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lozanna [386]
2 years ago
5

The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

excess static charge on a cockroach is modeled as point charges located at the end of each antenna, what magnitude of charge ???? would each antenna possess in order for each antennae to experience a force of magnitude 5.50 μN from the external electric field? Calculate ???? in units of nanocoulombs (nC) .
Physics
1 answer:
otez555 [7]2 years ago
8 0

Answer:

0.647 nC

Explanation:

The force experienced by a charge due to the presence of an electric field is given by

F=qE

where

q is the charge

E is the magnitude of the electric field

In this problem, each antenna is modelled as it was a single point charge, experiencing a force of

F=5.50\mu N = 5.50\cdot 10^{-6} N

Therefore, if the electric field magnitude is

E=8.50 kN/C = 8500 N/C

Then the charge on each antenna would be

q=\frac{F}{E}=\frac{5.50\cdot 10^{-6} N}{8500 N/C}=6.47\cdot 10^{-10} C = 0.647 nC

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Answer:

4.325\times10^6J

Explanation:

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Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.

Hence, total change in energy = mgh = 5800\times9.81\times76.0166J=4.325\times10^6J

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4 0
2 years ago
What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22
SIZIF [17.4K]

Answer:

5.45\times 10^{-4} W

Explanation:

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A = Surface area = 1.50 m²

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Rate of heat transfer is given as

R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})

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3 0
2 years ago
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GIVEN DATA:

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let force be F

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we know that

Change in kinetic energy = work done

kinetic energy = = \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})

kinetic energy = = \frac{1}{2}*4*(1.5^{2}-11^{2}) = -237.5 kg m/s2

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7 0
2 years ago
How much gravitational potential energy does a 45.2 kg object have when it is 21.9m above the ground?
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Answer:

Explanation:

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6 0
2 years ago
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QveST [7]

gravitational potential energy is given by formula

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here we need to compare the gravitational potential energy of stone 2 with respect to stone 1

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given that

m_2 = 4 m_1

now we have

\frac{U_2}{U_1} = 4

5 0
2 years ago
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