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4 years ago

7

why would granite have larger crystals than igneous rocks formed from lava cooled above Earth's surface

1 answer:

Oduvanchick [21]4 years ago

8 0

because crystals in granite had more time to form as magma cooled slowly. Crystals that form on the surface are smaller because they cooled really quickly

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Please help me. I have to submit this today.

Naily [24]

B I think hope this helps

3 0

3 years ago

I throw a ball off the edge of a 15.0m cliff. If I threw it horizontally at 8.0 m/s, how much time did it take to fall?

pickupchik [31]

Just ignore the horizontal component

if you have a vertical displacement of 15m, 0ms^1 initial velocity, end velocity is ignored, we know the acceleration due to gravity as 9.81ms^2 so we can work out the time using SUVAT

S=15
U=0
V=?
A=9.81
T=?

S=UT + 0.5 AT^2

UT=0

therefore,

S=0.5AT^2

rearrange to:

T=SQR (2S/A)

T = 1.75 seconds

3 0

3 years ago

Jamie dropped a water bottle from the top of an office building. The water bottle impacted the ground after falling for 5 second

PSYCHO15rus [73]

Answer:

<em>The speed of the water bottle at impact with the street is 49 m/s</em>

Explanation:

<u>Free Fall Motion </u>

A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not experience air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is $g = 9.8 m/s^2$ .

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The water bottle dropped by Jamie takes t=5 seconds to impact the ground, thus its speed at that moment is:

vf= 9.8*5

vf = 49 m/s

The speed of the water bottle at impact with the street is 49 m/s

4 0

3 years ago

Equations 22-8 and 22-9 are approximations of the magnitude of the electric field of an electric dipole, at points along the dip

OleMash [197]

On axis of the dipole the electric field is given by

$E = \frac{kq}{(r - \frac{d}{2})^2} - \frac{kq}{(r + \frac{d}{2})^2}$

$E = \frac{kq(r + d/2)^2 - (r - d/2)^2}{(r^2 - \frac{d^2}{4})^2}$

$E_{act} = \frac{kq*2rd}{(r^2 - \frac{d^2}{4})^2}$

now if we approximate above equation

$E_{eppr} = \frac{kq*2d}{r^3}$

now we will find the ratio of these two

$\frac{E_{eppr}}{E_{act}} = \frac{(r^2 - \frac{d^2}{4})^2}{r^4}$

put r = 3d

$\frac{E_{eppr}}{E_{act}} = \frac{(9d^2 - \frac{d^2}{4})^2}{81d^4}$

$\frac{E_{eppr}}{E_{act}} = \frac{(9- \frac{1}{4})^2}{81}$

$\frac{E_{eppr}}{E_{act}} = 0.945$

<em>so above is the ratio of two field</em>

6 0

4 years ago

Read 2 more answers

B observes that galaxies A and C are moving in opposite directions away from her, A with β = −0.8 and C with β = +0.6 . What are

irga5000 [103]

Answer: Speed of galaxy B with respect to galaxy A = 0.8

Speed of galaxy C with respect to galaxy A = 1.4

Explanation:

Let the speed of galaxy A with respect to galaxy B be Vab & it is equal to Va - Vb = -0.8

Let the speed of galaxy C with respect to galaxy B be Vcb = Vc - Vb = 0.6

Speed of galaxy B with respect to galaxy A = Vba = Vb - Va = -Vab = -(-0.8) = 0.8

Speed of galaxy C with respect to galaxy A = Vca = Vc - Va

But from above, Vc - Vb = 0.6; Vc = 0.6 + Vb

And Va - Vb = -0.8; Va = -0.8 + Vb

Vca = Vc - Va = 0.6 + Vb - (-0.8 + Vb) = 0.6 + Vb + 0.8 - Vb = 0.6 + 0.8 = 1.4

(Solved)

3 0

3 years ago