What is the energy of the photon that could cause (a) an electronic transition from then= 4 state to the n 5 state of hydrogen and (b) an electronic transition from the n 5 state to the n 6 state? What is the energy of the photon that could cause (a) an electronic transition from then= 4 state to the n 5 state of hydrogen and (b) an electronic transition from the n 5 state to the n 6 state?

Answer:

6593.4

Explanation:

wavelength, λ = 910 nm = 910 x 10^-9 m

Speed of laser = 3 x 10^8 m/s

t = 20 pico seconds

Distance traveled by laser in this time, d = c x t

d = 3 x 10^8 x 20 x 10^-12 = 6 x 10^-3 m

number of wavelengths, n = d / λ

n = (6 x 10^-3) / (910 x 10^-9)

n = 6593.4

Answer:

157.8 J

Explanation:

m = mass of the cylinder = 7 kg

h = height difference in top and bottom of the incline = 2.3 m

g = acceleration due to gravity = 9.8 m/s²

TE = Total Energy at the bottom

PE = Gravitational potential energy at the top

Using conservation of energy

Total Energy at the bottom = Gravitational potential energy at the top  

TE = PE

TE = m g h

TE = (7) (9.8) (2.3)

TE = 157.8 J

Explanation:

It is given that,

Mass of the passenger, m = 75 kg

Acceleration of the rocket, [tex]a=49\ m/s^2[/tex]

(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :

F = m a

[tex]F=75\ kg\times 49\ m/s^2[/tex]

F = 3675 N

Ratio, [tex]R=\dfrac{F}{W}[/tex]

[tex]R=\dfrac{3675}{75\times 9.8}=5[/tex]

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

[tex]F'=\sqrt{F^2+W^2}[/tex]

[tex]F'=\sqrt{(3675)^2+(75\times 9.8)^2}[/tex]

F' = 3747.7 N

The direction of force is calculated as :

[tex]\theta=tan^{-1}(\dfrac{W}{F})[/tex]

[tex]\theta=tan^{-1}(\dfrac{1}{5})[/tex]

[tex]\theta=11.3^{\circ}[/tex]

Hence, this is the required solution.

The horizontal component of the force the seat exerts against the passenger's body is 3675 N. The ratio of this force to the passenger's weight is 5. The total force the seat exerts has a magnitude of 3793 N.

(a) To calculate the horizontal component of the force the seat exerts against the passenger's body, we can use Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the passenger is 75.0 kg and the acceleration of the rocket sled is 49.0 m/s2. So the force exerted by the seat is:

Force = mass * acceleration

Force = 75.0 kg * 49.0 m/s2

Force = 3675 N

Now let's compare this force to the passenger's weight. The weight of an object is given by the formula:

Weight = mass * gravitational acceleration

Weight = 75.0 kg * 9.8 m/s2

Weight = 735 N

To find the ratio, we divide the force exerted by the seat by the weight of the passenger:

Ratio = Force / Weight

Ratio = 3675 N / 735 N

Ratio = 5

(b) The total force the seat exerts against the passenger's body has both a horizontal and vertical component. The direction of the total force is the same as the direction of the acceleration of the rocket sled. The magnitude of the total force can be found using the Pythagorean theorem:

Total Force = √(horizontal component2 + vertical component2)

Total Force = √(36752 + 7352)

Total Force = 3793 N

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Answer:

Torque is 93 Nm anticlockwise.

Explanation:

We have value of torque is cross product of position vector and force vector.

A force of 38 N, directed 30° above the x axis in the x-y plane.

        Force, F = 38 cos 30 i + 38 sin 30 j = 32.91 i + 19 j

A particle is located on the x axis 4.9 m and we have to find torque about the origin on the particle.

Position vector, r = 4.9 i

Torque, T = r x F = 4.9 i x (32.91 i + 19 j) = 4.9 x 19 k = 93.1 k Nm

So Torque is 93 Nm anticlockwise.

Final answer:

The angle a vector makes with the positive x-axis can be found using the cosine ratio; the angle is approximately 58.2 degrees for the vector with a magnitude of 25 m and an x-component of 12 m.

Explanation:

To determine the angle a vector makes with the positive x-axis, we can use trigonometry since we have the magnitude of the vector and the length of the x-component. Specifically, the cosine of the angle (θ) that the vector makes with the x-axis is equal to the x-component divided by the magnitude of the vector.

Here is the formula:

cos(θ) = (x-component) / (magnitude)

By substituting the given values:

cos(θ) = 12 m / 25 m

Now calculate the angle:

θ = cos⁻¹(12/25)

θ = 58.2° (approx.)

Therefore, the vector makes an angle of approximately 58.2 degrees with the positive x-axis.

Answer:

The energy transferred is 2.93 MJ.

Explanation:

Given that,

Time = 1 h = 3600 s

Area[tex]A = 1.6\times4.00[/tex]

Width [tex]\Delta x=0.20 m[/tex]

Temperature [tex] T_{1}=20^{\circ}\ C[/tex]

[tex]T_{2}=5^{\circ}\ C[/tex]

For concrete wall,

The thermal conductivity

k = 1.7 W/mK

Using Fourier's law

[tex]Q=kA\dfrac{dt}{\Delta x}[/tex]

Where, A = area

[tex]\Delta x[/tex] = width

[tex]\Delta t[/tex] = change in temperature

k = thermal conductivity

Put the value into the formula

[tex]Q=1.7\times 1.6\times 4.00\times\dfrac{15}{0.20}[/tex]

[tex]Q=816\ W[/tex]

The energy transferred in 1 hour

[tex]E = Q\times t[/tex]

[tex]E=816\times3600[/tex]

[tex]E=2937600[/tex]

[tex]E = 2.93\times10^{6}\ J[/tex]

[tex]E = 2.93\ MJ[/tex]

Hence, The energy transferred is 2.93 MJ.

Answer:

[tex]N = 8.1 \times 10^{10}[/tex]

Explanation:

Initial charge on the rod is

[tex]Q_i = 14 nC[/tex]

final charge on the rod is

[tex]Q_f = 1 nC[/tex]

now the charge transferred from to the sphere is given as

[tex]\Delta Q = Q_i - Q_f[/tex]

[tex]\Delta Q = 14 - 1 = 13 nC[/tex]

now we also know that

Q = Ne

so number of particles transferred is

[tex]N = \frac{\Delta Q}{e}[/tex]

[tex]N = \frac{13 \times 10^{-9}}{1.6 \times 10^{-19}}[/tex]

[tex]N = 8.1 \times 10^{10}[/tex]

When the charged plastic rod touches the metal sphere, it transfers around 13 nC of charge to the sphere. This amounts to about 8.125 x 10^10 electrons, as each electron carries a charge of approximately -1.6 x 10^-19 C.

When the charged plastic rod touches the metal sphere, charge transfer occurs until both objects have the same charge. Initially, the plastic rod had a charge of -14 nC and after touching it had a charge -1.0 nC, which means that 13 nC was transferred to the metal sphere.

The elementary charge (charge of an electron) is approximately -1.6 x 10^-19 C. Therefore, the number of electrons transferred would be the total transferred charge divided by the charge of one electron, which results in approximately 8.125 x 10^10 electrons.

The charge is negative indicating electrons, which have negative charge, were transferred.

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Answer:

4.08 x 10⁻¹⁵ m

Explanation:

r₀ = constant of proportionality = 1.3 x 10⁻¹⁵ m

r = radius of the nucleus 31P = ?

A = mass number of the nucleus 31P = 31

Radius of the nucleus is given as

[tex]r=r_{o}A^{\frac{1}{3}}[/tex]

Inserting the values

[tex]r=(1.3\times 10^{-15})(31)^{\frac{1}{3}}[/tex]

[tex]r=(1.3\times 10^{-15})(3.14)[/tex]

r = 4.08 x 10⁻¹⁵ m

Answer:

speed of boat as

[tex]v_b = 10 mph[/tex]

river speed is given as

[tex]v_r = 2 mph[/tex]

Explanation:

When boat is moving down stream then in that case net resultant speed of the boat is given as

since the boat and river is in same direction so we will have

[tex]v_1 = v_r + v_b[/tex]

Now when boat moves upstream then in that case the net speed of the boat is opposite to the speed of the river

so here we have

[tex]v_2 = v_b - v_r[/tex]

as we know when boat is in downstream then in that case it covers 24 miles in 2 hours

[tex]v_1 = \frac{24}{2} = 12 mph[/tex]

also when it moves in upstream then it covers same distance in 3 hours of time

[tex]v_2 = \frac{24}{3} = 8 mph[/tex]

[tex]v_b + v_r = 12 mph[/tex]

[tex]v_b - v_r = 8 mph[/tex]

so we have speed of boat as

[tex]v_b = 10 mph[/tex]

river speed is given as

[tex]v_r = 2 mph[/tex]

The rate of the boat in still water is 10 mph

The rate of the current is 2 mph

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

Given:

distance covered = d = 24 miles

time for driving down the river = td = 2 hours

time for driving up the river = tu = 3 hours

Unknown:

velocity of the boat in still water = vs = ?

velocity of the current = vc = ?

Solution:

When Marcus drive his boat down the river , the velocity of the boat is in the same direction to the velocity of the current.

[tex]v_s + v_c = \frac{d}{t_d}[/tex]

[tex]v_s + v_c = \frac{24}{2}[/tex]

[tex]v_s + v_c = 12[/tex]

[tex]v_s = 12 - v_c[/tex] → Equation 1

When Marcus drive his boat up the river , the velocity of the boat is in the opposite direction to the velocity of the current.

[tex]v_s - v_c = \frac{d}{t_d}[/tex]

[tex]v_s - v_c = \frac{24}{3}[/tex]

[tex]v_s - v_c = 8[/tex]

[tex]12 - v_c - v_c = 8[/tex] ← Equation 1

[tex]12 - 2v_c = 8[/tex]

[tex]2v_c = 12 - 8[/tex]

[tex]2v_c = 4[/tex]

[tex]v_c = 4 \div 2[/tex]

[tex]\large {\boxed {v_c = 2 ~ mph} }[/tex]

[tex]v_s = 12 - v_c[/tex]

[tex]v_s = 12 - 2[/tex]

[tex]\large {\boxed {v_s = 10 ~ mph} }[/tex]

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

Final answer:

Direct current (DC) is the flow of electric current in only one direction. According to the electron flow convention, the current flows from the positive terminal to the negative terminal.

Explanation:

Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant. The direction of current flow in a direct current circuit is from the positive terminal to the negative terminal. This is according to the electron flow convention, where the movement of negative charges (electrons) is considered as the flow of current.

Using a formula linking initial and final angular velocities, angular acceleration, and total angle, the final angular velocity can be obtained. Multiplying the final angular velocity and radius gives the final linear velocity. Angular acceleration and linear acceleration on the rim of a wheel are related by their radius.

Given the angular acceleration, a, the number of revolutions, and the radius of the wheel, r, you can determine the angular velocity at the end of 2 revolutions using the formula w_f = sqrt(w_i^2 + 2*a*θ), where w_i is the initial angular velocity, w_f is the final angular velocity, a is angular acceleration, and θ is the total angle swept out in radians. The angular velocity can then be used to find the final linear velocity, v = r*w_f.

Angular acceleration is directly given by a = α*r, where α is angular acceleration and r is the radius. Angular acceleration is also related to the linear (tangential) acceleration, a_t, by a = r*α, where α is angular acceleration and r is the radius. Tangential acceleration involves not only changes in speed (which causes tangential acceleration) but also changes in direction (which causes radial or centripetal acceleration).

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To solve this question, we need to use the concept of the conversion between potential and kinetic energy.

The roller coaster's initial potential energy before it goes down the hill can be calculated using the formula for gravitational potential energy: `PE = m * g * h`,

where `m` is the mass,

           `g` is the acceleration due to gravity, and

           `h` is the height.

Substituting the given values, where `m` is 1200 kg, `g` is 9.81 m/s² and the initial height `h` is 19 m,

`PE_initial = 1200 kg * 9.81 m/s² * 19 m = 223668 J`

This is the energy the roller coaster has due to its position at the top of the 19m high hill before it starts to move.

When the roller coaster reaches the hill that is 13 m high, we can calculate its potential energy at this point the same way we calculated the initial potential energy, with `h` being now the final height of 13 m,

`PE_final = 1200 kg * 9.81 m/s² * 13 m = 153036 J`

This is the energy the coaster has due to its position at the top of the 13m high hill.

The kinetic energy (KE) of the roller coaster at this point is gained by the conversion of some of the initial potential energy into kinetic energy. This conversion is equal to the difference between the initial potential energy and the final potential energy:

`KE = PE_initial - PE_final = 223668 J - 153036 J = 70632 J`

So, when the roller coaster goes over the hill that is 13 m high, its kinetic energy is 70632 J. This is the energy the roller coaster has due to its speed at this point.

Learn more about Conservation of Mechanical Energy here:

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Answer:

Sagittarius

Explanation:

The center of the Milky Way galaxy lies in the direction of the Sagittarius constellation, about 26,000 light-years away.

The center of the Milky Way galaxy is in the direction of the Sagittarius constellation and is about 26,000 light-years away from us.

The center of the Milky Way galaxy lies in the direction of the Sagittarius constellation and is approximately 26,000 light-years away. When we look towards Sagittarius, we are looking towards the densest part of the Milky Way, which is why it appears brighter in the night sky.

A light-year, by definition, is the distance that light can travel in one year. Hence, saying that the center of the Milky Way is 26,000 light-years away means it would take 26,000 years for the light from the center of the galaxy to reach us.

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Answer: A) Quenching

Hope this helps

Answer:

The pilot must keep the tip pointed at 12.1 degrees to the right with respect to the direction of the runway to align the flight path with the runway.

Explanation:

x= -10m/s

y= 47m/s

r= √(x²)+(y²)

r=48.05 m/s

β= tan⁻¹(y/x)

β=102.01°

the runway is at 90 degrees. Considering the wind, the airplane is flying at 102.01 º direction. Must fly at 12.1 degrees to the right with respect to the direction of the runway to contrarest the wind effect.

Final answer:

The question involves calculating the individual masses of two objects that collectively weigh 5.09 kg and attract each other with a gravitational force of 0.98 x 10^-8 N, separated by 19.5 cm. The gravitational formula F = G×(m1×m2)/r^2 and a system of equations are used to find the solution.

Explanation:

The student's question is about determining the mass of two objects that attract each other with a known gravitational force when separated by a certain distance. The given force is 0.98 x 10-8 N, and the total mass of the two objects is 5.09 kg. The distance between them is 19.5 cm (or 0.195 m in SI units).

To solve this problem, we'll use the formula for the gravitational force between two masses, which is:

F = G×(m1×m2)/r2,

where:

Answer:

here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field

Explanation:

As we know by Faraday's law of electromagnetic induction

Rate of change in magnetic flux will induce EMF in the coil

so here we will have

[tex]EMF = \frac{d\phi}{dt}[/tex]

here we know that

[tex]\phi = NB.A[/tex]

now if the magnetic flux will change with time then it will induce EMF in the coil

[tex]EMF = N\frac{d}{dt}(B.A)[/tex]

so here induced EMF will be zero in the coil if the flux linked with the coil will remain constant

so here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field

In such a way when coil will rotate then the flux linked with the coil will remains constant and there will be no induced EMF in it

To have zero induced emf in a circular loop rotating in a magnetic field, the loop's axis of rotation should be parallel to the magnetic field lines.

The orientation of the loop's axis of rotation must be parallel to the magnetic field lines in a region where a uniform magnetic field points straight down for the induced emf to be zero. When the loop's axis is perpendicular to the magnetic field, the maximum emf is induced. Thus, to minimize the induced emf, the axis should be parallel to the magnetic field.

Answer:

F = 0.7 N

Explanation:

As we know that the wire touch the ends of the diagonal of the rectangle

so here the length of the wire is given as

[tex]L = \sqrt{x^2 + y^2}[/tex]

[tex]L = \sqrt{0.6^2 + 0.8^2}[/tex]

[tex]L = 1 m[/tex]

now force due to magnetic field on current carrying wire is given as

[tex]F = iLB[/tex]

now we have

[tex]F = (7)(1)(0.1)[/tex]

[tex]F = 0.7 N[/tex]

Answer:

t=4.2 years

Explanation:

velocity v= 0.96c

destination star distance = 14.4 light year

According to the theory of relativity length contraction

[tex]l= \frac{l_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

[tex]l_0= l{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

putting values we get

[tex]l_0= 14.4{\sqrt{1-\frac{(0.96c)^2}{c^2}}}[/tex]

[tex]l_0= 4.032 light years[/tex]

now distance the trip covers

D= vt

[tex]4.032\times c= 0.96c\times t[/tex]

[tex]t= \frac{4.032c}{0.96c}[/tex]

t= 4.2 years

so the trip will take 4.2 years

Answer:

6.44 × 10^10 N/C

Explanation:

Electric field due to the ring on its axis is given by

E = K q r / (r^2 + x^2)^3/2

Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.

r = 0.25 m, x = 0.5 m, q = 5 C

K = 9 × 10^9 Nm^2/C^2

E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2

E = 6.44 × 10^10 N/C

Answer:

a)

492 kJ

b)

Consistent

Explanation:

Q = Heat stored by woman from food = 600 k J

η = Efficiency of woman = 18% = 0.18

Q' = heat transferred to the environment

heat transferred to the environment is given as

Q' = (1 - η) Q

Inserting the values

Q' = (1 - 0.18) (600)

Q' = 492 kJ

b)

Yes the amount of heat transfer is consistent. The process of sweating produces the heat and keeps the body warm  

A woman climbing the Washington Monument metabolizes food energy with 18% efficiency, meaning 82% of the energy is lost as heat. When we calculate this value, we find that 492 kJ of energy is released as heat, which is consistent with the fact that people quickly warm up when exercising.

The woman climbing the Washington Monument metabolizes 6.00×10² kJ of food energy with an efficiency of 18%. This implies that only 18% of the energy consumed is used for performing work, while the remaining (82%) is lost as heat to the environment.

To calculate the energy lost as heat:

The released heat of 492 kJ is consistent with the fact that a person quickly warms up when exercising, because a significant portion of the body's metabolic energy is lost as heat due to inefficiencies in converting energy from food into work.

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Answer:

Time of race  = 10.18 s

Explanation:

She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash

Time to travel 17 m can be calculated

         s = ut + 0.5at²

         17 = 0 x t + 0.5 x 3.89 x t²

          t = 2.96 s

Velocity after 2.96 seconds

          v = 3.89 x 2.96 = 11.50 m/s

Remaining distance = 100 - 17 = 83 m

Time required to cover 83 m with a speed of 11.50 m/s

          [tex]t=\frac{83}{11.50}=7.22s[/tex]

Time of race = 2.96 + 7.22 = 10.18 s

Answer:

4.665×10⁻¹¹ N

Explanation:

Intensity of solar radiation=I=1.4×10³ W/m²

Area of panel=A=5.0 m²

speed of sound=c=3×10⁸ m/s

[tex]P_t=\text {Total\ Solar\ Radiation\ Pressure}[/tex]

[tex]P_t=2\frac{I}{c}\\\Rightarrow P_t=2\frac{1.4\times 10^{-3}}{3\times 10^{8}}\\\Rightarrow P_t=0.933\times 10^{-11}[/tex]

∴Total Solar Radiation Pressure=0.933×10⁻¹¹ Ws/m³

Force= Pressure×Area

Force=0.933×10⁻¹¹×5

Force=4.665×10⁻¹¹ Ws/m=4.665×10⁻¹¹ N

∴Force is exerted by solar radiation impinging normally on a perfectly reflecting panel of an artificial satellite orbiting the earth is 4.665×10⁻¹¹ N

The force exerted by solar radiation on the panel is 7.0 × 10³ N.

The force exerted by solar radiation can be calculated using the formula: force = power/area. In this case, the power is 1.4 × 10³ W/m² and the area is 5.0 m². Substituting these values in, the force exerted by solar radiation on the panel is: force = (1.4 × 10³ W/m²) x (5.0 m²) = 7.0 × 10³ N.

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I = 28.6mA.

The magnetic field in the center of a solenoid is given by:

B = μ₀NI/L

Clear I from the equation above, we obtain:

I = BL/μ₀N

With B = 1.00 x 10⁻⁴T, L = 0.42m, μ₀ = 4π x 10⁻⁷T.m/A  and N = 1170turns

I = [(1.00 x 10⁻⁴T)(0.42m)]/[(4π x 10⁻⁷T.m/A)(1170turns)]

I = 0.0286A

I = 28.6mA

Answer:

Part a)

[tex]Q = 320 \mu C[/tex]

Part b)

[tex]t = 8.52 \times 10^{-5} s[/tex]

Part c)

Rate of energy = 301.5 J/s

Explanation:

Part a)

Since energy is always conserved in LC oscillating system

So here for maximum charge stored in the capacitor is equal to the magnetic field energy stored in inductor

[tex]\frac{1}{2}Li^2 = \frac{Q^2}{2C}[/tex]

now we have

[tex]Q = \sqrt{LC} i[/tex]

[tex]Q = \sqrt{(3.76 \times 10^{-3})(3.13 \times 10^{-6})} (2.95)[/tex]

[tex]Q = 320 \mu C[/tex]

Part b)

Energy stored in the capacitor is given as

[tex]U = \frac{q^2}{2C}[/tex]

now rate of energy stored is given as

[tex]\frac{dU}{dt} = \frac{q}{C}\frac{dq}{dt}[/tex]

so here we also know that

[tex]q = Q sin(\omega t)[/tex]

[tex]\frac{dq}{dt} = Q\omega cos(\omega t)[/tex]

now from above equation

[tex]\frac{dU}{dt} = \frac{Qsin(\omega t)}{C} (Q\omega cos\omega t)[/tex]

so maximum rate of energy will be given when

[tex]sin\omega t = cos\omega t[/tex]

[tex]\omega t = \frac{\pi}{4}[/tex]

[tex]t = \frac{\pi}{4}\sqrt{LC}[/tex]

[tex]t = 8.52 \times 10^{-5} s[/tex]

Part c)

Greatest rate of energy is given as

[tex]\frac{dU}{dt} = \frac{Q^2\omega}{C}[/tex]

[tex]\frac{dU}{dt} = \frac{(320 \mu C)^2 \sqrt{\frac{1}{(3.76 mH)(3.13 \mu C)}}}{3.13 \mu C}[/tex]

[tex]\frac{dU}{dt} = 301.5 J/s[/tex]

Answer:

The current in wire 3 is 0.25 A.

Explanation:

It is given that, three wires meet at a junction.

Current in wire 1, I₁ = 0.4 A

Current in wire 2, I₂ = -0.65 A  (out of the junction)

We need to find the magnitude of the current in wire 3 (I₃). Applying Kirchhoff's current law which states that the sum of current in the circuit at a junction is equal to zero.

[tex]I_1+I_2+I_3=0[/tex]

[tex]I_3=-(I_2+I_1)[/tex]

[tex]I_3=-(-0.65+0.4)\ A[/tex]

I₃ = 0.25 A

So, the current in wire 3 is 0.25 A. Hence, this is the required solution.

Using the principle of Kirchhoff's junction rule, we calculate the current in Wire 3 as being 0.25 amperes.

The understanding of this question relies on the principle of Kirchhoff's junction rule in Physics, which states the sum of currents entering a junction equals the sum of currents leaving the junction. In this case scenario, Wire 1 has a current of 0.40 A into the junction and wire 2 has a current of 0.65 A out of the junction. The current of Wire 3 can be solved via the equation: I1 + I3 = I2, where I1, I2, and I3 are the currents on Wires 1, 2, and 3 respectively. Plugging in the given values, we get: 0.40 A + I3 = 0.65 A. Solving the equation for I3, we find that I3 = 0.25 A. Hence, the magnitude of the current in Wire 3 is 0.25 A.

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(a) The frequency from the stationary whistle as heard by the listener is approximately [tex]\(409.62 \, \text{Hz}\)[/tex].

(b) The frequency from the moving whistle as heard by the listener is approximately [tex]\(454.55 \, \text{Hz}\).[/tex]

(c) The beat frequency detected by the listener is approximately [tex]\(44.93 \, \text{Hz}\).[/tex]

To solve this problem, we can use the Doppler effect formula for sound frequencies.

The formula for the apparent frequency [tex](\(f'\))[/tex] heard by an observer when the source and/or observer is in motion relative to the medium is given by:

[tex]\[ f' = \frac{f \cdot (v + v_o)}{(v + v_s)} \][/tex]

where:

- [tex]\( f \)[/tex] is the frequency emitted by the source,

- [tex]\( v \)[/tex] is the speed of sound in the medium (assumed to be constant),

- [tex]\( v_o \)[/tex] is the speed of the observer relative to the medium,

- [tex]\( v_s \)[/tex] is the speed of the source relative to the medium.

Given:

- [tex]\( f = 392 \, \text{Hz} \)[/tex] (frequency emitted by both whistles),

-[tex]\( v = 343 \, \text{m/s} \)[/tex] (speed of sound in air),

- [tex]\( v_o = 15.0 \, \text{m/s} \)[/tex] (speed of the listener),

- [tex]\( v_s = 35.0 \, \text{m/s} \)[/tex] (speed of the moving whistle).

Let's calculate each part of the problem:

(a) Frequency from the stationary whistle [tex](\(f_1'\))[/tex] as heard by the listener:

[tex]\[ f_1' = \frac{f \cdot (v + v_o)}{(v + v_s)} \]\[ f_1' = \frac{392 \, \text{Hz} \cdot (343 \, \text{m/s} + 15.0 \, \text{m/s})}{(343 \, \text{m/s} + 0 \, \text{m/s})} \]\[ f_1' = \frac{392 \, \text{Hz} \cdot (358.0 \, \text{m/s})}{343 \, \text{m/s}} \]\[ f_1' \approx 409.62 \, \text{Hz} \][/tex]

(b) Frequency from the moving whistle [tex](\(f_2'\))[/tex] as heard by the listener:

[tex]\[ f_2' = \frac{f \cdot (v + v_o)}{(v - v_s)} \]\[ f_2' = \frac{392 \, \text{Hz} \cdot (343 \, \text{m/s} + 15.0 \, \text{m/s})}{(343 \, \text{m/s} - 35.0 \, \text{m/s})} \]\[ f_2' = \frac{392 \, \text{Hz} \cdot (358.0 \, \text{m/s})}{308 \, \text{m/s}} \]\[ f_2' \approx 454.55 \, \text{Hz} \][/tex]

(c) Beat frequency detected by the listener:

The beat frequency is the difference in frequencies between the two whistles as heard by the listener:

[tex]\[ \text{Beat frequency} = |f_1' - f_2'| \]\[ \text{Beat frequency} = |409.62 \, \text{Hz} - 454.55 \, \text{Hz}| \]\[ \text{Beat frequency} \approx 44.93 \, \text{Hz} \][/tex]

So, the answers are:

(a) The frequency from the stationary whistle as heard by the listener is approximately [tex]\(409.62 \, \text{Hz}\)[/tex].

(b) The frequency from the moving whistle as heard by the listener is approximately [tex]\(454.55 \, \text{Hz}\).[/tex]

(c) The beat frequency detected by the listener is approximately [tex]\(44.93 \, \text{Hz}\).[/tex]

Answer :

The time is [tex]1\times10^{5}\ yrs[/tex]

(B) is correct option.

Explanation :

Given that,

Diameter = 100000 light year

Velocity = 0.99 c

We need to calculate the time

Using formula of time

[tex]t = \dfrac{d}{v}[/tex]

Put the value into the formula

[tex]t =\dfrac{100000 c}{0.99c}[/tex]

[tex]t =1\times10^{5}\ yrs[/tex]

Hence, The time is [tex]1\times10^{5}\ yrs[/tex]

Answer:

Wavelength, [tex]\lambda=3.01\ m[/tex]

Explanation:

It is given that,

Frequency, f = 99.5 MHz = 99.5 × 10⁶ Hz

We need to find the wavelength of the radio waves from an FM station operating at above frequency. The relationship between the frequency and the wavelength is given by :

[tex]c=f\lambda[/tex]

[tex]\lambda=\dfrac{c}{f}{[/tex]

c = speed of light

[tex]\lambda=\dfrac{3\times 10^8\ m/s}{99.5\times 10^6\ Hz}[/tex]

[tex]\lambda=3.01\ m[/tex]

So, the wavelength of the radio waves from an FM station is 3.01 m. Hence, this is the required solution.

Given:

applied tensile stress, [tex]\sigma[/tex] = 170 MPa

radius of curvature of crack tip,  [tex]r_{t}[/tex] =  [tex]3.5\times 10^{-4}[/tex] mm

crack length = [tex]4.5\times 10^{-2}[/tex] mm

half of internal crack length, a = [tex]\frac{crack length}{2} = \frac{4.5\times 10^{-2}}{2}[/tex]

a =  [tex]2.25\times 10^{-2}[/tex]

Formula Used:

[tex]\sigma _{max} =  2\times\sigma \sqrt{\frac{a}{r_t}}[/tex]

Solution:

Using the given formula:

[tex]\sigma _{max} = 2\times170 \sqrt{\frac{2.25\times 10 ^{-2}}{3.5\times 10^{-4}}}[/tex]

[tex]\sigma _{max}[/tex] = 2726 MPa (395372.9 psi)

The magnitude of the maximum stress at the tip of an internal crack can be determined using the stress concentration factor formula.

The magnitude of the maximum stress at the tip of an internal crack can be determined using the formula for the stress concentration factor, which is the ratio of the maximum stress to the applied stress. The stress concentration factor (Kt) for an internal crack can be calculated using the following equation:

Where Kt is the stress concentration factor, a is the crack length, and r is the radius of curvature of the crack.

Plugging in the given values:

Kt = 1 + 2 * (4.5 x 10-2 mm) / (3.5 x 10-4 mm) = 127

The magnitude of the maximum stress can be calculated by multiplying the stress concentration factor by the applied stress:

Maximum Stress = 127 * 170 MPa = 21,590 MPa