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3 years ago

Two springs have different spring constants. How could you identify the spring with the greater spring constant value?

Physics

2 answers:

Anni [7]3 years ago

7 0

Answer:

B.It will be stiffer than the other. - I JUST DID THE QUIZ ON PLATO AND IT WAS CORRECT.

Explanation:

sladkih [1.3K]3 years ago

3 0

You could attach both springs to the ceiling and hang the same mass from both. The one that stretches less has the larger spring constant value.

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In an inverse relationship, when one variable increases, the other___

eimsori [14]

Answer:

In an inverse relationship, when one variable increases, the other variable decreases.

Explanation:

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2 years ago

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Elena-2011 [213]

Answer:

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2 years ago

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When a car crashes, do you think its speed can make a difference in the<br> amount of damage done?

Oduvanchick [21]

Answer:

Yes

Explanation:

An increased speed will result in an increased amount of energy, so when it crashes some of that energy will bounce back and crumple the car.

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3 years ago

On an amusement park ride passengers accelerate straight downward from rest to 22.9 m/s in 2.2 s.

Kamila [148]

Given,

Initial Velocity (u) = 0 m/s

Final Velocity (v) = 25.9 m/s

Time (t) = 2.5 sec

avg. acceleration = ?

avg. acceleration = v-u/t

avg = 25.9/2.5 = 10.36 m/s²

avg = 10.36 m/s²

8 0

3 years ago

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Particle A has charge qA and particle B has charge qB. When they are separated by a distance ri, they experience an attractive f

kumpel [21]

Answer:

$r_{final}=r_{i}/4$

Explanation:

We know force between 2 charges is given by

$F=\frac{1}{4\pi \varepsilon }\frac{q_{1}q_{2}}{r^{2}}$

It is given initial force is attractive of magnitude $F_{i}$

Let $r_{i}$ be the initial separation

Thus

$F_{i}=\frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{i}^{2}}$

Now

Let $r_{f}$ be the final separation

Thus final force becomes

$\frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{f}^{2}}$

It is given $F_{f}$ = $16F_{i}$ ...................(i)

Using values of $F_{f}$ and $F_{i}$ in equation i we have

$16\times \frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{i}^{2}}$ = $\frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{f}^{2}}$

thus

$thus\\\\(\frac{r_{i}}{r_{f}})^{2}=16\\\\\frac{r_{i}}{r_{f}}=4\\\\\therefore r_{f}=r_{i}/4$

Thus the final separation is one-fourth of the initial separation

7 0

3 years ago