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Ksivusya [100]
3 years ago
14

An eccentric inventor attempts to levitate a cork ball by wrapping it with foil and placing a large negative charge on the ball

and then putting a large positive charge on the ceiling of his workshop. Instead, while attempting to place a large negative charge on the ball, the foil flies off. Explain.
Physics
1 answer:
djverab [1.8K]3 years ago
7 0

Answer:

because of the idea that like charges get repulsion as a force.

Explanation:

because you wrap the ball with foil, the negative charges will leave the foil and go into the ball by induction. This leaves the foil as a positively charged particle since its electrons left it for the ball, making the ball a negatively charged particle. but if you bring the negative charge near the foil, the electrons will transfer from that and go into the foil, making it negatively charged. Now, because the ball and the foil have the same charge, they repel. the foil flies off.

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The power in an electrical circuit is given by the equation P = I2R, where I is the current flowing through the circuit and R is
kiruha [24]

Answer:

Explanation: Here we have given a direct equation . There for no need to worry .

P = I²×R

P = (12)² ×100

P = 14400W = 14.4 kW

For second one

P = I² ×R

200 = I²×150

I = √200/150

I = 1.15 A

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A gymnast of mass 70.0 kgkg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume
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Explanation:

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4 years ago
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2 years ago
What type of image can be larger or smaller than the object?
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An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)
weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767

T₂ = 290.15 × 3.17767 = 922.00139

\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3}  = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12

\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702

Therefore;

T_4 = \frac{1383.002}{2.702} =511.859 \ k

Q_1 = c_p(T_3-T_2)

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

c_v(T_4-T_1)  = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1  \right )r_{v}^{k-1}}

Where:

β = Cut off ratio

Plugging in the values, we get;

\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1  \right )18^{1.4-1}}= 0.5191

Therefore;

\eta = \frac{\sum Q}{Q_1}

\therefore 0.5191 = \frac{150}{Q_1}

Heat supplied = \frac{150}{0.5191}  = 288.978 \ hp

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

7 0
4 years ago
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