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3 years ago

What type of rock forms most often near an active volcano?

Physics

2 answers:

Lina20 [59]3 years ago

5 0

So the answer will be

*Intrusive Igneous Rock

hope you enjoyed the answer

BaLLatris [955]3 years ago

3 0

Extrusive Igneous Rock

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g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

sdas [7]

Answer:

The vertical distance is $d = \frac{2}{k} *[mg + f]$

Explanation:

From the question we are told that

The mass of the cylinder is m

The kinetic frictional force is f

Generally from the work energy theorem

$E = P + W_f$

Here E the the energy of the spring which is increasing and this is mathematically represented as

$E = \frac{1}{2} * k * d^2$

Here k is the spring constant

P is the potential energy of the cylinder which is mathematically represented as

$P = mgd$

And

$W_f$ is the workdone by friction which is mathematically represented as

$W_f = f * d$

$\frac{1}{2} * k * d^2 = mgd + f * d$

=> $\frac{1}{2} * k * d^2 = d[mg + f ]$

=> $\frac{1}{2} * k * d = [mg + f ]$

=> $d = \frac{2}{k} *[mg + f]$

5 0

3 years ago

Consider a situation where the acceleration of an object is always directed perpendicular to its velocity. This means that

Bond [772]

Answer:

this situation would not be physically possible

7 0

3 years ago

If you are following a car and also being tailgated by another vehicle, your best option would be to.. a. Increase the following

Pie

Answer:

Your answer here is D

Explanation:

Slowly pressing your breaks will help ensure you are not hit by the other car. If they hit you its their fault. Hope this helps :)!

7 0

2 years ago

Read 2 more answers

Find the direction and magnitude of Ftot, the total force exerted on her by the others, given that the magnitudes F1 and F2 are

Mars2501 [29]

Answer:

θ = 36°

Explanation:

given,

F₁ = 22.8 N

F₂ = 16.6 N

magnitude of force = ?

direction of force = ?

$F = \sqrt{F_1^2 + F_2^2}$

$F = \sqrt{22.8^2 + 16.6^2}$

$F = \sqrt{795.4}$

F = 28.20 N

direction

$\theta = tan^{-1}(\dfrac{F_2}{F_1})$

$\theta = tan^{-1}(\dfrac{16.6}{22.8})$

$\theta = tan^{-1}(0.728)$

θ = 36°

5 0

3 years ago

A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box

forsale [732]

We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.

From N-Gcosα=0 we get:

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.

8 0

3 years ago