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2 years ago

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A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2.00 m making 0.700 revoluti

ons per second. What is the centripetal acceleration of the ball?

2 answers:

Pavlova-9 [17]2 years ago

8 0

Centripetal acceleration = (speed)²/(radius) .

We know the radius. We have to find the speed.

Speed around a circle = (circumference) / (time to go around)

The circumference of the circle is (2 π) (radius) = 4 π meters.

We don't exactly know the time to go around.
We know that the ball goes around 0.7 times/second.
Flip that over, and you have time to go around = second/0.7 .

So now, the centripetal acceleration is

       (speed)²/(radius) .

   = (4π meters · 0.7/sec)² / (2 meters)

   = (4π · 0.7 / 2) m/s²

   = about 4.4 m/s²

Ira Lisetskai [31]2 years ago

6 0

1) find speed (8.8 m/s)
2) find acceleration (38.7 m/s^2)
answer is about 38.7 m/s^2

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How do you put 0.0015kg in scientific notation

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0.0015 kg would be 1.5*10^-2 kg in scientific notation

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2 years ago

A 0.01-kg object is initially sliding at 9.0 m/s. It goes up a ramp (increasing its elevation by 1.5 m), and then moves horizont

barxatty [35]

Answer:

During this motion, 0.133 J of heat energy was created

Explanation:

Hi there!

Let´s calculate the energy of the object in each phase of the motion.

At first, the object has only kinetic energy (KE):

KE = 1/2 · m · v²

Where:

m = mass of the object.

v = velocity.

KE = 1/2 · 0.01 kg · (9 m/s)²

KE = 0.405 J

When the object goes up the ramp, it gains some gravitational potential energy (PE). Due to the conservation of energy, the object must convert some of its kinetic energy to obtain potential energy. By calculating the potential energy that the object acquires, we can know the loss of kinetic energy:

PE = m · g · h

Where:

m = mass of the object.

g = acceleration due to gravity (9.81 m/s²)

h = height.

PE = 0.01 kg · 9.81 m/s² · 1.5 m

PE = 0.147 J

The object "gives up" 0.147 J of kinetic energy to be converted into potential energy.

Then, after going up the ramp, the kinetic energy of the object will be:

0.405 J - 0.147 J = 0.258 J

When the object reaches the spring, kinetic energy is used to compress the spring and the object obtains elastic potential energy (EPE). Let´s calculate the EPE obtained by the object:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compression of the spring

EPE = 1/2 · 100 N/m · (0.05 m)² = 0.125 J

Then, only 0.125 J of kinetic energy was converted into elastic potential energy. The object is at rest at the end of the motion, i.e., the object does not have kinetic energy when it compresses the spring by 5.0 cm. Since energy can´t be lost, the rest of the kinetic energy, that was not used to compress the spring, had to be converted into heat energy:

Heat energy = initial kinetic energy - obtained elastic potential energy

Heat energy = 0.258 J - 0.125 J = 0.133 J

During this motion, 0.133 J of heat energy was created.

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3 years ago

A car is moving in a straight line with the same speed of 100 m/s ,the acceleration in this

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Answer:

a = 0m/s²

Explanation:

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2 years ago

As air pressure decreases, what happens

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Answer:

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2 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:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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2 years ago