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2 years ago

5

A pasta factory made 3.84 pounds of pasta in 4 minutes. How much pasta did the factory make in each minute? plzz help me

1 answer:

den301095 [7]2 years ago

5 0

${\underline{\green{\textsf{\textbf{ Answer : }}}}}$

➡ 0.96 pounds in a minute.

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Alexandra [31]

The object takes 0.5 seconds to complete one rotation, so its rotational speed is 1/0.5 rot/s = 2 rot/s.

Convert this to linear speed; for each rotation, the object travels a distance equal to the circumference of its path, or 2<em>π</em> (1.2 m) = 2.4<em>π</em> m ≈ 7.5 m, so that

2 rot/s = (2 rot/s) • (2.4<em>π</em> m/rot) = 4.8<em>π</em> m/s ≈ 15 m/s

thus giving it a centripetal acceleration of

<em>a</em> = (4.8<em>π</em> m/s)² / (1.2 m) ≈ 190 m/s².

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<em>T</em> = (50 kg) <em>a</em> ≈ 9500 N.

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2 years ago

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2 years ago

A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J

Lyrx [107]

Explanation:

Given that,

Wavelength of the light, $\lambda=4170\ A=4170\times 10^{-10}\ m$

Work function of sodium, $W_o=4.41\times 10^{-19}\ J$

The kinetic energy of the ejected electron in terms of work function is given by :

$KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J$

The formula of kinetic energy is given by :

$KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s$

Hence, this is the required solution.

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2 years ago

True or false cells of humans and plants are same.

timama [110]

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1 year ago

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th

tatuchka [14]

<u>Answer:</u> The final temperature of the solution is $80.14^oC$

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

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3 years ago