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2 years ago

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A 0.0250 kg sock spins in a dryer of radius 0.225 m once every 0.304s. how much centripetal force acts on the sock?(unit=N)

1 answer:

shutvik [7]2 years ago

5 0

Sorry for bad handwriting but I hope it helps :)

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Answer:

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Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p

ira [324]

Answer:

The force exerted on the $q_1$ is $F = 2.25*10^{3} \ N$

Explanation:

From the question we are told that

The area is $A = 2.34*10^{-3} \ m^2$

The magnitude of charge placed on them is $q = 7.07 * 10^{-7} C$

The charge placed between the plate is $q_1 = 6.62 *10^{-5} C$

The electric field generated around the plate is mathematically represented as

$E = \frac{q}{A \epsilon_o}$

Substituting values

$E = \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}$

$E = 34*10^{6} \ V/m$

The force exerted the charge $q_1$ is mathematically represented as

$F = q_1 * E$

Substituting values

$F = 6.62 *10^{-5} * 34*10^{6}$

$F = 2.25*10^{3} \ N$

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2 years ago

Given that the atomic weight of hydrogen is approximately 1 gram per mole, use the method above to estimate the average molecula

denpristay [2]

Answer:

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Explanation:

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2 years ago

A Carnot heat engine has an efficiency of 0.400. If it operates between a deep lake with a constant temperature of 298.0k and a

tatuchka [14]

Answer:

496.7 K

Explanation:

The efficiency of a Carnot engine is given by the equation:

$\eta = 1 - \frac{T_H}{T_L}$

where:

$T_H$ is the temperature of the hot reservoir

$T_C$ is the temperature of the cold reservoir

For the engine in the problem, we know that

$\eta = 0.400$ is the efficiency

$T_C = 298.0 K$ is the temperature of the cold reservoir

Solving for $T_H$ , we find:

$\frac{T_C}{T_H}=1-\eta\\T_H = \frac{T_C}{1-\eta} =\frac{298.0}{1-0.400}=496.7 K$

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2 years ago