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3 years ago

HELP ASAP Describe when contact metamorphism occurs?

Physics

2 answers:

iVinArrow [24]3 years ago

6 0

Answer:

Contact metamorphism occurs adjacent to igneous intrusions and results from high temperatures associated with the igneous intrusion. Since only a small area surrounding the intrusion is heated by the magma, metamorphism is restricted to the zone surrounding the intrusion, called a metamorphic or contact aureole.

Explanation:

diamong [38]3 years ago

5 0

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What kind of model is this?

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Answer:

I think it's Experimental

Explanation:

it has atoms on it.

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2 years ago

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A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 73° above the hor

Brrunno [24]

Answer:

7 s

Explanation:

u = Initial velocity of rock = 12.2 m/s

$\theta$ = Angle of throw = $73^{\circ}$

x = Displacement in x direction = 25 m

Displacement in x direction is given by

$x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{25}{12.2\times \cos73^{\circ}}\\\Rightarrow t=7\ \text{s}$

Time taken to reach the ground is 7 s.

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3 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

Now the resultant of the two velocities perpendicular to each other:

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

7 0

4 years ago

The pressure in an automobile tire is 198 kpa at 300.0 k. at the end of a trip on a hot, sunny day, the pressure has risen to 22

Kamila [148]

We can use the ideal gas equation which is expressed as PV = nRT. At a constant volume and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

T1/P1 = T2/P2

T2 = P2 x T1 / P1

T2 = 225 x 300 / 198

T2 = 340.91 K

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4 years ago

Jack is falling down the hill at a speed of 12 m/s. If Jack has a mass of 45 kg, what is his kinetic energy?

Troyanec [42]

Answer:

K=3240J

Explanation:

speed=12m/s

m=45kg

K=½mv²

K=½(45•144)J=3240J

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4 years ago