Two satellites, one in geosynchronous orbit (T = 24 hrs) and one with a period of 12 hrs, are orbiting Earth. How many times larger than the radius of Earth is the distance between the orbits of the two satellites.

Answer:

240 m

120 m

Explanation:

d = Path difference = 120 m

For destructive interference

Path difference

[tex]d=\dfrac{\lambda}{2}\\\Rightarrow \lambda=2d\\\Rightarrow \lambda=2\times 120\\\Rightarrow \lambda=240\ m[/tex]

The longest wavelength is 240 m

For constructive interference

[tex]d=\lambda\\\Rightarrow 120\ m=\lambda[/tex]

The longest wavelength is 120 m

Final answer:

The longest wavelength for destructive interference at point Q, where the path difference is 120 m, is 240 m.

Explanation:

The longest wavelength for which there will be destructive interference at point Q can be found by considering the path difference between the waves from the two antennas at point Q. The waves must differ by an odd multiple of ½ wavelengths for destructive interference to occur. The distance from antenna A to point Q is 120 m + 40 m = 160 m, and the distance from antenna B to point Q is 40 m. Therefore, the path difference is 120 m.

For the first occurrence of destructive interference, the path difference should be ½ wavelength, so we calculate the longest wavelength (λ) as:

½ λ = 120 m

λ = 240 m

This is the longest wavelength that causes destructive interference at point Q.

Answer:

Complete question

In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.095 T magnetic field.

a. What electric field strength, in volts per mater, is needed to select a speed of 4.2 x 10^6 m/s?

b. What is the voltage, in kilovolts, between the plates if they are separated by 0.95 cm?

Explanation:

Given that,

magnetic field B = 0.095T

Speed of particle v = 4.2 ×10^6m/s

Separation between plate d = 0.95cm

d = 0.95/100 = 0.0095m

a. Using the mass spectrometer velocity selector relationship between the electric field and magnetic field.

v = E/B

Where

v is the speed selector

B is magnetic field

E is electric field

Therefore, E = vB

E = 4.2 × 10^6 × 0.095

E = 0.399× 10^6

E = 3.99 × 10^5 V/m

b. Voltage?

The relationship between electric field and potential difference between the two plates is given as

V = Ed

V = 3.99 × 10^5 × 0.0095

V = 3790.5 V

To kV, 1kV = 1000V

Then, V = 3.7905kV

V ≈ 3.791 kV

The electric field strength needed in a mass spectrometer to select a velocity of 4.00 × 106 m/s with a 0.100-T magnetic field is 400,000 N/C. The voltage required across plates separated by 1.00 cm is 4,000 V.

In a velocity selector within a mass spectrometer, a charged particle remains undeflected when the electric force equals the magnetic force. This condition is given by qE = qvB, where q is the charge, E is the electric field strength, v is the velocity of the particle, and B is the magnetic field strength. The question asks for the electric field strength needed to select particles with a velocity of 4.00 × 106 m/s when subjected to a 0.100-T magnetic field. Using the condition for balance, E = vB, we can substitute the given values to find E = 4.00 × 106 m/s × 0.100 T = 400,000 N/C. For part (b), voltage (V) can be determined by the relation V = Ed, where d is the separation between the plates. For a plate separation of 1.00 cm, or 0.01 m, the voltage is V = 400,000 N/C × 0.01 m = 4,000 V.

Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light [tex]\lambda =550nm=550\times 10^{-9}m[/tex]

Distance D = 8 m

Distance between first minimum dark fringe and the central maximum is 2 mm

So [tex]x=3\times 10^{-3}m[/tex]

We have to find the width of the slit

For the first order wavelength is equal to [tex]\lambda =\frac{x}{D}\times a[/tex], here a width of slit

So [tex]a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm[/tex]

So width of the slit will be equal to 1.47 mm

Answer:

a)   E = -4.8 10⁻⁵ N / C , b) E = 0 , c) E = 0

Explanation:

For this exercise let's use Gauss's law

         Ф = E. dA = [tex]q_{int}[/tex] / ε₀

As a Gaussian surface we use a sphere, whereby the electric field lines are parallel to the normal area, and the scalar product is reduced to the algebraic product

a) the field for r = 30 cm

At this point we place our Gaussian surface and see what charge there is inside, which is the charge of the solid sphere (r> 20cm), the charge on the outside does not contribute to the flow

           E = q_{int} / A ε₀

           q_{int} = -4.8 10-16 C

The area of ​​a sphere is

              A = 4π R²

We replace

             E = -4.8 10⁻¹⁶ / 4π 0.30² 8.85 10⁻¹²

             E = -4.8 10⁻⁵ N / C

b) r = 45 cm

 This point is inside the spherical conductor shell, as in an electric conductor in electrostatic equilibrium the charges are outside inside the shell there is no charge for which the field is zero

        E = 0

c) R = 60 cm

This part is outside the two surfaces

   The chare inside is

            q_{int} = -4.8 10⁻¹⁶ + 4.8 10⁻¹⁶

            q_{int} = 0

Therefore the electric field is

         E = 0

Answer:

(a) With a short line, the A,B,C,D parameters are:

    A = 1pu    B = 1.685∠60.8°Ω    C = 0 S    D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 [tex]KV_{LL}[/tex]

(c) The sending-end voltage for 0.9 leading power factor is 33.40 [tex]KV_{LL}[/tex]

Explanation:

(a)

Considering the short transition line diagram.

Apply kirchoff's voltage law to the short transmission line.

Write the equation showing the relations between the sending end and the receiving end quantities.

Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

Determine the line-to-neutral receiving end voltage.

Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

Determine the sending-end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

Answer: 10^-3 V^2/Hz

Explanation:

1 Hz:

Su(f) = No * |H(f)|^2

= 10^-3 * 1/(1+(2*pi*f*R*C)^2)

= 10^-3 V^2/Hz

Answer:

1) [tex]v_{A} \approx 8.272\,\frac{m}{s}[/tex]

Explanation:

1) Let assume that the campion begins running at a height of zero. The movement of the Olympic champion is modelled after the Principle of Energy Conservation:

[tex]K_{A} = K_{B} + U_{g,B}[/tex]

[tex]\frac{1}{2}\cdot m \cdot v_{A}^{2} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + m \cdot g \cdot h_{B}[/tex]

[tex]\frac{1}{2} \cdot v_{A}^{2} = \frac{1}{2} \cdot v_{B}^{2} + g \cdot h_{B}[/tex]

The minimum speed is obtained herein:

[tex]v_{A}=\sqrt{v_{B}^{2} + 2 \cdot g \cdot h}[/tex]

[tex]v_{A} = \sqrt{(6.7\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)}[/tex]

[tex]v_{A} \approx 8.272\,\frac{m}{s}[/tex]

The minimum speed he need at launch point is [tex]8.27m/s[/tex]

Energy is neither be created nor be destroyed just change into one form to another form.

              [tex]\frac{1}{2}mv^{2} =\frac{1}{2}mv_{a}^{2} +mgh\\ \\v=\sqrt{v_{a}^{2}+2gh }[/tex]

Where,

Given that, [tex]v_{a}=6.7m/s,h=1.2m,g=9.8m/s^{2}[/tex]

Substitute all values in above relation.

             [tex]v=\sqrt{(6.7)^{2}+2*9.8*1.2 } \\\\v=\sqrt{68.41}=8.27m/s[/tex]

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The fundamental frequency of the string will remain at 250 Hz, even when the string is doubled in length and the tension is quadrupled, due to the relationship between string length, tension, and frequency.

When a transverse standing wave on a string is vibrating at its fundamental frequency, the length of the string and the tension within it play crucial roles in determining that frequency. The fundamental frequency (f) is given by f = (1/2L) * sqrt(T/μ), where L is the length of the string, T is the tension, and μ is the linear mass density of the string.

In the original scenario, the string vibrates at a fundamental frequency of 250 Hz. If the same piece of string is made twice as long and the tension is increased fourfold, the new fundamental frequency F' can be found using the formula mentioned above.

Considering the string is now twice as long (L' = 2L), and the tension is four times greater (T' = 4T), we can substitute these into the original equation for fundamental frequency: f' = (1/2L') * sqrt(T'/μ). By inserting these new values, we can determine that the new fundamental frequency f' will be the same as the original frequency f, because the doubling of the string length and the quadrupling of the tension will have cancelling effects.

Final answer:

The fundamental frequency of a string that is twice as long and with four times the tension would be half the original frequency; therefore, the new fundamental frequency would be 125 Hz.

Explanation:

The student is asking about how the fundamental frequency of a string changes when its length is doubled and the tension is increased by four times. The fundamental frequency f of a string fixed at both ends can be described by the equation f = √(T/μ)/(2L), where T is the tension, μ (mu) is the mass per unit length, and L is the length of the string. When we double the length of the string, the new length is 2L. When we quadruple the tension, the new tension is 4T. Substituting these values into the equation, the new frequency f' becomes f' = √(4T/μ)/(2 × 2L) = √(T/μ)/2L = f/2.

Therefore, the new fundamental frequency of the longer string with four times the tension is half the original frequency, which is 125 Hz in this case.

Answer: a) vox = vo × cos θ, b) voy =vo× sin θ,

c) H=2.94 m, d) t = vo sinθ / g, e) R = 38.57 m

Explanation:

A)

The velocity v0 is at angle θ to the horizontal.

The horizontal component of vo (vox), vo and the vertical component of vo (voy) all form a right angle triangle.

With vo as the hypotenus, vox as the adjacent and voy as the opposite.

To get vox, we relate vo and vox ( hypotenus and adjacent)

From trigonometry

Cos θ relates hypotenus and adjacent, hence we have that

Cos θ = vox/vo

vox = vo × cos θ

B)

To get the vertical component of vo, we relate vo and voy ( hypotenus and opposite).

According to trigonometry, sin θ relates hypotenus and opposite, hence we have that

Sin θ = voy/vo

voy =vo× sin θ

C)

The formulae for the maximum height of a projectile motion is given as

H = vo² (sin θ)²/2g

Where g = acceleration due to gravity = 9.8 m/s²

By substituting the parameters, we have that

H = 26² × (sin 17)²/2(9.8)

H = 676 × 0.0854/19.6

H = 57.7304/ 19.6

H = 2.94 m

D)

This is the motion of a projectile and the conditions at maximum height are vy = 0 and ay = - g

From the equation of motion

vy = voy - gt

0 = voy - gt

But voy = vo sinθ

0 = vo sinθ - gt

gt = vo sinθ

t = vo sinθ / g

E)

The horizontal distance covered formulae is given by

R = u² sin2θ/g

R = 26² × sin 2(17)/9.8

R = 676 × sin 34/ 9.8

R = 378.014/ 9.8

R = 38.57 m

The correct Answer is:

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Answer:

The reason the bird is not electrocuted is due to some facts about circuit:

1. Completeness of circuit- This circuit needs to be

complete in order for current to flow. The bird standing on only one wire has not completed the circuit.

2. A potential difference: Another factor deciding

the direction of flow of current is (electric)

potential. Current always flows from a higher

potential to a lower potential. In other words it

can be said that electrons flow from lower

potential to higher one. (the direction of electric

current is opposite to that of the electrons). So

we need the potential difference for current to

flow. The bird standing on only one wire has no potential difference.

3. Path of least Resistance- Factor that decides

the path a current will flow in case of parallel

paths is the (electric) resistance offered by the

path. Current will always flow in the path that

offers least resistance. The leg of a bird has high resistance.

Explanation:

It has no potential difference as both the legs of bird are touching the same wire at same constant potential. ... If the bird would touch the ground while sitting on the wire or flap its wings and touch another electric wire with a different voltage, then it would get shocked and likely die by electrocution.

while in the other hand, the Bird that touches two wires with it wings at the same time will get electrocuted because it has completed a circuit and the its feathers created a potential difference .

A bird is not electrocuted on a high-voltage power line because it's not completing a circuit. By contrast, if a large bird touches two wires simultaneously, it creates a closed circuit, allowing electricity to flow through its body, leading to electrocution. Circuit breakers serve to prevent excess current flow.

A bird sitting on a high-voltage power line does not get electrocuted because it is not completing a circuit. The bird is not touching the ground or another line, so the electricity stays within the conducting wire and doesn't pass through the bird. The bird is safe as long as it touches only one wire.

However, in contrasting scenarios, a larger bird could span the distance between two wires with its wings. If this happens, the bird would close the circuit, allowing the electricity to pass through its body, which would lead to electrocution. This is due to the significance of voltage difference between two given points to create an electric shock, as there is a potential difference between the two wires.

Also, the role of circuit breakers is important in the context of high voltage power supply. Circuit breakers prevent excess current flow and accidental contact with the line. They interrupt the electricity flow anytime there's a fault, safeguarding anyone who may come in contact with the cable.

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Final answer:

The acceleration of the child in the wagon can be calculated using Newton's Second Law of Motion. The sum of the forces acting on the system is calculated by adding the forces exerted by the children and subtracting the force of friction. Using the given values, the acceleration can be determined by dividing the sum of the forces by the mass of the system. If the force of friction is increased to 20.0 N, the acceleration can be recalculated using the new value.

Explanation:

To calculate the acceleration of the child in the wagon, we can use Newton's Second Law of Motion:

ΣF = ma

Where ΣF is the sum of all the forces acting on the system, m is the mass of the system, and a is the acceleration.

In this case, the forces acting on the system are the forces exerted by the two children and the force of friction.

Using the given values:

Force1 = 75.0 N

Force2 = 95.0 N

Friction = 12.0 N

Mass = 21.0 kg

The sum of the forces acting on the system is:

ΣF = Force1 + Force2 - Friction

Substituting the values:

ΣF = 75.0 N + 95.0 N - 12.0 N

ΣF = 158.0 N

Now we can calculate the acceleration:

a = ΣF / m

Substituting the values:

a = 158.0 N / 21.0 kg

a = 7.52 m/s²

Therefore, the acceleration of the child in the wagon is 7.52 m/s².

(d) If friction is 20.0 N, we can recalculate the acceleration using the new value:

ΣF = 75.0 N + 95.0 N - 20.0 N

ΣF = 150.0 N

a = ΣF / m

a = 150.0 N / 21.0 kg

a = 7.14 m/s²

Therefore, if friction is 20.0 N, the acceleration of the child in the wagon would be 7.14 m/s².

Answer:

a. 55.6v

b. [tex]1.61*10^{-7}J[/tex]

Explanation:

Data given

charge 1=+4.10nC

charge 2 =-2.40nC

distance, r= 55cm =0.55m

a. the electric potential at the mid point is the sum of the potential due to individual charge.

The electric potential is expressed as  

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

since we are interested in the electric potential at the mid point, we have

[tex]V=\frac{kq_1}{r/2}-\frac{kq_2}{r/2}\\ V=\frac{2k}{r}(q_1-q_2)\\ V=\frac{2*9*10^9}{0.55}(4.1-2.4)*10^{-9}\\ V=55.6v[/tex]

Hence the electric potential at the mid-point is 55.6v

b. to calculate the potential energy, we use the formula below

[tex]U=\frac{kq_1q_2}{r} \\U=\frac{9*10^9 *4.10*10^{-9}*2.4*10^{-9}}{0.55}\\ U=1.61*10^{-7}J[/tex]  

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron=[tex]v=6.1\times 10^7m/s[/tex]

Charge on electron, [tex]q=e=1.6\times 10^{-19} C[/tex]

Mass of electron,[tex]m_e=9.1\times 10^{-31} kg[/tex]

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

[tex]V=\frac{v^2m_e}{2e}[/tex]

Where V=Potential difference

[tex]m_e=[/tex]Mass of electron

v=Velocity of electron

Using the formula

[tex]V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}[/tex]

[tex]V=10581.59 V[/tex]

Hence, the potential difference=10581.59 V

Final answer:

To accelerate electrons to a speed of 6.1 × 10^7 m/s in a constant magnetic field with a strength of 0.65 T, the potential difference required in the first part of the experiment is approximately 88.6 volts.

Explanation:

To accelerate electrons to a speed of 6.1 × 10^7 m/s in a constant magnetic field with a strength of 0.65 T, we need to calculate the potential difference required in the first part of the experiment. The formula for the potential difference is given by:

V = (1/2)m*(v^2)/(q * B)

Where V is the potential difference, m is the mass of the electron (9.11 × 10^-31 kg), v is the velocity of the electron (6.1 × 10^7 m/s), q is the charge of the electron (-1.6 × 10^-19 C), and B is the magnetic field strength (0.65 T).

Plugging in the values into the formula, we get:

V = (1/2)(9.11 × 10^-31 kg)(6.1 × 10^7 m/s)^2/(-1.6 × 10^-19 C)(0.65 T)

Simplifying the expression, we find that the potential difference required is approximately 88.6 volts.

Answer:

[tex]6.65\times 10^5 m[/tex]

Explanation:

We are given that

Magnetic field=B=[tex]4\times 10^{-8} T[/tex]

[tex]v=2.7\times 10^7 m/s[/tex]

We have to find the height of proton from the surface of the Earth.

Mass of proton,[tex]m_p=1.67\times 10^{-27} kg[/tex]

Charge on proton,[tex]q=1.6\times 10^{-19} C[/tex]

Radius of Earth, r=[tex]6.38\times 10^6 m[/tex]

Centripetal force due to rotation of proton=[tex]\frac{mv^2}{r+h}[/tex]

Magnetic force,F=[tex]qvB[/tex]

[tex]\frac{mv^2}{r+h}=qvB[/tex]

[tex]\frac{mv}{r+h}=qB[/tex]

Substitute the values

[tex]\frac{1.67\times 10^{-27}\times (2.7\times 10^7)}{6.38\times 10^6+h}=1.6\times 10^{-19}\times 4\times 10^{-8}[/tex]

[tex]6.38\times10^6+h=\frac{1.67\times 10^{-27}\times 2.7\times 10^7}{1.6\times 10^{-19}\times 4\times 10^{-8}}[/tex]

[tex]6.38\times10^6+h=7.045\times 10^6[/tex]

[tex]h=7.045\times 10^6-6.38\times 10^6[/tex]

[tex]h=0.665\times 10^6=6.65\times 10^5 m[/tex]

Approximately 1.6 million joules of work is required to pump the liquid out of the spherical tank through a 1 m long spout at the top.

The work done to pump the liquid out of a tank can be calculated using the formula W = ρgVh, where ρ is the density of the liquid, g is the acceleration due to gravity, V is the volume of the liquid, and h is the height up to which the liquid is pumped. Since the tank is spherical and half full, the volume of the liquid is ½(4/3πr³), or 2πr³. Substituting the given values: ρ = 900 kg/m³, g = 9.8 m/s², r = 3 m, and h is the radius of the sphere plus the length of the spout (3 m + 1 m = 4 m), we get W = 900 kg/m³ * 9.8 m/s² * 2π(3 m)³ * 4 m ≈ 1.6 * 10⁶ J. Therefore, approximately 1.6 million joules of work is required to pump the liquid out of the tank.

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Answer: h = 0.52m

Explanation:

Using the equation of out flow;

A1 × V1 = A2 ×V2

Where A1 = area of the first nozzle

A2 = area of the second nozzle

V1= velocity of flow out from the first nozzle

V2 = velocity of flow out from 2nd nozzle

But AV= area of nozzle × velocity of water = volume of water per second(m³/s).

Now we can set A×V = Area of nozzle × height of rise.

Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)

D1 = 20mm= 0.02m; h1 = 0.13m

D2 = 10mm = 0.01m; h2= ?

h2 = π(D1/2)²× h1 /π(D2/2)²

h2 = (0.02/2)² × 0.13/(0.01/2)²

= (0.01)² ×0.13 /(0.005)²

= 1.3 × 10^-5/(5 × 10^-3)²

= 1.3 × 10^-5/25 × 10^-6

= (1.3/25) 10^-5 × 10^6

= 0.052× 10

= 0.52m

Answer:

(a) 21.44 ft/s

(b) 0 ft/s

(c) 19.51 ft/s

Explanation:

2 in = 2/12 ft = 0.167 ft

For steady laminar flow, the function of the fluid velocity in term of distance from center is modeled as the following equation:

[tex]v(r) = v_c\left[1 - \frac{r^2}{R^2}\right][/tex]

where R = 0.167 ft is the pipe radius and [tex]v_c[/tex] is the constant fluid velocity at the center of the pipe.

We can integrate this over the cross-section area of the in order to find the volume flow

[tex]\dot{V} = \int\limits {v(r)} \, dA \\= \int\limits^R_0 {v_c\left[1 - \frac{r^2}{R^2}\right]2\pi r} \, dr\\ = 2\pi v_c\int\limits^R_0 {r - \frac{r^3}{R^2}} \, dr\\ = 2\pi v_c \left[\frac{r^2}{2} - \frac{r^4}{4R^2}\right]^R_0\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^4}{4R^2}\right)\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^2}{4}\right)\\= 2\pi v_c R^2/4\\=\pi v_c R^2/2\\A = \pi R^2\\\dot{V} = Av_c/2\\[/tex]

So the average velocity

[tex]v = \dot{V} / A = v_c/2 = 10.72[/tex]

[tex]v_c = 10.72*2 = 21.44 ft/s[/tex]

b) At the wall of the pipe, r = R so [tex]v(R) = v_c(1 - 1) = 0 ft/s[/tex]

c) At a distance of 0.6 in = 0.6/12 = 0.05 ft

[tex]v(0.05) = v_c(1 - 0.05^2/0.167^2) = 0.91v_c = 0.91*21.44 = 19.51 ft/s[/tex]

Answer:

The answers to the questions are;

(a) The velocity at the center of the pipe is 21.44 ft/s

(b) The velocity at the wall of the pipe is 0 ft/s

(c) The velocity at a distance of 0.6 inches from the center-line is 19.63 ft/s.

Explanation:

To solve the question, we note that

The velocity profile in the cross section of a circular pipe with laminar flow is given by

U = 2×v×[1 - (r/r₀)²]

Where

U = The sought velocity at a point

r = Pipe radius where velocity is sought

r₀ = Internal radius of pipe = for  2-in Schedule 40 steel pipe =  2.067 in 52.6 mm

v = Average velocity of flow = 10.72 ft/s = 3.2675 m/s

Therefore we have

(a) The velocity at the center of the pipe

At the center r = 0 so we have

U = 2×v×[1 - (r/r₀)²]

At center U = 2×10.72 ft/s×[1 - (0/2.067 in)²] =  2×10.72 ft/s = 21.44 ft/s

(b) The velocity at the wall of the pipe is given by

r = r₀ ⇒ U = 2×v×[1 - (r/r₀)²] ⇒ U = 2×v×[1 - (r₀/r₀)²]

= U = 2×v×[1 - (1)²] =  2×v×0 = 0

The velocity at the wall of the pipe is 0 ft/s

(c) The velocity at a distance of 0.6 inches from the center-line is given by

U = 2×v×[1 - (r/r₀)²] = 2×10.72 ft/s×[1 - (0.6/2.067)²]  =  19.63 ft/s.

Answer:

2s

Explanation:

The position function of the motion can be expressed as:

[tex]s = s_0 + v_0t + at^2/2[/tex]

where [tex]s_0 = <0,0,0>[/tex] is the origin where the particle starts off, [tex]v_0 = <0,0,2> m/s[/tex] and a = <1,2,0> m/s2. In the 3-coordinate system it can be written as

[tex]s = <0 + 0t+ t^2/2, 0 + 0t + 2t^2/2, 0 + 2t + 0t^2/2> = <t^2/2, t^2, 2t>[/tex]

For the particle to hit the 2x+y-z=4 plane then its coordinates must meet that criteria

[tex]2t^2/2 +t^2-2t = 4[/tex]

[tex]2t^2 - 2t -4 = 0[/tex]

[tex]t^2 - t - 2 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{1\pm \sqrt{(-1)^2 - 4*(1)*(-2)}}{2*(1)}[/tex]

[tex]t= \frac{1\pm3}{2}[/tex]

t = 2 or t = -1

Since t can only be positive we will pick t = 2s

Answer:

19.6 m

Explanation:

The work-energy theorem applies here,

The theorem states that the change in momentum of a particle between two points is equal to the work done in moving the force between the two distance.

ΔK.E = W

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Kinetic energy = (1/2)(m)(v²)

m = 3.5 kg, v = 1.6 m/s

Final kinetic energy = 0 J, since the block of mass comes to rest.

Initial kinetic energy = (1/2)(3.5)(1.6²) = 4.48 J

ΔK.E = - 4.48 J

The workdone on the block of mass is done by the frictional force, F, which acts opposite to the direction of the displacement.

W = - Fd = - 1.6 F

ΔK.E = W

- 4.48 = - 1.6 F

F = 2.8 N

when the initial velocity is increased by a factor 3.5,

v = 1.6×3.5 = 5.6 m/s

Final kinetic energy = 0 J, since the block of mass comes to rest.

Initial kinetic energy = (1/2)(3.5)(5.6²) = 4.48 J

ΔK.E = - 54.88 J

The workdone on the block of mass is done by the frictional force, F, which acts opposite to the direction of the displacement. The frictional force is the same as above.

W = - Fd = - (2.8) d

ΔK.E = W

- 54.88 = - 2.8 d

d = 19.6 m

Compared to 1.6 m, 19.6 m is an increase by a factor 12.25.

The given question is incomplete. The complete question is as follows.

Two hollow conducting spheres (radius r) with a uniformly distributed charge are placed a distance d apart center to center. A thin wire with a switch S is connected to the surface of each sphere. The switch is initially open.

a. What is the potential between points a and b?

b. If the switch is then closed, what is the charge on each sphere at time [tex]t \rightarrow \infty[/tex].

c. What is the potential between points a and b after the sphere reaches its steady state?

Explanation:

(a) In order to bring a positive test charge from infinity to a point 'a', the work done is equal to the potential energy of the charge at point 'a'.

Hence,      [tex]V_{a} = \frac{1}{4 \pi \epsilon_{o}} \frac{q}{a}[/tex]

Now, work done in bringing a positive test charge from infinity to point 'b' is equal to the potential energy of the charge at point 'b'.

      [tex]V_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{-q}{b}[/tex]      

     [tex]V_{a} - V_{b} = \frac{q}{4 \pi \epsilon_{o}}(\frac{1}{a} + \frac{1}{b})[/tex]

Therefore, the potential between points a and b is as follows.

  [tex]V_{a} - V_{b} = \frac{q}{4 \pi \epsilon_{o}}(\frac{1}{a} + \frac{1}{b})[/tex]

(b)   As the spheres are connected through a conducting wire then charges will flow from one sphere to another unless and until the charge on both the sphere will become equal. In this case, it is equal to zero.

(c)   Since, the charge of both the spheres is equal to zero so, no work is necessary to bring another charge to a and b. Therefore, potential difference between the points will also become equal to zero.    

Answer: e

Explanation:

Argument against the person, circumstantial.

Answer:

2. Microwaves have lower frequency, same speed, and longer wavelength than visible light.

Explanation:

Microwaves are a form of electromagnetic radiation. Most people are familiar with this type of waves because they are used in microwave ovens. When compared to visible light, microwaves have lower frequency, same speed and longer wavelength than visible light. The prefix "micro" is used to indicate that microwaves are smaller (shorter wavelengths) than radio waves.

"The correct option is 2. Microwaves have lower frequency, same speed, and longer wavelength than visible light.

To understand why this option is correct, let's consider the relationship between frequency, wavelength, and speed for electromagnetic waves, which is given by the equation:

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

 1. Speed of light (in a vacuum) is constant for all electromagnetic waves, including microwaves and visible light. Therefore, the speed of microwaves and visible light is the same.

2. Microwaves have a lower frequency than visible light. The frequency of microwaves typically ranges from 300 MHz to 300 GHz, while the frequency of visible light ranges from approximately 430 THz to 750 THz. Since microwaves have a lower frequency, they also have a longer wavelength according to the equation [tex]\( c = f \times \lambda \)[/tex].

3. Since the speed of all electromagnetic waves is the same in a vacuum, and microwaves have a lower frequency, they must have a longer wavelength to maintain the constant speed. This is consistent with the relationship , where a lower frequency  requires a longer wavelength  to keep the speed constant.

In summary, microwaves have a lower frequency and longer wavelength than visible light, but they travel at the same speed. This makes option 2 the correct choice."

Answer:

Explanation:

Battery voltage is 6V

A current of 0.361A is draw the voltage reduces to, 5.07V

This shows that the appliances resistance that draws the currents is

Using KVL

The battery has an internal resistance r

V=Vr+Va

Vr is internal resistance voltage

Va is appliance voltage

6=5.07+Va

Va=6-5.07

Va=0.93

Using ohms law to the resistance of the appliance

Va=iR

R=Va/i

R=0.93/0.361

R=2.58ohms

Then if the circuit draws a current of 0.591A

Then the voltage across the load is

V=iR

Va=0.591×2.58

Va=1.52V

Then the voltage drop at the internal resistance is

V=Vr+Va

Vr=V-Va

Vr=6-1.52

Vr=4.48V

Answer:

V = 4.48 V

Explanation:

• As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.

• We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

 [tex]V_{rint} = I* r_{int}[/tex]

• The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

   [tex]V = V_{b} - V_{rint} = 5.07 V = 6.00 V - 0.361 A * r_{int}[/tex]

• We can solve for rint, as follows:

 [tex]r_{int} =\frac{V_{b}- V_{rint} }{I} = \frac{6.00 V - 5.07 V}{0.361A} = 2.58 \Omega[/tex]

• When the circuit draws from battery a current I of 0.591A, we can find the potential difference between the terminals of the battery, as follows:

 [tex]V = V_{b} - V_{rint} = 6.00 V - 0.591 A * 2.58 \Omega = 4.48 V[/tex]

• As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.  

Answer:

Zero work done,since the body isn't acting against  or by gravity.

Explanation:

Gravitational force is usually  considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.

Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.

The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.

The electromagnetic force generates the heat and light produced by the sun and other stars.

The force that generates the heat and light produced by the sun and other stars is the electromagnetic force. This force is responsible for holding atoms together and producing electromagnetic radiation, which includes heat and light. It is much stronger compared to the weak force and gravity.

Answer:

a) [tex]v(2) = 3.9\,\frac{m}{s}[/tex], b) [tex]v(4) = -15.7\,\frac{m}{s}[/tex]

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

[tex]v(t) = 23.5 -9.8\cdot t[/tex]

The velocities at given instants are, respectivelly:

[tex]v(2) = 3.9\,\frac{m}{s}[/tex]

[tex]v(4) = -15.7\,\frac{m}{s}[/tex]

Answer:

0.26086 Hz

Explanation:

solution:

The relative speed of the wave is:

v = 4.9+1.1 = 6 m/s

the frequency is:

f = v/λ

 = 6/23

 = 0.26086 Hz

Answer:

Explanation:

1 )

Charge density of left plate

= - Q / A

Charge density of right plate

= + Q / A

2 )

capacitance c = ε₀ A / d

potential difference = charge / capacitance

= Q / [ ε₀ A / d ]

= Q d  / ε₀ A

Final answer:

The charge density on each plate is ±Q/A, the electric potential difference between the plates is calculated using the electric field and the separation distance, and the capacitance C=ε₀A/d demonstrates that the size of the plates does not affect their capacitance as long as their proportion remains constant.

Explanation:

When considering two parallel plates each with area A and charges of +Q and -Q respectively, separated by a distance d, we can address the posed questions systematically.

Finding the Charge Density on the Plates

The surface charge density σ is defined as charge per unit area. Given the total charge +Q or -Q and the area A of each plate, the charge density on each plate is σ = ±Q/A. This is a direct result of the uniform distribution assumption of the charges across the plates.

Finding the Electric Potential Difference Between the Plates

The electric field E created between the plates by the charge distribution is uniform and can be represented as E = σ/ε₀, where σ is the surface charge density and ε₀ is the vacuum permittivity. Consequently, the electric potential difference V between the plates can be derived from the relation V = Ed, linking the electric field and the separation of the plates.

Demonstrating the Capacitance of Enlarged Plates Remains Constant

The capacitance C of a parallel-plate capacitor is given by C = ε₀A/d, which is independent of the charge on the plates. This formula illustrates that the capacitance is solely dependent on the physical characteristics of the capacitor (i.e., the area of the plates A, the distance between them d, and the permittivity of free space ε₀), and does not change with the amount of charge nor with the size of the plates as long as their proportional relationship remains constant.

Given Information:

Velocity of water flow = 2 m/s

Temperature of water = 17° C

Heat dissipation = 2500 W

Area of copper plate = 0.04 m²

Required Information:

Temperature of copper plate  = ?

Answer:

[tex]T_{p} = 27[/tex]° [tex]C[/tex]

Explanation:

Each chip dissipates 25 W so 100 chips will dissipate 25*100 = 2500 W

Area of copper plate = 0.2*0.2 = 0.04 m²

According to the convection rate equation

[tex]T_{p}= T_{w} + \frac{q}{hA}[/tex]

Where Tp is the temperature of copper plate, Tw is the temperature of water, q is the the heat dissipation of chips, A is the area of copper plate and h is the convection coefficient

The convection coefficient is given by turbulent flow correlation

[tex]h = Nu_{L}(k/L) =0.037Re_{L}^{4/5}P_{r}^{1/3}(k/L)[/tex]

Where Nu is Nusselt number, Re is Reynolds number, Pr = 5.2 is Prandtl number and k = 0.620 W/m.K

[tex]Re_{L}= uL/v[/tex]

Where u = 2 m/s and L = 0.2 m and  v = 0.96x10⁻⁶m² /s

[tex]Re_{L}= 2*0.2/0.96x10^{-6}[/tex]

[tex]Re_{L}= 416666.66[/tex]

[tex]h = 0.037(416666.66)^{4/5}(5.2)^{1/3}(0.620/0.2)[/tex]

[tex]h = 6223.89[/tex] [tex]W/m^{2}K[/tex]

[tex]T_{p}= 17 + \frac{2500}{(6223.89)0.04}[/tex]

[tex]T_{p} = 27[/tex]° [tex]C[/tex]

Therefore, the temperature of the copper plate is 27° C

Explanation:

The given data is as follows.

       Speed of the bullet (v) = 500 m/s

     Distance during one RC time constant (d) = 1 mm = [tex]1 \times 10^{-3} m[/tex]

      Capacitance (C) = 600 [tex]\mu F[/tex] = [tex]600 \times 10^{-6} F[/tex]

Hence, formula for speed of the bullet is as follows.

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

or,           t = [tex]\frac{d}{v}[/tex]

Time constant for RC circuit is as follows.

          t = RC

        R = [tex]\frac{t}{C}[/tex]

            = [tex]\frac{d}{vC}[/tex]

            = [tex]\frac{1 \times 10^{-3}}{500 \times 600 \times 10^{-6}}[/tex]

            = [tex]3.33 \times 10^{-3} ohm[/tex]

Thus, we can conclude that resistance in the flash tube is [tex]3.33 \times 10^{-3} ohm[/tex].