Ask question

Ask question

All categories

3 years ago

8

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

You might be interested in

John walks 1.00 km north, then turns right and walks 1.00 km east. His speed is 1.50 m/s during the entire stroll.a) What is the

avanturin [10]

Answer:

(a) 1.414 km

(b) 1.06 m/s

Explanation:

(a) For John:

Distance = 1 km north and then 1 km east

Speed = 1.5 m/s

total distance traveled = 1 + 1 = 2 km = 2000 m

Time taken to travel = Distance / speed

t = 2000 / 1.5 = 1333.3 seconds

Displacement = $\sqrt{1^{2}+1^{2}}=1.414 km$

(b) For jane :

Time is same as john = 1333.33 second

Distance = 1.414 km = 1414 m

Speed = distance / time = 1414 / 1333.33 = 1.06 m/s

3 0

3 years ago

A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a

Stels [109]

The power dissipated across a component can be calculated through the formula P=I^2xR

Substituting the values in we get P=(0.5)^2x10=2.5W

4 0

3 years ago

the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr

Dmitry [639]

Answer:

The approximate number of decays this represent is $N= 23*10^{10}$

Explanation:

From the question we are told that

The amount of Radiation received by an average american is $I_a = 2.28 \ mSv$

The source of the radiation is $S = 5.49 MeV \ alpha \ particle$

Generally

$1 \ J/kg = 1000 mSv$

Therefore $2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg$

Also $1eV = 1.602 *10^{-19}J$

Therefore $2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg$

An Average american weighs 88.7 kg

The total energy received is mathematically evaluated as

$1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x$

Cross-multiplying and making x the subject

$x = 88.7 * 1.423*10^{16} eV$

$x = 126.2*10^{16}eV$

Therefore the total energy deposited is $x = 126.2*10^{16}eV$

The approximate number of decays this represent is mathematically evaluated as

N = $\frac{x}{S}$

Where n is the approximate number of decay

Substituting values

$N = \frac{126 .2*10^{16}}{5.49*10^6}$

$N= 23*10^{10}$

7 0

3 years ago

If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun

inessss [21]

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

$s=-8t^2+32t$ ...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

$\dfrac{ds}{dt}=0$

$\dfrac{d(-8t^2+32t)}{dt}=0$

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

$s=-8(2)^2+32(2)$

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.

6 0

3 years ago

An umbrella tends to move upward on a windy day because _____.

masha68 [24]

E. all of the above

An umbrella tends to move upward on a windy day because _<span>A. buoyancy increases with increasing wind speed </span>
<span>B. air gets trapped under the umbrella and pushes it up </span>
<span>C. the wind pushes it up </span>
<span>D. a low-pressure area is created on top of the umbrella </span>

3 0

3 years ago