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3 years ago

3. In 1989, Michel Menin of France walked on a tightrope suspended under a

Physics

1 answer:

Tamiku [17]3 years ago

6 0

Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

Let the distance of Menin's pocket to the ground be x

Let the distance between Menin's pocket to the balloon be y

Hence, x=3150-y------1

Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

3070m

Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

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A 1.15 kg book is at rest on the table. What is the magnitude of the normal force that the table is exerting on the book?

Agata [3.3K]

Answer:

11.27N

Explanation:

Given parameters:

Mass of the book = 1.15kg

Unknown:

Magnitude of the normal force = ?

Solution:

The normal force is the vertical force exerted by a body on an object.

It can be described as the weight of an object.

Normal force = mass x acceleration due to gravity

Normal force = 1.15 x 9.8 = 11.27N

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3 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

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2 years ago

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sukhopar [10]

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3 years ago

a backpack has a mass of 8 kg. it is lifted and given 54.9 J of gravitational potential energy. how high is is lifted? accelerat

dalvyx [7]

P=mgh
h=P/mg
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8 0

3 years ago

The disk that BTK sent to the television station contained just one valid file. What was the name of the file?

icang [17]

Answer:

The name of the file is Floppy.

Explanation:

Given that,

The disk that BTK sent to the television station contained just one valid file.

We need to find the name of the file

According to given data,

The disk that BTK sent to the television station contained just one valid file.

This file is called floppy.

Floppy :

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Floppy disk are read and written to by floppy drive.

Hence, The name of the file is Floppy.

3 0

3 years ago