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Lelechka [254]
3 years ago
13

g Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass

gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 46.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.834 m/s2. Calculate her mass.
Physics
1 answer:
katrin [286]3 years ago
3 0
Newton’s second law is F = mass times acceleration.

F divided by acceleration equals her mass.

46/0.834 = 55.156 kg
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What is the answer to this problem 2.40e+3kg
8090 [49]

3/5•(4e+5kg)




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Factor out 3/5 from the expression

3/5•(4e+5k)

Use the commutative property to reorder the terms

3/5(4e+5k)
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What portion of the electromagnetic spectrum is most absorbed by the chlorophyll pigments in the thylakoid membranes?
nadya68 [22]

Answer: The visible part of the electromagnetic spectrum, <u>specifically blue and red wavelength</u>

Explanation:

Let's begin by explaining that pigments generally absorb more light than they reflect (they absorb certain wavelengths and reflect others). Therefore, the color that a given object seems to have depends on which parts of the visible electromagnetic spectrum are reflected and which parts are absorbed.

In this sense, in the thylakoid membranes there are two types of <u>chlorophyll pigments</u>:

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4 0
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Which of the following changes will increase the frequency of an oscillating pendulum?
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6 0
3 years ago
A) A wire made from iron with a cross-section of diameter 0.800 mm carries a current of 14.0 A. Calculate the "areal current den
Veseljchak [2.6K]

Answer:

A) ρ=1.74x10^{26}

B) μ=1.68x10^{29}\frac{electron}{m^3}

C) v=1.03x10^{-3}\frac{m}{s}

D)e=8.99x10^-9

Explanation:

A)

The magnetic field can be find knowing the current is the charge per second

β= \frac{14eC*s}{1.6x10^{-19}C}\\

β= 8.75x10^{19}e*s

Electron density

ρ=\frac{8.75x10^{19}}{\pi*0.400x10^{-3}m} = 1.74x10^{26}

B)

μ= \frac{7.86}{56.2}\frac{g}{cm^3} \frac{mol}{g}*6.022x10^{23}\frac{molecules}{mol} *2 \frac{electron}{molecule}

μ=1.68x10^{23} \frac{electron}{cm^3}= 1.68x10^{29} \frac{electron}{m^3}

C)

The drift speed using last information found

v= \frac{J}{n*q} \\v= \frac{14A}{\pi*((0.40x10^-3)^2)*1.68x10^29*1.60x10^-19)} = 1.03x10^-3(\frac{m}{s} )

D)

To compared the random thermal motion and the current's drift speed

e=\frac{1.03x10^-3}{1.15x10^5} = 8.99x10^-9

8 0
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Importance of SI system in points​
monitta

Answer:

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4 0
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