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2 years ago

I need help with the 1st and the 3rd one plzzzzzzzzzzzzz

Physics

1 answer:

Viktor [21]2 years ago

7 0

Answer for the first one is 2. The answer for the 3rd one is 8

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2) If the current in any circuit reach to infinity then its resistance becomes

In-s [12.5K]

Answer:

Zero according to ohm's law R= V / i

4 0

3 years ago

What is the avarage velocity of a car that travels 30 kilometers due west in 0.5 hours

Fed [463]

Velocity is distance over time $v= \frac{d}{t}$ . If it wants it in km/h you're good to go otherwise if its m/s you need to convert km to m by multiplying 30 by 1000 and t in hours to seconds by multiplying 0.5 by 3600.

6 0

2 years ago

Read 2 more answers

N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2

liberstina [14]

Answer:

Explanation:

The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.

So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.

Using the formula for the sum of the infinite terms of a geometric series, we have:

Sum = First term / (1 - rate)

Sum = C / (1 - 0.5)

Sum = C / 0.5 = 2C

So the equivalent capacitance of this parallel connection is 2C.

5 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0

2 years ago

Calculate the average speed in metres per second from Glasgow to Edinburgh

mariarad [96]

This is the same question as the one previously but with more details, so I will just use my previous answer.

1800 to 1820 is 20 minutes.1830 to 1838 is 8 minutes.1840 to 1905 is 25 minutes.
The total time travelled is 20+8+25 = 53 minutes = 3180 seconds.
The distance between Glasgow and Edinburgh is 28 + 12 + 34 = 74 km = 74000 m.

So, the average speed is 74000m/3180s = 23.27 m/s (4 s.f.)

5 0

2 years ago