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4vir4ik [10]
3 years ago
11

A truck ttaveling with an initial velocity of 60.0m/s comes to a stop in 12.93 secs. What is the accelerationof the truck?

Physics
1 answer:
anastassius [24]3 years ago
3 0

When a car travelling at an initial velocity of 10 m/s applies the brakes and bring ... accelerates from rest for a time of 8 seconds with an acceleration of 3.2m/s^2?


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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:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
The mass of the earth is 5.98 1024 kg, the mass of the moon is 7.36 1022 kg, and the distance between the centers of the earth a
Setler [38]

Answer:

 x_{cm} = 4.644 10⁶ m

Explanation:

The center of mass is given by the equation

         x_{cm} = 1 / M_{total}  ∑ x_{i}  m_{i}

Where M_{total} is the total masses of the system, x_{i} is the distance between the particles and m_{i} is the masses of each body

Let's apply this equation to our problem

        M = Me + m

        M = 5.98 10²⁴ + 7.36 10²²

        M = 605.36 10²² kg

Let's locate a reference system located in the center of the Earth

Let's calculate

       x_{cm} = 1 / 605.36 10²²   [Me 0 + 7.36 10²² 3.82 10⁸]

       x_{cm} = 4.644 10⁶ m

4 0
3 years ago
A car runs for about 10 years during its average useful life. If the odometer reads 120,000 miles at the end, what was the avera
Eva8 [605]

Answer:

12000 mph

Explanation:

Given that,

Distance read by the odometer = 120,000 miles

Duration of car, t = 10 years

We need to find the average speed of the car. Speed of an object is equal to the total distance covered divided by total time taken. So,

v=\dfrac{120000}{10}\\\\=12000\ mph

So, the average speed of the car is 12000 mph.

7 0
3 years ago
The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the ampli
Setler79 [48]

Answer:

0.72

Explanation:

T = Time period of oscillation = 1.5 s

Angular frequency is given as

w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s

A = Amplitude of oscillation = 40 cm = 0.40 m

\mu = Coefficient of static friction = ?

a = acceleration of the block

m = mass of the block

Maximum acceleration of the block is given as

a = Aw^{2}

frictional force is given as

f = \mu mg

As per newton's second law

f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72

8 0
2 years ago
I'm willing to give a lot of points if u can help me out.
Anettt [7]

Answer:

i have absolutly no idea how to do it but i looked it up and your answer should be B. i could be wrong but thats what the web told me

7 0
2 years ago
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