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Irina18 [472]
2 years ago
15

Why does it take more time for a larger sample of water to freeze?

Physics
2 answers:
Stels [109]2 years ago
3 0

Answer:The greater the amount of water that there is it will take longer for the water to freeze because more heat has to be dissipated into the environment

Explanation:

exis [7]2 years ago
3 0

Answer:

The greater the mass of water, the more heat has to be dissipated into the environment and the longer it will take to freeze. The greater the surface area of the water, the faster the water will lose its heat and freeze.

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agasfer [191]

Answer:

i/f = i/o + i/i       f = focal, o = object, i = image

1 / i = 1 / f - 1 / o  =    (o - f) / o f

i = o * f / ( o - f)      image distance

i = 12.5 * 22 / (12.5 - 22) = -28.9 cm

Image is real

Image is 28.9 cm to left of lens

M = - i / o = = 28.9 / 12.5 = 2.3     magnification (convex lens)

8 0
3 years ago
Please need help on this one
jolli1 [7]

Answer:

The second answer from the top, no the energy in the wave pushed the water particles from above the earthquake in the opposite direction.

Explanation:

I believe this is the correct answer. Hope you do well

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3 years ago
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andrew-mc [135]
Whoever scores the highest
8 0
3 years ago
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What is the displacement of an object during a specific unit of time.
strojnjashka [21]

Answer:

velocity

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the displacement of an object during a specific unit of time.

7 0
3 years ago
You are observing traffic in a single lane of a highway at a specific location. You measure the average headway and average spac
HACTEHA [7]

Answers:

1) flow of traffic =   1198.8 veh/h

2) average speed = 34.09 mi/h

3)density of traffic = 34.34 veh/mi

Explanation:

1) to find flow of traffic

we use the relation

q=1/h

where q is the traffic flow and h is average time headway.

h= 3s       (given)

insert the value

q=1/3=0.333veh/s=1198.8veh/h

2) to find average traffic speed

use the relation

u=S/h

where u is average speed S is average spacing and h average time

S= 150 ft   (given)

so inserting the values

u= 150/3*3600/5280=34.09 mi/h

3) density of traffic

K=q/u

where K is density of traffic q is flow of traffic and u is average speed

inserting values from above solved parts

K=1198.8/34.09=34.34 veh/mi

6 0
3 years ago
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