m2 × c2 × ΔT2

Substituting the given values, we can solve for c2 to find the specific heat capacity of the alloy:

c2 = (m1 × c1 × ΔT1) / (m2 × ΔT2)

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Answer:

a) 0.0693 m

b) Work done = 8.644 J

Explanation:

Given:

Spring constant, k = 3600 N/m

Radius of the piston, r = 0.028 m

Now, we know that the atmospheric pressure at STP = 1.01325 × 10⁵ Pa  = 101325 Pa

Now,

The force ([tex]F_P[/tex]) due to the atmospheric pressure on the piston will be:

[tex]F_P[/tex] = Pressure × Area of the piston

on substituting the values we get,

[tex]F_P[/tex] = 101325 × πr²

F = 101325 × π × (0.028)² = 249.56 N

also,

Force on spring is given as:

F = kx

where,

x is the displacement in the spring

 on substituting the values we get,

 249.56 N = 3600N/m × x

or

x = 0.0693 m

thus, the compression in the spring will be = 0.0693 m

b) Applying the concept of conservation of energy

we have,

Work done by the atmospheric pressure in compressing the spring = Potential energy gained  by the spring

mathematically,

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

 on substituting the values we get,

[tex]W = \frac{1}{2}\times 3600\times (0.0693)^2[/tex]

W = 8.644 J

a) x = 0.0693 m

b) W = 8.644 J

Given :

Spring constant, K = 3600 N/m

Radius of the piston, r = 0.028 m

Solution :

Now the atmospheric pressure at STP = 1.01325 × 10⁵ Pa  = 101325 Pa

Force due to the atmospheric pressure on the piston is,

Force = Pressure × Area of the piston

on substituting the values we get,

[tex]\rm F_P = 101325\times \pi r^2[/tex]

[tex]\rm F_P = 249.56\;N[/tex]

a) We know that the force on spring is given by,

F = Kx

where, k is spring constant and x is the displacement in the spring.

[tex]249.56 = 3600\times x[/tex]

[tex]\rm x = 0.0693\;m[/tex]

b) We know that the Work Done is given by,

[tex]\rm W= \dfrac{1}{2} k x^2[/tex]

[tex]\rm W = 0.5\times 3600\times (0.0693)^2[/tex]

W = 8.644 J

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Answer:

To the nearest integer, the penny's speed in m/s be right as it reaches the ground if it was dropped from rest = 91 m/s

Explanation:

We have equation of motion

S = ut + 0.5at²

Here u = 0, a = g  and t = 9.3 s

We have equation of motion v = u +at

Substituting

       v = u +at

       v = 0 + 9.8 x 9.3 = 91.14 m/s

To the nearest integer, the penny's speed in m/s be right as it reaches the ground if it was dropped from rest = 91 m/s

Final answer:

The rate constant for this reaction at 384.7 K is 7.945 x 10^-4 cm^3/(molecule·s).

Explanation:

The rate constant, denoted by k, can be determined using the Arrhenius equation: k = Ae^-Ea/RT, where A is the frequency factor, Ea is the activation energy, R is the ideal gas constant, and T is the temperature in Kelvin. To find the rate constant at 384.7 K, we first need to convert Ea from kJ/mol to J/mol by multiplying it by 1000, giving us 12,900 J/mol. Plugging in the values A = 4.23 x 10^-12 cm^3/(molecule·s), Ea = 12,900 J/mol, and R = 8.314 J/mol/K, into the Arrhenius equation, we can calculate k as follows:

k = (4.23 x 10^-12 cm^3/(molecule·s)) * e^(-12,900 J/mol / (8.314 J/mol/K * 384.7 K))

k = 7.945 x 10^-4 cm^3/(molecule·s)

C. [tex]E_{k}=2.03x10^{11}J[/tex]

The kinetic energy of a body is the ability to perform work due to its movement given by the equation [tex]E_{k}=\frac{1}{2}mv^{2}[/tex].

To calculate the kinetic energy of a rocket with mass 15000kg and speed of 5200m/s:

[tex]E_{k}=\frac{1}{2}(15000kg)(5200m/s)^{2}=202800000000J=2.03x10^{11}J[/tex]

Answer:

b) Projectile MOTION

Explanation:

SHM is periodic motion or to and fro motion of a particle about its mean position in a straight line

In this type of motion particle must be in straight line motion

So here we can say

a) Simple Pendulum : it is a straight line to and fro motion about mean position so it is a SHM

b) Projectile motion : it is a parabolic path in which object do not move to and fro about its mean position So it is not SHM

d) Spring Motion : it is a straight line to and fro motion so it is also a SHM

So correct answer will be

b) Projectile MOTION

Final answer:

Projectile motion is not a simple harmonic motion because it does not meet the conditions for SHM.

Explanation:

Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is proportional to the displacement. The three conditions that must be met to produce SHM are: a linear restoring force, a constant force constant, and no external damping forces. Based on these conditions, the answer to the question is (b) Projectile motion, as it does not meet the conditions for SHM. A projectile follows a parabolic path and does not have a linear restoring force.

Answer:

k = 212.55 newton per meter

Explanation:

A girl is standing on a trampoline. Her mass is 65 kg and she is able to jump 3 meters.

We have to find the spring constant.

Since by Hooke's law,

F = -kx

Where F = force applied by the spring

k = spring constant

x = displacement

And we know force applied by the spring will be equal to the weight of the girl.

So, F = mg

Therefore, (-mg) = -kx

65×(9.81) = k×(3)

k = [tex]\frac{(65)(9.81)}{3}[/tex]

k = 212.55 N per meter

Therefore, spring constant of the spring is 212.55 Newton per meter.

Answer:

The distance of bob when Alice starts running is 50 m.

Explanation:

Given that,

Speed v = 10.0 m/s

Time t = 5 sec

We need to calculate the distance

Using formula of distance

[tex]D=v\times t[/tex]

[tex]D=10\times5[/tex]

[tex]D=50\ m[/tex]

Hence, The distance of bob when Alice starts running is 50 m.

Answer:

60°

Explanation:

In the given question it is given that the two mirrors perpendicular to each other.

The light ray will reflect off the first mirror with an angle of 30° (∠i= ∠r) and then arrive at the second mirror. Using geometry of the two mirrors and the fact that the angle between the two of them is 90°, the incident angle for second mirror is 60°, which will again equal to the angle of reflection.

hence, angle of reflection of second mirror is 60°

The angle of reflection from the second surface is (b) 60◦

Two mirrors are at right angles to one another. A light ray is incident on the first at an angle of 30° with respect to the normal to the surface. What is the angle of reflection from the second surface?

(a) 30◦

(b) 60◦

(c) 45◦

(d) 53◦

(e) 75◦

Reflection is the direction change of a wavefront at interface between two different media. The light ray (it is a line (straight or curved) that is perpendicular to the light's wavefronts and its tangent is collinear with the wave vector) will reflect off of the first mirror with an equal angle  of 30◦  then it arrives at the second mirror. By using the geometry (branch of mathematics that concerned with questions of shape, size, relative position of figures, and the properties of space) of the two  mirrors and the angle between the two of them is 90◦ , the  incident angle (the angle between a ray incident on a surface and the line perpendicular to the surface at the point of incidence called the normal) for the second mirror is 60◦ that equal to the final angle of  reflection.

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To allow the electron to pass through the region without being deflected, the magnetic field should be equal and opposite to the electric field with a magnitude of 5.0 x 10^3 T in the -i^ direction.

The force experienced by an electron moving in a magnetic field is given by the formula F = qvB sin(θ), where q is the charge of the electron, v is its velocity, B is the magnetic field, and θ is the angle between the direction of velocity and the magnetic field. To allow the electron to pass through the region without being deflected, the magnetic force should be equal and opposite to the electric force. Since the electric field is in the j^ direction, the magnetic field should be in the -i^ direction with a magnitude of 5.0 x 10^3 T.

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To prevent the electron from being deflected, the magnetic field should be [tex]2.0 \times 10{^-4} T[/tex]. This is calculated by balancing the electric and magnetic forces acting on the electron.

To determine the magnetic field that allows an electron to pass through a region with perpendicular electric and magnetic fields without being deflected, we use the concept of force balance. When the forces from the electric field and the magnetic field are equal and opposite, the electron will move in a straight line without deflection. Given the electron's velocity[tex]v = 5.0 \times 10^7 m/s \hat i[/tex] and the electric field [tex]E = 10^4 V/m \hat j[/tex], we can use the formula:

[tex]qE = qvB[/tex]

Here, q is the charge of the electron, E is the magnitude of the electric field, v is the electron's velocity, and B is the magnetic field strength. Solving for B, we get:

[tex]B = E/v[/tex]

Plug in the given values:

[tex]B = 10^4 V/m / 5.0 \times 10^7 m/s[/tex]

This simplifies to:

[tex]B = 2.0 \times 10^{-4} T[/tex]

Therefore, the magnetic field required is [tex]2.0 \times 10^{-4} T[/tex].

•If a system gains 757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ. How much work is done is  - 581 kJ

•The correct statement is: Work was done by  the system

Let Change in internal energy ΔU = 176 kJ

Let Heat gained by the system (q) = 757 kJ

Using the  First law of thermodynamics

ΔU = q + w

Where:

ΔU  represent  change in internal energy

q represent  heat added to system and w is work done.

Let plug in the formula

176 kJ = 757 kJ + w

w = 176 kJ - 757 kJ

w= - 581 kJ

Based on the above calculation the negative sign means  that work is done by the system

Inconclusion:

•If a system gains 757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ. How much work is done is  - 581 kJ

•The correct statement is: Work was done by  the system

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The amount of work done by the system, based on given heat gain and change in internal energy is 581 kJ, meaning the work was done by the system.

The question asks about the amount of work done by or on a system in the field of thermodynamics. According to the first law of thermodynamics, the change in internal energy of a system (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W), or written as ΔU = Q - W. In this case, the heat added to your system was 757 kJ and the change in internal energy of the system was +176 kJ.

So we have: 176 kJ = 757 kJ - W. Subtracting 757 kJ from both sides of the equation would give us W = 757 kJ - 176 kJ. This results in the value of W = 581 kJ. Conclusively, since W is positive, we say that work was done by the system.

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Answer:

The dose absorbed by the person's body is 52600 mrem.

Explanation:

Given that,

Radiation = 52.6 rad

We need to calculate the absorbed dose

We know that,

The equivalent dose is equal to the absorbed radiation for beta source.

So, The patient is exposed 52.6 rad of radiation.

Therefore, The absorbed radiation is equal to the exposed 52.6 rad of radiation.

[tex]1 rad = \dfrac{1\ rem}{1000\ mrem}[/tex]

So, radiation absorbed = 52.6 rad

[tex]radiation\ absorbed =52.6\times1000\ mrem[/tex]

[tex]radiation\ absorbed = 52600\ mrem[/tex]

Hence, The dose absorbed by the person's body is 52600 mrem.

Answer:

The second natural frequency an for this system is 15710 rad/s.

(e) is correct option.

Explanation:

Given that,

Length = 2 m

Wave speed = 5000 m/s

We need to calculate the second natural frequency

Using formula of Time period

[tex]T = \dfrac{L}{v}[/tex]

[tex]T=\dfrac{2}{5000}\ s[/tex]

We know that,

The frequency is the reciprocal of time period.

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

[tex]f=\dfrac{5000}{2}[/tex]

[tex]f=2500\ Hz[/tex]

We know that,

[tex]1\ rad/s =\dfrac{1}{2\pi}\ Hz[/tex]

So, The frequency is in rad/s

[tex]f= 15710\ rad/s[/tex]

Hence, The second natural frequency an for this system is 15710 rad/s.

Answer:

Electric force between the charges, [tex]F=1.35\times 10^{10}\ N[/tex]

Explanation:

It is given that,

Charge 1, q₁ = 3 C

Charge 2, q₂ = 2 C

Distance between them, r = 2 m

We need to find the electric force between them. The formula for electric force is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

k is the electrostatic constant

[tex]F=9\times 10^9\times \dfrac{3\ C\times 2\ C}{(2\ m)^2}[/tex]

[tex]F=1.35\times 10^{10}\ N[/tex]

So, the force between the charges is [tex]1.35\times 10^{10}\ N[/tex]. Hence, this is the required solution.

Explanation:

It is given that,

Momentum of the photon, [tex]p=5.55\times 10^{-27}\ kg-m/s[/tex]

(a) We need to find the wavelength of this photon. It can be calculated using the concept of De-broglie wavelength.

[tex]\lambda=\dfrac{h}{p}[/tex]

h is the Planck's constant

[tex]\lambda=\dfrac{6.67\times 10^{-34}\ Js}{5.55\times 10^{-27}\ kg-m/s}[/tex]

[tex]\lambda=1.2\times 10^{-7}\ m[/tex]

or

[tex]\lambda=120\ nm[/tex]

(b) The wavelength lies in the group of ultraviolet rays. The wavelength of UV rays lies in between 400 nm to 10 nm.