what is ex(p), the value of the x-component of the electric field produced by by the line of charge at point p which is located at (x,y) = (a,0), where a = 9.7 cm?

Answer:

An anode is an electrode through which conventional current (positive charge) flows into the device from the external circuit, while a cathode is an electrode through which conventional current flows out of the device.

Explanation:

The equation that describes the graph is y = 16.7x + 150.

We have been given the graph of a certain function data set in physics. We know that the graph is the representation of data on cartesian coordinates. In this case, we are asked to find the equation of the graph in the form; y=mx+b

m = slope of the graph

b = intercept of the graph.

To obtain the slope;

m = y2 - y1/ x2 - x1

m = 400 - 200/ 16 - 4

m = 200/ 12

m = 16.7

Then we can see from the graph that the y - intercept is 150. Having these data, the equation that could describe the graph is now;

y = 16.7x + 150

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Myelin and larger diameter will lower the speed of propagation of an action potential .Myelin is a fatty substance that wraps around and insulates the axons of neurons. It acts as an electrical insulator, preventing the leakage of ions and enhancing the speed of signal conduction. When an axon is myelinated, the action potential "jumps" from one node of Ranvier (unmyelinated region) to the next, a process known as saltatory conduction. This allows for faster propagation of the action potential along the axon. Therefore, the presence of myelin increases the speed of propagation.

Low Temperature: Temperature can have an impact on the speed of propagation of an action potential. In general, lower temperatures tend to slow down the rate of chemical reactions and electrical conductivity. In nerve cells, lower temperatures can decrease the efficiency of ion channel function and ion movement across the cell membrane, leading to slower propagation of the action potential.
Larger Diameter: The diameter of the axon also plays a role in the speed of propagation. A larger diameter allows for a greater surface area, facilitating faster movement of ions and reducing resistance to electrical flow. This results in faster propagation of the action potential compared to axons with smaller diameters.
High Temperature: As mentioned earlier, temperature affects the rate of chemical reactions and electrical conductivity. Higher temperatures can increase the speed of ion channel function and ion movement, leading to faster propagation of the action potential.
In summary, myelination and larger axon diameter increase the speed of propagation, while low temperatures and high temperatures can decrease or increase the speed, respectively.

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Linear momentum is a physical quantity that describes the motion of an object in a straight line. It is the product of an object's mass and its velocity, and it has both magnitude and direction. The direction of the linear momentum is the same as the direction of the object's velocity. In other words, linear momentum is a vector quantity.

Angular momentum, on the other hand, is a physical quantity that describes the rotational motion of an object. It is the product of an object's moment of inertia and its angular velocity, and it also has both magnitude and direction. The direction of the angular momentum is perpendicular to the plane of rotation, and it is defined by the right-hand rule. In other words, angular momentum is also a vector quantity.

In summary, the main difference between linear momentum and angular momentum is the type of motion that they describe. Linear momentum describes the motion of an object in a straight line, while angular momentum describes the rotational motion of an object.

Answer:

because if someone close to u gets sick you would know how to cure it

The heat loss per hour from the top surface of the plate is 20kW when an air at 20°C blows over a heated steel plate with its surface maintained at 200°C.

Given the temperature of air (Ta) = 20°C = \(273 + 20 = 293K\)

The temperature of heated steel surface plate (Ts) = 200°C = \(273 + 200 = 473K\)

The dimensions of the plate are = 50 x 40 cm such that the area of the plate =\(2000cm^2 = 0.2m^2\)

the thickness of the plate is = 2.5cm

The coefficient of convective heat-transfer(h) = \(20 W/(m^2*K)\)

The thermal conductivity of steel is(α) = 45 W/(m*K)

Let the heat loss per hour from the top surface = Q

The heat loss per hour from the top surface of the plate can be calculated using the following equation:

\(Q = h A (T1 - T2):\)

\(Q = 20 W/(m^2K) * 0.2 m^2 * (473 K - 293 K)\)

Q = 20000 W = 20 kW

Hence, the heat loss per hour from the top surface of the plate is 20 kW.

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According to the following formula, the intensity upon an angle with the axis Intensity is defined as (Central Highest Intensity) * (sinc (angle * /w))2.

in which the slit's width (w) and also the light's wavelength () are both given.

Accordingly, the concentration perpendicular towards the axis is equal to the intensity of something like the central maximum and is as follows:

Intensity is measured in terms: (Centre Maximum Intensity) * (sinc (angle * 589 nm/w))2

The the light's intensity at such an angles towards the center maxima can indeed be computed utilizing diffraction equation  theory and presuming the singular slits is just a rectangle split according follows:

I() = I0 sin2 (Bsin/I)

Here B is the slit's width, I0 is the center maximum's intensity, & seems to be the light's wavelengths.

Substituting the given values we get:

I(θ) = I₀ sin²(πBsinθ/589 x 10⁻⁹ m)

For sodium yellow light (589 nm), the intensity of the central maximum is I₀.

In addition to highlighting of the center maxima, the intensity at such an angle with the axis is nonetheless:

I(θ) = I₀ sin²(πBsinθ/589 x 10⁻⁹ m)

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Explanation:

mass = newton ÷ g

m = 1025 ÷ 10

mass= 102.5kg

Since Intensity depends on power and area, star b will have the same brightness as star a but look smaller

Luminous Intensity is the magnitude or quantity of the brightness of light emitted by a luminous object.

If two stars have the same luminosity, but star b is three times farther away from us than star a. If we are to compare star a with star b, star b will look bright but small.

The reason is due to the fact that Intensity is directly proportional to power and inversely proportional to area and not to the distance or length.

Therefore, star b will look bright but small.

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\( \sf{(i)  \: We \:  are \:  given  \: a  \: figure \:  of  \: Series \:  Circuit. }\)

\( \sf{Here \:  Resistances  \: are ,}\)

\( \sf• \:  R_{1}  =3  Ω\)

\( \sf•  \: R_{2}  =3Ω\)

\( \sf•  \: R_{3}  =3Ω\)

\( \sf{We \:  know  \: the  \: formula  \: of  \: the  \: Equivalent \:  Resistance \:  for  \: Series  \: Circuit, }\)

\( \sf ⇒ R_{s} =(3 + 3 + 3)Ω\)

\( \sf \therefore R_{s}  =9Ω\)

\( \sf  \pink{ \boxed{Answer : 9 Ω.}}\)

\( \\  \\ \)

\( \sf{(ii)  \: We \:  are \:  given  \: a  \: figure \:  of  \: Parallel \:  Circuit. }\)

\(\sf{Here \:  Resistances  \: are ,}\)

\( \sf• \:  R_{1}  =3  Ω\)

\( \sf• \:  R_{2}  =3  Ω\)

\( \sf• \:  R_{3}  =3  Ω\)

\( \sf{We \:  know  \: the  \: formula  \: of  \: the  \: Equivalent \:  Resistance \:  for  \: Parallel  \: Circuit, }\)

\( \sf⇒ \frac{1}{ R_{p} }  =  \frac{1}{3}  +  \frac{1}{3}  +  \frac{1}{3} \)

\( \sf⇒ \frac{1}{ R_{p} }  =  \frac{1 + 1 + 1}{3} \)

\( \sf⇒ \frac{1}{ R_{p} }  =  \frac{3}{3} \)

\( \sf⇒ \frac{1}{ R_{p} }  =  1\)

\( \sf \therefore R_{p}   =  1Ω\)

\( \sf  \pink{ \boxed{Answer : 1 Ω.}}\)

Answer:

Vf = 11.04 m/s

Explanation:

First, we consider free fall motion with zero initial velocity. Using 2nd equation of motion:

h = Vi t + (1/2)gt²

where,

h = height of bridge = ?

Vi = Initial  Speed  = 0 m/s

t = time taken = 4.5 s

g = 9.8 m/s²

Therefore,

h = (0 m/s)(4.5 s) + (1/2)(9.8 m/s²)(4.5 s)²

Therefore,

h = 99.225 m

Now, we consider the forced motion when the youth throws the ball with some negative initial energy:

Vi = - Vi

t = 3 s

h = 99.225 m

Therefore,

99.225 m = - Vi(3 s) + (1/2)(9.8 m/s²)(3 s)²

Vi = (- 99.225 m + 44.145 m)/3 s

Vi = - 18.36 m/s

Now, we use this in 1st equation of motion for final velocity:

Vf = Vi + gt

Vf = -18.36 m/s + (9.8 m/s²)(3 s)

Therefore,

Vf = 11.04 m/s

Answer:

3.75m/s²

Explanation:

g= GM/r²

For planet 1

\(g_{1}\)= GM/r²                   (i)

\(g_{1}\) = 15m/s²

for planet 2

radius= 2*r= 2r

g= GM/r

\(g_{2}\)= GM/(2r)²

\(g_{2}\)= GM/4r²

\(g_{2}\)=  GM/r² *1/4

from (i)

\(g_{2}\)=  \(g_{1}\) *1/4

\(g_{2}\) = 15/4

\(g_{2}\) = 3.75m/s²

Answer:

potential energy is transformed into kinetic energy

Answer:

Explanation:

Welcome to Pagasi b,  new recruit.  You will enjoy your new, fresh , home planet.  Your new home world is more relaxed than earth, days are longer by about quadruple, as are the nights.   Also seasons are also about 4 times  as long on Pagasi b. one year on Pagasi b is a bit over 4 years on earth.  You'll get great harvests and plenty of time to rest up during the mild, yet cool winters.  If the equatorial region is too warm for you during summer you can always travel north as the planet is at a 79 degree axis tilt , and will provide much cooler climes as you travel north for summers.  Have a great day, day, day, day   :P

Dr. Hewitt's experiment using the bowling ball to showcase the relationship between momentum, energy and speed of a body in motion. In his first attempt, the ball returns and stops almost exactly at the point it was launched.

Therefore, the ball stopped at the position of initial launch during the first trial.

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Hi there!

Both bullets will land at the same time.

The bullet's velocity in the HORIZONTAL direction has no impact on its vertical velocity since the horizontal and vertical components of its velocity are SEPARATE.

Thus, the bullet fired horizontally still has an initial VERTICAL velocity of 0 m/s, just like the bullet that was dropped from rest.

Since both bullets have no initial VERTICAL velocity, both will land at the same time.

Answer:

9.99

Explanation:

The value of (997)^1/3

(997)^1/3

997 = (1000 - 3)

(1000 - 3)^1/3

Expanding :

[1000(1 - 3/1000)]^1/3

1000^1/3 * (1 - 3/1000)^1/3

Cube root of 1000

10 * (1 - 3/1000 * 1/3)

10 * (1 - 1/1000)

10 * (1 - 0.001)

10(0.999)

= 9.99

Hence, the value of (997)^1/3 according to binomial theorem is 9.99

The value of (997)^(¹/₃) according to binomial theorem is;

9.99

The value of

(997)^(¹/₃)

⇒ (1000 - 3)^(¹/₃)

Factorizing out common terms gives;

⇒ (1000)^(¹/₃) * (1 - 3/1000)^(¹/₃)

Using the given approximations from binomial theorem, we have;

⇒ 10(1 + (¹/₃ × -³/₁₀₀₀) ....

⇒ 10(1 - 0.001)

⇒ 10 × 0.999

⇒ 9.99

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The specific heat of the reddish-brown metal is 0.092 cal/(g*°C).

We can use the formula Q = m * c * ΔT to solve this problem, where Q is the heat absorbed or released by the metal, m is the mass of the metal, c is its specific heat, and ΔT is the change in temperature.

In this problem, the metal gave off 71.7 cal of heat as its temperature decreased from 97.5°C to 22.0°C. Therefore, ΔT = (22.0°C - 97.5°C) = -75.5°C. Since the metal gave off heat, Q is negative, so Q = -71.7 cal. The mass of the metal is 10.3 g.

Substituting these values into the formula, we get:

-71.7 cal = (10.3 g) * c * (-75.5°C)

Solving for c, we get:

c = -71.7 cal / (10.3 g * (-75.5°C))

c = 0.092 cal/(g*°C)

Therefore, the specific heat of the reddish-brown metal is 0.092 cal/(g*°C).

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The system has a steady-state response of 1. The response of the system oscillates before reaching the steady-state. The overshoot of the system is approximately 5%.

The given transfer function is of third-order. We can reduce the transfer function to an approximate second-order system by ignoring the pole at s = -50. Here, the transfer function is given as:

RC​=(s+1) (s+5) (s+50) 50​ To obtain the approximate second-order transfer function,

we will ignore the pole at s = -50. RC​=1(s+1) (s+5)50​

The transfer function of a second-order system is given by:

G(s)= ωn2s2+2 ωns+ωn2

For the given transfer function, ωn 2=1/RC50= 1/5C Ωn=0.45ωn=0.15

For the given transfer function, we can take ζ = 0.45 which will provide a closed-loop response with an overshoot of 5%.

The transfer function of the approximate second-order system will be: G(s)=0.0225s2+0.0675s+0.0225

We can find the roots of the characteristic equation using the quadratic formula:

s=− ωn± ωn2−4 ωn 2 2s=-0.0342 ± j0.1508, -0.0342 - j0.1508 Using the MATLAB software, we can plot the unit step response and the closed-loop poles. We can obtain the following plot:

Unit Step Response: The plot of the unit step response is given below:

The approximate second-order transfer function of the system is G(s)=0.0225s2+ 0.0675s+ 0.0225. The closed-loop poles are located at -0.0342 ± j0.1508. The plot of the unit step response is shown below.

The system has a steady-state response of 1. The response of the system oscillates before reaching the steady-state. The overshoot of the system is approximately 5%.

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The distance traveled by the wheel edge by one rotation of the wheel, if the radius is 22 cm, is  138.16 cm.

The circular movement of an object about a primary axis is referred to as rotation or spin. A revolving object in two dimensions only has one conceivable central axis and can revolve either clockwise or counterclockwise. There are countless conceivable central axes and rotating orientations for a three-dimensional object.

Given:

The radius of the wheel, r = 22 cm,

The distance traveled by the point on the wheel's edge in one rotation will be equal to the circumference of the wheel.

Calculate the circumference of the circle by the formula given below,

Circumference of circle = \(2\pi \ r\)

Substitute the values,

C = 2 × 3.14 × 22

C = 138.16 cm

Therefore, the distance traveled by the wheel edge by one rotation of the wheel, if the radius is 22 cm, is  138.16 cm.

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Answer:

The trebuchet was invented in France and was first reported to be used in 1124AD in the siege of Tyre (in present-day Lebanon) during the Crusades. As it was much more powerful than a catapult, a trebuchet became the siege weapon of choice.

Explanation:

Most people think that the trebuchet is a medieval weapon, but actually, it has its origins in ancient China. The ancient Chinese trebuchet could throw huge boulders up to 250 pounds, a distance of 200 feet. In addition, it was lightweight and mobile which was crucial as it could be moved around the battlefield with ease. The thing I love about the chinese trebuchet is that it pushes the boundaries of ancient engineering to create a ballistics revolution - a flexible, portable, killing machine.

The correct answer of Brewster’s angle is 48.8 degrees.

What is light's Brewster angle as it travels through water?

Brewster's law is the name of this equation, and the angle it defines is known as Brewster's angle. Brewster's angle for visible light is around 56° for a glass medium in air (n2 = 1.5), while it is roughly 53° for an air-water interface (n2 = 1.33)

The inverse tan of n 2 over n 1 is what is known as Brewster's Angle. Therefore, 48.8 degrees is the inverse tan of 1.52 divided by 1.33.

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Because r is given, we can use the second expression in the equation ac=v2r;ac=rω2 a c = v 2 r ; a c = r ω 2 to calculate the centripetal acceleration. Solution.

1.3 m/s is the average speed of runner. Option D is the correct answer.

The total distance covered by the runner in one lap is equal to the distance around the two semicircles plus the distance along the two parallel lines. The distance around one semicircle is πr = π(49 m) and since there are two semicircles, the total distance around the semicircles is 2π(49 m). The distance along each parallel line is 96 m, and since there are two parallel lines, the total distance along the parallel lines is 2(96 m). Therefore, the total distance covered by the runner in one lap is:

Total distance = 2π(49 m) + 2(96 m) = 2π(49 m + 48 m) = 2π(97 m)

The time taken by the runner to complete one lap is given as 100 seconds. Therefore, her average speed is:

Average speed = Total distance / Time taken = [2π(97 m)] / (100 s) = 1.93 m/s

Rounding to two significant figures, the average speed of the runner is 1.9 m/s, which corresponds to option (D) 1.3 m/s.

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