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MA_775_DIABLO [31]
3 years ago
8

URGENT!!!!! I WILL MARK U AS BRAINLIEST!!!!!!! A step-down transformer converts 120 V to 5 V in a phone charger. If the primary

coil has 600 turns, how many turns are in the secondary coil? SHOW ALL WORK WITH CORRECT ANSWER WITH CORRECT UNIT AS WELL.
Physics
1 answer:
Sloan [31]3 years ago
6 0

Answer:

  25 turns

Explanation:

The voltage ratio is the turns ratio:

  secondary turns / primary turns = secondary voltage / primary voltage

  secondary turns / 600 = 5/120

  secondary turns = 600(5/120) = 25 . . . turns

The secondary coil has 25 turns.

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When is the kinetic energy of an electron transformed into potential energy?
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Answer:

when it interacts with other electrons without changing its speed

Explanation:

without changing its speed

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3 years ago
1 nanometer is equals to?
netineya [11]

1×10^_19 meters

hope it helps you!!!!!

3 0
4 years ago
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A 2.64-kg copper part, initially at 400 K, is plunged into a tank containing 4 kg of liquid water, initially at 300 K. The coppe
marin [14]

Answer:

a) T_f=305.7049\ K

b) \Delta S=313.51\ J.K^{-1}

Explanation:

Given:

  • mass of copper, m_c=2.64\ kg
  • initial temperature of copper, T_{ic}=400\ K
  • specific heat capacity of copper, c_c=385\ J.kg^{-1}.K^{-1}
  • mass of water, m_w=4\ kg
  • initial temperature of water, T_{iw}=300\ K
  • specific heat capacity of water, c_w=4200\ J.kg^{-1}.K^{-1}

a)

<u>∵No heat is lost in the environment and the heat is transferred only between the two bodies:</u>

Heat rejected by the copper = heat absorbed by the water

2.64\times 385\times (400-T_f)= 4\times 4200\times (T_f-300)

T_f=305.7049\ K

b)

<u>Now the amount of heat transfer:</u>

Q=m_c.c_c.(T_{ic}-T_{f})

Q=2.64\times 385\times (400-305.7049)

Q=95841.5841\ J

∴Entropy change

\Delta S=\frac{dQ}{T}

\Delta S=\frac{95841.5841}{305.7049}

\Delta S=313.51\ J.K^{-1}

5 0
4 years ago
Please help! Question is attached.​
il63 [147K]

Answer:

a. 6

b. 6 m/s²

c. 300 m to the right

d. 30 secs

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slope = rise /run

60-0/10-0

= 6

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c. d=ut+1/2at²

t=10 (segment A last for 10 secs)

u - initial velocity = 0

so d = 0(10)+1/2*6*10²

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6 0
3 years ago
Plutonium-241 is an isotope of plutonium that is highly radioactive and is used in some nuclear reactors and many nuclear weapon
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We use the radioactive decay equation for this problem which is expressed as:

An = Aoe^-kt

An is the remaining amount after time t, Ao is the initial amount and k is a constant.

First, we determine the k from the half life as follows:

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Then, we can calculate An after 28.8 yr.

An = 1000 e^-0.04814(28.8)
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3 years ago
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