In need of help!!!!!!

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.

Force exerted by the Earth on a human being

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

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]

Now replacing we have

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

Force exerted by a building on a human being

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

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

3 0

3 years ago

What is the necessary conditions for the production of sound?

TEA [102]

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

Answer: Something that's vibrating, and you also need medium for those vibrations to start in.

I hope this helped!

<!> Brainliest is appreciated! <!>

- Zack Slocum

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

8 0

3 years ago

Your physics textbook is sliding to the right across the table draw the vectors starting at the black dot. the location and orie

liberstina [14]

You would take a black dot at the top right of the y axis and dray it to the the far right of your x axis. hope this helps have a nice day and God bless

6 0

3 years ago

A roller coaster uses 800 000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potentia

Zarrin [17]

Answer:

Good Luck!

Explanation:

6 0

3 years ago