For the purpose of calculating the electric field strength by means of Gauss’s law, determine whether approximate cylindrical symmetry holds in each of the following situations.
Part (a) We have a 3.6-m long copper rod of radius 1 cm, carrying a charge of 1.5 nC distributed uniformly along the rod’s length. We want to calculate the electric field strength at a point 4.9 cm from the rod near its center.
TRUE FALSE
Part (b) We have a 8.9-cm long copper rod of radius 1 cm, carrying a charge of 1.5 nC distributed uniformly along the rod’s length. We want to calculate the electric field strength at a point 4.9 cm from rod near its center.
TRUE FALSE
Part (c) A 1.9-m long wooden rod is glued end-to-end to a 1.9-m long plastic rod, both of radius 1 cm. The combined rod is then painted with an electrically charged paint so that it is covered with a uniform charge density, giving it total charge of 1.5 nC. We want to calculate the electric field strength at a point 4.9 cm from the rod near its center.
TRUE FALSE
Part (d) For the same charged rod as in part (c) we want to calculate the electric field strength at a point 8.9 m from each end of the rod.
TRUE FALSE

Answer: They’re the most reactive metals

Explanation: They have larger ionic radii and low ionization energies

A car is moving at 16 m/s at a time t = 0. With an acceleration of 0.50t, in m/s², for t, in seconds, it slows down. At t = 8 seconds, it stops. Option D is correct.

The change of displacement with respect to time is defined as the velocity.  velocity is a vector quantity.

Given data;

t = 0 ,u=16 m/s

acceleration = −0.50t m/s²

v=0,t=?

The acceleration is found as;

[tex]\rm a= \frac{dv}{dt} \\\\\ dv=adt \\\\ dv=-0.50 \ t dt[/tex]

Intregating both sides we get;

[tex]\rm [v]_0^{16}=-0.50\frac{t^2}{2} \\\\ 16=0.25 t^2\\\\\ t^2=\frac{16}{0.25} \\\\\ t= 8 \ sec[/tex]

Hence, the car stops at t=8sec. Option D is correct.

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Answer:

Im pretty sure they're equal, i could be wrong though. Hope i helped have an amazing day :D

Explanation:

Answer:

B: equal

Explanation:

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: [tex]\frac{∆v}{t}[/tex]

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 [tex]\frac{feet}{second}[/tex] then sped up to 5 [tex]\frac{feet}{second}[/tex], then the change in my speed would be 2 [tex]\frac{feet}{second}[/tex], because I started walking 2 [tex]\frac{feet}{second}[/tex] faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 [tex]\frac{meters}{second}[/tex] to 16 [tex]\frac{meters}{second}[/tex]  in 4 seconds. What would the acceleration be? Let's set up an equation:

a = [tex]\frac{∆v}{t}[/tex]

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 [tex]\frac{meters}{second}[/tex] by 4 [tex]\frac{meters}{second}[/tex]. That makes sense, right? Back to the equation.

a = [tex]\frac{∆v}{t}[/tex]
a = [tex]\frac{16-4}{4}[/tex]

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = [tex]\frac{12}{4}[/tex]

a = 3 ([tex]\frac{meters}{second^{2}}[/tex])

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit [tex]\frac{\frac{meters}{second}}{second}[/tex], which can be simplified down to [tex]\frac{meters}{second^{2} }[/tex])

Let me know if you need clarification on anything I explained here!
- breezyツ

Answer:

354875 J

Explanation:

The equation for Work is:

[tex]Work=Force*Distance[/tex]

We can plug in the given values into the equation:

[tex]Work=4250N*83.5m=354875J[/tex]

The speed of the plane wave defined by sin(ωt−kx) is 2.864 × 10⁸ m/s

the speed of the plane wave defined by sin(ωt−kx) is v = ω/k where

Now ω = 3376.7 Thz = 3376.7 × 10¹² Hz and k = 11.79μm⁻¹ = 11.79 × 10⁶ m⁻¹

So, substituting the values of the variables into v, we have

v = ω/k

v = 3376.7 × 10¹² Hz/11.79 × 10⁶ m⁻¹

v = 286.4 × 10⁶ m/s

v = 2.864 × 10⁸ m/s

The speed of the plane wave defined by sin(ωt−kx) is 2.864 × 10⁸ m/s

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Answer:

average speed= total distance/total time

Explanation:

Related to the distance traveled by the object which means, it is only represented by the magnitude and direction of travel is not important. so average speed= total distance/total time

Answer:b

Explanation: If you look at the line on the graph, you can see that it is going downward, meaning it has a negative slope, and choice b is the only one that has a negative slope

Answer:

150 watts

Explanation:

300/2 = 150 watts

Answer:

150 watts

Explanation:

The explanation is in the picture

PLEASE MARK BRAINLIEST

Answer:

10m-2m = 8m

8

Hope this helps!

Answer:

I don't know what are you saying but I don't have any results

Explanation:

[tex]kai6417[/tex]

Hi there!

We can use the following equation for the period of a pendulum:

[tex]\large\boxed{T = 2\pi \sqrt{\frac{L}{g}}}[/tex]

T = Period (s)
L = length of simple pendulum (.75 m)

g = acceleration due to gravity (9.8 m/s²)

Plug in the values:

[tex]T = 2\pi \sqrt{\frac{.75}{9.8}} = \boxed{1.738 s}[/tex]

Peak expiratory flow readings are higher when patients are well, but lower when a  patient's airway is constricted.

This term is used in medicine to refer to the maximum rate of a person's exhalation or to how fast and how long can they exhale.

This rate is measured by using the peak expiratory flow test that requires you to deeply inhale and then exhale as fast as hard as possible.

This rate is closely related to the amount of air flowing out of the lungs. This implies the peak expiratory flow is lower if there is a disease such as asthma that can constrict the airways.

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Answer:

30N

Explanation:

30N of force was applied

quizlet

In celestial mechanics, escape velocity or escape speed is the minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a primary body, thus reaching an infinite distance from it.

PLEASE MARK AS BRAINLIEST

Answer:

B

Explanation:

From Kepler's Third Law, T^2 is proportional to r^3.

Now r of B is larger than r of A,

so T of B is larger than T of A.

Then B takes the longest time.

Dwarf Galaxy B takes the longest time to orbit the large galaxy. This can be explained using Kepler's Third Law of Planetary Motion, also known as the Law of Periods.

Kepler's Third Law states that the square of the orbital period (T) of a celestial body is directly proportional to the cube of the semi-major axis (a) of its elliptical orbit. Mathematically, it can be expressed as:

[tex]\[ T^2 \propto a^3 \][/tex]

Where:

T is the orbital period (time taken for one complete orbit), and

a is the semi-major axis of the elliptical orbit.

In the case of dwarf galaxies A and B revolving around the large galaxy, we can use the distances from the large galaxy to each of the dwarf galaxies as the semi-major axes (a) of their orbits.

Let's compare the values of [tex]\( a^3 \)[/tex] for the two dwarf galaxies:

For Dwarf Galaxy A:

[tex]\( a_{\text{A}} = 70,000 \)[/tex] light years

For Dwarf Galaxy B:

[tex]\( a_{\text{B}} = 500,000 \)[/tex] light years

Now, let's calculate [tex]\( a^3 \)[/tex] for each galaxy:

For Dwarf Galaxy A:

[tex]\( a_{\text{A}}^3 = (70,000)^3 = 343,000,000,000 \)[/tex] cubic light years

For Dwarf Galaxy B:

[tex]\( a_{\text{B}}^3 = (500,000)^3 = 125,000,000,000,000,000 \)[/tex] cubic light years

Since [tex]\( a_{\text{B}}^3 \)[/tex] is significantly larger than [tex]\( a_{\text{A}}^3 \)[/tex], it means that Dwarf Galaxy B has a larger a value and therefore a longer orbital period according to Kepler's Third Law. In other words, Dwarf Galaxy B takes the longest time to orbit the large galaxy.

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Answer:

The noble gases are located in the 18th group of the periodic table.

Answer:

The inner planets are less dense than the outer planets

Please brainliest my answer iam begging you for god sake

Answer:

The inner planets are less dense than the outer planets .

Explanation:

Answer:

Net Force. when all combinations of force acting on an object. Force. A push or pull

So it's the third answer

( the combination of all forces acting on it )

Answer:

5.3 hours

Explanation:

(3180 km) / (600 km/h) = 5.3 hours

Given :-

To Find :-

Solution :-

Here it's given that a car of mass 1000kg moving with velocity of 25m/s North . Another car of mass 1200kg is moving with velocity 30m/s East.

So that ,

[tex]v_{1}= 25m/s \ E= 25 \hat{i} m/s [/tex]

And ,

[tex]v_{2}= 30m/s \ N = 30 \hat{j} m/s [/tex]

See attachment .

From Triangle law of vector addition , the resultant of both vectors will be given by third side of the triangle taken in opposite direction.

So , if the resultant velocity makes an angle [tex]\theta[/tex] . Then ,

[tex] tan\theta = \dfrac{v_2}{v_1} [/tex]

Substituting the respective values,

[tex]tan\theta =\dfrac{30m/s}{25m/s} [/tex]

Simplify ,

[tex]tan\theta =\dfrac{6}{5}[/tex]

Take arctan both sides ,

[tex]\theta = tan^{-1}\bigg(\dfrac{6}{5}\bigg)[/tex]

Hence the cars move at an angle of tan-¹(6/5) NE after the collision .

As the angle the plane makes with the horizontal is increased, the magnitude of the force of static friction between the block and the plane (2) increases

Since the static frictional force between the block and the plane is given by F = μN where

Now, F = μN

F = μWcosθ

Since μ and W are constant, we see that

F ∝ cosθ and cosθ ∝ θ

So, F ∝ θ

Since the static frictional force is directly proportional to the angle the plane makes with the horizontal, when the angle the plane makes with the horizontal is increased, the static friction between the block and the plane increases.

So, As the angle the plane makes with the horizontal is increased, the magnitude of the force of static friction between the block and the plane (2) increases

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Answer:

Force can be defined as the product of mass and acceleration.

Explanation:

~Cornasha_Weeb

When you double the speed of your car, your braking distance quadruples. every time you double your speed, you multiply your braking distance by four.

The stopping distance of a vehicle is the sum of breaking distance and thinking distance. The stopping distance of a vehicle when we increase the velocity from 10mph to 30 mph, increases from30 feet to 90 feet.

Stopping distance is the distance covered by a vehicle when we stops. Distance is the product of time and velocity. As the velocity increases then, the time taken to cover the distance will be lower.

Stopping distance mathematically is the sum of breaking distance and thinking distance .Thinking distance is simply the speed and it is in fact means how long it takes on average to think for a break.

Breaking distance is means how long the driver takes to break or stop in actual. In general, the breaking distance is speed multiplied by 2.

Thus, for 10 mph, the thinking distance is 10 feet and breaking distance is 10 × 2 = 20 feet and therefore, the stopping distance is 10 + 20 =30 feet.

Similarly for 30 mph, breaking distance is 60 feet and thinking distance is 30 feet thus, having 60 + 30 = 90 feet of stopping distance.

Therefore, when the speed increased from 10 mph to 30 mph, stopping distance increases from 30 feet to 90 feet.

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Answer:

because they have opposite charges

Explanation:

If you're talking about magnetism and electricity, the laws are that like charges attact and and unlike charges repel.

Charged particles interact with each other because they have opposite charges.

When charged particles are close to each other, their electric fields interact. So particles with opposite charges attract each other whereas particles with similar charges repel each other.

So we can conclude that Charged particles interact with each other because they have opposite charges.

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oh ok where is it bring it