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Sveta_85 [38]
1 year ago
10

A child is playing with a spring toy, first stretching and then compressing it.

Physics
1 answer:
Reika [66]1 year ago
5 0
Sorry, I’m only in 6th Grade, I don’t know the answer to this question.
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Why doesn't a ball roll on forever after being kicked at a soccer game?
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Because of gravity and friction. 
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2 years ago
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Mark and Balthazar are preparing to conduct neutralization reactions in which they add a base to two different solutions, citric
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Lab safety equipment prevents damage from accidents and helps keep the people working in the lab safe. The equipment goes hand in hand with the clothing of the person. The first step would be to wear closed shoes and a lab coat.
The equipment that must be worn are goggles to protect the eyes from irritants and latex gloves to protect the skin on the hands.
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A certain signal molecule S in heart tissue is degraded by two different biochemical pathways: when only Path 1 is active, the h
Misha Larkins [42]

Answer:

Half life of S = 3.76secs

Explanation:

The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.

For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564

for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615

Simply say = 1/3.9 + 1/104 = 0.266015

So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active

6 0
2 years ago
To show that displacement current is necessary to make Ampère's law consistent for a charging capacitor Ampère's law relates the
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Answer: The Ampère -Max-well law

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7 0
1 year ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
2 years ago
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