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Luba_88 [7]
3 years ago
7

How many 1/3 servings are available in 6 ounces

Mathematics
1 answer:
Paraphin [41]3 years ago
8 0

I'm going to assume that you're asking, "How many 1/3-oz servings are available in 6 oz?"

Divide 6 oz by 1/3 oz:

         6 oz

----------------------- = 18 servings

(1/3) oz/serving

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If the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb, how much work is needed to stretch it 6 in.
mel-nik [20]

Answer:

0.375 feet-lb

Step-by-step explanation:

We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.

We can represent our given information as:

6=\int\limits^2_0 {F(x)} \, dx

We will use Hooke's Law to solve our given problem.

F(x)=kx

Substituting this value in our integral, we will get:

6=\int\limits^2_0 {kx} \, dx

Using power rule, we will get:

6=\left[ \frac{kx^2}{2} \right ]^2_0

6=\frac{k(2)^2}{2}-\frac{k(0)^2}{2}

6=\frac{4k}{2}-0\\\\k=3

We know that 6 inches is equal to 0.5 feet.

Work needed to stretch it beyond 6 inches beyond its natural length would be \int\limits^{0.5}_0 {kx} \, dx =\int\limits^{0.5}_0 {3x} \, dx

Using power rule, we will get:

\int\limits^{0.5}_0 {3x} \, dx = \left [\frac{3x^2}{2}\right]^{0.5}_0

\frac{3(0.5)^2}{2}-\frac{3(0)^2}{2}\Rightarrow \frac{3(0.25)}{2}-0=\frac{0.75}{2}=0.375

Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.

3 0
3 years ago
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Answer:

\frac{12}{5}

Step-by-step explanation:

\sqrt{\frac{144}{25}}=\frac{\sqrt{144}}{\sqrt{25}}=\frac{12}{5}

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