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Ivan
3 years ago
5

The brightness of a star depends on what

Physics
1 answer:
Zinaida [17]3 years ago
7 0
It’s D size and temperature
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3. A crew team rows a boat at a rate of 20 km/h in still water. Beginning at time = 0 minutes, the team rows for 30 minutes up a
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Going upstream against the current gives a net speed equivalent to the speed at still water minus the speed of the current. Consequently, the speed downstream gives a net speed equal to the speed at still water plus the speed of the current, making it travel faster. The solution is:

UPSTREAM

v = 20 - 1.5 = 18.5 km/h
t = 30 mins or 0.5 hours
distance = 18.5km/h (0.5 h) = 9.25 km

DOWNSTREAM
for the same distance of 9.25 km:

v = 20 + 1.5 = 21.5 km/h
t = 9.25km / 21.5 km/h = 0.43 hours or 25.8 mins  =  26 mins --> FINAL ANS.
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Q2. A certain machine is used to lift a load of 500N when an effort of 50N is applied to the machine the load is raised by 10m a
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Answer:

2%

Explanation:

E=l/E x r/R x100%

=5000/50 x 0.1/0.6 x100

=2% or 166.7

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2 years ago
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Most of the mass in an atom is in the nucleus.<br> True<br> False
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There are "Protons" & "Neutrons" in an atom inside the nucleus. They are very massive as compared to "Electrons" (which reside outside the nucleus).So, the sentence is correct.

In short, Your Final answer would be "True"

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4 years ago
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Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85% of the of the entering water is e
Natalka [10]

Answer:

a)

i) v'=\frac{17}{20}                                   ii) \frac{m_v}{m_f-m_v} =\frac{17}{83}

b) m_r=752963.55\ kg

Explanation:

Given:

fraction of water in wet sugar of m kg by mass, m'_w=\frac{1}{5} \times m

% of water evapourated from the total water after passing through the evapourator, m'_v=85\%

a)

amount of wet sugar fed to the evapourator, m_f=100\ kg

Now the mass of water present in the fed amount of sugar:

m_w=\frac{1}{5} \times m_f

m_w=\frac{100}{5}

m_w=20\ kg

Now the amount of water leaving from this total amount of water after passing through the evapourator:

m_v=m_v' \times m_w

m_v=\frac{85}{100} \times 20

m_v=17\ kg

i)

So, the fraction of of water leaving the evapourator:

v'=\frac{m_v}{m_w}

v'=\frac{17}{20}

ii)

Now the ratio of kg water vaporized/kg wet sugar leaving the evaporator.:

\frac{m_v}{m_f-m_v} =\frac{17}{100-17}

\frac{m_v}{m_f-m_v} =\frac{17}{83}

b)

amount of sugar fed per day, m_f=907185\ kg

<u>Now the mass of water in the given amount of sugar per day:</u>

m_w=\frac{m_f}{5}

m_w=\frac{907185}{5}

m_w=181437\ kg

Mass of water vapourized after passing through the evaporator:

m_v=\frac{85}{100}\times m_w

m_v=\frac{85}{100}\times 181437

m_v=154221.45\ kg

Now the mass of water still remaining in the sugar:

m_r=m_w-m_v

m_r=907185-154221.45

m_r=752963.55\ kg

is the mass of extra water to be evapourated.

8 0
3 years ago
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