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riadik2000 [5.3K]

3 years ago

9

VTheStampy is this right and what do I do next?

1 answer:

koban [17]3 years ago

3 0

That is correct. Based on the listed requirements, that is all you have to do.

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Which of the following represents the greatest energy transition from a higher energy level to a lower one?

prohojiy [21]

You need to use Planck's law:
E = h·υ = (h·c)/λ

Without making all the calculations, a fraction is bigger than another when the denominator is smaller. Therefore you need to find the smallest wavelength (λ) which is 450nm.

You could also be helped by colors: in order of decreasing energy, you have blue - green - yellow - red.

In any case, the correct answer is a).

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Kruka [31]

Answer:

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3 years ago

Using the formula F = M* A. What is the acceleration of a .5 kg

Katyanochek1 [597]

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7 0

3 years ago

A boy slides a book across the floor using a force of five and over a distance of 2M what is the kinetic energy of the book afte

Alinara [238K]

Answer: 10Nm or 10J

Explanation:

Given the following :

Force (f) = 5

Distance (d) = 2m

Calculate the kinetic energy assuming no friction

Work done = force × distance

Work done = 5N × 2m = 10Nm

Recall :

Work done = ΔK.E ( change in kinetic energy)

Therefore, kinetic energy of the book after sliding = ΔK. E, which is equal to work done.

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5 0

3 years ago

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

$b = 1795.4 kg/s$

7 0

4 years ago