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2 years ago

A cellphone charger has a transformer with

Physics

1 answer:

Evgen [1.6K]2 years ago

6 0

Answer:

A.C → Alternating current.

D.C → Direct current

Explanation:

A.C is an abbreviation for Alternating current.

Alternating current is a type of current that flows sinusoidally or current that changes direction with time.

D.C is an abbreviation for Direct current.

This type of current flows only in one direction, i.e it doesn't change direction with time.

A rectifier can be used to convert A.C to D.C, allowing it to flow in one direction.

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Answer:

1 st question: Control variable

2nd question: random variable

3rd question: if the two objects are dropped from the same height they will hit the ground at the same time

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1 year ago

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Which term describes the number of joules used per second by a circuit?

jasenka [17]

Answer:

B) Power

Explanation:

The power is defined by the following equation:

P = W / t

where:

W = work = Force * Distance = [Newton] * [meter]

t = time = seconds

The units for work are give en Newton per second, which is equal to Joules

And for power the unit used commonly is Watts, therefore:

Watts = (Joule/second)

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2 years ago

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Rita jeptoo of kenya was the first female finisher in the 110th boston marathon. she ran the first 10.0 km in a time of 0.5689 h

stira [4]

Part a

Answer: 17.58 km/h

$Average speed=\frac{Total\hspace{1mm}Distance}{Total\hspace{1mm}Time}$

Total Distance =10 km

Total time =0.5689 h

$\Rightarrow Average speed=\frac{10\hspace{1mm}km}{0.5689\hspace{1mm}h}=17.6 \hspace{1mm}km/h$

Part b

Answer: 17.626 km/h

$Average speed=\frac{Total\hspace{1mm}Distance}{Total\hspace{1mm}Time}$

Total Distance =42.195 km

Total time =2.3939 h

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2 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

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2 years ago

Which one of the following distances is the shortest?

Kitty [74]

I think the correct answer is C

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2 years ago