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insens350 [35]
3 years ago
13

a length 650 cm of thin thread wraps around a cylinder exactly 30 times calculate the circumference and the radius of the cylind

er (take)​
Physics
1 answer:
Eduardwww [97]3 years ago
5 0

Answer:

circumference= 65/3  cm = 21.67 cm

radius R = 3.45 cm

Explanation:

To calculate the length of the circumference of the cylinder, we divide 650 cm by 30 (the number of times it wrapped exactly around it)

length of circumference= 65/3  cm = 21.67 cm

now use the formula of the circumference length to find the radius (R):

circumference length = 2 * pi * R

65/3 = 2 * pi * R

R = 65 / (6 pi)

R = 3.45 cm

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A laptop can convert 400J of electrical energy into a 240J of light and sound. What is the efficiency? Where does the rest of th
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Efficiency = (Wanted) energy out ÷ energy in × 100

Energy in = 400J
Wanted Energy out = 240J

Energy cannot be used up, only transferred, so the remaining energy is most likely to be transferred into unwanted energy (loss of energy) such as heat energy.

Efficiency = 240 ÷ 400 × 100
Efficiency = 0.6 × 100
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3 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
3 years ago
A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric
MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

J = \frac{I}{\pi R^2}

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}

This enclosed current is now to be used in Ampere’s Law.

\mu_o I_{enc} = \int {B} \, dl

Here, \int \, dl represents the circular path of radius r. So we can replace the integral with the circumference of the path, 2\pi r.

As a result, the magnetic field is

B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}

5 0
3 years ago
Based on Newton’s 3rd law if you were to push on the wall with a force of 100 N, how much force would the wall push back towards
rusak2 [61]
The answer is 100N. Look up the definition of Newton's third law.
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4 years ago
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