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lidiya [134]
3 years ago
10

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted s

o that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 12 cm. When the cylinder is rotating at 2.0 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?a. 4.8 m/s²b. 19 m/s²c. 27 m/s²d. 9.8 m/s²e. 6.0 m/s²
Physics
1 answer:
lukranit [14]3 years ago
7 0

Answer:

Option B

a=19 m/s^{2}

Explanation:

Given information

Radius of container, r=12cm=12/100=0.12m

Angular velocity= 2 rev/s, converted to rad/s we multiply by 2π

Angular velocity, \omega=2*2\pi =12.56637061

We know that speed, v=r\omega

Centripetal acceleration, a=\frac {v^{2}}{r} and substituting v=r\omega we obtain that

a=r\omega^{2}

Substituting \omega for 12.56637061  and r for 0.12

a=0.12*(12.56637061)^{2}=18.94964045 m/s^{2}

Rounded off, a=19 m/s^{2}

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Answer:240 meters

Explanation:60×4

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3 years ago
The position equation for a particle is s of t equals the square root of the quantity 2 times t plus 1 where s is measured in fe
loris [4]

Given that,

s = √(2t + 1)

Time, t = 4 s

Acceleration , a = ??

Since,

Acceleration = velocity / time

Velocity = distance/ time

Acceleration = distance/ time²

s/t² = √(2t+1)/t²

putting t = 4 sec, we have

a = √(2*4+1)/4²

a = √(5)/16

a= 0.139 ft/s²

Therefore, acceleration of the given particle will be 0.139 feet/ second².

8 0
3 years ago
When the hydrogen atom makes the transition from the n=2 to the n=1 energy level, it emits a photon. This photon can be absorbed
bazaltina [42]

Answer:

n_{fn}= 4

Explanation:

To solve this exercise we will use Bohr's atomic model

               E_{n} = - 13.606 / n²     [eV]

The transition from level n = 2 to level n = 1 is valid

               E_{21} = - 13.606 [¼ -1/1]

               E_{21} = 10.2045 eV

Bohr's model for atoms with only one electron is

               E_{n} = -13.606 Z² / n²

Where Z is the atomic number of the atom.

In this case the helium atom has an atomic number of Z = 2 from the level n₀ = 2 let's look up to what level it reaches

         ΔE = -13.606 [4 /  n_{fn}² - 4/4]

         4 / n_{fn}² = -ΔE / 13.606 + 1

         4 / n_{fn}² = -10.2045 / 13.606 +1 = -0.75 +1

         4 / n_{fn}² = 0.25

        n_{fn} = √ 4 / 0.25

        n_{fn}= 4

8 0
3 years ago
A large body of air hats has similar properties of a temperature and moisture is a
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6 0
4 years ago
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
3 years ago
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