A cupcake recipe designed to produce 28 cupcakes calls for 360 grams of flour Determine the quantity of flour that would be

2 answers:

4 0

Since the 360 grams of flour are used to produce 28 cupcakes , thus the quantity (mass) of flour is said to be distributed equally on all 28 cupcakes thus, every cupcake requires 360/28 grams of flour which is equal to 12.8571 grams.

By the same logic the 32 cupcakes would have a different amount of flour distributed equally amongst all 32 cupcakes thus the quantity required is equal to 32 times required to prepare 1 cupcake , we conclude that the mass needed is equal to 32 x 12.8571 = 411.4272 grams of flour.
If you still have any questions i’ll be there to answer. ♥️♥️

Novosadov [1.4K]2 years ago

3 0

Answer:24888.8g

Explanation: 28cupcakes calls 360g

32 cupcakes calls ???

28x32=896 cupcakes

896/360= 2.4888888g

In four decimal place = 24888.8grams

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jarptica [38.1K]

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a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$