The atomic radii of a divalent cation and a monovalent anion are 0.35 nm and 0.129 nm, respectively.(a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another).Enter your answer for part (a) in accordance to the question statement N(b) What is the force of repulsion at this same separation distance?

Applying the concept of scalar product. We know that vectors must be multiplied in their respective corresponding component and then add the magnitude of said multiplications. That is, those corresponding to the [tex]\hat {i}[/tex] component are multiplied with each other, then those corresponding to the [tex]\hat {j}[/tex] component and so on. Finally said product is added.

The scalar product between the two vectors would be:

[tex]\vec{A} \cdot \vec{B} = (4.2\hat{i}+7\hat{j})\cdot (5.7\hat{i}-2.6\hat{j})[/tex]

[tex]\vec{A} \cdot \vec{B} = (4.2*5.7) +(7*(-2.6))[/tex]

[tex]\vec{A} \cdot \vec{B} = 5.74[/tex]

Therefore the scalar product between this two vectors is 5.74

Final answer:

The scalar product of vectors A = 4.20 i + 7.00 j and B = 5.70 i - 2.60 j is calculated by multiplying corresponding components and adding them up, resulting in a scalar product of 5.74.

Explanation:

To find the scalar product (also known as the dot product) of two vectors, you multiply the corresponding components of the vectors and then add these products together. Given two vectors A = 4.20 i + 7.00 j and B = 5.70 i - 2.60 j, the scalar product A cdot B is calculated as follows:

Multiply the x-components together: (4.20)(5.70)

Multiply the y-components together: (7.00)(-2.60)

Add these two products together to get the scalar product.

Now let's do the calculations:

(4.20)(5.70) = 23.94

(7.00)(-2.60) = -18.20

23.94 + (-18.20) = 5.74

Therefore, the scalar product of vectors A and B is 5.74.

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = [tex]v_{oy}[/tex] t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = [tex]v_{oy}[/tex] t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

The minimum muzzle velocity of a cannon to clear a 25.0 m tall cliff from 60.0 m away involves using projectile motion equations to find the initial velocity components, while ensuring the shell reaches the required height and distance.

To solve for the minimum muzzle velocity for a cannon to shoot a shell past a 25.0 m cliff from 60.0 m away, we can use projectile motion equations. First, we separate the initial velocity into its horizontal (vx) and vertical (vy) components:

The shell must reach a height of at least 25.0 m to clear the cliff. We use the equation of motion in the vertical direction, considering the initial vertical velocity (vy) and the displacement (s = 25 m), to find the time (t) it takes to reach the top of the cliff. Then, using the horizontal velocity (vx), we can calculate how far the shell would travel horizontally in that time, ensuring that it covers at least 60.0 m. With the horizontal distance (d) and time (t) determined, we can calculate the shell's trajectory past the cliff edge.

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Answer:

No.

Explanation:

Through uniform We can presume you mean constant in both  that is in magnitude as well as in direction ⠀

When you apply a steady, non-zero acceleration to a resting body, the velocity can be   parallel to the acceleration and not always perpendicular at any time later.

We can do the same for an object with an initial velocity in a direction which is different than that of acceleration, and the direction of its velocity will asymptotically approach to that of acceleration over time.

In both cases  the motion of a object undergoing a non-zero persistent acceleration can not be uniform That is the concept of acceleration: the velocity change over time.

Answer:

Explanation:

a ) Position of the plane with respect to observer ( origin ) is

R = 2.71 i + 1.37 j + 4.65 k

magnitude of R = √ (2.71² + 1.37² + 4.65²)

√(7.344 + 1.8769 + 21.6225)

=√30.8434

= 5.55 km

b ) angle with north

cos Ф = 1.37 / 5.55

= .2468

Ф = 75°

c )

R = 2.71 i + 1.37 j + 4.65 k

=

It is not possible to find a vector quantity of magnitude zero but components different from zero

The magnitude can never be less than the magnitude of any of its components

Answer:

6.38 V

Explanation:

Note: Since the capacitors are identical, The same Charge flows through them.

For the first capacitor,

The Energy stored in a capacitor is given as

E = 1/2qv ............... Equation 1

Where E = Energy stored by the capacitor, q = charge in the capacitor v = Voltage.

make q the subject of the equation

q = 2E/v ................ Equation 2

Given: E = 8 J, v = 51 v.

Substitute into equation 2

q = 8/51

q = 0.3137 C.

For the second capacitor,

v = 2E/q ................... Equation 3

Given: q = 0.3137 C, E = 2 J.

Substitute into equation 3

v = 2/0.3137

v = 6.38 V

Hence the voltage across the second capacitor = 6.38 V

The honeybee lost about 112 million electrons to acquire a charge of +18pC. This calculation involves converting the charge from picocoulombs to coulombs and then dividing by the charge of a single electron.

To calculate the number of electrons a honeybee lost to acquire a charge of +18pC, we first need to understand that the fundamental unit of charge, often represented as e, is +1.602 x 10-19 C for a proton and -1.602 x 10-19 C for an electron.

The number of electrons lost (n_e) is the total charge divided by the charge per electron. Therefore, we convert the charge of the honeybee from picocoulombs (pC) to coulombs (C) by multiplying by 10-12, because 1pC = 10-12C. The +18pC charge is thus equivalent to 18 x 10-12 C.

In relation to the charge of an electron, the honeybee's charge is -18 x 10-12 C / -1.602 x 10-19 C/e-, which gives approximately 1.12 x 108 or 112,000,000 electrons.

So, the honeybee lost about 112 million electrons to get a positive charge of +18pC.

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Answer:f1 = 7.90Hz, f2= 15.811Hz, f3 = 23.71Hz.

Explanation: length of string = 10m, mass of string = 100g = 0.1kg, T= 250N

We need the velocity of sound wave in a string when plucked with a tension T, this is given below as

v = √T/u

Where u = mass /length = 0.1/ 10 = 0.01kg/m

Hence v = √250/0.01, v = √25,000 = 158.11

a) at the lowest frequency.

At the lowest frequency, the length of string is related to the wavelength with the formulae below

L = λ/2, λ= 2L.

λ = 2 * 10

λ = 20m.

But v = fλ where v = 158.11m/s and λ= 20m

f = v/ λ

f = 158.11/ 20

f = 7.90Hz.

b) at the first frequency.

The length of string and wavelength for this case is

L = λ.

Hence λ = 10m

v = 158.11m/s

v= fλ

f = v/λ

f = 158.11/10

f = 15.811Hz

c) at third frequency

The length of string is related to the wavelength of sound with the formulae below

L =3λ/2, hence λ = 2L /3

λ = 2 * 10 / 3

λ = 20/3

λ= 6.67m

v = fλ where v = 158.11m/s, λ= 6.67m

f = v/λ

f = 158.11/6.67

f = 23.71Hz.

Answer:

72.3 MeV

Explanation:

E = Total energy of muon = 178 MeV

[tex]E_0[/tex] = Rest energy of a muon = 105.7 MeV

Kinetic energy of the muon at the surface of the Earth is given by the difference of the total energy and the rest energy of the muon

[tex]K=E-E_0\\\Rightarrow K=178-105.7\\\Rightarrow K=72.3\ MeV[/tex]

The kinetic energy of a muon at the surface is 72.3 MeV

Answer:

θ=12.7°

Explanation:

Lets take the mass of the ball = m

The acceleration due to gravity = g = 9.81 m/s²

The acceleration of the block = a

a= 2.2  m/s²

Lets take angle of incline surface = θ

When block slide down :

The gravitational force on the block =  m g sinθ

By using Newton's second law

F= m a

F=Net force ,a acceleration ,m=mass

m g sinθ =  ma

a= g sinθ

Now by putting the values in the above equation

2.2 = 9.81 sinθ

[tex]sin\theta =\dfrac{2.2}{9.81}\\sin\theta=0.22\\\theta = 12.7\ degrees[/tex]

θ=12.7°

To solve this problem we will apply the concepts related to the deformation of a body and the normal effort, from which we will obtain the area. From this area applied to the geometric concept of a circular bar we will find the diameter.

The deformation equation in a rod tells us that

[tex]\delta = \frac{PL}{AE}[/tex]

Here,

P = Load

L = Length

A = Cross-sectional area

E = Elastic Modulus

Rearranging the Area,

[tex]A = \frac{PL}{\delta E}[/tex]

Replacing we have that the area is,

[tex]A = \frac{(109*10^{3})(6.1)}{(10*10^{-3})(150*10^9)}[/tex]

[tex]A = 0.000443266m^2[/tex]

[tex]A = 44.32*10^{-6}m^2[/tex]

Using the geometric expression for the Area we have,

[tex]A = \frac{\pi}{4} d^2[/tex]

[tex]d = \sqrt{\frac{4A}{\pi}}[/tex]

[tex]d = \sqrt{\frac{4(44.32)}{\pi}}[/tex]

[tex]d = 7.51mm[/tex]

Therefore the smalles diameter rod is 7.51mm

Final answer:

By using the equations of motion under constant acceleration, we can calculate the additional time and distance required for a truck, decelerating on a ramp, to come to a complete stop after having its speed reduced to half of its initial value.

Explanation:

A truck enters a 750-ft ramp at high speed v0 and travels 540 ft in 6 s at constant deceleration before its speed is reduced to v0/2. To solve for both the additional time required for the truck to stop and the additional distance traveled, we use the equations of motion under constant acceleration.

Given:

Initial distance traveled: 540 ft

Time taken: 6 seconds

Initial speed: v0

Final speed at this stage: v0/2

Solution:

Calculate the constant deceleration using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Using the determined deceleration, calculate the additional time required for the truck to stop using the formula: t = (v - u) / a.

To find the additional distance traveled, we use the formula: s = ut + 0.5at2.

Through these calculations, we can determine the additional time required for the truck to stop and the additional distance it will travel.

Answer:

when the Sun illuminates half of the Moon it must be at 90°

Explanation:

The Moon has a circular motion around the Earth and the relative position of the sun, the earth and the moon create the lunar phases.

For this case when the Sun illuminates half of the Moon it must be at 90°, this angle changes with the movement of the moon, it is zero degree for the new moon and 180° for the full moon

Answer:

(a). Vf = 7.14 m/s

(b). Vf = 7.14 m/s

(c). same answer

Explanation:

for question (a), we would be applying conservation of energy principle.

but the initial height is h = 1.5 m

and the initial upward velocity of the ball is Vi =  10 m/s

Therefore

(a). using conservation law

Ef = Ei

where Ef = 1/2mVf² + mghf  ........................(1)

also Ei = 1/2mVi² + mghi  ........................(2)

equating both we have

1/2mVf² + mghf = 1/2mVi² + mghi

eliminating same terms gives,

Vf = √(Vi² + 2g (hi -hf))

Vf = √(10² + -2*9.8*2.5) = 7.14 m/s

Vf = 7.14 m/s

(b). Same process as done in previous;

Ef = Ei

but here the Ef = mghf ...........(3)

and Ei = 1/2mVi² + mghi ...........(4)

solving for the final height (hf) we relate both equation 3 and 4 to give

mghf = 1/2mVi² + mghi ..............(5)

canceling out same terms

hf = hi + Vi²/2g

hf = 1.5 + 10²/2*9.8 = 6.60204m ............(6)

recalling conservation energy,

Ef = Ei

1/2mVf² + mghf = mghi

inputting values of hf and hi we have

Vf = √(2g(hi -hf)) = 7.14 m/s

Vf = 7.14 m/s

(c). From answer in option a and c, we can see there were no changes in the answers.

Answer:

Short circuit

Explanation:

In an ideal inductor circuit with constant current and voltage, it implies that the voltage drop in the circuit is zero (0).

Also, In circuit analysis, a short circuit is defined as a connection between two nodes that forces them to be at the same voltage.

In an ideal short circuit, this means there is no resistance and thus no voltage drop across the connection. That is voltage drop is zero (0).

Therefore, the circuit element is short circuit.

For circuits with constant currents and voltages, an ideal inductor is equivalent to a short circuit or a piece of wire with no resistance, as it does not affect the circuit voltage or resistance.

An ideal inductor in a circuit with constant currents and voltages acts as if it is effectively a wire with no resistance, since an ideal inductor does not dissipate energy when the current is constant. However, if we consider the energy transfer in a circuit with an inductor and another circuit element, often referred to as a 'black box', which could be a resistor, we notice energy is exchanged only when there is a change in current. Using the lumped circuit approximation, the inductor's magnetic fields are assumed to be completely internal, meaning it only interacts with other components via the current flowing through the wires, not by overlapping magnetic fields in space.

Given the Kirchhoff's voltage law, which states that the sum of the voltage drops in a closed loop must be zero, an inductor with a constant current will have a voltage drop that is zero, making it equivalent to a short circuit in terms of its impact on the voltage in the circuit. Therefore, for circuits with constant currents and voltages, an ideal inductor is equivalent to a short circuit or a piece of wire with no resistance, as it does not contribute any voltage drop or generate heat like a resistor would.

Answer:

Explanation:

Given

volume of Tank [tex]V=0.25\ m^3[/tex]

Specific gravity [tex]=2[/tex]

specific gravity is the defined as the ratio of density of fluid to the density of water

Density of water [tex]\rho _w=1000\ kg/m^3[/tex]

Density of Fluid [tex]\rho =2\times 1000=2000\ kg/m^3[/tex]

We know mass of a fluid is given by the product of density and volume

[tex]m=\rho \times V[/tex]

[tex]m=2000\times 0.25[/tex]

[tex]m=500\ kg[/tex]    

Answer:

Option A

Explanation:

This can be explained based on the conservation of energy.

The total mechanical energy of the system remain constant in the absence of any external force. Also, the total mechanical energy of the system is the sum of the potential energy and the kinetic energy associated with the system.

In case of two stones thrown from a cliff one vertically downwards the other vertically upwards, the overall gravitational potential energy remain same for the two stones as the displacement of the stones is same.

Therefore the kinetic energy and hence the speed of the two stones should also be same in order for the mechanical energy to remain conserved.

Answer:

b. A, because it travels a longer path.

Explanation:

This can be justified by the equation of motion given below:

[tex]v^2=u^2+2\times a\times s[/tex]

where:

[tex]v=[/tex] final velocity

[tex]u=[/tex] initial velocity

[tex]a=[/tex] acceleration = g (here)

[tex]s=[/tex] displacement of the body

Explanation:

The temperature in the inner solar system was too high for light gases to condense, while in the outer solar system, the temperature was much lower, which allowed the Jovian planets to form, which grew enough to accumulate and retain the hydrogen gas that remained in the solar nebula, which led to its high levels of hydrogen and large size.

Answer:

Acceleration of the proton will be equal to [tex]4.79\times 10^{14}m/sec^2[/tex]

Explanation:

We have given electric field [tex]E=5\times 10^6N/C[/tex]

Mass of proton is equal to [tex]m=1.67\times 10^{-27}kg[/tex]

And charge on proton is equal to [tex]e=1.6\times 10^{-19}C[/tex]

Electrostatic force will be responsible for the motion of proton

Electrostatic force will be equal to [tex]F=qE=1.6\times 10^{-19}\times 5\times 10^6=8\times 10^{-13}N[/tex]

According to newton law force on the proton will be equal to F = ma, here m is mass of proton and a is acceleration

This newton force will be equal to electrostatic force

So [tex]1.67\times 10^{-27}\times a=8\times 10^{-13}[/tex]

[tex]a=4.79\times 10^{14}m/sec^2[/tex]

So acceleration of the proton will be equal to [tex]4.79\times 10^{14}m/sec^2[/tex]

Final answer:

The initial acceleration of a proton in a [tex]5.00 * 10^6 N/C[/tex] electric field is approximately [tex]4.79 * 10^1^5 m/s^2[/tex], calculated by applying the electric force formula and Newton's second law.

Explanation:

Calculating Proton Acceleration in an Electric Field

To calculate the initial acceleration of a proton in a [tex]5.00 * 10^6 N/C[/tex] electric field, we follow the steps in the Problem-Solving Strategy for electrostatics:

Identify the known quantities. The electric field (E) is [tex]5.00 * 10^6 N/C[/tex], and the charge of a proton (q) is approximately [tex]1.60 * 10^-^1^9[/tex] C.

Write down the formula for electric force (F = qE).

Calculate the force on the proton: F = [tex](1.60 * 10^-^1^9 C)(5.00 * 10^6 N/C) = 8.00 * 10^-^3 N.[/tex]

Use Newton's second law (F = ma) to find the acceleration (a), knowing the mass of a proton (m) is approximately [tex]1.67 * 10^-^2^7 kg[/tex].

Solve for acceleration: a = F/m =[tex](8.00 * 10^-^3 N) / (1.67 * 10^-^2^7 kg) = 4.79 * 10^1^5 m/s^2[/tex].

Thus, the initial acceleration of the proton is approximately[tex]4.79 * 1015 m/s^2.[/tex]

Answer:

The SI units of A, B and C are :

[tex]m,\ m/s\ and\ m/s^2[/tex]                  

Explanation:

The position x, in meters, of an object is given by the equation:

[tex]x=A+Bt+Ct^2[/tex]

Where

t is time in seconds

We know that the unit of x is meters, such that the units of A, Bt and [tex]Ct^2[/tex] must be meters. So,

[tex]b=\dfrac{m}{s}=m/s[/tex]

[tex]C=m/s^2[/tex]

So, the SI units of A, B and C are :

[tex]m,\ m/s\ and\ m/s^2[/tex]

So, the correct option is (B).

The SI units of A, B, and C is option B [tex]m, m/s, m/s^2[/tex]

The position x, in meters, of an object is provided by the equation:

[tex]x = A + Bt + Ct^2[/tex]

Here

t should be t in seconds

As We know that the unit of x should be in meters, in such a way that the units of A, Bt and  must be meters.

So,

A = m

bt = m

[tex]b = m\div s = m/s\\\\ Ct^2 = m\\\\ C = m/s^2[/tex]

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Answer : The volume of ethyl alcohol overflow will be, 7.49 mL

Explanation :

To calculate the volume of ethyl alcohol overflow we are using formula:

[tex]\Delta V=V_o(\alpha \Delta T)\\\\\Delta V=V_o\times \alpha \times (T_2-T_1)[/tex]

where,

[tex]\Delta V[/tex] = volume expand = ?

[tex]\alpha[/tex] = volumetric expansion coefficient = [tex]0.00109/^oC[/tex]

[tex]V_o[/tex] = initial volume = 430 mL

[tex]T_2[/tex] = final temperature = [tex]22.0^oC[/tex]

[tex]T_1[/tex] = initial temperature = [tex]6.00^oC[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta V=(430mL)\times (0.00109/^oC)\times (22.0-6.00)^oC[/tex]

[tex]\Delta V=7.49mL[/tex]

Thus, the volume of ethyl alcohol overflow will be, 7.49 mL

Answer:

7.3094 ml

Explanation:

Initial volume of the glass, Vo = 430 ml

Initial temperature, T1 = 6°C

final temperature, T2 = 22°C

Temperature coefficient of glass, γg = 27.6 x 10^-6 /°C

Temperature ethyl alcohol, γa = 0.00109 /°C

Use the formula of expansion of substances

Expansion in volume of glass

ΔVg = Vo x γg x ΔT

ΔVg = 430 x 27.6 x 10^-6 x 16 = 0.1898 ml

Expansion in volume of ethyl alcohol

ΔVa = Vo x γa x ΔT

ΔVa = 430 x 0.00109 x 16 = 7.4992 ml

The amount of volume over flow is

ΔV = ΔVa - ΔVg

ΔV = 7.4992 - 0.1898

ΔV = 7.3094 ml

Thus, the amount of ethyl alcohol over flow is 7.3094 ml.

Answer:

6698.03 J

Explanation:

Kinetic Energy: This is a form of mechanical energy that is due to a body in motion. The S.I unit of Kinetic Energy is Joules (J).

The formula for kinetic energy is given as

Ek = 1/2mv².......................... Equation 1

Where Ek = kinetic Energy, m = mass of the object, v = velocity of the object.

Given: m = 49.9 pounds, v = 54.4 miles per hours.

Firstly, we convert pounds to kilogram.

If 1 pounds = 0.454 kg,

Then, 49.9 pounds = (0.454×49.9) kg = 22.655 kg.

Secondly, we convert miles per hours to meters per seconds.

If 1 miles per hours = 0.447 meter per seconds,

Then, 54.4 miles per hours = (0.447×54.4) = 24.3168 meters per seconds.

Substitute the value of m and v into equation 1

Ek = 1/2(22.655)(24.3168)²

Ek = 6698.03 J.

Thus the Kinetic energy of the object = 6698.03 J

Answer:

14.715 m/s

11.03625 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

Time taken to go up will be [tex]\dfrac{3}{2}=1.5\ s[/tex]. This is also equal to the time taken to go down.

[tex]v=u+at\\\Rightarrow v=0+9.81\times 1.5\\\Rightarrow v=14.715\ m/s[/tex]

The speed of the ball when it reaches the player is 14.715 m/s

This is equal to the speed at which the player threw the ball

[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{14.715^2-0^2}{2\times 9.81}\\\Rightarrow s=11.03625\ m[/tex]

The ball reached a height of 11.03625 m

Answer:

Extrasolar solar system differ from our solar system in many ways such as of mass, size and shape of the planet, as well as temperature or amount of heat received in each planet.

Explanation:

An extrasolar planet is a planet outside the Solar System, while the Solar System orbit around the sun  as a result of the gravitational pull of the sun.

Thus, we can say that the major difference between extrasolar planetary systems and solar system is that in solar system, planets orbit around the Sun, while in extrasolar planetary systems, planets orbit around other stars.

All of the planets in our solar system orbit around the Sun. Planets that orbit around other stars are called exoplanets or extrasolar.

Extrasolar solar system differ from our solar system in many ways such as of mass, size and shape of the planet. They also differ in terms of temperature, because the temperature in each planet in solar system depends on its distance from the sun while that of the extrasolar depends on the activities of the star.

The speed of transverse waves on the steel piano wire can be calculated using the formula √(Tension / (Mass per unit length)). To reduce the wave speed by a factor of 2 without changing the tension, we can solve for the new wave speed, and then calculate the difference in mass per unit length with the copper wire.

The speed of transverse waves on a steel piano wire can be calculated using the formula:

Speed = √(Tension / (Mass per unit length))

Where the tension in the wire is 500 N and the mass per unit length is calculated by dividing the mass of the wire by its length. Therefore, the mass per unit length is 5 g / 0.7 m = 7.14 g/m.

To reduce the wave speed by a factor of 2 without changing the tension, we can use the equation:

New wave speed = √(New tension / (Mass per unit length))

We can solve this equation for the new tension by rearranging it as:

New tension = (New wave speed)^2 * (Mass per unit length)

Since we want to reduce the wave speed by a factor of 2, the new wave speed is half the original speed. Substituting these values into the equation, we have:

(0.5 * Old wave speed)^2 * (Mass per unit length) = 500 N

Solving for the new mass per unit length gives:

New mass per unit length = 500 N / (0.5 * Old wave speed)^2

The difference between the new mass per unit length and the mass per unit length of copper wire is the mass of copper wire that needs to be wrapped around the steel wire. However, we would need additional information to calculate this difference.

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A) The work done by the electric field is zero

B) The work done by the electric field is [tex]9.1\cdot 10^{-4} J[/tex]

C) The work done by the electric field is [tex]-2.4\cdot 10^{-3} J[/tex]

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

[tex]W=Fd cos \theta[/tex]

where

F is the magnitude of the force

d is the displacement of the particle

[tex]\theta[/tex] is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

This means that force and displacement are perpendicular to each other, so

[tex]\theta=90^{\circ}[/tex]

and [tex]cos 90^{\circ}=0[/tex]: therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

[tex]\theta=0^{\circ}[/tex]

and

[tex]cos 0^{\circ}=1[/tex]

Therefore, the work done by the electric force is

[tex]W=Fd[/tex]

and we have:

[tex]F=qE[/tex] is the magnitude of the electric force. Since

[tex]E=4.30\cdot 10^4 V/m[/tex] is the magnitude of the electric field

[tex]q=32.0 nC = 32.0\cdot 10^{-9}C[/tex] is the charge

The electric force is

[tex]F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N[/tex]

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

[tex]W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J[/tex]

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

[tex]\theta=90^{\circ}+45^{\circ}=135^{\circ}[/tex]

Moreover, we have:

[tex]F=1.38\cdot 10^{-3} N[/tex] (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

[tex]W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J[/tex]

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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Answer :

(a) [tex]1.00\text{ mile}=1.61km[/tex]

(b) [tex]1.00km=3.28\times 10^3ft[/tex]

Explanation :

As are given:

1 inch = 2.54 cm

(a) Now we have to convert the number of kilometers in 1.00 mile.

Conversions used:

1 mile = 5280 ft

1 ft = 12 inch

1 inch = 2.54 cm

1 cm = 10⁻⁵ km

Thus,

[tex]1.00\text{ mile}=5280ft\times \frac{12in}{1ft}\times \frac{2.54cm}{1in}\times \frac{10^{-5}km}{1cm}[/tex]

[tex]1.00\text{ mile}=1.61km[/tex]

(b) Now we have to convert the number of feet in 1.00 km.

[tex]1.00km=10^5cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12inch}[/tex]

[tex]1.00km=3.28\times 10^3ft[/tex]

The unit 1 mile is equivalent to 1.61 km.

The unit 1 km is equivalent to 3280 ft.

Part A

The unit 1 miles will convert into km. We know that 1 mile = 5280 ft and 1 ft = 12 inch.

Given that 1 inch = 2.54 cm. Hence

1 mile = 5280 ft.

1 mile = [tex]5280 \times 12[/tex] inch

1 mile = [tex]63360 \times 2.54[/tex] cm

1 mile = 160934.4 cm

Now, 1 cm = 10^-5 km. Then

1 mile = [tex]160934.4 \times 10^{-5}[/tex] km

1 mile = 1.61 km

The unit 1 mile is equivalent to 1.61 km.

Part B

The unit 1 km will convert into feet. We know that 1 km = 10^5 cm

Given that 1 inch = 2.54 cm

1 km = [tex]10^5[/tex] cm

1 km = [tex]10^5 \times \dfrac {1}{2.54}[/tex] inch

1 km = [tex]10^5 \times \dfrac {1}{2.54}\times \dfrac{1}{12}[/tex] ft.

1 km = 3280 ft.

The unit 1 km is equivalent to 3280 ft.

To know more about the inch, km and feet, follow the link given below.

https://brainly.com/question/9198905.

Final answer:

If the width of both slits in a double-slit interference experiment is doubled without changing the distance between their centers, the spacing of the fringes decreases and the intensity of the bright fringes decreases.

Explanation:

In a double-slit interference experiment, if the width of both slits is doubled without changing the distance between their centers:

a. The spacing of the fringes will decrease.

b. The intensity of the bright fringes will decrease.

When the width of the slits is increased, the interference fringes become wider and less compact. This means that the spacing between the fringes becomes smaller. Additionally, the intensity of the bright fringes decreases because spreading out the light over a wider area results in less light reaching a specific point on the screen.

Final answer:

The jet fighter pilot will black out during the acceleration period as it lasts longer than 5.0 s. The greatest speed the pilot can reach with an acceleration of 5g before blacking out is 245 m/s.

Explanation:

To determine whether the jet fighter pilot will black out, we need to calculate the time of acceleration. The speed of sound is v = 331 m/s. The speed the pilot wants to reach is 3 times the speed of sound, so the desired speed is 3 * 331 m/s = 993 m/s. We can use the equation:

v = u + at

Where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration (5 g = 5 * 9.8 m/s^2 = 49 m/s^2), and t is the time.

Substituting the values, we get:

993 m/s = 0 m/s + 49 m/s² x t

Rearranging this equation to solve for t, we get:

t = 993 m/s / 49 m/s^2 = 20.265 s

Since the acceleration lasts for longer than 5.0 seconds, the pilot will black out during the acceleration period.

To calculate the greatest speed the pilot can reach before blacking out, we can use the same equation, but rearrange it to solve for v:

v = u + at

Substituting the values, we get:

v = 0 m/s + 49 m/s^2 x 5.0 s = 245 m/s

Therefore, the greatest speed the pilot can reach with an acceleration of 5 g before blacking out is 245 m/s.

Answer:

Current, [tex]I=6.78\times 10^{-11}\ A[/tex]

Explanation:

In this case, we need to find the current in amperes if 1400 Na+ ions flow across a cell membrane in 3.3 μs.

Charge, [tex]q=1400\times 1.6\times 10^{-19}=2.24\times 10^{-16}\ C[/tex]

Time taken, [tex]t=3.3\ \mu s=3.3\times 10^{-6}\ s[/tex]

Let I is the current. It is given by total charge per unit time. It is given by :

[tex]I=\dfrac{q}{t}[/tex]

[tex]I=\dfrac{2.24\times 10^{-16}}{3.3\times 10^{-6}}[/tex]

[tex]I=6.78\times 10^{-11}\ A[/tex]

So, the current of [tex]6.78\times 10^{-11}\ A[/tex] is flowing across a cell membrane. Hence, this is the required solution.