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rjkz [21]
3 years ago
15

One ounce of a well-known breakfast cereal contains 103 Calories (1 food Calorie = 4186 J). If 1.6% of this energy could be conv

erted by a weight lifter's body into work done in lifting a barbell, what is the heaviest barbell that could be lifted a distance of 1.8 m?
Physics
1 answer:
faust18 [17]3 years ago
5 0

Answer:

the heaviest barbell that could be lifted  is 390.6kg

Explanation:

Hello!

To solve this question you must follow the following steps

1. Find the amount of energy that can transform the body into motion, this is achieved by multiplying the breakfast consumed by the percentage of energy conversion.

W=(103Calories)(0.016)\frac{4186J}{1calorie} =6898.5J

2. We use the equation that defines the work done by a body that has weight when it is lifted, this is defined by the product of mass by gravity by height.

W=mgh

where

W=work=6898.5J

m=mass

g=gravity=9.81m/s^2

h=height=1.8m

now we solve for mass, and use the values.

m=\frac{W}{hg} =\frac{6898.5}{(1.8)(9.81)} =390.6kg

the heaviest barbell that could be lifted  is 390.6kg

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3 years ago
What is the value of n in the balmer series for which the wavelength is 410.2 nm.?
m_a_m_a [10]

The answer is n= 6.

What is Balmer series?

The Balmer series is the portion of the emission spectrum of hydrogen that represents electron transitions from energy levels n > 2 to n = 2. These are four lines in the visible spectrum. They are also known as the Balmer lines. The four visible Balmer lines of hydrogen appear at 410 nm, 434 nm, 486 nm and 656 nm.

For the Balmer series, the final energy level is always n=2. So, the wavelengths 653.6, 486.1, 434.0, and 410.2 nm correspond to n=3, n=4, n=5, and n=6 respectively. Since the last wavelength, 410.2 nm, corresponds to n=6, the next wavelength should logically correspond to n=7.

To solve for the wavelength, calculate the individual energies, E2 and E7, using E=-hR/(n^2). Then, calculate the energy difference between E2 (which is the final) and E7 (which is the initial). Finally, use lamba=hc/E to get the wavelength.

To learn more about emission spectrum click on the link below:

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