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3 years ago

Which describes how sliding friction affects pushing a cereal box across a tabletop?

2 answers:

6 0

The term sliding friction refers to the resistance created by two objects sliding against each other. This can also be called kinetic friction. Sliding friction is intended to stop an object from moving.

Understanding Sliding Friction

The amount of sliding friction created by objects is expressed as a coefficient which takes into consideration the various factors that can affect the level of friction. These various factors that can impact sliding friction include the following:

The surface deformation of objects

The roughness/smoothness of the surface of the objects

The original speed of either object

The size of object

The amount of pressure on either object

The adhesion of the surface

So from the above options correct answer must be

<u>It acts in the direction opposite of motion.</u>

Vinil7 [7]3 years ago

5 0

I think your best bet would be.

It acts in the direction opposite of the motion

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A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.

Ksju [112]

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

$v = \sqrt{\frac{2GM}{R}}$

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

$v = 14.9km/h = 14900m/s$

Through this equation we can find the mass of the Planet in function of the distance, therefore

$M = \frac{v^2R}{2G}$

$M = \frac{14900^2R}{2(6.67*10^{-11})}$

$M = 16.64*10^{17}R$

The orbital velocity is

$v_o = \sqrt{\frac{GM}{R+h}}$

$9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}$

$11.1*10^7R = (R+15000*10^3)(9200)^2$

$2.64*10^7R = 12.69*10^{13}$

$R = 4.81*10^6m$

The time period of revolution is,

$T = \frac{2\pi(R+h)}{v_o}$

$T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}$

$T = 4307s$

$T = 72min = 1hour12min$

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

3 0

3 years ago

The brief wave of positive electrical charge that sweeps down the axon is _____.

kobusy [5.1K]

I believe Action potential is the brief wave of positive charge that sweeps down the axon. Axon is part of the neuron that conducts impulses from the dendrites towards the cell body along the neuron. The action potential is brief since the sodium channels can only stay open for a very brief amount of time. As it travels along the neuron there is a change in polarity across the membrane of the axon .

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3 years ago

Can you help and explain these for me?

ollegr [7]

Yes I can do you want me to

8 0

3 years ago

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

madreJ [45]

Answer:

$Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4orbits}$