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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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asambeis [7]

This question is incomplete, the complete question is;

The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answer:

the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Explanation:

Given that;

P₁ = 1.00 atm

P₂ = ?

V₁ = 1 L

V₂ = 1.60 L

the temperature of the gas is kept constant

we know that;

P₁V₁ = P₂V₂

so we substitute

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

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The ball drop 2kms in the air
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Answer:

N(80.8^{\circ})E

Explanation:

Bearing from control tower= N(32^{\circ}+\theta})E

But tan \theta=\frac {opposite}{adjacent}

tan \theta=\frac {88}{77}

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Answer:

A force does work on an object if a component of the force is parallel to the displacement of the object.

Explanation:

Work, a measurement of energy is said to be done when a force applied to an object results in the movement of that object to a certain distance and direction. Force is the act of push or pulls occurs on an object as a result of the interaction between that object with another one and displacement is the distance and direction covered by that object as a result of the force applied on it.

The work done (W) by a constant force (F) is equal to the product of the force in the direction of displacement of the object and the distance (d) moved by the object i.e., W = F * d.

The angle between the displacement and the force is θ, then the work done, W = Fd cos θ  ........ (1)

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Negative work - Force acts in the opposite direction with respect to the displacement of the object.  Here, θ is 180°, so cos θ i.e., cos 180° is -1. Therefore, from the equation (1), W = -Fd (i.e., work done by the force is negative).

If a force is applied to an object and it does not move, then the work done is zero i.e., W = F * 0 = 0. Also, if the force and displacement are at right angle to each other, then θ is 90°. Therefore, from the equation (1), W = 0 since cos 90° is zero.

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In (A. Dalton's) atomic model, negative electrons orbit the positively charged nucleus.
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