An 1,840 W toaster, a 1,460 W electric frying pan, and a 50 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.) (a) What current (in A) is drawn by each device?

Answer:

The y-component of the car's position vector is 670m/s.

The x-component of the acceleration vector is -3, and the y-component is 40.

Explanation:

The displacement vector of the car with velocity

[tex]\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s[/tex]

is the integral of the velocity.

Integrating [tex]\boldsymbol{v}[/tex] we get the displacement vector [tex]\boldsymbol{d}[/tex]:

[tex]\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j} )[/tex]

Now if the initial position if the car is

[tex]\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})[/tex]

then the displacement of the car at time [tex]t[/tex] is

[tex]\boldsymbol{d(t)}= \boldsymbol{r+d}[/tex]

[tex]\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j} )}[/tex]

Now at [tex]t=10s[/tex], we have

[tex]\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j} )}m[/tex]

The y-component of the car's position vector is 670m/s.

The acceleration vector is the derivative of the velocity vector:

[tex]\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})[/tex]

and at [tex]t=10s[/tex] it is

[tex]\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2[/tex]

The x-component of the acceleration vector is -3, and the y-component is 40.

The x and y components of the car's position at 10 s are -147.0 m and 668.67 m, respectively. The x and y components of the car's acceleration at 10 s are -3 m/s² and 40 m/s², respectively.

The problem involves determining the position and acceleration components of a remote-controlled car from given velocity functions over time.

1.) X Component of Car's Position at 10 s:

Given the velocity component, vx = -3t m/s, we need to integrate it with respect to time to find the position (x). The initial position x0 is 3.0 m.

When t = 10 s:

2.) Y Component of Car's Position at 10 s:

Given the velocity component, vy = 2t² m/s, integrating it with respect to time gives the position (y). The initial position y0 is 2.0 m.

When t = 10 s:

3.) X Component of Car's Acceleration at 10 s:

Given the velocity component, vx = -3t m/s, the acceleration is the time derivative of velocity.

Hence, at t = 10 s:

4.) Y Component of Car's Acceleration at 10 s:

Given the velocity component, vy = 2t² m/s, the acceleration is the time derivative of velocity.

Hence, at t = 10 s:

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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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]

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.

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.

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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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

In a single phase, the addition or removal of energy changes Kinetic Energy not Potential Energy. However, during a phase change, this energy addition or subtraction results in a change in Potential Energy, not Kinetic Energy.

The subject of this question is energy and phase, particularly in the context of Potential Energy (PE) and Kinetic Energy (KE). When only one phase is present, adding or removing energy will mainly change the KE, not the PE. This is because the energy is utilized to speed up or slow down the particles, thus changing their kinetic energy. However, during a phase change, adding or removing energy changes PE but not KE as it alters the state rather than the speed of the particles. Statement B is the one that is accurate while only one phase is present, whereas the correct option for the phase change scenario is option D.

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To produce a second tone or the first overtone in a tube closed at one end, the length of the tube required is three times the length used for the fundamental frequency, resulting in a length of 1.008 m.

To understand the length required to produce a second tone or the first overtone in a tube closed at one end, it's essential to grasp the concept of harmonics in sound resonance. In such a tube, the resonant frequencies occur in odd multiples of the fundamental frequency. The first resonance the students observed, with the fundamental frequency of 256 Hz at a length of 0.336 m, corresponds to a quarter wavelength of the sound wave in the tube.

For the first overtone (second resonance), the air column in the tube must accommodate three-quarters of a wavelength, meaning the effective length will be three times larger than that of the fundamental. Thus, if the fundamental resonance occurs at a length of 0.336 m, the length for the second resonance will be:

This calculation is based on the understanding that the second tone or first overtone in a closed tube happens at three times the length necessary for the fundamental frequency, leading to the described increase in the length of the air column.

To find the length of tube for the second resonance, halve the initial length where the first resonance occurred at a fundamental frequency of 256 Hz.

The length required to produce a second tone under the same experimental conditions can be calculated based on the concept of resonance in a closed tube.

To find the length for the second resonance (first overtone), we know that the first resonance occurs at 0.336m for a fundamental frequency of 256 Hz. The second resonance, in this case, would occur at half the wavelength of the fundamental frequency, so the length would be half of the initial length: 0.168m.

Answer:

maximum width of the doorway = 35.77ft

Explanation:

The detailed calculation and derivation from first principle is as shown in the attachment

Answer:

the maximum width is x= 4√2 ft = 5.656 ft

Explanation:

for the parabola

y= a*x² + b*x + c

where y= height and x= width

an aircraft hangar should be symmetric with respect to the y axis , then

y(-x)=y(x) → a*x² + b*x + c = a*x² - b*x + c →-2*b*x =0 → b=0

it also should be pointing downwards → a is negative

, then the parabola would be

y= c- a*x²

since c= maximum height = 18 ft

then for y=0 , x= 48 ft/2 = 24 ft  →  0 = 18 ft - a*(24 ft)² → a= 1/32 ft⁻¹

then

y= 18 ft- 1/32 ft⁻¹ *x²

since the doorway cannot go beyond the parabola , the maximum possible doorway is obtained when the doorway touches the parabola.

then for a height y= 8 ft

8 ft = 18 ft- 1/32 ft⁻¹ *x²

x= 4√2 ft = 5.656 ft

Final Answer:

The speed at which the car hits the water is approximately 75.2 feet per second.

Explanation:

To find the speed at which the car hits the water, we can use one of the kinematic equations that relates the initial velocity, acceleration due to gravity, the height it fell from, and the final velocity. The kinematic equation that we need is:

[tex]\[ v^2 = u^2 + 2gh \][/tex]

Where:
-  v  is the final velocity,
-  u  is the initial velocity,
-  g  is the acceleration due to gravity (which we will use  [tex]\( 32.174 \, \text{ft/s}^2 \)[/tex] for since we are dealing with feet),
-  h  is the height.

Here, we are given:
-  [tex]\( u = 8 \, \text{ft/s} \)[/tex] (initial velocity)
- [tex]\( h = 87 \, \text{ft} \)[/tex] (height)
- [tex]\( g = 32.174 \, \text{ft/s}^2 \)[/tex] (acceleration due to gravity)

Let's find the final velocity \( v \) using these values.

[tex]\[ v^2 = u^2 + 2gh \][/tex]

[tex]\[ v^2 = (8 \, \text{ft/s})^2 + 2 \cdot 32.174 \, \text{ft/s}^2 \cdot 87 \, \text{ft} \][/tex]

[tex]\[ v^2 = 64 \, \text{ft}^2/\text{s}^2 + 2 \cdot 32.174 \, \text{ft/s}^2 \cdot 87 \, \text{ft} \][/tex]
[tex]\[ v^2 = 64 \, \text{ft}^2/\text{s}^2 + 5591.148 \, \text{ft}^2/\text{s}^2 \][/tex]
[tex]\[ v^2 = 5655.148 \, \text{ft}^2/\text{s}^2 \][/tex]

Now we take the square root of both sides to solve for the final velocity \( v \):

[tex]\[ v = \sqrt{5655.148} \, \text{ft/s} \][/tex]

Performing the square root calculation, we get:

[tex]\[ v \approx 75.2 \, \text{ft/s} \][/tex]

So, the speed at which the car hits the water is approximately 75.2 feet per second.

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

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.

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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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:

(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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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:

Explanation:

False

Electric conductivity in metals is the result of the motion of charged particles called electrons. Atoms of metallic elements are distinguished by the availability of valence electrons. Valence electrons are the electrons in the outer shell of an atom and free to move around.

These free electrons are free to move in the lattice of metal and result in electric current when a voltage is applied.

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.

Answer: Asteroids, meteoroids, and comets are remnants of the early solar system. This Statement is TRUE.

Explanation:

METEOROID: these are small rocky or metallic objects found in outer space.

ASTEROIDS: these are also known as minor planets of the inner solar system. They are irregularly shaped object in space that orbits the Sun.

COMETS: these are dusty chunk of ice, that moves in a highly elliptical orbit about the sun.

Asteroids, meteoroids, and comets as remnants of the early solar system was further proved in nebular hypothesis

initially proposed in the eighteenth century by German philosopher Immanuel Kant and French mathematician Pierre-Simon Laplace. (The word nebula means a gaseous cloud.) According to the modern version of the theory, about 4.5 to 5 billion years ago the solar system developed out of a huge cloud of gases and dust floating through space. These materials were at first very thin and highly dispersed.

Answer:

The correct answer is:

d) A and B have the same magnitude flux

Explanation:

Electric flux is the property of electric field that measures the electric field lines, passing through a surface and electric flux is also directly proportional to the number of electric field lines passing through a surface.  

The formula of electric flux is:

Φ = E A Cos θ

(where E is the electric field, A is the area of face and θ is the angle between the face and the electric field).

Since, faces A and B are perpendicular to the electric field and the electric field lines passing through face A also passes through face B therefore, both of these faces have larger and same magnitude of electric flux.

Since, face C is parallel to the electric field so, the electric flux is smaller at face C, because the magnitude of Cos 180 (when face is parallel) is smaller than the magnitude of Cos 90 (when face is perpendicular).

Face A, which is perpendicular to the uniform electric field, has the largest magnitude electric flux through it because the angle between the field lines and the normal to the surface is zero, maximizing the electric flux.

The question revolves around calculating the electric flux through different faces of a prism in a uniform electric field. Electric flux (Φ) is given by the equation Φ = E ⋅ A ⋅ cos(θ), where E is the magnitude of the electric field, A is the area through which the field lines pass, and θ is the angle between the field lines and the normal (perpendicular) to the surface.

Face A is perpendicular to the electric field, which means the angle θ is 0 degrees and cos(θ) is 1. Thus the flux through Face A is maximum. For Face B, the leaning face, θ is greater than 0 degrees but less than 90 degrees, thus cos(θ) will be less than 1. Hence, flux through Face B will be less than through Face A. Face C, being parallel to the electric field, has θ as 90 degrees, and cos(90) is 0, so the flux through Face C is zero. Therefore, in comparison, Face A has the largest magnitude electric flux through it.

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:

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.

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.

Read more: https://brainly.com/question/15121836?referrer=searchResults

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:

To find the car's acceleration, use the kinematic equation v = u + at. The car's acceleration is 2.5 m/s^2. To find the car's speed as it passes the 0.14 km post, plug the values into the kinematic equation v = u + at. The car's speed is 20 m/s.

Explanation:

To find the car's acceleration, we can use the kinematic equation:
v = u + at
where v is the final velocity, u is the initial velocity (which is 0 m/s since the car starts from rest), a is the acceleration, and t is the time. We are given that the car passes the 0.14 km post at 8.0 s after starting, which means the car travels a distance of 0.14 km in 8.0 s. Converting 0.14 km to meters gives us 140 m. Plugging the values into the equation, we have:
20 = 0 + a * 8.0
Simplifying, we find that the car's acceleration is 2.5 m/s^2.

To find the car's speed as it passes the 0.14 km post, we can use the kinematic equation:
v = u + at
Since the car starts from rest (u = 0 m/s) and the car's acceleration is 2.5 m/s^2 (which we just found), we can plug these values into the equation along with the time (8.0 s) to find the car's speed:
v = 0 + 2.5 * 8.0
Simplifying, we find that the car's speed as it passes the 0.14 km post is 20 m/s.

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:

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.

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.

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]