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My name is Ann [436]

3 years ago

6

Two students noted that the acceleration of a car was - 40 km/s^2. The first student said, “That can’t be right. You can’t have

a negative acceleration.” The second student said, “That can’t be right. You can’t have s^2.” What would you say to each student to help them better understand how acceleration is described? pls answer

1 answer:

GuDViN [60]3 years ago

4 0

The 40 cause ion the air is not gone be in madden

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Identification: This refers to the speed of a chemical process.

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<span>The speed that reactants are converted into products is Reaction Rate. Reaction Rates describe slow reactions like rusting of an iron bar or fast reactions like combustion of gasoline in a car engine. This is extensively studied in the field of Chemical k</span>

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A bowling ball with a mass of 7.0kg strikes a pin that had a mass of 2.0kg the pin flies forward with a velocity of 6.0m/s, and

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The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.

We don't have external forces, so:

$P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\$

Where:

mb is the mass of the bowling ball
mp the mass of the pin
$v_{0b}\quad and\quad v_{0p}$ the initial velocities of the bowling ball and the pin.
$v_{1b}\quad and\quad v_{1p}$ the final velocities of the bowling ball and the pin.

Solving for v0b:

$v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}$

<h2>R/ The original velocity of the ball was 5.71 m/s.</h2>

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3 years ago

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

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3 years ago

What is the connection between our sense of hearing and our balance?

Vesna [10]

Answer:

When the head moves, fluid within the labyrinth moves and stimulates nerve endings that send impulses along the balance nerve to the brain. Those impulses are sent to the brain in equal amounts from both the right and left inner ear.

Explanation:

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A toy car of mass 8 kg initially moves with a speed of 5 m/s. How much work must be done on the car to increase its speed to 10

sukhopar [10]

ANSWER:

300 J

STEP-BY-STEP EXPLANATION:

To know the work required, we must calculate the work in both cases, the difference would be the amount of work necessary for the speed to increase. The work done is the same as the amount of energy increase. The formula for kinetic energy is:

$\frac{1}{2}m\cdot v^2$

We calculate in each case:

$\begin{gathered} \frac{1}{2}\cdot8\cdot10^2=400\text{ J} \\ \frac{1}{2}\cdot8\cdot5^2=100\text{ J} \end{gathered}$

We calculate the difference between the two to find out the work required:

$400-100=300\text{ J}$

The work required is 300 J

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2 years ago