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3 years ago

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A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What can you conclude about the box

?
A. An unbalanced force of 25N is acting on the box.
B. The net horizontal force acting on the box is 50N.
C. Friction between the box and the floor is 25N to the left.
D. Friction between the box and the floor is 25N to the right.

2 answers:

kotykmax [81]3 years ago

7 0

Answer:

C. Friction between the box and the floor is 25N to the left

Explanation:

dolphi86 [110]3 years ago

4 0

Answer:

Since the forces exerted on the box are equal and opposite in direction, the net force will be 0, so the box will not move.

Explanation:

UwU

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4 years ago

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

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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.

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3 years ago