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tino4ka555 [31]
3 years ago
6

What force does the gravitational attraction of earth exert on a 12.8kg object, such as a toolbox loaded with tools

Physics
1 answer:
Hatshy [7]3 years ago
5 0

Answer:

16

Explanation:

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Gravity pulls the moon towards the earth.
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One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra
sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration (a_{r}), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

a_{r} = 4\pi^{2}\cdot f^{2}\cdot R (1)

Where:

f - Frequency, measured in hertz.

R - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension (T), measured in newtons, through the string, then we derive the following model:

\Sigma F = T = m\cdot a_{r} (2)

Where m is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right) (4)

\frac{T_{2}}{T_{1}} = 4\cdot 2

\frac{T_{2}}{T_{1}} = 8

If both the radius and frequency are doubled, then the tension is increased 8 times.

5 0
3 years ago
A person standing 1.20 m from a portable speaker hears its sound at an intensity of 5.50 ✕ 10−3 W/m2. HINT (a) Find the correspo
iris [78.8K]

Answer:

PART A)

L = 97.4 dB

PART B)

I = 6.11 \times 10^{-6} W/m^2

PART C)

L = 67.9 dB

Explanation:

PART A)

level of sound is given as

L = 10 Log\frac{I}{I_o}

now we have

L = 10 Log\frac{5.50\times 10^{-3}}{10^{-12}}

L = 97.4 dB

PART B)

Since source is a spherical source

so here the intensity of sound is inversely depends on the square of the distance from the source

\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2}

\frac{I_2}{5.50 \times 10^-3} = \frac{1.20^2}{36^2}

I_2 = 6.11 \times 10^{-6} W/m^2

PART C)

level of sound is given as

L = 10 Log\frac{I}{I_o}

now we have

L = 10 Log\frac{6.11\times 10^{-6}}{10^{-12}}

L = 67.9 dB

3 0
3 years ago
How can you find the reading of main scale and vernier scale​
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Answer:

this pdf should help you out

Explanation:

Download pdf
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