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Answer:
thanks for 50 points
Explanation:
!am sorry
Related Questions
determine the frequency bandwidth (in hz) and the high cut-off frequency (in hz) of a first- order thermal sensor having a time constant of 0.1s when subjected to the sinusoidal temperature variation, t(t)
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The frequency band width is 1.59 Hz and high cut-οff frequency is 1.59 Hz.
What is frequency?The frequency οf a repeated event is its number οf instances per unit οf time. It differs frοm angular frequency and is sοmetimes referred tο as tempοral frequency fοr clarificatiοn. The unit οf frequency is hertz (Hz), οr οne οccurrence per secοnd. The time elapsed between events is measured by the periοd, which is the reciprοcal οf the frequency.
Given data:-
The temperature [tex]$\mathrm{T}(\mathrm{t})=20 \sin \omega t$[/tex] And the time constant [tex]$\mathbf{T}=0.1 \mathrm{sec}$[/tex] The general equation of first order [tex]$\mathbf{H}(\mathrm{s})=\frac{\mathrm{K}}{T \mathrm{~T}+1}$[/tex] and [tex]$\mathrm{s}=\mathrm{j} \omega$[/tex]. Now the general first-order equation:
[tex]$\begin{aligned}\mathbf{H}(\mathrm{s}) & =\frac{\mathrm{K}}{(\mathrm{Ts}+1)} \\\mathrm{H}(\mathrm{j} \omega) & =\frac{\mathrm{K}}{(0.1 \mathrm{j} \omega)+1}\end{aligned}[/tex]
The magnitude of the [tex]$\mathrm{H}(\mathrm{j} \omega)$[/tex]
[tex]$|H(j \omega)|=\frac{K}{\sqrt{(0.1 \omega)^2+1}}$[/tex]
Now for the higher and lower frequency, i.e. [tex]$\mathrm{f}_{\mathrm{L}}$[/tex] and [tex]$\mathrm{f}_{\mathrm{h}}$[/tex]The upper frequency when [tex]$\omega=\frac{1}{\mathrm{~T}}=\frac{1}{0.1}=10 \mathrm{rad} / \mathrm{sec}$[/tex] the magnitude [tex]$|\mathrm{H}(\mathrm{j} \omega)|=\frac{\mathrm{K}}{\sqrt{2}}$[/tex] and a lower frequency when [tex]$\mathrm{\omega=0\ magnitude}\ |\mathrm{H}(\mathrm{j} \omega)=\mathrm{K}|$[/tex]
The upper frequency [tex]$\omega=2 \pi f$[/tex]
[tex]$\mathrm{f}_{\mathrm{h}}=\frac{10}{2 \pi}=1.59 \mathrm{~Hz}$$And lower frequency$$\mathrm{f}_{\mathrm{L}}=0$[/tex]
Now the frequency band width [tex]$\mathrm{B}=\mathrm{f}_{\mathrm{h}}-\mathrm{f}_{\mathrm{L}}=1.59-0=1.59 \mathrm{~Hz}$[/tex]
Thus, The frequency band width is 1.59 Hz and high cut-off frequency is 1.59 Hz.
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A car is driving counterclockwise on a circular path, moving at non-uniform speed. The small black circles in the figure show the position of the object in equal intervals of time.
a- The x component of the instantaneous acceleration at B is negative, positive or zero? Why?
b- The y component of the instantaneous acceleration at B is negative, positive or zero? Why?
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a- The instantaneous acceleration at B has a negative x component.
b- The instantaneous acceleration at B has a positive y component.
Explain about the uniform circular motion?We refer to a particle as being in uniform circular motion when it is traveling in a circular path (and perhaps a portion of one) at a constant speed.
The object is accelerating despite the fact that its speed is remaining constant because of a change in direction in its velocity (v).We refer to a particle moving uniformly around a circle as having a centripetal acceleration owing to the direction of the acceleration, which is toward the center.It is useful to discuss the time T required for the particle to accomplish one full trip around the circle if the particle consistently follows a complete circular path. This is referred to as the motion's period.On a round, counterclockwise route, an automobile is traveling at an irregular speed.
The position of a object is depicted in the figure's tiny black circles at regular intervals of time.
a- The instantaneous acceleration at B has a negative x component.
Because the x-component of the velocity will be be along the negative x axis, which will be directed towards point D.
b- The instantaneous acceleration at B has a positive y component. Because, the y-component of the velocity will be be along the positive y-axis, which will be directed towards point C.
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1 /B 1 B 1 1/ Q B A A space ship, initially traveling at constant velocity along a straight line, decides to switch course. It first fires it's engine at full power for a very short time in the direction perpendicular to its trajectory at point A, giving it an instantaneous large boost. Then at point B, it fires its engine continuously at a much lower power in the opposite direction until it reaches point C. Which of the following trajectories most closely represent the trajectory of the ship? R Search listening t... KWFinder: Keywor... A space ship, initially traveling at constant velocity along a straight line, decides to switch course. It first fires it's engine at full power for a very short time in the direction perpendicular to its trajectory at point A, giving it an instantaneous large boost. Then at point B, it fires its engine continuously at a much lower power in the opposite direction until it reaches point C. Which of the following trajectories most closely represent the trajectory of the ship? Pick the correct answer 0 1 g 2 O 3 o 4 o 5 Submit
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The spaceship undergoes an instantaneous change in velocity at point A and then undergoes a continuous change in velocity at point B until it reaches point C. The trajectory of the spaceship is therefore curved and can be represented by one of the given options.
What is Instantaneous Velocity?
Instantaneous velocity is the velocity of an object at a specific moment in time or at a specific point in its motion. It is the limit of the average velocity as the time interval over which it is calculated approaches zero. In other words, it is the slope of the tangent to the position-time graph at a particular point.
To calculate the instantaneous velocity of an object, you need to determine the object's position at a specific moment in time and then take the derivative of the position function with respect to time. The resulting derivative gives you the object's instantaneous velocity at that specific moment in time.
Instantaneous velocity is important in physics and many other fields because it allows us to understand the behavior of objects in motion and to calculate important quantities such as acceleration, momentum, and energy.
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A small 8.00 kg rocket burns fuel that exerts a time-varying upward force on the rocket (assume constant mass) as the rocket moves upward from the launch pad. This force obeys the equation F=A+Bt2. Measurements show that at t=0, the force is 100.0 N, and at the end of the first 2.00 s, it is 162.0 N.
A. Find the net force on this rocket at the instant after the fuel ignites.
B. Find the acceleration of this rocket at the instant after the fuel ignites.
C. Find the net force on this rocket 3.00 ss, after the fuel ignites.
D. Find the acceleration of this rocket 3.00 ss, after fuel ignition.
E. Suppose that you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 ss, after fuel ignition?
Answers
a) The net force at the point of fuel ignition is F = A = 100 N.
b) Acceleration after fuel ignites is a = 12.5 m/s^2
c) The net force on the rocket at t=3.00 s is 403 N.
d) Acceleration of rocket 3.00 ss after fuel ignites is a = 50.4 m/s^2
e) Acceleration in outer space is a = 50.4 m/s^2
How the solution was obtainedA. The force at t=0 is given as 100 N, so A = 100 N. We can use the given information to find B:
F = A + Bt^2
162 N = 100 N + B(2.00 s)^2
B = (162 N - 100 N) / (2.00 s)^2
B = 31 N/s^2
Therefore, the net force on the rocket at t=0 is:
F = A = 100 N.
B. The acceleration of the rocket is given by Newton's second law:
F_net = ma
where F_net is the net force acting on the rocket, and a is the acceleration of the rocket. At t=0, the net force on the rocket is 100 N. Therefore, the acceleration of the rocket at t=0 is:
a = F_net / m
a = 100 N / 8.00 kg
a = 12.5 m/s^2
C. To find the net force on the rocket at t=3.00 s, we can simply plug in t=3.00 s into the force equation:
F = A + Bt^2
F = 100 N + 31 N/s^2 (3.00 s)^2
F = 403 N
Therefore, the net force on the rocket at t=3.00 s is 403 N.
D. To find the acceleration of the rocket at t=3.00 s, we can use the same equation as in part B:
F_net = ma
At t=3.00 s, the net force on the rocket is 403 N. Therefore, the acceleration of the rocket at t=3.00 s is:
a = F_net / m
a = 403 N / 8.00 kg
a = 50.4 m/s^2
E. In outer space, far from all gravity, the only force acting on the rocket is the force from the burning fuel. Therefore, the net force on the rocket is simply the force from the burning fuel:
F = A + Bt^2
F = 100 N + 31 N/s^2 (3.00 s)^2
F = 403 N
Using the same equation as in part B, the acceleration of the rocket is:
a = F_net / m
a = 403 N / 8.00 kg
a = 50.4 m/s^2
Therefore, the acceleration of the rocket in outer space would be the same as in part D.
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It takes tiffany 0. 25 hours to get to school in the mornings. She lives 4. 5 miles away from the school. At what speed (in miles per hour) is she traveling to get there?.
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If It takes Tiffany zero.25 hours to get to school in the mornings. She lives 4.5 miles far away from the faculty, then the speed of traveling off the Tiffany would be 28.96 kilometers per hour.
velocity of touring = general distance traveled by means of the Tiffany / general time
velocity of touring = 4.5 × 1.609 / 0.25
= 28.96 kilometers per hour
Speed in physics is a scalar quantity that refers to the rate at which an object moves, usually expressed in units of meters per second (m/s) or kilometers per hour (km/h). It is defined as the distance an object travels per unit of time, which means that it only takes into account the magnitude of the object's motion and not its direction.
Instantaneous speed refers to the speed of an object at any given moment, while average speed is calculated over a period of time. The speed of an object can be influenced by several factors, such as the forces acting upon it and the medium through which it is moving.
In addition to speed, there are other related concepts in physics, such as velocity, which includes both the speed and direction of an object's motion, and acceleration, which describes the fee at which an item's speed modifications over time.
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Quantity with kgm^3/s^1
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The quantity cannot be expressed in [tex]kgm^3/s^1[/tex] as this is not a valid unit of measurement.
What is a quantity?A quantity is a measurable amount of something, such as length, weight, time, or area. It is usually expressed as a numerical value, such as 3 meters, 5 kilograms, or 10 minutes. Quantities can be compared and combined to form new quantities. For example, adding two lengths of 3 meters each yields a new quantity of 6 meters.Quantities are used to describe physical phenomena and to measure the results of experiments.
A unit of quantity is a standard measurement used to measure the amount of a particular item. This can be a single unit, such as a kilogram, or a group of units, such as a dozen eggs. Units of quantity are used to help standardise measurements and allow for easier comparison between different products.
The quantity cannot be expressed in [tex]kgm^3/s^1[/tex] as this is not a valid unit of measurement. The correct unit for this quantity is kg/s.
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Can quantity be expressed by [tex]kgm^3/s^1[/tex]?
The modern atomic theory has been updated over the years as new observations of the atom have been made. What is likely to happen in the future?.
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The modern atomic theory is likely to continue evolving as new discoveries are made, particularly in areas such as dark matter, quantum mechanics, computational power and simulation techniques
It is likely that the modern atomic theory will continue to be updated as new observations and discoveries are made in the field of atomic and subatomic particles.
One area where there is ongoing research is in the study of dark matter and dark energy, which make up a large portion of the universe but cannot be directly observed. Understanding the nature of these phenomena could lead to new insights into the behavior of particles at the atomic level.
Another area of ongoing research is in the study of quantum mechanics and its application to atomic and subatomic particles. As technology advances and scientists are able to study these particles in greater detail, it is likely that our understanding of quantum mechanics will continue to evolve.
Additionally, advancements in computational power and simulation techniques may allow scientists to simulate and predict the behavior of atoms and molecules with greater accuracy, leading to further refinements of the modern atomic theory.
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3. What is the electrostatic force of attraction between a -5.8 x 10-7C charge and a 3.8 x 10-8C charge if they are separated by a distance of 4.4 meters?
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Answer:
1.73 x 10-7 N
Explanation:
3. What is the electrostatic force of attraction between a -5.8 x 10-7C charge and a 3.8 x 10-8C charge if they are separated by a distance of 4.4 meters?
The electrostatic force of attraction between two charges can be calculated using Coulomb's law, which states that the force of attraction is inversely proportional to the square of the distance between them. In this case, the two charges are -5.8 x 10-7C and 3.8 x 10-8C and they are separated by a distance of 4.4 meters. Plugging these values into Coulomb's law, the electrostatic force of attraction is 1.73 x 10-7 N.
