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3 years ago

Explain what happens at a continental continental convergent

1 answer:

7 0

A subduction zone is formed at a convergent plate boundary when one or both of the tectonic plates is composed of oceanic crust. The denser plate, made of oceanic crust, is subducted underneath the less dense plate, which can be either continental or oceanic crust.

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Anna Litical is practicing a centripetal force demonstration at home. She fills a bucket with water, ties it to a strong rope, a

dlinn [17]

Answer:

Explanation:

Suppose when bucket is half full it has a mass of 2 m rotating in a circle of radius r

When Bucket is quarter full then it has a mass of m rotating in a circle of radius r.

When an object moves in a circular path then it experiences an inward force which is given by

$F_1=\frac{2mv^2}{r}$

where v=velocity of bucket

Force in case 2 is given by

$F_2=\frac{mv^2}{r}$

Thus $F_1=2F_2$

therefore force required in half bucket is more than force required in quarter bucket full.

6 0

4 years ago

How many cranial nerves do we have

maxonik [38]

12.

The twelve cranial nerves are: olfactory nerve, optic nerve, oculomotor nerve, trochlear nerve, trigeminal nerve, abducens nerve, facial nerve, vestibulocochlear nerve, glossopharengeal nerve, vagus nerve, spinal accessory nerve, and hypoglossal nerve.
:)

8 0

3 years ago

A student pulls on a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185N at

Vitek1552 [10]

Answer:

$2.75 m/s^2$

Explanation:

We can solve the problem by writing the equations of motion along the horizontal and vertical direction.

Along the horizontal direction we have:

$T cos \theta - \mu N = ma$ (1)

where

$T cos \theta$ is the horizontal component of the tension, where

T = 185 N is the tension in the rope

$\theta=25^{\circ}$ is the angle between the rope and the horizontal

$\mu N$ is the force of friction, where

$\mu=0.27$ is the coefficient of friction

N is the normal reaction of the floor on the box

m = 35.0 kg is the mass of the box

a is the acceleration

Along the vertical direction we have:

$N+T sin \theta-mg=0$ (2)

where

N is the normal force (upward direction)

$T sin \theta$ is the vertical component of the tension in the rope (upward direction)

$mg$ is the weight of the box (downward direction), where

m = 35.0 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

From eq.(2) we get:

$N=mg-T sin \theta$

And substituting into (1), we can find the acceleration:

$T cos \theta - \mu (mg-T sin \theta) = ma\\Tcos \theta -\mu mg + \mu T sin \theta = ma\\a=\frac{T cos \theta- \mu mg + \mu T sin \theta}{m}=\\=\frac{(185)(cos 25^{\circ})-(0.27)(35.0)(9.8)+(0.27)(185)(sin 25^{\circ})}{35.0}=2.75 m/s^2$