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3 years ago

Is the force of friction the sum of all the microscopic contact forces between the bottom of the block and the surface?

1 answer:

6 0

The force of friction is that force that tends to oppose the motion of the body with the surface in contact... are you clear with it?

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Calculate the acceleration of a turtle going from 0.3 m/s to 0.7 m/s in 30 seconds.

Hoochie [10]

Acceleration = velocity/ time
Acceleration = 0.7-0.3 /30= 0.01 m/s^2
Notice that velocity is calculated the final speed minus the initial !

6 0

2 years ago

Which determines the reactivity of an alkali metal?

shutvik [7]

Answer:

Explanation:

I'm pretty sure it's A because when you look up the properties of alkali metal it states stuff about its melting points and what to expect. (correct me if I'm wrong but I had this same question on a test and I'm sure this is the answer lol)

8 0

3 years ago

Which statement correctly describes gravity?

Bess [88]

The second one is best. Gravity is an attractive force that acts between all objects, whether they're in contact or not.

3 0

3 years ago

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What are the atmospheric conditions over Minneapolis,Minnesota? Check all that apply

Masja [62]

Answer:

the answer is A & C & E just did the lab

Explanation:

3 0

3 years ago

Read 2 more answers

Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of

Radda [10]

According to the information provided to define an average density, it is necessary to use the concepts related to mass calculation based on gravitational constants and radius, as well as the calculation of the volume of a sphere.

By definition we know that the mass of a body in this case of the earth is given as a function of

$M = \frac{gr^2}{G}$

Where,

g= gravitational acceleration

G = Universal gravitational constant

r = radius (earth at this case)

All of this values we have,

$g = 9.8m/s^2\\G = 6.67*10^{-11} m^3/kg*s^2\\r = 6378*10^3 m$

Replacing at this equation we have that

$M = \frac{gr^2}{G} \\M = \frac{(9.8)(6378*10^3)^2}{6.67*10^{-11}} \\M = 5.972*10^{24}kg$

The Volume of a Sphere is equal to

$V = \frac{4}{3}\pi r^3\\V = \frac{4}{3} \pi (6378*10^3)^3\\V = 1.08*10^{21}m^3$

Therefore using the relation between mass, volume and density we have that

$\rho = \frac{m}{V}\\\rho = \frac{5.972*10^{24}}{1.08*10^{21}}\\\rho = 5.52*10^3kg/m^3$

6 0

3 years ago