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Misha Larkins [42]
3 years ago
12

Which of the following best describes the circuit shown below?

Physics
2 answers:
Allisa [31]3 years ago
8 0

i think it,s going to be B

Stels [109]3 years ago
8 0

Answer:

Short for Appex

Explanation:

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The rainbow is a result of a. Diffraction of the light.
suter [353]

Rainbows are caused by the dispersion of light, which itself consists of a combination of refraction and reflection of light around little droplets of water.

Choice C

5 0
3 years ago
A cue stick hits a cue ball with an average force of 22 N for a duration of 0.029 s. If the mass of the ball is 0.15 kg, how fas
Likurg_2 [28]

Answer:

4.25 m/s

Explanation:

Force, F = 22 N

Time, t = 0.029 s

mass, m = 0.15 kg

initial velocity of the cue ball, u = 0

Let v be the final velocity of the cue ball.

Use newton's second law

Force = rate of change on momentum

F = m (v - u) / t

22 = 0.15 ( v - 0) / 0.029

v = 4.25 m/s

Thus, the velocity of cue ball after being struck is 4.25 m/s.

8 0
3 years ago
Read 2 more answers
the statement that current is equal to the voltage difference divided by the resistance is known as what
dangina [55]

Answer:it is known as (Ohm's law)

Explanation:

3 0
3 years ago
At what position in its elliptical orbit is the speed of a planet a maximum? when it is closest to the sun when it is farthest f
Phantasy [73]

Answer:

1.when it is closest to the sun

2.when it is midway between its farthest

Explanation:

According to the law of  Kepler's

T ² ∝ r³

T=Time period

r=semi major axis

We also know that time period T given as

T=\dfrac{2\pi r}{v}

v=Speed

v=\dfrac{2\pi r}{T}

v\alpha \dfrac{r}{T}

v\alpha \dfrac{r}{T}

v^2\alpha \dfrac{r^2}{T^2}

v^2\alpha \dfrac{r^2}{r^3}

v^2\alpha \dfrac{1}{r}

v\alpha \dfrac{1}{\sqrt{r}}

So we can say that ,when r is more then the speed will be minimum and when r is low then speed will be maximum.

7 0
3 years ago
Electric power companies measure energy consumption in kilowatt-hours, denoted kWh. One kilowatt-hour is the amount of energy tr
Natali [406]

The amount of energy used in the billing period is 5,400,000,000 joules.

One kWH is the amount of energy transferred in one hour, there 1 kWh is equal to

1 kWh=1*1000*(J/s)*3600 s

=3600000 J

Thus the amount of energy in joules consumed by the user for the billing period is =1500*3600000=5,400,000,000 J.

The amount of energy used in the billing period is 5,400,000,000 joules.

7 0
3 years ago
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