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fenix001 [56]
3 years ago
7

2. In the activities below indicate if friction is useful or not useful? Explain your answer. i. Writing yes it is useful becaus

e when your writing because friction helps you see what your writing ii. Rubbing iii. Skiing iv. Rotating a wheel
Physics
1 answer:
Gemiola [76]3 years ago
8 0

Answer:

Please find the answer in the explanation

Explanation:

Friction is a force that opposes motion. One or two of the advantages of friction are break and ability of an object to walk.

Writing yes it is useful because when your writing because friction helps you see what your writing

ii. Rubbing. Yes, it is useful.

friction make it possible for two object to rub each other

iii. Skiing. No. It is not useful because With presence of friction, skiing will not be possible.

iv. Rotating a wheel No. It is not useful because Friction will oppose the rotation of the wheel.

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Sound level B in decibels is defined as
Lelechka [254]

Answer:

The approximate combined sound  intensity is I_{T}=1.1\times10^{-4}W/m^{2}

Explanation:

The decibel  scale intensity for busy traffic is 80 dB. so intensity will be

10log(\frac{I_{1}}{I_{0}} )=80, therefore I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}

In the same way for the loud conversation having a decibel intensity of 70 dB.

10log(\frac{I_{2}}{I_{0}} )=70, therefore I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}

Finally we add both of them I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}, is the approximate combined sound  intensity.

3 0
3 years ago
An elevator is moving is an upwards
Minchanka [31]

Answer:

The elevator's free-body diagram has three forces, the force of gravity, a downward normal force from you, and an upward force from the tension in the cable holding the elevator. The combined system of you + elevator has two forces, a combined force of gravity and the tension in the cable.

Explanation:

6 0
3 years ago
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.
elena-s [515]

Answer:

g=13.42\frac{m}{s^2}

Explanation:

1) Notation and info given

\rho_{center}=13000 \frac{kg}{m^3} represent the density at the center of the planet

\rho_{surface}=2100 \frac{kg}{m^3} represent the densisty at the surface of the planet

r represent the radius

r_{earth}=6.371x10^{6}m represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of  the radius

r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}

r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}

So we can create a linear model in the for y=b+mx, where the intercept b=\rho_{center}=13000 \frac{kg}{m^3} and the slope would be given by m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}

So then our linear model would be

\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be dV=r^2 sin\theta d\phi d\theta dr.

And the total mass would be given by the following integral

M=\int \rho (r) dV

Replacing dV we have the following result:

M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)

We can solve the integrals one by one and the final result would be the following

M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})

Simplyfind this last expression we have:

M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]

And replacing the values we got:

M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg

And now that for any shape the gravitational acceleration is given by:

g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}

4 0
3 years ago
How would the attractive force between two spheres change if the mass of one sphere was doubled?
Dominik [7]
The force of attraction between two objects can be illustrated using Newton's Law of Universal Gravitation.
The relation between the different parameters is shown in the attached image.

Now, from the relation, we can deduce that the force between the two objects is directly proportional to the masses of the two objects.

This means that, if the mass of one object is doubled, then the force between the two objects will also be doubled.

8 0
3 years ago
What change will always result in an increase in gravitational force between two objects
tekilochka [14]

The gravitational force between two object depends on their masses and on their distance.


Since the formula is


F = G\frac{m_1m_2}{d^2}


If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.


So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.


So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase

6 0
4 years ago
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