in curvilinear particle motion which of the following can be negative? group of answer choices normal component of velocity tangential component of velocity tangential component of acceleration normal component of acceleration

The power of the pump is its work done by time. The power used  for a work done of 800 Nm in 50 seconds is 16 J/s or W. The force acting here is 40 N. Then the mass of the body is 4 kg.

Power of an object is the rate of work done or energy used.

power = work done/ time.

It has the unit watt which is equal to J/s.

a. Given work done by the pump  = 800 Nm

time = 50 s

then power = 800 Nm or J / 50 s  = 16 Watt.

b. Force F = Work done/ distance

F = 800 Nm/20 m = 40 N

c. Then force = m a

given acceleration a = 10 m/s²

m = f/ a

   = 40 N/10 m/s²

   = 4 kg.

Therefore, mass of the pump is 4 kg.

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(a) The initial velocity of the projectile is 42.23 m/s.

(b) The maximum height reached by the projectile is 91 m.

The initial velocity of the projectile is calculated by applying the following kinematic equation.

v = ( 24.0 î - 8.00j ) m/s

The angle of projection is calculated as;

θ = arc tan ( Vy / Vx )

θ = arc tan ( 8 / 24 )

θ = 18.43⁰

h = ( u² sin²θ ) / 2g

u² = 2gh / (sin²θ)

u² = ( 2 x 9.8 x 9.1 ) / ( sin 18.43 )²

u² = 1783.6

u = √1783.6

u = 42.23 m/s

The maximum height reached by the projectile is calculated as follows;

v² = u² - 2gh

where;

2gh = u²

h = ( u² ) / ( 2g)

h = ( 42.23² ) / ( 2 x 9.8 )

h = 91 m

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The final speed is 58.8 m/s.

We have to note that if we are to solve the problem that we have here, we have to look at the equations of motion and this is how we can be able to get the final velocity of the object.

Thus we are going to have that;

v = u + gt

v = final speed

u =- initial speed

g = acceleration

t = time

v = gt

v = 9.8 * 6

= 58.8 m/s

The object is going to have a final speed of about 58.8 m/s when we look at the calculations above here.

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Can you ensure that perhaps the bicycle's speedometer must have displayed 11 mph at least once for a rider who covers 33 miles. 

A speedometer is what?

Speedometer is a computer benchmark that evaluates how quickly Web applications respond. It simulates user behaviors like adding to-do items using demo web applications. The window of your browser is too small. Lets create the view port at least 850px x 650px in size to get the most accurate results. The current size is 1024 by 569 pixels.

Which speedometer works best with a Royal Enfield motorcycle?

Version 2.0 of the BRPEARl Heavy Digital Speedometer 5.5" Screen A8 automobile hud Head up Displays OBD2 Speedometer D'Mega Shop RX 100 135 RXG Instrument panel Mount Bracket An... Dig... Bullkartzone Speedo Dial for Royal Honda Cycle (W... bike rpm, tachometer, speedometer, tacho gauge

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Answer: The given time period of oscillation would be around 0.9515 seconds.

Explanation:

The very first 54.0 km were driven at an average speed, while the final 86.0 km were driven at in a speed of (43.0 b) km/h. She travels at an average speed of 43.75 km/h for the entire journey.

Velocity is the pace and velocity of an object's movement, whereas speed is the forecasting at which an objects is travelling along a path. In other words, velocity is a vector, whereas quickness is a scalar value. As a result, the fundamental unit of time and the basic element of distance are combined to form the Special name of speed. Thus, the metre per second (m/s) is the Substrate concentration of speed.

a = 19 and b = 7

Average speed, 54 km = 48+a km/h

Plug a = 19

54 km = 48+19

= 67 km/h

remaining speed

86 km = 43-b km/h

where b = 7

Average speed = 36 km/h

Average speed = total distance / total time taken

speed = distance/time

time = distance/speed

To cover 54 km = 54/67 hr

To cover 86 km = 86/36 hr

Total time = 3.2 hours.

Total distance = 86+54

= 140 km

Average speed 140/3.2

= 43.75 km/ h

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The answer is closest to option e, 1.97 m/s. To solve this problem, we need to use conservation of energy. The initial potential energy of the system is converted into kinetic energy as the string unwinds.

We can find the potential energy using the equation U = Iω²/2, where I is the moment of inertia of the pulley, and ω is its angular velocity.

The potential energy of the system is given by U = Iω²/2 = (0.0352 kg m²)(ω²)/2. Initially, the potential energy is all stored in the string, so we can set U equal to the work done by the constant force to find the final kinetic energy:

U = Fd = (5.00 N)(2.31 m) = 11.55 J

Equating this to the kinetic energy, we get:

K = U = (0.0352 kg m²)(ω²)/2

Solving for ω, we get:

ω = √(2K/I) = √(2U/I) = √(2(11.55 J)/(0.0352 kg m²)) = 14.44 rad/s

Now we can find the final speed of the string by multiplying the final angular velocity by the radius of the pulley:

v = ωr = (14.44 rad/s)(0.125 m) = 1.805 m/s

Therefore, the answer is closest to option e, 1.97 m/s.

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The following values are given: m1 = € kg, resistor = y1 Means 3.00 m, sq.m = 1.75 kg, r2 / y2. = 1.75 m, m3 = 3.00 metric tons, r3 = restrictions that limit. = 4.00 m, and = 1.75 rad/s about x-axis = m1r. 2.

The combined translational and rotational kinetic energies of an object in motion determine its total kinetic energy. 12 mvCM2 is the translational kinetic energy. Rotational kinetic energy is equal to 12 I2. Total kinetic energy is 12mvCM2 plus 12I2.

E s = x 2 x I ω 3 . . This equation shows that its kinetic energy of such a rotating solid body is inversely related to the cube of the angular acceleration and the moment of inertia. Flywheel energy-storage systems, which are intended to, take advantage.

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Answer:

Explanation:

when gravity pulls the water molecules downwards, they will fall. But if they're in a container, the container will keep them from spreading out completely. This is what's called 'taking the shape of the container'

The magnitude of the applied force is 14.59 N.

Newton's second law, which states that the net force exerted on an item is equal to its mass times its acceleration, can be used to address this issue.

Let's use the letters m1 and m2 to represent the masses of the left and right blocks, respectively. The following equations are therefore able to be written:

m1 * a = F - f

m2 * a = f

where F is the size of the applied force, f is the frictional force acting on the blocks, and an is the acceleration of both blocks. Because a string connects the two blocks, it should be noted that the acceleration is the same for both. The frictional force can be expressed as

f = * N,

where is the coefficient of friction and N is the normal force, can be used to express the frictional force. The weight of the blocks, which can be represented as m1 * g and m2 * g, respectively, determines the normal force acting on them.

When we solve for F and replace the frictional force expression in the first equation, we obtain:

F = m1 * a + μ * m1 * g

Inputting the values provided yields:

F = (m1 + m2) * a + μ * (m1 + m2) * g

F = (m1 + m2) * 5.2 + 0.4 * (m1 + m2) * 9.8

F = 5.2 * (m1 + m2) + 3.92 * (m1 + m2)

F = 9.12 * (m1 + m2)

We now need to determine what m1 Plus m2 equals. Although their specific masses are withheld from us, we do know that the ratio of their masses is 2:3. Let's write m for the smaller mass and 1.5m for the larger value. Next, we have:

m + 1.5m = 2.5m

1.5m/m = 3/2

So the masses are m = 2/5 and 1.5m = 6/5.

Inputting these values into the F equation yields the following results:

F = 9.12 * (2/5 + 6/5)

F = 9.12 * 8/5

F = 14.59 N

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If Earth were 9.5 times farther away from the Sun than it is now, 95 times weaker would be the gravitational force between the Sun and Earth.

Any two things with mass are drawn together by the gravitational force. The gravitational force is known as. The formula F = Gm1m2/r2 states that the direction of force will always point in the direction of the other mass along the line connecting the two bodies.

In contrast to gravity, which is the force of attraction between any two bodies, gravitation is the pull of any body towards the earth.

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The minimum magnitude of the acceleration of the car is 0.

The maximum acceleration will be 30m/s².

The distance traveled is 55m.

Minimum magnitude of acceleration of the car is 0. This is because in a V-t graph slope of tangent on the of car and at t=us slope is zero.

Maximum acceleration of car is from o tois where slope is steepest.

= V - u / t

= 30 - 0 / 1

= 30m/s²

The distance will be:

d1 = 15

d2 = 1/2(30 + 50) = 40

Distance = 15 + 40 = 55m

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Answer:

Approximately [tex]5800\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

Let [tex]m[/tex] denote the initial mass of the rocket and the fuel. The question states that the mass of the fuel is [tex]65\%[/tex] of the total mass, which is [tex]0.65\, m[/tex]. The mass of the rocket without the fuel would be [tex](1 - 0.65)\, m[/tex].

When an object of mass [tex]m[/tex] travels at a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of that object would be [tex]p = m\, v[/tex].

Initial momentum of the rocket and the fuel, combined, would be [tex]0\; {\rm kg \cdot m\cdot s^{-1}}[/tex] since initial velocity was [tex]0\; {\rm m\cdot s^{-1}}[/tex].

Let [tex]v(\text{rocket})[/tex] and [tex]v(\text{fuel})[/tex] denote the final velocity of the rocket and the fuel. Final momentum of the rocket would be [tex]((1 - 0.65)\, m)\, (v(\text{rocket}))[/tex]. Final momentum of the fuel would be [tex](0.65\, m)\, (v(\text{fuel}))[/tex].

The total final momentum of the rocket and the fuel, combined, woul dbe:

[tex]((1 - 0.65)\, m)\, (v(\text{rocket})) + (0.65\, m)\, (v(\text{fuel}))[/tex].

Under the assumptions, momentum would be conserved. In other words, the total momentum of the rocket and the fuel would stay the same:

[tex](\text{total final momentum}) = (\text{total initial momentum})[/tex].

[tex]((1 - 0.65)\, m)\, (v(\text{rocket})) + (0.65\, m)\, (v(\text{fuel})) = 0[/tex].

Given that [tex]v(\text{fuel}) = 3100\; {\rm m\cdot s^{-1}}[/tex]:

[tex]\begin{aligned}v(\text{rocket}) &= \frac{0.65}{1 - 0.65}\, v(\text{fuel}) \\ &= \frac{0.65}{1 - 0.65}\, (3100)\; {\rm m\cdot s^{-1}} \\ &\approx 5800\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

The uniform line of charge and [tex]\rho[/tex] for the uniformly charged sphere. Using slope the value of [tex]\lambda[/tex] is  [tex]\\\frac{4\pi\epsilon_0\times slope}{2}$[/tex].

For the uniform line of charge, the electric field $E$ is inversely proportional to the distance $r$, so $Er$ should be constant. For the uniformly charged sphere, the electric field $E$ is inversely proportional to the square of the distance $r$, so $Er^2$ should be constant. Therefore, we can use the graphs of $Er^2$ versus $r$ and $Er$ versus $r$ to determine which data set is for the uniform line of charge and which set is for the uniformly charged sphere.

For data set A, the graph of $Er^2$ versus $r$ is a straight line, which means that $Er^2$ is constant. Therefore, data set A is for the uniformly charged sphere. For data set B, the graph of $Er$ versus $r$ is a straight line, which means that $Er$ is constant. Therefore, data set B is for the uniform line of charge.

For the uniform line of charge, the electric field $E$ is given by

E =[tex]\frac{2\lambda}{4\pi\epsilon_0r}$,[/tex]

where $\lambda$ is the charge per unit length and $\epsilon_0$ is the permittivity of free space. Since $Er$ is constant, we can write $Er = \frac{2\lambda}{4\pi\epsilon_0}$. From the graph of $Er$ versus $r$ for data set B, we can find the slope of the straight line, which is equal to $Er$. Therefore, we can use the slope to calculate $\lambda$:

[tex]$\lambda = \frac{4\pi\epsilon_0Er}{2} = \frac{4\pi\epsilon_0\times slope}{2}$[/tex]

For the uniformly charged sphere, the electric field $E$ is given by $E = \frac{\rho r}{3\epsilon_0}$, where $\rho$ is the volume charge density. Since $Er^2$ is constant, we can write $Er^2 = \frac{\rho r^3}{3\epsilon_0}$. From the graph of $Er^2$ versus $r$ for data set A, we can find the slope of the straight line, which is equal to $Er^2$. Therefore, we can use the slope to calculate [tex]$\rho$[/tex]

[tex]$\rho = \frac{3\epsilon_0Er^2}{r^3} = \frac{3\epsilon_0\times slope}{r^3}$[/tex]

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Both a prism and a diffraction grating can be used to disperse white light into its component colors, which are a range of wavelengths of visible light.

θ = sin^(-1) (-1)

[n avg] = [(n blue - n red) sin()]

Assuming an angle and the values given as n red = 1.52, n blue = 1.53, and n avg = (n red + n blue)/2 = 1.525

of incidence of 45 degrees, we can calculate the angle of deviation to be:

θ = sin^(-1)[(1.53 - 1.52) sin(45°) / 1.525] = 0.34°

Therefore, the angular separation between red and blue light leaving the prism is 0.34 degrees.

nλ = d(sin θ_i + sin θ_d)

where n is the order of the diffraction (usually n=1), λ is the wavelength of the light, d is the spacing between the grating lines, θ_i is the angle of incidence of the light on the grating, and θ_d is the angle of diffraction.

θ_red = sin^(-1)[(nλ_red / d) - sin(θ_i)] = sin^(-1)[(1650 nm / 500 nm) - sin(0°)] = 0.837°

θ_blue = sin^(-1)[(nλ_blue / d) - sin(θ_i)] = sin^(-1)[(1450 nm / 500 nm) - sin(0°)] = 0.579°

Therefore, the angular separation between red and blue light leaving the grating is:

θ_separation = θ_red - θ_blue = 0.837° - 0.579° = 0.258°

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The velocity of the runner at t=π/2 seconds is 15 miles/hour.

The directional speed of an item in motion, as measured by a specific unit of time and viewed from a certain point of reference, is what is referred to as velocity.

The function is as follows:

f(t)=-10cos(t) + 15

The time in this case is shown as t.

We must change t=/2 in the given function to obtain the runner's speed at t=/2 seconds:

f(π/2) = -10cos(π/2) + 15

f(π/2) = -10(0) + 15

f(π/2) = 15

As a result, the runner's speed at t=/2 seconds is 15 miles per hour.

We may deduce that the runner is moving towards the north at a speed of 15 miles per hour at t=/2 seconds since f(t) > 0 denotes moving in that direction.

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1) The overall potential difference across the capacitor is 1.66 V.

2) The charge on each capacitor is 1.66 C.

3) The energy stored in the combination​ will be 1.3778 J.

A capacitor is a device that stores electrical energy in an electric field by accumulating electric charges on two insulated surfaces. It is a two-terminal passive electronic component. Capacitance is the effect of a capacitor.

Given that three capacitors of capacitance 1 micro Farad, 2 microfarads and 2 microfarads are each charged to 250 volts and then connected in series.

Q₁ = C₁Vt

Q₁ = 1 x 5 = 5 μF

Q₂ = C₂Vt

Q₂ = 2 x 5 = 10 μF

The equivalent capacitance will be,

1 / Ct = 1 / C₁ + 1 / C₂

Ct = 2 / 3 μF

Vt = 10 / 3 ÷ 2

Vt = 1.66 V

The charge will be,

q =CV

q = 1 x 1.66

q = 1.66 C

The energy will be calculated as:-

U = 1/2 x C  xV

U = 1 / 2 x 1 x 1.66²

U = 1.3778 J

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The motion that could be represented by the given position-versus-time graph is "A car drives toward the right at first slowly, then faster."

We can see that the graph has a positive slope, which indicates a motion in the positive direction. The fact that the slope is increasing over time means that the object is accelerating, which is consistent with the car starting off slowly and then getting faster.

phase 1 : object moves towards its mean position, velocity must be slow.

phase 2 : object moved to its initial position taking much less time compared to phase 1 but in the same direction to reach its mean position.

Therefore, the correct option is "A car drives toward the right at first slowly, then faster."

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density is the mass of a specific material per unit volume. d = M/V, in which d is densities, M is mass, as well as V is volume, is the formula for density. Common units for expressing density are grams per cubic inch.

What is the mass unit?

 There are several ways to measure mass, including kilograms, grams, pounds, and pounds, but the SI unit de mass is the "kilogram" or kg. Using the right conversion formula, any unit of weight can be changed into another unit without changing the significance or meaning of the quantity being measured.

Describe mass.

Definition, Units, Formula, and Examples of Mass The quantity of matter that makes up every object or body is the greatest way to understand mass. Everything that we can see has mass. Examples of objects with mass include a table, an chair, your mattress, a ball, a tumbler, and even air. The mass of a thing determines whether it is light or heavy.

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The given statement " plato's apology tells the story of how sorry socrates was for having upset his fellow athenians with his persistent questioning." is False

Plato's Apology is not about how sorry Socrates was, but rather a speech he delivered during his trial in which he defended his philosophical beliefs and actions, and refused to apologize for them. Socrates argued that his questioning was necessary to improve Athens and its citizens, and that his death would only harm the city. The Apology is often seen as a key text in the development of Western philosophy, and has been interpreted in many different ways.

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The amount of force after 0.05 seconds will be 528000 N.

A force is an influence that can cause an object's motion to change. A force can cause a mass object to change its velocity, or accelerate. Intuitively, a force can be described as a push or a pull.

Given that a car of mass 1500kg collides with a wall. the initial and final speed of the car is -15m/s and 2.6m/s. if the collision lasts for 0.05 seconds.

The force will be calculated as:-

Mc x V₁ + F x t = Mc x V₂

Here, V₁ = -15 m /s and V₂ = 2.6 m /s.

Mc x V₁ + F x t = Mc x V₂

-( 1500 x 15 ) + F x 0.05 = 1500 x 2.6

F x 0.05 = 22500 + 3900

F = 26400 / 0.05

F = 528000 N

Therefore, the force after the collision will be 528000 N.

The complete question is given below.

A car of mass 1500kg collides with a wall. the initial and final speed of the car is -15m/s and 2.6m/s. if the collision lasts for 0.05 seconds. find the impulse caused by the collision and the average force on the car​.

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The reaction rate constant and activation energy have an exponential connection, which is described by the Arrhenius equation, which may be used to predict reaction rate constants for chemical reactions as a function of temperature.

Define Q, ko, and R, as well as the other specified constants.

The "linspace" function in Matlab can be used to create a vector of temperature values ranging from 300 K to 1000 K.

Use the Arrhenius equation and the listed constants to determine the value of k for each temperature value.

Plot temperature values on the x-axis and k values on the y-axis using the Matlab "plot" function. Plot 1/T values on the x-axis and the log10 of k values on the y-axis using the same Matlab "plot" function.

Add titles to the figure window and label the x- and y-axes on both graphs.

% Define constants

Q = 1000; % J/mol

ko = 10; % s^-1

R = 8.314; % J/mol-K

% Create a vector of temperature values

T = linspace(300, 1000);

% Calculate k values using the Arrhenius equation

k = ko * exp(-Q./(R.*T));

% Create a subplot with two graphs

subplot(1, 2, 1)

plot(T, k)

xlabel('Temperature (K)')

ylabel('Reaction Rate Constant (s^-1)')

title('Arrhenius Plot')

subplot(1, 2, 2)

plot(1./T, log10(k))

xlabel('1/Temperature (1/K)')

ylabel('log10(Reaction Rate Constant)')

title('Modified Arrhenius Plot')

The Arrhenius plot and the modified Arrhenius plot are two subplots that should appear in a single figure window created by this code. The Arrhenius plot displays the correlation between temperature and the reaction rate constant, whereas the modified Arrhenius plot displays the correlation between 1/T and the log10 of the reaction rate constant.

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The activation energy for the reverse (backward) reaction would be 20 + 60 = 80 kJ/mol.

From the question, we are told that the activation energy of the reaction is 20 kJ and its E is -60 kJ/mol. A negative number for E means that it is an exothermic reaction, a reaction in which energy is released instead of absorbed.

The energy graph for an exothermic reaction is attached below. You can see that the activation energy for a backward reaction is EQUAL to the activation energy for the forward reaction + energy released.

Since we know the activation energy and E, the activation energy for the reverse (backward) reaction would be 20 + 60 = 80 kJ/mol.

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Answer:

Explanation:

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

Work done = 335-125=110J

Answer:

3N

Explanation:

yeah just that is it you don't need calculation to solve it

The worker can load {500/ρ} number of bricks into the van.

Mass is the total amount of matter present inside a body.

Given is that a worker wants to load some bricks in to his van. There are 1000 bricks and when stacked neatly they measure 2m by 1m by 1m.

Assume the density of the brick to be {ρ}. Let's assume that he can load {n} bricks inside the van. So, we can write -

n x mass of 1 brick = maximum load

n x ρ x V = L{max}

nρV = L{max}

n = L{max}/ρV

n = 1000/(2 x 1 x 1 x ρ)

n = 500/ρ

Therefore, the worker can load {500/ρ} number of bricks into the van.

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The speed of the cars right after the collision is 1/3 m/s.

To solve this problem, we can apply the law of conservation of momentum, which states that the total momentum of a system remains constant if there are no external forces acting on it. In this case, the two railroad cars form a closed system, and there are no external forces acting on them during the collision. Therefore, we can write:

(m1 * v1) + (m2 * v2) = (m1 + m2) * v

where m1 and v1 are the mass and velocity of the first railroad car, m2 and v2 are the mass and velocity of the second railroad car, m1 + m2 is the total mass of the system after the collision, and v is the velocity of the system after the collision.

Plugging in the given values, we get:

(1500 kg * 4 m/s) + (3000 kg * (-3 m/s)) = (1500 kg + 3000 kg) * v

Simplifying and solving for v, we get:

v = (1500 kg * 4 m/s - 3000 kg * (-3 m/s)) / (1500 kg + 3000 kg)

v = (6000 kgm/s + 9000 kgm/s) / 4500 kg

v = 15/45 m/s = 1/3 m/s

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The current flowing through potentiometer R is 0.855mA, the current flowing through potentiometer Rį for an RL 5KΩ and the current flowing through potentiometer R is 854μA,

The potentiometer is an instrument used for measuring the unknown voltage by comparing it with the known voltage. It can be used to determine the emf and internal resistance of the given cell and also used to compare the emf of different cells. The relative system is used by the potentiometer. The reading is more accurate in a potentiometer.

1) Rl = 1kΩ

Applying nodal analysis at V:

V/1K + V/3.3K + V-12/1K

3.3V+V+3.3V-39.6/3.3K = 0

7.6V = 39.6

V = 39.6/7.6 = 5.21 V

Il = v/Rl = 5.21/1K

Il = 5.21 mA

For Rl = 1K and Il = 5.21mA

Rl = 5 kΩ

Applying nodal analysis at V:

V/5 K + V/3.3K + V-12/1K

3.3V+5V+16.5V-198/3.3K = 0

24.8V = 198

V = 198/24.8 = 7.98 V

Tl = 7.98/5 K = 1.596 mA

Rl = 10 kΩ

Applying nodal analysis at V:

V/10K + V/3.3K + V-12/1K

3.3V+10V+33V-396/3.3K = 0

46.3V = 396

V = 396/46.3 = 8.55 V

IL = V/Rl = 8.55/10 K = 0.855mA

2) IL = 12m x 0.767K/0.767K + 10K

= 12m x 0.767/10.767

= 0.854 mA

= 854 μA

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Correct, low, quiet voices have sound waves with low frequency and low amplitude. Instead of hearing, the vestibular system controls balance and spatial orientation.

In order to determine where a sound is coming from, humans rely on two key clues. These signals include (1) the ear the sound strikes first (interaural temporal differences) and (2) the volume of the sound when it reaches each ear (known as interaural intensity differences).

Sound waves enter the outer ear and travel to the eardrum via the auditory canal. The three tiny ossicles carry the ensuing vibrations into the cochlea, where hair cells are able to pick them up.

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The ranking of the energy that is requires is;

4 > 3 > 2> 1

Atomic energy levels refer to the specific energy states that an electron can occupy in an atom. Electrons in an atom can only occupy certain energy levels, and the energy levels are quantized, meaning that only specific values are allowed.

When an electron absorbs energy, it can jump to a higher energy level, and when it releases energy, it drops to a lower energy level. The energy levels are determined by the properties of the atom, such as the number of protons and neutrons in the nucleus and the electrons' distribution in the atom's electron shells.

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Part A

The angular separation of the two stars is 0.03 degrees

Part B

The angular separation of the two stars is 108 arcseconds.

Part C

Yes, the Hubble Space Telescope can resolve the two stars.

Angular separation is the angular distance between two objects as seen from a particular point in space. This is usually expressed in degrees, arcminutes, or arcseconds. It is a measure of the angular separation between two objects, such as stars, planets, galaxies, or other celestial objects, as seen from Earth. The angular separation of two objects is the angle between them as seen from the observer's point of view.

For example, the Moon and the Sun have an angular separation of 0.5 degrees when they are in opposition.

Part A:

To find the angular separation of the two stars, we can use the formula:

θ = arctan(d/D)

where θ is the angular separation, d is the physical distance between the two stars, and D is the distance to the stars from Earth.

Substituting the given values, we get:

θ = arctan(100,000,000 km / (200 x 9.461 trillion km))

θ = arctan(100,000,000 km / 1.8922 x 10^15 km)

θ = 0.03 degrees

Therefore, the angular separation of the two stars is 0.03 degrees.

Part B:

To convert degrees to arcseconds, we can use the formula:

1 degree = 3600 arcseconds

Substituting the value we found in part A, we get:

θ = 0.03 degrees x 3600 arcseconds/degree

θ = 108 arcseconds

Therefore, the angular separation of the two stars is 108 arcseconds.

Part C:

To determine if the Hubble Space Telescope (HST) can resolve the two stars, we need to compare the angular separation of the stars to the angular resolution of the HST. The angular resolution of a telescope is the smallest angle between two point sources that the telescope can distinguish.

The angular resolution of the HST is about 0.05 arcseconds. Since the angular separation of the two stars is 108 arcseconds, the HST can easily resolve the two stars as separate objects.

Therefore, the HST can resolve the two stars in the binary star system.

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