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1 year ago

How many years does it take for pluto to orbit the sun.

Physics

1 answer:

Ugo [173]1 year ago

7 0

Answer:

248 years

Explanation:

The last time this happened was from 1979 to 1999. It won't happen again until 2227.

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Which change would decrease the total current, I, flowing through this circuit?

iren [92.7K]

Answer:

A. Increasing the voltage of the battery

Explanation:

The relationship between voltage, V, current, I and resistance, R, is given as follows;

V = I × R

∴ I = V/R

From the above relationship, the current flowing in the circuit is directly proportional to the voltage of the battery, and inversely proportional to the resistance, 'R', of the circuit

Therefore, increasing the voltage, 'V', of the battery, increases the total current, 'I', flowing in the circuit.

7 0

2 years ago

I need to choose a theme for my physics assignment My experiment is finding g

Kobotan [32]

<h3>Question:</h3>

How to find g (acceleration due to gravity)

<h3>Solution:</h3>

We know,

Acceleration due to gravity (g)

$= \frac{GM}{ {R}^{2} }$

where, G = Gravitational constant

$= 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\$

M = Mass of the earth

$= 6 \times {10}^{24} \: kg$

R = Radius of the earth

$= 6.4 \times {10}^{6} m$

Putting these values of G, M and R in the above formula, we get

$g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2}$

So, the value of acceleration due to gravity is

$9.8m/s ^{2}$

Hope it helps.

Do comment if you have any query.

5 0

2 years ago

Select the correct answer

devlian [24]

Explanation:

please send full question....

7 0

2 years ago

Read 2 more answers

"a wind shift from the south or southwest to the northwest is commonly associated with the passage of which type of front

cluponka [151]

If i wouldve know i would tell you sorry

4 0

2 years ago

Suppose a 1300 kg car is traveling around a circular curve in a road at a constant

Scrat [10]

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :

$F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N$

So, the force has a magnitude of 4212 N

4 0

2 years ago