Which wave has the highest frequency when traveling through the indicated metal?
Wavelength of 2.1 m in copper
Wavelength of 1.4 m in lead
Wavelength of 0.9 m in brass
Wavelength of 1.8 m in gold

Answer:

Solar radiation may be converted directly into electricity by solar cells (photovoltaic cells). In such cells, a small electric voltage is generated when light strikes the junction between a metal and a semiconductor (such as silicon) or the junction between two different semiconductors.

Explanation:

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Solar radiation may be converted directly into electricity by solar cells or photovoltaic cells. This is how solar energy is used to generate electricity.

Solar energy is the radiant light and heat from the Sun that is captured and used in a variety of technologies, including solar power to generate electricity, solar thermal energy, and solar architecture.

When the sun shines on a solar panel, the photovoltaic cells in the panel absorb the energy from the sun. When light strikes the junction of a metal and a semiconductor (such as silicon) or the junction of two different semiconductors, a small electric voltage is generated. This energy generates electrical charges that move in response to an internal electrical field in the cell, resulting in the flow of electricity.

Therefore, solar energy is used to generate electricity.

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We know

[tex]\boxed{\sf Work\:Done=Force\times Displacement} [/tex]

[tex]\\ \rm\longmapsto Work\:done=0.420\times 4.60[/tex]

[tex]\\ \rm\longmapsto Work\:done=1.932J[/tex]

Answer:

Explanation:

( a ) At top position , vertical component of velocity will be zero .

So answer is position no (3)

b )

At position (3) which is the topmost position , acceleration is acting due to gravity , so it will be downwards.

c )

Horizontal component of acceleration at all points will be zero because gravity acts vertically downwards.

d )

Vertical displacement will be zero at position ( 1 ) and ( 5 )

e )

Displacement is maximum at extreme  position , ie at position ( 5 )

f )

Speed is highest at position (1) and it is lowest at position ( 3 )

From highest to lowest

( 1 ) , ( 2 ) , ( 3 )

Answer:

Explanation:

Bhjb

Explanation:

its because a concrete surface is a hard surface which doesn't absorb the energy of gravitation when we fall down so we get hurt more badly..

hope this helps

Answer:

Explanation:

Relative velocity is defined as the velocity of an object B in the rest frame of another object A.

Answer:

65.87 s

Explanation:

For the first time,

Applying

v² = u²+2as.............. Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

From the question,

Given:  u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m

Substitute these values into equation 1

v² = 0²+2(1.99)(60)

v² = 238.8

v = √238.8

v = 15.45 m/s

Therefore, time taken for the first 60 m is

t = (v-u)/a............ Equation 2

t = (15.45-0)/1.99

t = 7.77 s

For the final 40 meter,

t = (v-u)/a

Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²

Substitute into the equation above

t = (0-15.45)/-0.266

t = 58.1 seconds

Hence total time taken to cover the distance

T = 7.77+58.1

T = 65.87 s

Answer:

The energy stored in a body due to either it's position or change in shape is called gravitational potential energy.

Answer:

the amplitude of the oscillation of the given mass is 0.1 m.

Explanation:

The amplitude of an oscillation is the maximum displacement of the object from the equilibrium position.

The equilibrium position of the given mass in question is at the zero (0) mark.

The maximum displacement of the object from the equilibrium position is 0.1 m upwards or 0.1 m downwards.

Therefore, the amplitude of the oscillation of the given mass is 0.1 m.

Complete question:

The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?

Answer:

The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

Explanation:

Given;

combined mass of the shotgun and arm–shoulder, m₁ = 15 kg

mass of the projectile, m₂ = 0.04 kg

speed of the projectile, u₂ = 380 m/s

let the recoil velocity of the shotgun and arm–shoulder combination = u₁

Apply the principle of conservation of linear momentum;

m₁u₁  +  m₂u₂ = 0

m₁u₁ = - m₂u₂

[tex]u_1 = -\frac{m_2u_2}{m_1} \\\\u_1 = - \frac{0.04\times 380}{15} \\\\u_1 =-1.013 \ m/s\\\\u_1 = 1.013 \ m/s \ \ \ in \ opposite \ direction[/tex]

Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

Explanation:

[tex] \implies v_{av} = \dfrac{total \: displacement}{total \: time} [/tex]

[tex] \implies v_{av} = \dfrac{100}{10} [/tex]

[tex]\implies v_{av} =10 \: {ms}^{ - 1} [/tex]

Answer:

9.1875 Joules

Explanation:

The kinetic energy of an object can be given by the equation [tex]EK = \frac{1}{2} mv^{2}[/tex], in which m is the mass and v is the velocity. Plugging in the mass and velocity values into the equation, we get:

[tex]EK = \frac{1}{2} (6kg)(1.75m/s)^2\\EK = 9.1875 J[/tex]

Answer:

[tex]\boxed {\boxed {\sf 9.19 \ Joules}}[/tex]

Explanation:

Kinetic energy is the energy an object possesses due to motion. The following formula is used to calculated kinetic energy.

[tex]E_k= \frac{1}{2} mv^2[/tex]

In this formula, m is the mass and v is the velocity.

The car has a mass of 6.00 kilograms and a velocity of 1.75 meters per second.

Substitute the values into the formula.

[tex]E_k= \frac{1}{2} (6.00 \ kg )(1.75 \ m/s)^2[/tex]

Solve the exponent.

[tex]E_k= \frac{1}{2}(6.00 \ kg )(3.0625 \ m^2/s^2)[/tex]

Multiply the numbers together.

[tex]E_k=\frac {1}{2} (18.375 \ kg*m^2/s^2)[/tex]

[tex]E_k= 9.1875 \ kg*m^2/s^2[/tex]

The original measurements of mass and velocity both have 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredths place. The 7 in the thousandths place to the right (9.1875) tells us to round to 8 up to a 9.

[tex]E_k \approx 9.19 \ kg*m^2/s^2[/tex]

1 kilogram meter squared per second squared is equal to 1 Joule. Our answer of 9.19 kg*m²/² is equal to 9.19 Joules.

[tex]E_k \approx 9.19 \ J[/tex]

The kinetic energy of the toy care is approximately 9.19 Joules.

Answer:

ΔL = 3.12 x 10⁻⁴ m = 0.312 mm

Explanation:

First, we will find the stress applied by the mass on the rod:

[tex]\sigma = \frac{F}{A}[/tex]

where,

σ = stress = ?

F = Force applied by the mass = weight = mg = (10 kg)(9.81 m/s²) = 98.1 N

A = cross-sectional area = πr² = π(1 mm)² = π(0.001 m)² = 3.14 x 10⁻⁶ m²

Therefore,

[tex]\sigma = \frac{98.1\ N}{3.14\ x\ 10^{-6}\ m^2}\\\\[/tex]

σ = 3.12 x 10⁷ Pa

Now, we will find the value of strain:

[tex]E = \frac{\sigma}{\epsilon}\\\\\epsilon = \frac{\sigma}{E}\\\\\epsilon = \frac{3.12\ x\ 10^7\ Pa}{200\ x\ 10^9\ Pa}[/tex]

∈ = 1.56 x 10⁻⁴

Now, the change in length can be given by the formula of strain:

[tex]\epsilon = \frac{\Delta L}{L}\\\\\Delta L = (\epsilon)(L)[/tex]

where,

L = Original Length = 2 m

ΔL = Elongation = ?

Therefore,

ΔL = (1.56 x 10⁻⁴)(2 m)

ΔL = 3.12 x 10⁻⁴ m = 0.312 mm

Answer:

[tex]P=45.2W[/tex]

Explanation:

From the question we are told that:

Amplitude [tex]A=0.56m[/tex]

Period [tex]T=3.4s[/tex]

Wavelength [tex]\lambda=2.0[/tex]

Area [tex]a=0.14m^2[/tex]

Generally the equation for Power is mathematically given by

[tex]Power = 2 \pi ^2 \rho a(\frac{\lambda}{T})(\frac{1}{T})*A[/tex]

[tex]P= 2 3.142^2 1025 0.14(\frac{2.0}{3.4})(\frac{1}{3.4})^2*0.56^2[/tex]

[tex]P=45.2W[/tex]

Answer:

the work required to turn the crank at the given revolutions is 8,483.4 J

Explanation:

Given;

torque required to turn the crank, T = 4.50 N.m

number of revolutions, = 300 turns

The work required to turn the crank is given as;

W = 2πT

W = 2 x 3.142 x 4.5

W = 28.278 J

1 revolution = 28.278 J

300 revlotions = ?

= 300 x 28.278 J

= 8,483.4 J

Therefore, the work required to turn the crank at the given revolutions is 8,483.4 J

Answer:

256 ohms

Explanation:

Applying,

R = R'[1+α(T-T')]............. Equation 1

Where R = Final resistance of the wire, R' = Initial resistance of the wire, T = Final temperature, T' = Initial temperature, α = Temperature coefficient of resistance

From the question,

Given: R' = 200 ohms, T = 90 degrees, T' = 20 degrees, α = 0.004/degree

Substitute these values into equation 1

R = 200[1+0.004(90-20)]

R = 200[1+0.28]

R = 200(1.28)

R = 256 ohms

The resistance of the wire at 90 °C correct to the nearest ohm assuming the coefficient of resistance is 0.004 °C¯¹ is 256 ohm

α = R₂ – R₁ / R₁(T₂ – T₁)

0.004 = R₂ – 200 / 200(90 – 20)

0.004 = R₂ – 200 / 200(70)

0.004 = R₂ – 200 / 14000

Cross multiply

R₂ – 200 = 0.004 × 14000

R₂ – 200 = 56

Collect like terms

R₂ = 56 + 200

R₂ = 256 ohm

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The sign is held in equilibrium. Using Newton's second law, we set up the equations of the net forces acting on the sign in the horizontal and vertical directions:

∑ F (horizontal) = T₁ cos(29.5°) - T₂ cos(44.5°) = 0

(right is positive, left is negative)

∑ F (vertical) = T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N = 0

(up is positive, down is negative)

Solve the system of equations. I use elimination here:

• Multiply the first equation by sin(29.5°) and the second by cos(29.5°):

sin(29.5°) (T₁ cos(29.5°) - T₂ cos(44.5°)) = 0

cos(29.5°) (T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N) = 0

T₁ cos(29.5°) sin(29.5°) - T₂ cos(44.5°) sin(29.5°) = 0

T₁ cos(29.5°) sin(29.5°) + T₂ cos(29.5°) sin(44.5°) = (215 N) cos(29.5°)

• Subtract the first equation from the second to eliminate T₁ :

T₂ cos(29.5°) sin(44.5°) - (- T₂ cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)

• Solve for T₂ :

T₂ (cos(29.5°) sin(44.5°) + cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)

T₂ sin(74.0°) = (215 N) cos(29.5°)

… … … (using the fact that sin(x + y) = sin(x) cos(y) + cos(y) sin(x))

T₂ = (215 N) cos(29.5°) / sin(74.0°)

T₂ ≈ 195 N

• Solve for T₁ :

T₁ cos(29.5°) - T₂ cos(44.5°) = 0

T₁ cos(29.5°) = T₂ cos(44.5°)

T₁ = T₂ cos(44.5°) / cos(29.5°)

T₁ ≈ 160. N

Answer:

Creation of energy by joining the nuclei of two hydrogen atoms to form helium

Answer:

v(7) = 52.915 m/s

Explanation:

First, find the value for acceleration.

F = ma

100 = .5 * a

a = 200 m/s²

Next find the velocity at x = 7 using kinematic equations.

v² = v₀² + 2a(Δx)

v² = (0)² + 2(200)(7)

v = [tex]\sqrt{2800}[/tex]

v = 52.915 m/s

90 degrees - 30 = 60 degrees

Velocity = (5m/s - 4.35m/s x cos(30)) / cos(60)

Velocity = 2.47 m/s

The answer is D) 2.47 m/s at 61.9 degrees

The velocity and angle of the second ball after the collision are (A.) 1.25 m/s at  31.2° below the horizontal. Option A

Use the conservation of momentum and conservation of kinetic energy to solve this problem.

Let's denote the velocity and angle of the second ball after the collision as v₂ and θ₂, respectively.

Thus:

Conservation of momentum: [tex]m_1v_1 = m_1v_1'cos(30^o) + m_2v_2cos(\theta_2)[/tex]

Conservation of kinetic energy: [tex](1/2)m_1v_1^2 = (1/2)m_1v_1'^2 + (1/2)m_2v_2^2[/tex]

where m₁ and m₂ are the masses of the first and second balls, respectively.

Since the masses and initial velocity are the same, we can simplify the equations to:

m₁v₁ = m₁v₁'cos(30°) + m₂v₂cos(θ₂)

[tex]v_1^2 = v_1'^2 + v_2^2[/tex]

Substitute in the given values

[tex](1)(5) = (1)(4.35)cos(30^o) + (1)(v_2)cos(\theta_2)\\5^2 = 4.35^2 + v2^2\\v_2 = 1.25 m/s\\\theta2 = 31.2^o[/tex]

Therefore, the velocity and angle of the second ball after the collision are approximately 1.25 m/s at an angle of 31.2° below the horizontal.

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56.1∘

Question: A glass is half-full of water, with a layer of vegetable oil (n = 1.47) floating on top. A ray of light traveling downward through the oil is incident on the water at an angle of 56.1∘ .

A glass is half-full of water, with a layer of vegetable oil (n ...https://study.com › academy › answer › a-glass-is-half-ful...

Answer:

Explanation:

You final velocity is correct but not to the correct number of significant digits. The actual answer should be 18.1 m/s. We will use that to find the total time the stone was in the air in the equation:

v = v₀ + at

18.1 = 6.63 + (-9.81)t and

11.5 = -9.81t so

t = 1.17 seconds.

We know this was how long the stone was in the air (as compared to the time that the stone took to reach its max height or some other height) because we used the velocity with which the stone hit the ground to find the total time the rock was in the air before it hit the ground going at that velocity.

Complete Question

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the wheel starts turning, you are at the very top.

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.

Answer:

[tex]\phi=123.75[/tex]

Explanation:

From the question we are told that:

Height [tex]h=27m[/tex]

Period [tex]T=32sec[/tex]

Time [tex]t=75sec[/tex]

Generally the equation for angular velocity is mathematically given by

[tex]\omega=\frac{2 \pi}{T}[/tex]

[tex]\omega=\frac{2 \pi}{32}[/tex]

[tex]\omega=0.196rad/s[/tex]

Therefore

[tex]\theta=\omega t[/tex]

[tex]\theta=0.196rad/s*75sec[/tex]

[tex]\theta=843.75 \textdegree[/tex]

Therefore

[tex]\phi=\theta-2(360)[/tex]

[tex]\phi=123.75[/tex]

The electric force exerted by S1 on S2 is 21.58μN. The electric field created by S1 at the level of S2 is [tex]1.20 \frac{kN}{C}[/tex]. The electric potential created by S1 at the level or S2 is 360V. The electric force exerted by S1 on S2 is 21.58μN. The electric field generated by S1 in the middle of S1 and S2 is [tex]4.79 \frac{kN}{C}[/tex]. The electric field in the middle of S1 and S2 is [tex]11.99 \frac{kN}{C}[/tex]. The electric potential in the middle of S1 and S2 is 1.80 kV. The electric potential generated by S2 on the position of S1 is 539.4V

a) The electric force exerted by S1 on S2 is 21.58μN.

b) The electric field created by S1 at the level of S2 is [tex]1.20 \frac{kN}{C}[/tex]

c) The electric potential created by S1 at the level or S2 is 360V

d) The electric force exerted by S1 on S2 is 21.58μN.

e) The electric field generated by S1 in the middle of S1 and S2 is [tex]4.79 \frac{kN}{C}[/tex]

f)  The electric field in the middle of S1 and S2 is [tex]11.99 \frac{kN}{C}[/tex]

g) The electric potential in the middle of S1 and S2 is 1.80 kV

h) The electric potential generated by S2 on the position of S1 is 539.4V

The magnitude of the force is determined by using the following formula:

[tex]F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}[/tex]

where:

F= Electric force [N]

k = Electric constant ([tex]N\frac{m^{2}}{C^{2}}[/tex])

q1= First charge [C]

q2 = Second charge [C]

r =  distance between the two charges

So, in this case, the force can be calculated like this:

[tex]F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}[/tex]

So the force will be equal to:

[tex]F=21.58x10^{-6}N[/tex]

which is the same as:

[tex]F=21.58 \micro N[/tex]

b) The electric field created by S1 at the level of S2 is [tex]1.20 \frac{kN}{C}[/tex]

We can take the following formula for the electric field.

[tex]E_{S1}=\frac{F_{e}}{q_{2}}[/tex]

[tex]E_{S1}=\frac{1.20 x10^{-6}N}{18x10^{-9}C}[/tex]

which yields:

[tex]E_{S1}=1.20x10^{3} \frac{N}{C}[/tex]

[tex]E_{S1}=1.20 \frac{kN}{C}[/tex]

In this case, since S1 is positive, we can asume the electric field is in a direction away from S1.

c)

The electric potential created by S1 at the level of S2 is 360V

This electric potential can be found by using the following formula:

V=Er

Where V is the electric potential and it is given in volts.

So we can use the data found in the previous sections to find the electric potential:

[tex]V=(1.20x10^{3} \frac{N}{C})(30x10^{-2}m)[/tex]

V=360V

d)  The force exerted by S2 on S1 will be the same in magnitude as the force exerted by S1 on S2 but oposite in direction. This is because the force will depend on the two charges, and the distance between them, so:

The electric force exerted by S1 on S2 is 21.58μN.

The magnitude of the force is determined by using the following formula:

[tex]F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}[/tex]

[tex]F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}[/tex]

So the force will be equal to:

[tex]F=21.58x10^{-6}N[/tex]

which is the same as:

[tex]F=21.58 \micro N[/tex]

e) The electric field generated by S1 in the middle of S1 and S2 is [tex]4.79 \frac{kN}{C}[/tex]

In order to find the electric field generated by S1, we can make use of the following formula

[tex]E=k_{e} \frac{q_{1}}{r_{1}^{2}}[/tex]

[tex]E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}})[/tex]

which yields:

[tex]E=4.79 \frac{kN}{C}[/tex]

f)  The electric field in the middle of S1 and S2 is [tex]11.99 \frac{kN}{C}[/tex]

In order to find the electric field generated by two different charges at a given point is found by using the following formula:

[tex]E=k_{e} \sum \frac{q_{i}}{r_{i}^{2}}[/tex]

where:

[tex]q_{i}[/tex]= each of the charges in the system

[tex]r_{i}[/tex]= the distance between each of the charges and the point we are analyzing.

We'll suppose the electric fields will be positive then, so:

[tex]E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})[\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}}+\frac{18x10^{-9}C}{(15x10^{-2}m)^{2}}][/tex]

which yields:

[tex]E=11.99 \frac{kN}{C}[/tex]

g) The electric potential in the middle of S1 and S2 is 1.80 kV

Since we know what the electric field is from the previous question, we can make use of the same formula we used before to find the electric potential in the middle of S1 and S2

So let's take the formula:

V=Er

So we can use the data found in the previous sections to find the electric potential:

[tex]V=(11.99x10^{3} \frac{N}{C})(15x10^{-2}m)[/tex]

V=1.80kV

h)

The electric potential generated by S2 on the position of S1 is 539.4V and can be found by using the following formula:

[tex]V=k_{e}\frac{q_{2}}{r}[/tex]

So we can use the data provided by the problem to find the electric potential.

[tex]V=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{18x10^{-9}C}{30x10^{-2}m})[/tex]

V=539.4V

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1) F12 = 216* 10⁻⁷ [N]

b) E1 = 12*10⁻² [ N / C]

c) V(s1/s2) = 3.6 [Volts]

d)|F21| = 216* 10⁻⁷ [N]    For direction of F21  seeAnnex

e)Es₁/p = 0,48*10⁴ [ N / C]

f) Et = 1.2 *10⁴  [N/C]

g)V(p) = 1800 [V]

h) V(s2/s1)  = 540 [V]

The electric force due to the small sphere S1 over S2, is equal in module and of opposite direction at the force that S2 produces on S1. As the charges on both spheres are of opposite signs they attract each other.

See Annex: Notice directions of F1 and F2

Let´s call F1 the force exerted by S1 on S2 then according to Coulomb´s law

F1 = K * Q*q/d²

In that equation, K is the constant of Coulomb  K = 9*10⁹ N-m²/C²

Q = 12 nC  =  12 * 10⁻⁹ C

q = - 18 nC = - 18 *10⁻⁹ C

d = 30 cm = 0,3 m            ⇒  d² = 0.09 m²

By substitution we get:

F12 = 9*10⁹ *  12 * 10⁻⁹ *18 *10⁻⁹ / 0.09        N-m²/C² *C*C/m²

F12 = 9*12*18*10⁻⁹/ 0.09  [N]

F12 = 216* 10⁻⁷ [N]

According to what was explained at the beginning F2 ( the electric force exerted by S2 on S1 is equal in module and of opposite direction ( see Annex)  

b) The electric field created by S1 at the level of S2

Let´s call that electric field E1

E1  =  K * Q / d²

Then

E1 = 9*10⁹* 12 * 10⁻⁹/0.09      [ N-m²/C²*C/m²]

E1 = 12*10⁻² [ N / C]

c) Electric potential is created by S1 at S2 level

V  = K * Q/d

V(s1/s2) = 9*10⁹ * 12 * 10⁻⁹/ 0,3           [ N* m²/C² * C /m]

V(s1/s2) = 3*12 *10⁻¹ [N*m/C]    ⇒ V(s1/s2) = 3.6 [ J/C ]    ⇒  

V(s1/s2) = 3.6 [Volts]

d) Already explain in 1 then

|F21| = |F12|

|F21| = 216* 10⁻⁷ [N]    For direction of F21  seeAnnex

e)

Es₁/p = K * Q / d²

Es₁/p = 9*10⁹* 12 * 10⁻⁹/(0.15)²

Es₁/p = 9*10⁹* 12 * 10⁻⁹ / 0.0225

Es₁/p = 108 * 10⁴ / 225  [ N / C]

Es₁/p = 0,48*10⁴ [ N / C]

f) To determine the electric field in the middle ( point P)

We need to calculate the electric field due to charge S2 ( - 18 *10⁻⁹ C)

Then

Es2/p = K *q / d²

Es2/p =   9*10⁹* ( - 18 *10⁻⁹ )/ (0.15)²

Es2/p  = - 162* 10⁴ /225

Es2/p  = - 0.72 * 10⁴ [ N / C]

The electric field ( a vector )  and the force due to a charge ( a vector ) both have the same direction.

Assuming we place a test charge (+) in the middle the sphere S1 will reject such a charge, and S2will attract the test charge therefore the resultant field is the sum of both components, then:

Et = Es₁/p + Es2/p

Et = 0,48*10⁴ + 0,72*10⁴  

Et = 1.2 *10⁴  [N/C]

g) To calculate the electric potential at p.

Vp  = V s1/p  + V s2/p

V s1/p  = K * Q /d/2

V s1/p  = 9 * 10⁹ * 12*10⁻⁹ / 0,15

V s1/p  =  720 [V]

V s2/p =  9 * 10⁹ * 18*10⁻⁹ /0,15

V s2/p = 1080 [V]

Then

V(p) = 1080 + 720

V(p) = 1800 [V]

h) The electric potential at S1 level is due to charge in S2

V(s2/s1)  =  K * q / d

V(s2/s1)  = 9*10⁹ * 18 * 10⁻⁹ / 0,3

V(s2/s1)  = 540 [V]

Air pressure has no effect at all in an ideal gas approximation. This is because pressure and density both contribute to sound velocity equally, and in an ideal gas the two effects cancel out, leaving only the effect of temperature. Sound usually travels more slowly with greater altitude, due to reduced temperature.

When a liquid is cooled,the kinetic energy of the particles decreases. The force of attraction between the particles increases, the space between the particles decreases.

Answer:

Tangential

Explanation: This is a kind of force which act on a moving body in such a way that it is curved in the direction of the path of the body. This implies that when the velocity of the object is positive, the acceleration will be negative.

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

[tex]C = \frac{Q}{V} = e_0\frac{A}{d}[/tex]

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

[tex]V = \frac{Q}{e_0} *\frac{d}{A}[/tex]

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

[tex]V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V[/tex]

So, if we double the distance between the plates, the potential difference will also be doubled.

Answer:

(a) 0.277 m

(b) 0.553 m

Explanation:

(a) For the closed cylinder,

Applying,

f' = v/4l............. Equation 1

Where f' = fundamental frequency, v = velocity of sound, l = lenght of the cylinder

make l the subject of the equation

l = v/4f'............. Equation 2

From the question,

Given: v = 343 m/s, f' = 310 Hz

Substitute these values into equation 2

l = 343/(4×310)

l = 0.277 m

(b) For the open cylinder,

f' = v/2l

l = v/2f'................ Equation 3

Substitute into equation 3

l = 343/(2×310)

l = 0.553 m

Answer:

a)  [tex]l=0.276m[/tex]

b)  [tex]l'=1.11m[/tex]

Explanation:

From the question we are told that:

Frequency [tex]F=310Hz[/tex]

Speed [tex]V=343m/s[/tex]

a)

Generally the equation for Frequency is mathematically given by

The cylinder is closed at one end

[tex]F=\frac{V}{4*l}[/tex]

[tex]310=\frac{343}{4*l}[/tex]

[tex]l=0.276m[/tex]

b)

Generally the equation for Frequency is mathematically given by

the cylinder is open at both ends.

[tex]F=\frac{V}{l'}[/tex]

[tex]310=\frac{343}{l'}[/tex]

[tex]l'=1.11m[/tex]

Answer:

The resulting pressure is 3 times the initial pressure.

Explanation:

The equation of state for ideal gases is described below:

[tex]P\cdot V = n \cdot R_{u}\cdot T[/tex] (1)

Where:

[tex]P[/tex] - Pressure.

[tex]V[/tex] - Volume.

[tex]n[/tex] - Molar quantity, in moles.

[tex]R_{u}[/tex] - Ideal gas constant.

[tex]T[/tex] - Temperature.

Given that ideal gas is compressed isothermally, this is, temperature remains constant, pressure is increased and volume is decreased, then we can simplify (1) into the following relationship:

[tex]P_{1}\cdot V_{1} = P_{2}\cdot V_{2}[/tex] (2)

If we know that [tex]\frac{V_{2}}{V_{1}} = \frac{1}{3}[/tex], then the resulting pressure of the system is:

[tex]P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)[/tex]

[tex]P_{2} = 3\cdot P_{1}[/tex]

The resulting pressure is 3 times the initial pressure.