All categories

3 years ago

How long would it take for a person to sprint from the 0m line to the 100m line if they are traveling forward at 6m/s?

Physics

1 answer:

valina [46]3 years ago

8 0

Answer:

<em><u>This section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Average and Instantaneous Acceleration we introduced the kinematic functions of velocity and acceleration using the derivative. By taking the derivative of the position function we found the velocity function, and likewise by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, we can work backward and calculate the velocity function from the acceleration function, and the position function from the velocity function.</u></em>

Explanation:

<h3>Derive the kinematic equations for constant acceleration using integral calculus.</h3><h3>Use the integral formulation of the kinematic equations in analyzing motion.</h3><h3>Find the functional form of velocity versus time given the acceleration function.</h3><h3>Find the functional form of position versus time given the velocity function.</h3>

You might be interested in

Can someone help me???

LenaWriter [7]

Answer:

heliocentric is when the earth and the other planets revolves around the sun.

Explanation:

I'm not sure about which one but I do know what heliocentric is.

6 0

3 years ago

Four students made a graphic organizer describing the parts of the atom.

Nikitich [7]

Answer:

I believe the answer is D.

Explanation:

Protons are found inside the nucleus so are neutrons. Electrons are found outside the nucleus.

8 0

3 years ago

Read 2 more answers

I need help with this someone help me?

Brilliant_brown [7]

Graph A so answer B also why isn’t answer a with graph A and B with graph B etc like that’s just confusing lol

3 0

3 years ago

What are conductors, semiconductors, non-conductors, dielectrics?

elena55 [62]

Conductors are substances that pass an electrical charge.

Semiconductors are substances whose electrical conductivity is lower than that of metals and greater than that of dielectrics.

Electricity nonconductors or insulators - in the terminology of Faraday - dielectrics (see). N. perfect does not exist; they represent only a large resistance to galvanic current and then different bodies in varying degrees (see Galvanic current), so that between poor and good conductors there are many bodies of average conductivity. N. The galvanic current is also the best insulators of static electricity. N. Heat or its bad conductors are at the same time electrical insulators (see Thermal Conductivity).

A dielectric (insulator) is a substance that is poorly conducting or not conducting at all. The concentration of free charge carriers in a dielectric does not exceed 108 cm-3. The main property of the dielectric is the ability to polarize in an external electric field. From the point of view of the band theory of a solid body, a dielectric is a substance with a band gap greater than 3 eV.

5 0

3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm , giving it a ch

vladimir1956 [14]

A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is $3.2\cdot 10^7 N/C$ (radially inward)

C) The electric field at 6.00 cm from the surface is $1.2\cdot 10^7 N/C$ (radially inward)

Explanation:

We can solve the problem by applying Gauss Law, which states that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the magnitude of the electric field

dS is the element of the surface

q is the charge contained within the surface

$\epsilon_0$ is the vacuum permittivity

By taking a sphere centered in the origin,

$\int E dS = E \cdot 4\pi r^2$

where $4\pi r^2$ is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

$R=9.0 cm = 0.09 m$ (radius of the plastic sphere is half of the diameter)

Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:

$q=0$

And therefore,

$E4\pi r^2 = 0\\\rightarrow E = 0$

So, the electric field inside the plastic sphere is zero.

Here we apply again Gauss Law:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

$r=R=0.18 m$

The charge contained within the Gaussian sphere is therefore

$q=-29.0 \mu C = -29.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C$

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

$E=3.2\cdot 10^7 N/C$

For this part again, we apply Gauss Law:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be

r = 9 cm + 6 cm = 15 cm = 0.15 m

While the charge contained within the sphere is again

$q=-29.0 \mu C = -29.0\cdot 10^{-6}C$

Therefore, the electric field in this case is

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C$