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3 years ago

Add these measurements, using significant figure rules 1.0090cm + 0.02cm

Physics

1 answer:

Harlamova29_29 [7]3 years ago

5 0

Answer:

1.03

Explanation:

0.02 has two significant decimal figures, so start by shortening 1.0090 to 2 decimal places (3 significant figures). Round up the decimal to become 1.1 and add .02 to get 1.3.

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The de Broglie relation λ=h/p can be rewritten in terms of the wave number k as p=kℏ. Recall that wave number is defined by k=2π

garik1379 [7]

Answer:

k₁ = 2πf₁/v

k₂ = 2πf₂/v

Explanation:

Since the de Broglie relation λ=h/p (1 ) and momentum, p =kℏ (2)

From (1) p = h/λ = kℏ

So, h/λ = kℏ

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k = h/[(h/2π)(v/f)]

k = 2πf/v where k = wave number, f = frequency of wave and v = velocity of wave.

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3 years ago

A force of 25 N applied to a spring stretches it by 5 cm. What is the potential energy of the spring in the compressed position?

melamori03 [73]

Answer:

Explanation:

<u>Elastic Potential Energy </u>

We find objects like springs that hold potential energy when stretched and they have the capacity to release it when left return to its equilibrium position. The elastic potential energy of a spring of constant k when is stretched a distance x is

$\displaystyle PE=\frac{kx^2}{2}$

The spring constant can be obtained if we know the force N needed to stretch the spring a distance x, by using the Hook's Law

$F=kx$

The question provides the information that a force of F = 25 N stretches a spring x= 5 cm = 0.05 m. Using the formula F=kx we can compute the value of k.

$\displaystyle k=\frac{F}{x}$

$k=\displaystyle \frac{25}{0.05}$

$k=500\ N/m$

The potential energy of the spring in the compressed position (assumed 5 cm as well) if

$\displaystyle PE=\frac{(500)(0.05)^2}{2}$

$PE=0.625\ J$

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3 years ago