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N76 [4]
2 years ago
8

0.75 km expressed in centimeters

Physics
1 answer:
disa [49]2 years ago
7 0
75000 lol enjoy..............using up 20 characters 
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Why don’t we feel the gravitational force of a large object such as a skyscraper semi-truck?
Kobotan [32]

Answer:

Se the explanation below

Explanation:

We do not feel these forces of these bodies, because they are very small compared to the force of Earth's attraction. Although its mass is greater than that of a human being, its mass is not compared to the Earth's mass. In order to understand this problem we will use numerical data and the universal gravitation formula, to give validity to the explanation.

<u>Force exerted by the Earth on a human being</u>

<u />

F=G*\frac{m_{1}*m_{2}}{r^2}

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 5.97*10^24[kg]

r = distance from the center of the earth to the surface or earth radius = 6371 *10^3 [m]

<u />

Now replacing we have

F = 6.673*10^{-11} *\frac{80*5.97*10^{24}}{(6371*10^{3})^{2}  } \\F = 785[N]

<u>Force exerted by a building on a human being</u>

<u />

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 300000 [ton] = 300 *10^6[kg]

r = distance from the building to the person = 2[m]

F = 6.673*10^{-11}*\frac{80*300*10^6}{2^{2} }  \\F= 0.4 [N]

As we can see the force exerted by the Earth is 2000 times greater than that exerted by a building with the proposed data.

8 0
3 years ago
Light of wavelength 505 nm passes through a single slit of width 4.32 x 10-5 m. At what angle does the first interference minimu
Nataly_w [17]

Answer:

0.665

Explanation:

I did the work. Just plug everything in from the formula. Look at the lesson manual.

3 0
3 years ago
Ohms law is A.(R=E/W). B.(R=E/1). C.(E/Z). D.none of them
nikklg [1K]

Answer:

D. none of them.

Explanation:

This is because Ohm's law is:

Voltage = Current × Resistance

or,

V = IR

6 0
2 years ago
A volleyball player bumps a ball across a net with the velocity and angle shown below. What is the maximum height of the ball?
Marrrta [24]

Answer:

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

                             h = u² sin²θ/2g  m

Where,

                           g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

                            h = 18² · sin²60 / 2 x 9.8

                               = 18² x 0.75 / 2 x 9.8

                               = 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

6 0
3 years ago
The sound intensity at a distance 2.00 m from a sound source is 5.00 Find the total sound energy emitted by the source in each s
notka56 [123]

Answer:

     P = 251, 3 W

Explanation:

The intensity is defined as the power emitted per unit area

           I = P / A

Since sound is distributed in all directions spherical shape, the area of ​​a sphere is

           A = 4π r²

let's clear the power and replace

         P = I A

         P = I (4π r²)

let's calculate

         P = 5.00 (4π 2²)

         P = 251, 3 W

6 0
3 years ago
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