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3 years ago

I need this done by tonight!! Can anyone help me please? Answer these 4 questions

Physics

1 answer:

VikaD [51]3 years ago

4 0

Answer:

1. 14 g of chocolate mixture.

2. 24 fl oz of chocolate milk

3. 10 cups of chocolate milk.

4. 12½ cups.

Explanation:

From the question given above, the following data were obtained:

1 TBSP = 7 g

1 Cup = 8 fl oz

2 Table spoons (TBSP) for 1 cup (8 fl oz) of milk.

1. Determination of the mass of chocolate mixture in 1 cup of chocolate milk.

From the question given above,

1 Cup required 2 Table spoons (TBSP)

But

1 TBSP = 7 g

Therefore,

2 TBSP = 2 × 7 = 14 g

Thus, 1 Cup required 14 g of chocolate mixture.

2. Determination of the number fl oz of chocolate milk in 3 cups

1 Cup = 8 fl oz

Therefore,

3 Cups = 3 × 8

3 Cups = 24 fl oz

Thus, 24 fl oz of chocolate milk are in 3 cups.

3. Determination of the number of cups of chocolate milk produce from 20 TBSP.

2 TBSP is required to produce 1 cup.

Therefore,

20 TBSP will produce = 20/2 = 10 Cups.

Thus, 10 cups of chocolate milk produce from 20 TBSP.

4. Determination of the number of cups obtained from 100 fl oz chocolate milk.

8 fl oz is required to produce 1 cup.

Therefore,

100 fl oz will produce = 100 / 8 = 12½ cups.

Thus, 12½ cups is obtained from 100 fl oz chocolate milk.

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2 years ago

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Answer:

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Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

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Substituting values

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$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

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So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

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Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

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