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3 years ago

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The number of kilograms of water in a human body varies directly as the mass of the body. Upper AA 9393-kg person contains 6262

kg of water. How many kilograms of water are in anan 8181-kg person?

1 answer:

Pie3 years ago

3 0

Answer:

54 kg

Explanation:

Mass of person = 93 kg = $m_p$

Mass of water = 62 kg = $m_w$

Dividing the above two masses we get

$\frac{m_p}{m_w}=\frac{93}{62}\\\Rightarrow \frac{m_p}{m_w}=1.5\\\Rightarrow m_p=1.5m_w$

Hence, the mass of the person is 1.5 times the mass of the water in them

Now, Mass of person = 81 kg = $m_p$

$m_p=1.5m_w\\\Rightarrow m_w=\frac{1}{1.5} m_p\\\Rightarrow m_w=\frac{1}{1.5} \times 81\\\Rightarrow m_w=54\ kg$

So, the mass of water in a person that has mass of 81 kg is 54 kg

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Assume that Parker Company will receive SF200,000 in 360 days. Assume the following interest rates:

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Answer:

b. $96,914

Explanation:

360-day borrowing rate = 5%

spot rate = 0.48

360-day deposit rate = 6%

Borrow at the rate of 5% to get

SF200,000/1.05 = $190,476.19

Convert at the spot rate of $0.48 to get

190,476.19*0.48 = $91,428.57

Invest at the interest rate of 6% to get

91,428.57/1.06 = 96,914.28

Therefore, Parker Company will receive $96,914 in 360 days.

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2 years ago

What did Thomson contribute to the model of atom

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<h3><u>Answer;</u></h3>

<em>Electrons </em>

<h3><u>Explanation;</u></h3>

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Read 2 more answers

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

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