The destination airport has one runway, 08-26, and the wind is calm. The normal approach in calm wind is a left hand pattern to runway 08. There is no other traffic at the airport. A thunderstorm about 6 miles west is beginning to develop to its mature stage, and rain is starting to reach the ground. The pilot decides to a. Fly normal approach to runway 8b. Fly a right hand approach to runway 8c. Fly the approach to runway 26d. Fly to the west for a fun ride

To calculate the magnitude of the total acceleration of the car 11 seconds after it starts, consider both tangential and centripetal accelerations. The tangential acceleration is 2.457 m/s², and the centripetal acceleration is 8.63 m/s², resulting in a total acceleration of 8.97 m/s².

To determine the magnitude of the total acceleration of the car 11 seconds after the start, we need to consider both the tangential acceleration (uniform acceleration as it speeds up) and the centripetal acceleration (due to the car's circular motion). To calculate the tangential acceleration ([tex]a_t[/tex]), we use the final speed (v) the car reaches in 13 seconds, which is 115 km/h, and convert it to meters per second:

v = 115 km/h = (115 * 1000 m) / (3600 s) = 31.94 m/s

Since the car accelerates uniformly from rest, the tangential acceleration is given by:

[tex]a_t[/tex] = v / t = 31.94 m/s / 13 s = 2.457 m/s²

After 11 seconds, the speed of the car (v₁₁) is:

v₁₁ = at * 11 s = 2.457 m/s² * 11 s = 27.03 m/s

The centripetal acceleration ([tex]a_c[/tex]) is calculated using the formula:

[tex]a_c[/tex] = v₁₁² / r = (27.03 m/s)² / 85 m = 8.63 m/s²

To find the total acceleration, we use the Pythagorean theorem because the tangential and centripetal accelerations are perpendicular to each other:

a = √([tex]a_t[/tex]² + [tex]a_c[/tex]²) = √((2.457 m/s²)² + (8.63 m/s²)²) = √(6.036 + 74.516) = √(80.552) = 8.97 m/s².

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:

Explanation:

V = Deltax/Deltat

V = 15.0 m/s

Displacement:

(a) Vf = Vi + adeltat

Vf = 15.0m/s - 9.8m/s^2 x 0.500s = 10.1m/s

Displacement = (15.0m/s x 0.500s) - (0.5)(9.8m/s^2)(0.500s)^2 = 6.275m

(b) Vf = 15.0m/s - 9.8m/s^2 x 1.00s = 5.2m/s

Displacement = (15.0m/s x 1.00s) - (0.5)(9.8m/s^2)(1s)^2 = 10.1m

(c) Vf = 15.0m/s - 9.8m/s^2 x 1.50s = 14.7m/s

Displacement = (15.0m/s x 1.50s) - (0.5)(9.8m/s^2)(1.5s)^2 = 11.475m

(d) Vf = 15.0m/s - 9.8m/s^2 x 2.00s =  19.6m/s

Displacement = (15.0m/s x 2.00s) - (0.5)(9.8m/s^2)(2s)^2 = 10.4m

Answer:

(a) Melting point is 136.8°C

(b) Melting point is 278.24°F

Boiling point is 832.28°F

(c) Melting point is 409.8K

Boiling point is 717.6K

Explanation:

(a) 586.1°F = 5/9(586.1 - 32)°C = 307.8°C

Melting point = 444.6°C - 307.8°C = 136.8°C

(b) Melting point = 136.8°C = (9/5×136.8) + 32 = 278.24°F

Boiling point = 444.6°C = (9/5×444.6) + 32 = 832.28°F

(c) Melting point = 136.8°C = 136.8 + 273 = 409.8K

Boiling point = 444.6°C = 444.6 + 273 = 717.6K

To determine the melting point of sulfur in degrees Celsius and Fahrenheit, and in Kelvin.

a. To determine the melting point of sulfur in degrees Celsius, we can subtract the Fahrenheit value from the boiling point of sulfur. Given that the melting point of sulfur is 586.1 Fahrenheit lower than its boiling point, we can calculate the melting point as follows:

Melting point in degrees Celsius = Boiling point in degrees Celsius - 586.1

b. To convert the melting and boiling points from degrees Celsius to Fahrenheit, we can use the formula:

Degrees Fahrenheit = (Degrees Celsius × 9/5) + 32

c. To convert the melting and boiling points from degrees Celsius to Kelvin, we can use the formula:

Kelvin = Degrees Celsius + 273.15

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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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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.

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.

Answer:

A)The time taken for the car to come to a full stop is:

[tex]t=\displaystyle \frac{v_0}{a_0}[/tex]

b) The time taken for the car to accelerate from full stop to  its original cruising speed is:

[tex]t=\displaystyle \frac{v_0}{a_0}[/tex]

c) The separation distance between the car and the train is:

[tex]d=v_0^2/a_0[/tex]

Completed question:

a) How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0.

b) How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

c) The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed again? Express the separation distance in terms of v0 and a0. Your answer should be positive.

Explanation:

a) The car slows down with constant acceleration (-a₀). Therefore, this movement is a linearly accelerated motion:

[tex]v(t)=v_0+t\cdot (-a_0)\\0=v_0+t\cdot (-a_0)\\t=v_0/a_0[/tex]

b) for the acceleration process we use the same equation than before:

[tex]v(t)=v_{orig}+t\cdot (a_0)\\v_0=0m/s+t\cdot (a_0)\\t=v_0/a_0[/tex]

c) To determine the separation distance between the car and the train we can observe how much distance each of them travels in the time spend for the car to deaccelerate and accelerate again.

For the train:

[tex]x(t_{tot})=x_0+v_0\cdot(t_{tot})\\x(t_{tot})=v_0\cdot(t_{dea}+t_{acc})\\x(t_{tot})=2v_0^2/a_0[/tex]

For the car:

[tex]x(t_{tot})=x(t_{dea})+x(t_{acc})\\x(t_{tot})=v_0\cdot(t_{dea})+0.5(-a)(t_{dea})^2+0.5a(t_{acc})^2\\x(t_{tot})=v_0\cdot(t_{dea})\\x(t_{tot})=v_0^2/a_0[/tex]

Therefore the separation distance between the car and the train is:

[tex]d=|x_{car}-x_{train}|=v_0^2/a_0[/tex]

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

If a sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz then the needle move in one period = 0.0508 m

If the displacement of a particle undergoing simple harmonic motion of amplitude A at a time is [tex]→x=Asin[/tex]ω[tex]t^i[/tex]

=  [tex](+A^i)+ (-2A^i)+(+A^i)[/tex]

= [tex]0^i.[/tex]

Given:

Amplitude = 0.0127 m

frequency = 2.55 Hz

Solution:

To find the displacement of the needle in one period we need to put value in the formula:

The total distance traveled = 4A

= 4*0.0127 m

= 0.0508 m

Thus, If a sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz then the needle move in one period = 0.0508 m

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Final answer:

The distance moved by the sewing machine needle in one period of its simple harmonic motion is four times the amplitude, which equals 0.0508 meters.

Explanation:

The simple harmonic motion of a sewing machine needle is defined by its amplitude and frequency. The amplitude (0.0127 m) is the maximum displacement from its rest position, and the needle's frequency (2.55 Hz) indicates how often it repeats this motion in one second. To determine the distance the needle moves in one period, we need to consider that it moves from the rest position to the maximum amplitude, back through the rest position to the negative maximum amplitude, and back to rest again, completing one full cycle.

In simple harmonic motion, the distance moved in one period is four times the amplitude. So, the needle moves a distance of 4 x 0.0127 m in one period. Therefore, the needle moves 0.0508 meters in one cycle.

The velocity of the boat relative to the earth is 7.2 m/s, 32.0° south of east. It will take the boat 0.118 seconds to cross the river. The boat will reach a point 0.236 meters south of the starting point.

To find the velocity of the boat relative to the earth, we need to find the resultant of the boat's velocity relative to the water and the river's velocity relative to the earth. Using vector addition, we can find that the magnitude of the total velocity is 7.2 m/s and the direction is 32.0° south of east.

To find the time required to cross the river, we can use the formula t = d/v, where d is the width of the river and v is the horizontal component of the boat's velocity relative to the earth. Plugging in the values, we find that the time required is 0.118 seconds.

To find how far south of the starting point the boat will reach the opposite bank, we can use the formula d = v*t, where v is the vertical component of the boat's velocity relative to the earth and t is the time. Plugging in the values, we find that the boat will reach a point 0.236 meters south of the starting point.

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We use vector addition to find the magnitude of 4.65 m/s at an angle of 25.4° south of east. The time required to cross the river is approximately 119.05 seconds, and the boat will reach a point around 238.1 meters south of its starting point.

The question is about calculating the motion of a boat in a river flowing in a specific direction. Let's solve each part step-by-step.

(a) The boat's velocity relative to the water is 4.2 m/s due east, and the river flows due south at 2.0 m/s. We can use the Pythagorean theorem to find the magnitude:

[tex]Magnitude = \sqrt{(velocity_x^2 + velocity_y^2)}\\Magnitude = \sqrt{(4.22 + 2.02)}\\Magnitude = \sqrt{(17.64 + 4.00)}\\Magnitude = \sqrt{21.64}\\Magnitude = 4.65 m/s[/tex]

To find the direction, we need to calculate the angle θ relative to the east:

[tex]tan(\theta) = velocity_y / velocity_x = 2.0 / 4.2\\\theta = tan-1(2.0 / 4.2)\\\theta \approx 25.4\textdegree south\right \left of\right \left east[/tex]

(b) Given the river's width is 500 m and the boat's velocity relative to the water is 4.2 m/s, we use the formula:

Time = Distance / Velocity
Time = 500 m / 4.2 m/s
Time ≈ 119.05 s

(c) To find the southward distance, we use the river's flow velocity and the time calculated in part (b):

[tex]Distance_{south} = Velocity_{river} \times Time\\Distance_{south} = 2.0 m/s \times 119.05 s\\Distance_{south} \approx 238.1 m[/tex]

Answer:  An emission line occurs when an electron drops down to a lower orbit around the nucleus of an atom and looses energy.

An absorption line also occurs when any electron move to a higher orbit by absorbing energy.

Each atom has a unique way of using some space of the orbits and can absorb only certain energies or wavelengths.

Explanation:

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.

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Answer:

1. True

2. True

3. False

Explanation:

According to Ohm's law, V = IR where V is the voltage, I is the current and R is the resistance. So if V is constant, R is inversely proportional to I

1. The resistivity of metal wires would depend on the temperature, this is true.

2. When length L of a wire decreases, the resistance R decreases as well, this would make I increase.

3. When the resistivity increases, resistance R increases as well, this would make I decrease.

The resistivity of a wire is affected by temperature, with most metals increasing in resistivity with higher temperatures. Current increases when the length of a wire decreases at constant voltage. However, current decreases when resistivity increases at constant voltage.

When examining the resistance of a wire, there are several factors to consider that impact how the resistance changes under various conditions. The statements about resistance of a wire need to be evaluated for their truthfulness.

When an object is dropped from a certain height with no air resistance, it takes a specific time to reach the ground. The time it takes to reach the ground is proportional to the square root of the height. If the object is dropped from three times the original height, it will take T * sqrt(3) time to reach the ground.

When an object is dropped from a certain height with no air resistance, it takes a specific time (T) to reach the ground. If the object is dropped from three times that height, it will take a longer time to reach the ground. The relationship between the height and the time taken is linear, assuming no air resistance.

To determine the time it would take to reach the ground from three times the original height, we need to consider that the time is proportional to the square root of the height. Let's represent the original time as T and the original height as H. Therefore, the time it would take to reach the ground from three times the original height (3H) would be:

t = T * sqrt(3)

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Explanation:

A nucleus that absorbs 11.45δ is less shielded than a nucleus that absorbs at 4.13δ.

the nucleus that absorbs at 11.45δ requires weaker applied field strength to come into resonance than the nucleus that absorbs at 4.13δ.

A nucleus absorbing at 4.13 δ is more shielded than one absorbing at 11.45 δ because it is in a weaker local magnetic field and resonates at a lower frequency.

In the context of nuclear magnetic resonance (NMR) spectroscopy, the chemical shift is given in units of delta (δ) which represents the resonance frequency of a nucleus relative to a standard reference compound. When a nucleus is surrounded by a dense cloud of electrons, it is considered to be ‘shielded’. A shielded nucleus is influenced by a smaller local magnetic field because the electrons repel some of the external magnetic field. As a result, shielded nuclei resonate at a lower frequency (higher δ values) when compared to de-shielded nuclei. Therefore, a nucleus that absorbs at 4.13 δ is more shielded than a nucleus that absorbs at 11.45 δ because the latter is in a stronger local magnetic field and is, hence, de-shielded.

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:

R=250 N

α = 38.86⁰

Explanation:

Given that

Force ,F₁ = 150 N (Towards north  )

Force ,F₂ = 200 N ( Towards east )

We know that the angle between north and east direction is 90⁰ .

The resultant force R is given as

[tex]R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta\\[/tex]

We know that

cos 90⁰ =  0

That is why

[tex]R=\sqrt{150^2+200^2}\\R=250\ N[/tex]

Therefore the magnitude of the forces will be 250  N.

The angle  from the horizontal of the resultant force = α

[tex]tan\alpha=\dfrac{150}{200}\\tan\alpha=0.75\\\alpha=38.86\ degrees[/tex]

R=250 N

α = 38.86⁰

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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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.

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.

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:

BOYLE'S Law

Explanation:

Boyle's law states that at constant temperature the volume of a fixed quantity of an ideal gas is inversely proportional to the pressure of the gas.

Mathematically:

From the universal gas law we have:

[tex]P.V=n.R.T[/tex]

where:

P = pressure of the gas

V = volume of the gas

n = no. of moles of gas

T = temperature of the gas

R = universal gas constant

when the mass of gas is fixed i.e. n is constant and temperature is also constant.

[tex]PV=constant[/tex]

[tex]P\propto\frac{1}{V}[/tex]

[tex]P_1.V_1=P_2.V_2[/tex]

here the suffix 1 and 2 denote two different conditions of the same gas.

Answer:

The mass of solid is 173.45 g.

Explanation:

Given that,

Total volume of solid and liquid = 84.0 mL

Mass of liquid = 26.5 g

Density of liquid = 0.865 g/mL

Density of solid = 3.25 g/mL

We need to calculate the volume of liquid

Using formula of density

[tex]\rho_{l}=\dfrac{m_{l}}{V_{l}}[/tex]

[tex]V_{l}=\dfrac{m_{l}}{\rho_{l}}[/tex]

Put the value into the formula

[tex]V_{l}=\dfrac{26.5}{0.865}[/tex]

[tex]V_{l}=30.63\ mL[/tex]

We need to calculate the volume of solid

Volume of solid = Total volume of solid and liquid- volume of liquid

[tex]V_{s}=84.0-30.63[/tex]

[tex]V_{s}=53.37\ mL[/tex]

We need to calculate the mass of solid

Using formula of density

[tex]\rho_{s}=\dfrac{m_{s}}{V_{s}}[/tex]

[tex]m_{s}=\rho_{s}\timesV_{s}[/tex]

Put the value into the formula

[tex]m_{s}=3.25\times53.37[/tex]

[tex]m_{s}=173.45\ g[/tex]

Hence, The mass of solid is 173.45 g.

To find the mass of the solid, subtract the volume of the liquid solvent from the total volume of the mixture. Then use the density of the solid to find its mass.The mass of the solid is 173.44 g.

To find the mass of the solid, we can first find the total volume of the solid by subtracting the volume of the liquid solvent from the total volume of the mixture.

Since the density of the liquid solvent is given, we can use it to calculate its volume.

The volume of the liquid solvent is found by dividing its mass by its density: 26.5 g / 0.865 g/mL = 30.64 mL. Therefore, the volume of the solid is 84.0 mL - 30.64 mL = 53.36 mL.

Next, we can use the density of the solid to find its mass.

The density of the solid is given as 3.25 g/mL.

We can use the formula mass = density * volume to find the mass of the solid.

Plugging in the values, we get:

mass = 3.25 g/mL * 53.36 mL = 173.44 g.

Therefore, the mass of the solid is 173.44 g.

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1) Potential difference: 1 V

2) [tex]V_b-V_a = -1 V[/tex]

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

[tex]\Delta U=q\Delta V[/tex]

where

q is the charge's magnitude

[tex]\Delta V[/tex] is the potential difference between the initial and final position

In this problem, we have:

[tex]q=4.80\cdot 10^{-19}C[/tex]is the magnitude of the charge

[tex]\Delta U = 4.80\cdot 10^{-19}J[/tex] is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

[tex]\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V[/tex]

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

[tex]V_b-V_a = - 1V[/tex]

Final answer:

The particle moves in the direction of decreasing potential and through a potential difference of 3.0 V.

Explanation:

The potential difference through which the particle moves can be calculated using the formula for work done by the electric force: W = q(Vb - Va), where W is the change in kinetic energy, q is the charge of the particle, and (Vb - Va) is the potential difference. Rearranging the formula, we have Vb - Va = W/q. Substituting the given values, Vb - Va = (4.80×10-19 J) / (-1.60×10-19 C) = -3.0 V.

This implies that the particle moves in the direction of decreasing potential. Therefore, the particle moves from point a to point b through a potential difference of 3.0 V in the direction of decreasing potential.

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:

42.5 Hz.

Explanation:

The fundamental frequency of a closed pipe is given as

f₀ = v/4l....................... Equation 1

Where f₀ = lowest frequency, v = speed of sound in air, l = length of the organ pipe

Given: v = 340 m/s, l = 2.00 m.

Substitute into equation 1

f₀ = 340/(4×2)

f₀ = 340/8

f₀ = 42.5 Hz.

Hence the smallest frequency that will resonant in the organ pipe = 42.5 Hz.

112.11 V

The electromotive force, e.m.f, (E) of the generator is related to its terminal voltage(V) and the current (I) it delivers as follows;

E = V + (I x r) -------------------------(i)

Where;

r = the internal resistance of the generator

Now, as stated in the question;

at V = 108V, I = 10.2A

Substitute the values into equation (i) as follows;

E = 108 + (10.2 x r)

E = 108 + 10.2r       ------------------(ii)

Also, as stated in the question;

at V = 93V, I = 47.4A

Substitute these values also into equation (i) as follows;

E = 93 + (47.4 x r)

E = 93 + 47.4r      ------------------------(iii)

Now solve equations (ii) and (iii) simultaneously;

Subtract equation (iii) from (ii)

  E = 108 + 10.2r

_

  E = 93 + 47.4r

______________

  0 =  15 - 37.2r           ----------------(iv)

_______________

Solve for r in equation (iv)

37.2r = 15

r = 15 / 37.2

r = 0.403Ω

The internal resistance is therefore 0.403Ω

Substitute r = 0.403Ω into equation (ii);

E = 108 + 10.2(0.403)

E = 108 + 4.11

E = 112.11

Therefore, the emf is 112.11 V

The electromotive force (emf) produced in the generator is 112.1126 Volts.

Given the following data:

To calculate the electromotive force (emf) produced in the generator:

Mathematically, the electromotive force (emf) is given by the formula:

[tex]E = V + Ir[/tex]

For the first instance A:

[tex]E = 108 + 10.2r[/tex]   .....equation 1.

For the first instance B:

[tex]E = 93 + 47.4r[/tex]   .....equation 2.

Next, we would equate eqn 1 and eqn 2:

[tex]108 + 10.2r = 93 + 47.4r\\\\108 - 93 = 47.4r - 10.2r\\\\15 = 37.2r\\\\r = \frac{15}{37.2}[/tex]

Internal resistance, r = 0.4032 Ohms

Now, we can calculate the electromotive force (emf) produced in the generator:

[tex]E = 108 + 10.2r\\\\E = 108 + 10.2(0.4032)\\\\E = 108 + 4.1126[/tex]

Emf, E = 112.1126 Volts.

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Explanation:

The relation between force, mass and acceleration is as follows.

               F = ma

or,          a = [tex]\frac{F}{m}[/tex]

As two blocks are in contact with each other. Hence, total mass will be as follows.

                mass = 5 kg + 2 kg

                          = 7 kg

Now, we will calculate the acceleration as follows.

             a = [tex]\frac{F}{m}[/tex]

                = [tex]\frac{20}{7}[/tex]

Hence, force exerted by mass of 2 kg on a mass of 5 kg will be calculated as follows.

               [tex]20 - F_{1} = 5 \times \frac{20}{7}[/tex]

                   [tex]F_{1}[/tex] = 5.714 N

Thus, we can conclude that magnitude of the force that the 5.00-kg block exerts on the 2.00-kg block is 5.714 N.