The element hydrogen has the highest specific heat of all elements. At a temperature of 25°C, hydrogen’s specific heat capacity is 14300J/(kg K). If the temperature of a .34kg sample of hydrogen is to be raised by 25 K, how much heat will have to be transferred to the hydrogen?

Answer:

[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]

Explanation:

Given,

current in the wire I₁ = 1 A

                               I₂ = 1 A

distance between them, r = 1 m

using Force per unit length formula

[tex]\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi r}[/tex]

[tex]\mu_0 = magnetic\ permeability\ of\ free\ space = 4\pi\times 10^{-7}[/tex]

[tex]\dfrac{F}{l}=\dfrac{4\pi \times10^{-7}\times 1\times 1}{2\pi\times 1}[/tex]

[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]

Hence, the magnitude of force per unit length is equal to [tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]

Final answer:

The magnitude of the force per unit length each 1.0 A current-carrying wire experiences when separated by a distance of 1.0 m is 2 x 10⁻⁷ N/m, calculated using the formula for magnetic force between parallel conductors.

Explanation:

When discussing two parallel current-carrying wires, we're addressing the magnetic force that arises between them due to their currents. Specifically, when each wire carries a current of 1.0 A and they are placed 1.0 m apart, the force per unit length that each wire experiences can be calculated using the formula for the magnetic force between two parallel conductors. The equation is FE = (µ0 / 2π) × (I1I2 / r), where µ0 is the magnetic constant (4π x 10-7 T·m/A), I1 and I2 are the currents in the wires, and r is the distance between the wires. Given that each wire carries 1.0 A of current and the separation is 1.0 m, we can plug these values into the formula to find that FE = (4π x 10-7 T·m/A / 2π) × (1.0 A2 / 1.0 m), resulting in a force per unit length of 2 x 10-7 N/m.

Answer:

The diameter is  0.8376 m

Explanation:

Magnetic force is the force that is associated with the magnetic field, the magnitude of the magnitude force can be obtained using equation 1;

F = q v B  .....................................1,

where q is the magnitude of the charge of the particle =;

v is the  velocity and;

B is the magnetic field = 1.833 T ;

Here, the path of the charge is circular so the  force can be also considered as the centripetal force  which is represented in equation 2;

[tex]F_{c}[/tex] = m [tex]v^{2}[/tex] / r ....................................2,

m is the particle's mass = 2.00 x[tex]10^{-26}[/tex] kg

v is the speed of ion =  2.05% of 3.00 x [tex]10^{8}[/tex] = 0.0205 x 3.00 x [tex]10^{8}[/tex]

                                    = 6.15 x  [tex]10^{6}[/tex] m/s ;

Singly charged ion has a charge equal to the electron charge and the magnitude = 1.60217646 ×[tex]10^{-19}[/tex] C and;

r is the radius of the circular path.

to get  the diameter of the orbit  we equate equation 1 to 2 and isolate r in equation 2.

q v B = m [tex]v^{2}[/tex] / r

r = m v/q B.....................................3

r = (2.00 x[tex]10^{-26}[/tex] kg) x (6.15 x  [tex]10^{6}[/tex] m/s)  / (1.60217646 ×[tex]10^{-19}[/tex]  C) x (1.833 T)

r = 0.4188 m

The diameter is r x 2

D = 0.4188 m x 2

D = 0.8376 m

Therefore the diameter is  0.8376

Answer:

 φ = √2/L sin (kx),   E = (h² / 8 mL²) n²  

Explanation:

The Schrödinger equation for a particle in a box is, described by a particle within a potential for simplicity with infinite barrier

      V (x) =   ∞            x <0

                    0      0 <x <L

                    ∞           x> L

This means that we have a box of length L

We write the equation

              (- h’² /2m  d² / dx² + V) φ = E φ

             h’= h / 2π

The region of interest is inside the box, since being the infinite potential there can be no solutions outside the box. The potential is zero

                - h’² /2m d²φ/ dx² = E φ

The solution for this equation is a sine wave,

Because it is easier to work with exponentials, let's use the reaction between the sine function and cook with the exponential

               [tex]e^{ikx}[/tex] = cos kx + i sin kx

Let's make derivatives

              dφ / dx = ika e^{ikx}

              d²φ / dx² = (ik) e^{ikx} = - k² e^{ikx}

Let's replace

            - h'² / 2m (-k² e^{ikx}) = E e^{ikx}

            E = h'² / 2m    k²

To have a solution this expression

Now let's work on the wave function, as it is a second degree differential bond, two solutions must be taken

             φ = A e^{ikx} + B e^{-ikx}

This is a wave that moves to the right and the other to the left.

Let's impose border conditions

         φ (0) = 0

         φ (L) = 0

For being the infinite potential

With the first border condition

         0 = A + B

         A = -B

They are the second condition

         0 = A e^{ikL}+ B e^{-ikL}

We replace

        0 = A (e^{ikL} - e^{-ikL})

We multiply and divide by 2i, to use the relationship

        sin kx = (e^{ikx} - e^{-ikx}) / i2

        0 = A 2i sin kL

Therefore kL = nπ

         k = nπ / L

The solution remains

         φ = A sin (kx)

        E = (h² / 8 mL²) n²

To find the constant A we must normalize the wave function

       φ*φ = 1

       A² ∫ sin² kx dx = 1

We change the variable

       sin² kx = ½ (1 - cos 2kx)

       A =√ 2 / L

The definitive function is

          φ = √2/L sin (kx)

Answer:

|Ax| =1.03 m  (directed towards negative x-axis)

|Ay|= 5.30 (directed towards negative y-axis)

Explanation:

Let A is a vector = 5.4 m

θ = 259°

to Find Ax, Ay

Sol:

according the condition it lies in 3rd quadrant

we know that Horizontal Component Ax = A cos θ

Ax = 5.4 Cos 259°

Ax = - 1.03 m

|Ax| =1.03 m  (directed towards negative x-axis)

Now Ay = A sin θ

Ay = 5.4 Sin 259°

Ay = -5.30

|Ay|= 5.30 (directed towards negative y-axis)

(a) -1.030m

(b) -5.301m

Given a vector F in the xy plane, of magnitude F and in a direction θ counterclockwise from the positive direction of the x-axis;

The x-component ([tex]F_{X}[/tex]) of vector F is given by;

[tex]F_{X}[/tex] = F cos θ    ---------------------(i)

And;

The y-component ([tex]F_{Y}[/tex]) of vector F is given by;

[tex]F_{Y}[/tex] = F sin θ            -----------------------(ii)

Now to the question;

Let the vector be A

Therefore;

The magnitude of vector A is A = 5.4m

The direction θ of A counterclockwise from the positive direction of the x-axis = 259°

(a) The x-component ([tex]A_{X}[/tex]) of the vector A is therefore given by;

[tex]A_{X}[/tex] = A cos θ       ------------------------(iii)

Substitute the values of θ and A into equation (iii) as follows;

[tex]A_{X}[/tex] = 5.4 cos 259°

[tex]A_{X}[/tex] = 5.4 x (-0.1908)

[tex]A_{X}[/tex] = -1.030

Therefore, the x-component of the vector is -1.030m

(b) The y-component ([tex]A_{Y}[/tex]) of the vector A is therefore given by;

[tex]A_{Y}[/tex] = A sin θ       ------------------------(iv)

Substitute the values of θ and A into equation (iv) as follows;

[tex]A_{Y}[/tex] = 5.4 sin 259°

[tex]A_{Y}[/tex] = 5.4 x (-0.9816)

[tex]A_{Y}[/tex] = -5.301m

Therefore, the y-component of the vector is -5.301m

Answer:

#Check explanation below for a detailed description.

Explanation:

A plot of absorbance vs temperature is used to determine  [tex]T_m[/tex] of a duplex DNA.

-The duplex DNA denatures.

-The denaturation creates two single strands which enhances Ultraviolet absorption ( increased temperatures increases absorption rates).

-Higher concentrations of [tex]Sodium \ Chloride[/tex] hampers stability of the DNA.

Answer:

687,500 N

Explanation:

Workdone = Force × Distance

Making force the subject of the formula; we have:

Force =[tex]\frac{workdone}{distance}[/tex]

Given that:

workdone  = 5.5×10⁵ J

Distance = 0.80 m

∴ Force = [tex]\frac{5.5*10^5}{0.8}[/tex]

Force = 687,500 N

Answer:

6.875×10⁵  N.

Explanation:

Force: This can be defined as the product of mass and acceleration or it can be defined as the ratio of work done and distance. The S.I unit of force is Newton.

W = F×d................. Equation 1

Where W = work done, F = force, d = distance.

make F the subject of the equation

F = W/d.................... Equation 2

Given: W = 5.5×10⁵ J, d = 0.8 m

Substitute into equation 2

F =  5.5×10⁵ /0.8

F = 6.875×10⁵  N.

Hence the force exerted on the truck by the sand = 6.875×10⁵  N.

The velocity of A is 5.16m/s²

Explanation:

Given-

mass of cylinder A, mₐ = 2kg

mass of cylinder B, mb = 10kg

Distance, s = 2m

Velocity of A, v = ?

Let acceleration due to gravity, g = 10m/s²

We know,

[tex]a = \frac{mb * g - ma * g}{ma + mb} \\\\a = \frac{10 * 10 - 2 * 10}{ 2 + 10} \\\\a = \frac{80}{12} \\\\a = 6.67m/s^2[/tex]

We know,

[tex]v = \sqrt{2as}[/tex]

[tex]v = \sqrt{2 X 6.67 X 2} \\\\v = \sqrt{26.68} \\\\v = 5.16m/s^2[/tex]

Therefore, the velocity of A is 5.16m/s²

In the scenario described, increasing the height of a mass M in a vertical spring-mass system will result in both gravitational and spring potential energy. Thus, the potential energy in Case 2 is the sum of gravitational potential energy (m*g*D) and the elastic potential energy of the spring (1/2*k*D^2).

The situation described in the question deals with principles of Physics, specifically potential energy and its application to the spring-mass system. Consider both cases with the mass M hanging from a spring with a constant k in a vertical orientation. In the first case, the system is at equilibrium and thus the potential energy is defined as zero.

In the second case, the mass M is lifted upward a vertical distance D from the equilibrium position, it now possesses gravitational potential energy in addition to the elastic potential energy of the spring. This total energy is preserved as long as no external force behaves on the system. The gravitational potential energy gained by the mass is equal to m*g*D where m is the mass, g is the acceleration due to gravity and D is the vertical distance lifted.

The potential energy stored in the spring when it is stretched or compressed by a distance 'x' is given by the formula U = 1/2*k*x^2, where 'k' is the spring constant and 'x' is the displacement from the equilibrium position (in this case is D).

So, combining both energies, the total potential energy in Case 2, when the mass has been lifted a vertical distance D is m*g*D + 1/2*k*D^2.

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Answer:

[tex]\dot V = 0.542 \frac{m^{3}}{s}[/tex]

Explanation:

The power needed for the pump to raise the pressure of gasoline is defined by following equation. The maximum possible volume flow rate is isolated and then calculated:

[tex]\dot W = \dot V \cdot \Delta P\\\dot V = \frac{\dot W}{\Delta P}\\\dot V = \frac{3.8 kW}{7 kPa}\\\dot V = 0.542 \frac{m^{3}}{s}[/tex]

Explanation:

Below is an attachment containing the solution.

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force [tex]F_p[/tex] by the worker must be equal to the friction force [tex]F_f[/tex] on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

[tex]F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N[/tex]

b. The work is done on the crate by this force is the product of its force [tex]F_p[/tex] and the distance traveled s = 4.35

[tex]W_p = F_ps = 79.1*4.35 = 344 J[/tex]

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

[tex]W_f = F_fs = -79.1*4.35 = -344 J[/tex]

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

[tex]W_p + W_f = 344 - 344 = 0 J[/tex]

Answer:

(A) 79N

(B) W = 344J

(C) Wf= -344J

(D) W = 0J

(E) W = 0J

Explanation:

Please see attachment below.

The resultant voltage depends on the arrangement of the batteries. For series configuration, the voltage sums up to 9V. For parallel, it remains 1.5V. And for combined series and parallel, it sums up to 3V.

The voltage across batteries depends on how they are connected.

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Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

Answer:

Explanation:

capacitance of parallel plate capacitor

c = ε A / d , d is distance between plates , A is surface area , ε is constant

As d becomes two times , Capacitance c = 1/ 2 times ie c / 2

potential V = Q / C

Q is constant , potential

v = Q / c /2

= 2 . Q / C

= 2 V

So potential difference becomes 2 times.

NEW P D = 398 X 2

= 796 V.

Option d is correct; relativity formulas are not needed because the Lorentz factor (γ-1) is much less than 1, indicating that the electron's velocity is non-relativistic after being accelerated through a potential difference of 0.200 kV.

We need to determine whether we should use relativity formulas for the electrons accelerated through a potential difference. When electrons gain kinetic energy (K) by acceleration through a potential difference (V), the energy they gain can be expressed as Ve=K=(γ-1)mc2. Here, γ stands for Lorentz factor, m is the mass of electron, and c is the speed of light. Given that the potential difference ('V') is 0.200 kV, and the rest mass energy ('Eo') of an electron is 0.511 MeV, we calculate:

γ-1= (0.200 keV) / (0.511 MeV) ≈ 0.3916 × 10-3, which implies that γ is approximately 1. So, γ-1 << 1, which in turn means that the electron's velocity is much less than the speed of light, and the relativistic effect can be neglected. Thus, we do not need to use special relativity.

The correct answer is d.

Explanation:

A trapeze is rotating with 1 rotation per second .

Thus its angular velocity ω = 2π n

here n is the number of rotations per second

Thus ω = 2π b because n = 1 in this case

Suppose the moment of inertia of his is = I

Then angular momentum L₁ = I  ω = 2 I π

In the second case , the moment of inertia becomes = 0.4 I

Let his angular velocity is  ω₀

Thus angular momentum L₂ = 0.4 I  ω₀

Because no external torque is applied , therefore angular momentum will remain constant .

Thus L₁ = L₂

Therefore  2 I π = 0.4 I x 2 n₀ π

here n₀ is the number of rotations per second

n₀ = [tex]\frac{1}{0.4}[/tex] = [tex]\frac{5}{2}[/tex] = 2.5

Explanation:

Below is an attachment containing the solution.

Final answer:

The skier is moving at 11.1 m/s when she gets to the bottom of the hill. This solution is derived using the principles of energy conservation and work done by friction.

Explanation:

A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high. To determine how fast the skier is moving when she gets to the bottom of the hill, we analyze the problem using principles of energy conservation and work done by friction.

We start with the equation that equates the final kinetic and potential energies to the initial energies along with the work done by friction. This equation is 0.5 mv² + 0 = 0.5 mu² + μmgl + mgh, where v is the final velocity, μ is the coefficient of friction, g is the acceleration due to gravity (9.8 m/s²), l is the length of the rough patch, and h is the height of the hill.

Plugging in the given values,

we have: 0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.50. S

implifying, we get v² = 123.07, which leads to v = 11.1 m/s.

Thus, the skier's speed at the bottom of the hill is 11.1 m/s.

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

[tex]V =\frac{Kq}{r}[/tex]

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

[tex]V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V[/tex]

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

[tex]V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V[/tex]

Total potential due to this charges  = 4500 V + 6750 V = 11250 V

The potential due to two 3.0 μC charges on the x-axis at different distances from a point can be calculated using Coulomb's Law.

The potential due to two point charges can be found using Coulomb's Law. The potential, V, at a point on the x-axis is the sum of the potentials from each charge. The potential due to a point charge can be calculated using the formula V = k * (Q / r), where k is the electrostatic constant, 9 x 10^9 Nm^2/C^2, Q is the charge, and r is the distance between the charge and the point. In this case, since the charges are on the x-axis, the distance between the origin and the point is x, and the distance between the other charge and the point is (6-x). So, the potential at the point is V = k * (3.0 x 10^-6 / x) + k * (3.0 x 10^-6 / (6-x)) relative to infinity.

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Answer:

W / n = - 9133 J / mol, W / n = 3653 J / mol , e = 0.600

Explanation:

The Carnot cycle is described by

      [tex]e= 1 - Q_{c} / Q_{H} = 1 - T_{c} / T_{H}[/tex]

In this case they indicate that the final volume is

         V = 3V₀

In the part of the heat absorption cycle from the source is an isothermal expansion

         W = n RT ln (V₀ / V)

         W / n = 8.314 1000 ln (1/3)

          W / n = - 9133 J / mol

During the part of the isothermal compression in contact with the cold focus, as in a machine the relation of volumes is maintained in this part is compressed three times

            W / n = 8.314 400 (3)

           W / n = 3653 J / mol

The efficiency of the cycle is

            e = 1- 400/1000

            e = 0.600

I have attached the circuit image missing in the question.

Answer:

A) The range of vo is; -6.6V≤ vo ≤-1V

B) σ = 0.1861

Explanation:

A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.

Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0

So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0

Simplifying further; -25 vg - vΔ = 0

From the question, vg = 40mV = 0.04 V

So - 25(0.04) = vΔ

So: vΔ = - 1 V

Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0

So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0

So RΔ = 100kΩ or 100,000Ω from the question.

So, substituting for RΔ, we get,

[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0

Let's put the value of - 1 for vΔ as gotten before.

So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0

Now let's make vo the subject of the equation to get;

-1 - vo = (1 - σ)[2 + (1/σ)]

-1 - vo = 2 - 2σ + (1/σ) - 1

-vo = 1 + 2 - 2σ + (1/σ) - 1

-vo = 2 - 2σ + (1/σ)

vo = - 1 (2 - 2σ + (1/σ))

When σ = 0.2; vo = - 1(2 - 0.4 + 5) =

- 1 x 6.6 = - 6.6V

Also when σ = 1;

vo = - 1(2 - 2 + 1) = - 1V

Therefore, the range of vo is;

- 6.6V ≤ vo ≤ - 1V

B) it will saturate at vo = - 7V

So, from;

vo = - 1 (2 - 2σ + (1/σ))

-7 = - 1 (2 - 2σ + (1/σ))

Divide both sides by (-1)

7 = (2 - 2σ + (1/σ))

Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)

Multiply each term by α to get;

5σ = - 2σ^(2) + 1

So 2σ^(2) + 5σ - 1 = 0

Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861

Final answer:

The question involves understanding an ideal operational amplifier's output behavior, focusing on its output voltage changes and saturation with respect to variations in σ (sigma). Calculations for the output voltage range given a specific input and understanding the conditions under which the op-amp will saturate are key.

Explanation:

The question revolves around an operational amplifier (op-amp) circuit, exploring its behavior under certain conditions, specifically examining output voltage (vO) variations and saturation point related to the parameter σ (sigma). The op-amp is assumed to be ideal, implying infinite gain, zero input current, and that its input terminals are at the same potential.

a. Range of vO if vI = 40 mV

Given that the op-amp is ideal, the output voltage will depend on the input voltage (vI), the gain settings (σ), and the feedback resistor (R3). In practical scenarios, the gain can be adjusted by changing the value of σ or R3. However, for an ideal op-amp, the input voltage is directly proportional to the output voltage, influenced by σ. With vI = 40 mV and σ ranging between 0.2 and 1.0, the output voltage will vary accordingly, directly proportional to these parameters.

b. Saturation point of the op-amp

Saturation in an op-amp occurs when the output voltage exceeds the power supply limits, meaning the op-amp can no longer amplify the input signal. The specific value of σ at which saturation occurs depends on the supply voltage, the op-amp's maximum output voltage capability, and the configuration of the feedback network. Without specific values for the power supply or feedback network, calculating the exact σ value for saturation is not possible. Yet, in theory, as σ approaches the op-amp's gain limit or if the gain results in an output voltage beyond what the op-amp can deliver based on its power supply, saturation will occur.

Answer:

Explanation:

net force exerted on charge Q₃, exerted by charges Q₁and Q₂, will be  zero

if net  electric field due to charges Q₁ and Q₂  at origin is zero .

electric field due to Q₂

= 9 X 10⁹ X 3 x10⁹ / .04²

electric field due to Q₁

= 9 X 10⁹ X Q₁ / .02²

For equilibrium

9 X 10⁹ X Q₁ / .02² = 9 X 10⁹ X 3 x10⁻⁹ / .04²

Q₁  = 3 X10⁻⁹ x .02² / .04²

= 3 / 4 x 10⁻⁹

.75 x 10⁻⁹  C

Answer:

3.41m

Explanation:

The following were obtained from the question:

f (frequency) = 8.81x10^7Hz

V (velocity of electromagnetic wave) = 3x10^8 m/s

λ (wavelength) =?

Velocity, frequency and wavelength of a wave are related with the equation below:

V = λf

λ = V/f

λ = 3x10^8 /8.81x10^7

λ = 3.41m

Therefore, the wavelength of the radio wave is 3.41m

Answer:

Answer: 3.41

Explanation:

Edge 2020 (E2020)

Answer:

Resistance increases with increase in temperature which depends on power supplied which also depends on voltage.

Thermal expansion will make resistance larger.

Explanation:

Light bulb is a good example of a filament lamp. If we plot the graph of voltage against current we will notice that resistance is constant at constant temperature.

The filament heats up when an electric current passes through it, and produces light as a result.

The resistance of a lamp increases as the temperature of its filament increases. The current flowing through a filament lamp is not directly proportional to the voltage across it.

tensile stress begins to appear in resistor as the temperature rises. Thus, the resistance value increases as the temperature rises. Resistance value can only decrease as the temperature rises in case of thin film resistor with aluminium substrate.

In case of a filament bulb, the resistance will increase as increase in length of the wire. The thermal expansion in this regard is linear expansivity in which resistance is proportional to length of the wire.

Resistance therefore get larger.

Final answer:

The equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler water due to the specific heat capacity of water.

Explanation:

The equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler water.

When different temperatures of water are mixed, heat is transferred between them until they reach a common equilibrium temperature. The amount of heat transferred is determined by the specific heat capacity of water, which is greater than most common substances. As a result, water undergoes a smaller temperature change for a given heat transfer. Therefore, the equilibrium temperature will be closer to the initially cooler water.

Answer:

H = 4500 m

Explanation:

        [tex]H = \frac{1}{2}*g*t^{2} = \frac{1}{2} * 10 m/s2*(30s)^{2} = 4500 m[/tex]

Answer:

Option as B is correct At the same speed as before

Explanation:

As we know the relation between speed of the wave and tension in string

The speed of wave in stretched string

ν = [tex]\sqrt{\frac{T}{\mu} }[/tex]  

speed of wave is the directly proportional to the square root of tension as mentioned in question tension of string is unaffected when in linear mass density is constant,  so we can say that the  speed of wave will  be the same  

Option as B is correct At the same speed as before  

If you suddenly begin to wiggle more rapidly without appreciably affecting the tension, you will cause the waves to move down the string at the same speed as before (Option b).

A wave can be defined as a type of disturbance that contains energy independently of particle motion.

In conclusion, if you suddenly begin to wiggle more rapidly without appreciably affecting the tension, you will cause the waves to move down the string at the same speed as before (Option b).

Learn more in:

https://brainly.com/question/12215474

Answer:

(a). The maximum height by first ball is [tex]1.8545\times10^{4}\ m[/tex]

The maximum height by second ball is [tex]5.1020\times10^{4}\ m[/tex]

(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is [tex]1.0\times10^{7}\ J[/tex]

Explanation:

Given that,

Mass of cannonball = 20.0 kg

Speed = 1000 m/s

Angle with horizontal= 37.08

Fired angle = 90.08

We need to calculate the speed of the ball

Using formula of speed

[tex]v_{y}=v\sin\theta_{H}[/tex]

Put the value into the formula

[tex]v_{y}=1000\times\sin37.08[/tex]

[tex]v_{y}=602.9\ m/s[/tex]

(a). We need to calculate the maximum height by first ball

Using conservation of energy

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

[tex]h= \dfrac{v^2}{2g}[/tex]

Put the value into the formula

[tex]h=\dfrac{(602.9)^2}{2\times9.8}[/tex]

[tex]h=1.8545\times10^{4}\ m[/tex]

We need to calculate the maximum height by second ball

Using conservation of energy

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

[tex]h= \dfrac{v^2}{2g}[/tex]

Put the value into the formula

[tex]h=\dfrac{(1000)^2}{2\times9.8}[/tex]

[tex]h=5.1020\times10^{4}\ m[/tex]

(b). We need to calculate the total mechanical energy of the ball–Earth system at the maximum height for each ball

Using formula of energy

[tex]E=\dfrac{1}{2}mv^2[/tex]

[tex]E=\dfrac{1}{2}\times20\times1000^2[/tex]

[tex]E=1.0\times10^{7}\ J[/tex]

Hence, (a). The maximum height by first ball is [tex]1.8545\times10^{4}\ m[/tex]

The maximum height by second ball is [tex]5.1020\times10^{4}\ m[/tex]

(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is [tex]1.0\times10^{7}\ J[/tex]

Answer: tensional force

Explanation:

Tension force on a material occurs when two equal forces act on a material in an opposite direction away from the ends of the material.

Pre-tensing a wire material increases its load bearing capacity and reduces its flexure.

Answer:

When light goes from one material into another material having a HIGHER index of refraction, its speed and wavelength decrease, but its frequency stays the same.

The correct option is E

Explanation:

Refraction is the bending of light ray as it crosses the boundary between the two media of different densities, thus causing a change in it's direction.

When light goes from one material of lower refractive to another material of higher refractive index, the speed of light reduces, because from law of refraction

aNg= (speed of light in air)/(speed of light in glass)

Snell's Law

So generally,

n1/n2 = v2/v1

Let n1 be incident index

Let n2 be refracted index

v2 is refracted speed

v1 is incident speed

Since given that, n1<n2

Then, the right side of the equation will always be less than 1 i.e n1/n2

Then, v2=v1• ( n1/n2)

Since n1/n2 is less that one this show that the speed v2 will reduce.

Since we know that the speed decrease,

We also know that speed, wavelength and speed is related by

v=fλ

Since, the speed is directly proportional to wavelength this shows that as the speed reduces the wavelength also reduces where frequency is the constant of proportionality

v∝λ. .

then f is constant of proportional

v=fλ

So as speed reduces, the wavelength reduces.

frequency of a light wave does not depend on the medium, while wavelength and speed do.

its speed and wavelength decrease, but its frequency stays the same.

Then the answer is E

Answer:

1)   v = 2.20 10⁵ m / s , 2)  r = 4.86 10⁻³ m,    r = 4.92 10⁻³ m

Explanation:

1) A speed selector is a section where the magnetic and electrical forces have opposite directions, so

           [tex]F_{m} = F_{e}[/tex]

           q v B = q E

           v = E / B

           V = E s

           E = V / s

            v = V / s B

            v = 148 / (0.0016 0.42)

            v = 2.20 10⁵ m / s

2) when the isotopes enter the spectrometer, we can use Newton's second law

             F = m a

Acceleration is centripetal

             a = v² / r

             q v B = m v² / r

              r = m v / qB

The mass of uranium 235 is

           m = 3.903 10⁻²⁵ kg

The radius of this isotope is

             r = 3,903 10⁻²⁵ 2.20 10⁵ / (92  1.6 10⁻¹⁹  1.2)

             r = 4.86 10⁻³ m

The mass of the uranium isotope 238 is

              m = 238 a = 238 1.66 10-27 = 395.08 10-27 kg

The radius is

             r = 395.08 10⁻²⁷ 2.20 10⁵ / (92 1.6 10⁻¹⁹  1.2)

             r = 4.92 10⁻³ m

Final answer:

The speed of the ions entering the mass spectrometer is 352.38 m/s. The difference in the diameters of the orbits of the 238U and 235U ions can be calculated using the formula for the radius of a charged particle's orbit in a magnetic field.

Explanation:

To find the speed of the ions entering the mass spectrometer, we can use the equation for the force on a charged particle in a magnetic field: F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field. The force is equal to the electric force, qE, so we can set these two equations equal to each other and solve for v: qvB = qE. Since q is the charge of the ion and E is the electric field strength, we can rearrange this equation to solve for v, giving us v = E/B.

Plugging in the values given in the question, we find that the speed of the ions entering the mass spectrometer is 148 V / 0.42 T = 352.38 m/s.

To find the difference in the diameters of the orbits of the 238U and 235U ions, we can use the formula for the radius of a charged particle's orbit in a magnetic field: r = mv / (qB), where m is the mass of the ion, v is its velocity, q is its charge, and B is the magnetic field. Since the ions are singly ionized, their charges are +e, where e is the charge of an electron. Plugging in the values given in the question, we can calculate the radius of the orbits of the 238U and 235U ions, and then subtract these values to find the difference in their diameters.