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2 years ago

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If a total force exerted by water in a container with a bottom area of 2 square meters is 450 newtons, what's the water pressure

at the bottom of the container? A. 0.900 kPa B. 0.225 kPa C. 0.575 kPa D. 0.300 kPa

2 answers:

m_a_m_a [10]2 years ago

7 0

its B 0.225kPa using the formula p=f/A then change the pascals into kpa

Ipatiy [6.2K]2 years ago

7 0

The Correct answer to this question for Penn Foster Students is: 0.225 kPa

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The position vector of a particle of mass 1.70 kg as a function of time is given by r with arrow = (6.00 î + 5.70 t ĵ), where r

ollegr [7]

Answer:

The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .

Explanation:

Given that,

Mass = 1.70 kg

Position vector $r=(6.00\hat{i}+5.70 t \hat{j})$

We need to calculate the angular velocity

The velocity is the rate of change of the position of the particle.

$v = \dfrac{dr}{dt}$

$v=\dfrac{d}{dt}(6.00\hat{i}+5.70 t \hat{j})$

$v=5.70\hat{j}$

We need to calculate the angular momentum of the particle

Using formula of angular momentum

$L=r\cdot p$

Where, p = mv

Put the value of p into the formula

$L=m(r\times v)$

Substitute the value into the formula

$L=1.70(6.00\hat{i}+5.70 t \hat{j}\times5.70\hat{j})$

$L=1.70\times34.2$

$L=58.14\ kgm^2/s$

Hence, The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .

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Calculating heat gained or lost requires mass, specific heat , and

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B. relative humidity

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3 years ago

What is magnetic resonance???

Lyrx [107]

Answer:

procedure in which radio waves and a powerful magnet linked to a computer

Explanation:

Listen to pronunciation. (mag-NEH-tik REH-zuh-nunts IH-muh-jing) A procedure in which radio waves and a powerful magnet linked to a computer are used to create detailed pictures of areas inside the body. These pictures can show the difference between normal and diseased tissue.

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What is the average velocity of a car starting from 0 and travels 7 meters in 9 seconds?

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Your question has been heard loud and clear.

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Average velocity= 0.77m/s

Thank you.

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3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

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0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

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$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

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$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

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2 years ago