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scZoUnD [109]
3 years ago
10

During which process does layer upon layer of sediment build up, exerting pressure on the layers below?

Physics
1 answer:
irinina [24]3 years ago
8 0
I think the answer is b. compaction
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What is the basis for rutherford's planetary model?
Olenka [21]
The basis for Rutherford's Planetary model, was the results he got from experiments.

He observed that most of the alpha particles he fired at a gold foil, passed through the foil, but only few were deflected back. So he concluded that most of the Atom would be empty space, with a positive entity at the center.
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3 years ago
A minivan sells for 33,200 dollars. Give the price of the minivan in a) kilodollars and megadollars
amm1812
The prefix of a unit could give you a hint on its extent with respect to the base unit. In this case, the base unit is money in terms of dollars. When you say kilodollars, that would be 1,000 times as large as a dollar. Hence,

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When you say megadollars, that would be a million times as large as a dollar. Hence,

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A cool down is considered the activity performed
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3 years ago
A 0.500-kg mass suspended from a spring oscillates with a period of 1.36 s. How much mass must be added to the object to change
lord [1]

The mass that must be added is 0.628 kg

Explanation:

The period of a mass-spring system is given by

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant

For the initial mass-spring system in the problem, we have

m = 0.500 kg

T = 1.36 s

Solving for k, we find the spring constant:

k=(\frac{2\pi}{T})^2 m = (\frac{2\pi}{1.36})^2 (0.500)=10.7 N/m

In the second part, we want the period of the same system to be

T = 2.04 s

Therefore, the mass on the spring in this case must be

m=(\frac{T}{2\pi})^2 k =(\frac{2.04}{2\pi})^2 (10.7)=1.128 kg

Therefore, the mass that must be added is

\Delta m = 1.128 - 0.500 = 0.628 kg

Learn more about period:

brainly.com/question/5438962

#LearnwithBrainly

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