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kkurt [141]
2 years ago
6

Revolution: One orit of an object in space around another_ days?​

Physics
2 answers:
stealth61 [152]2 years ago
6 0

Answer:

365 days..

Explanation:

Hope it may help you シ︎♡︎

AlladinOne [14]2 years ago
6 0

\huge\mathfrak\green{365  \: or \:  366 \:  Days.}

Hope \: it \: helps

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A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is
kenny6666 [7]

Answer:

f2/f1 = \sqrt{2}

Explanation:

From frequency of oscillation

f = 1/2pi *\sqrt{k/m}

Initially with the suspended string, the above equation is correct for the relation, hence

f1 = 1/2pi *\sqrt{k/m}

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

f2 = 1/2pi *\sqrt{2k/m}

Employing f2/ f1, we have

f2/f1 = \sqrt{2}

3 0
3 years ago
Convert 10 kilowatt to watt​
Nataly [62]

Answer:

0.01 kilowatts cubed

Explanation:

just ask alexa lol

3 0
3 years ago
Read 2 more answers
A solid concrete block weighs 150. N and is resting on the ground. Its dimensions are 0.400 m ✕ 0.200 m ✕ 0.100 m. A number of i
N76 [4]

Answer:

27 blocks

Explanation:

First, the expression to use here is the following:

P = F/A

Where:

P: pressure

F: Force exerted

A: Area of the block.

Now , we need to know the number of blocks needed to exert a pressure that equals at least 2 atm. To know this, we should rewrite the equation. We know that certain number of blocks, with the same weight and dimensions are putting one after one over the first block, so we can say that:

P = W/A

P = n * W1 / A

n would be the number of blocks, and W1 the weight of the block.We have all the data, and we need to calculate the area of the block which is:

A = 0.2 * 0.1 = 0.02 m²

Solving now for n:

n = P * A / W1

The pressure has to be expressed in N/m²

P = 2 atm * 1.01x10^5 N/m² atm = 2.02x10^5 N/m²

Finally, replacing all data we have:

n = 2.02x10^5 * 0.02 / 150

n = 26.93

We can round this result to 27. So the minimum number of blocks is 27.

5 0
3 years ago
How much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is
mafiozo [28]

1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

<h3>What is Speed ?</h3>

Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s

In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.

What are the parameters to consider ?

The parameters are;

  • The distance S = 3.85 × 10^{5} km
  • The Speed of Light C = 3 × 10^{8} m/s
  • The time taken t = ?

Speed = distance S ÷ Time t

Convert kilometer to meter by multiplying it by 1000

C = S/t

3 × 10^{8} =  3.85 × 10^{8} / t

Make t the subject of formula

t = 3.85 × 10^{8} / 3 × 10^{8}

t = 1.2833

t = 1.3 s

Therefore, 1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

Learn more about Speed here: brainly.com/question/4931057

#SPJ1

3 0
2 years ago
A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan
Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0
3 years ago
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