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3 years ago

What happens to a circuit's resistance (R), voltage (V), and current (1) when

Physics

2 answers:

klemol [59]3 years ago

7 0

Answer:

R and I will change, V will remain constant

Explanation:

:) I just did the quiz (and have screenshots for proof)

Naya [18.7K]3 years ago

4 0

Answer:

Explanation:

Hope this helps

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A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the

Gelneren [198K]

Answer:

Option B. 2.8 s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 27 m/s

Angle of projection (θ) = 30

Acceleration due to gravity (g) = 9.8 m/s²

Time of flight (T) =?

The time of flight of the ball can be obtained as follow:

T = 2uSineθ / g

T = 2 × 27 × Sine 30 / 9.8

T = 2 × 27 × 0.5 / 9.8

T = 27 / 9.8

T = 2.8 s

Therefore, time of flight of the ball is 2.8 s

7 0

3 years ago

Consumers that eat both producers and other consumers are known as

matrenka [14]

Omnivores...————Fhfhfjdhfh

3 0

3 years ago

A soccer player practices kicking the ball into the goal from halfway down the soccer field. The time it takes for the ball to g

12345 [234]

This is correct, I just did the test. Yes, displacement is 45 meters, elapsed time is three seconds, and the direction is toward the goal.

4 0

3 years ago

Read 2 more answers

What is the radius of the Earth?

ki77a [65]

<span>the answer would be 3,959 miles</span>

7 0

3 years ago

a 100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. When the mixture reaches equillibriu

Artyom0805 [142]

Answer:

The latent heat of fusion of water is 334.88 Joules per gram of water.

Explanation:

Let the latent heat of ice be 'x' J/g

1) Thus heat absorbed by 100 gram of ice to get converted into water equals

$Q_1=100\times x$

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

$Q_2=100\times 4.186\times 11=4604.6Joules$

Thus total energy needed equals $Q_1+Q_2=100x+4604.6$

3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

$Q_3=650\times 4.186\times (25-11)\\\\Q_{3}=38092.6Joules$

Now by conservation of energy we have

$Q_1+Q_2=Q_3\\\\100x+4604.6=38092.6\\\\\therefore x=\frac{38092.6-4604.6}{100}=334.88J/g$

6 0

3 years ago