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3 years ago

Grace drives her car 168 km in 2 hours. What is her average speed in kilometers per hour?

Physics

1 answer:

wlad13 [49]3 years ago

5 0

Answer:

84kliometers

Explanation:

divide one hundred and sixty eight kilo meters by two hours

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Body types can be described as _____ (check all that are correct)Immersive Reader

rewona [7]

Apple, Pear, hourglass, bodybuilder, and rectangle.

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3 years ago

A 50 cm^3 block of iron is removed from an 800 degrees Celsius furnance and immediately dropped into 200 mL of 20 degrees Celsiu

NNADVOKAT [17]

Answer:

% of water boils away= 12.64 %

Explanation:

given,

volume of block = 50 cm³ removed from temperature of furnace = 800°C

mass of water = 200 mL = 200 g

temperature of water = 20° C

the density of iron = 7.874 g/cm³ ,

so the mass of iron(m₁) = density × volume = 7.874 × 50 g = 393.7 g

the specific heat of iron C₁ = 0.450 J/g⁰C

the specific heat of water Cw= 4.18 J/g⁰C

latent heat of vaporization of water is L_v = 2260 k J/kg = 2260 J/g

loss of heat from iron is equal to the gain of heat for the water

$m_1\times C_1\times \Delta T = M\times C_w\times \Delta T + m_2\times L_v$

$393.7\times 0.45\times (800-100) = 200\times 4.18\times(100-20) + m_2\times 2260$

m₂ = 25.28 g

25.28 water will be vaporized

% of water boils away = $\dfrac{25.28}{200}\times 100$

% of water boils away= 12.64 %

5 0

3 years ago

Calculating average speed

djverab [1.8K]

Average speed =

(distance covered during some period of time)
divided by
(length of time to cover that distance).

7 0

3 years ago

Another name for the water cycle is the

atroni [7]

Answer:

<h2>Hydrologic cycle</h2>

Explanation:

Stay safe, stay healthy and blessed.

Have a good day !

Thank you

5 0

2 years ago

Read 2 more answers

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:

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, $m'_v=85\%$

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:

$m_v=m_v' \times m_w$

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

$m_v=17\ kg$

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.:

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

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

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