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Automobile engines AND steam engines are examples of ______. a. internal combustion engines b. heat engines c. external combusti

uranmaximum [27]

Answer: B. Heat engines

Explanation:

8 0

3 years ago

Read 2 more answers

1. Suppose you're working at a constant temperature , when you exert more pressure on an object

vova2212 [387]

Answer:

Decrease

Explanation:

Working at a constant temperature when more pressure is exerted, the volume decreases.

This is known as Boyle's law.

According to Boyle's law;

"the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".

Mathematically;

P₁V₁ = P₂V₂

P and V are pressure and volume

1 and 2 are initial and final states.

8 0

3 years ago

According to the law of conservation of energy, when a car uses 10,000 J of chemical energy from gasoline, how much of this ener

Dmitriy789 [7]

The answer is A because its asking how much of all 3 energies combined will it give off.

6 0

3 years ago

A mass of 4.10 kg is suspended from a 1.69 m long string. It revolves in a horizontal circle as shown in the figure.

nikklg [1K]

The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R$

where $m=4.10\,\rm kg$ , $v=2.85\frac{\rm m}{\rm s}$ , and $R$ is the radius of the circular path.

As shown in the diagram, we can see that

$\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)$

where $r=1.69\,\rm m$ , so that

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}$

The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have

• net vertical force

$F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}$

Solve for $\theta$ :

$\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0$

Complete the square:

$\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}$

Plugging in the known quantities, we end up with

$\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27$

The second case has no real solution, since $-1\le\cos(\theta)\le1$ for all $\theta$ . This leaves us with

$\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}$

7 0

2 years ago

Placing a greater value on a research project that included 10,000 participants, as opposed to one investigating the same hypoth

fgiga [73]

Answer:

Sample size determination

Explanation:

Sample size determination is the process of determining the number of observations to be used in a research hypothesis.

The larger the sample size, the more precise and accurate the results of the hypothesis will be. Bearing this in mind, the researcher will prefer to place a greater value on the research project with 10,000 participants.

3 0

3 years ago