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3 years ago

If 1.4% of the mass of a human body is calcium, how many kilograms of calcium are there in a 195-pound man?

Physics

2 answers:

Mamont248 [21]3 years ago

8 0

Answer:

1.241 kg

Explanation:

The mass of the man in kg is 195 pound * (1/2.2) kg/pound = 195/2.2 = 88.64 kg

If 1.4% of the mass of a human body is calcium, then the mass of calcium inside the 88.64 kg man is

1.4% * 88.64 = 88.64*1.4/100 = 1.241 kg

So there's 1.241 kg of calcium inside a 195-pound man

Degger [83]3 years ago

3 0

<h2>Answer:</h2>

1.24kg

<h2>Explanation:</h2>

Given;

Mass of the man (M) = 195 pound

<em>First, convert the value to kilograms (kg) as follows;</em>

Note that;

1 pound = 0.453592kg

Therefore,

195 pounds = 195 x 0.453592kg = 88.45kg

<em>Now, since the mass of calcium (say m) in a human body is 1.4% of the mass (M) of the body;</em>

=> m = 1.4% of M

=> m = 1.4% x M ------------------(i)

Where;

M = 88.45kg

<em>Substitute the value of M into equation (i) as follows;</em>

=> m = 1.4 % x 88.45

=> m = (1.4 / 100) x 88.45

<em>Solve for m;</em>

m = 0.014 x 88.45

m = 1.24kg

Therefore, the mass of calcium in the 195-pound man is 1.24kg.

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One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a

Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

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3 years ago

A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in

katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= $10+1.50$

= $11.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v = \frac{d}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{11.5}$

$=26.08 \ s$

Throughout the opposite direction, when the boat seems to be travelling then,

= $10-1.50$

= $8.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v=\frac{v}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{8.5}$

$=35.29 \ s$

hence,

The time taken by the boat will be:

= $26.08+35.29$

= $61.37 \ s$

8 0

3 years ago

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is . Find the time i

a_sh-v [17]

Answer:

a)<em> It took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves. </em>

b) <em>The minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds. </em>

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensitie. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies:

Gamma rays

X-rays

Ultraviolet rays

Visible region

Infrared

Microwave

Radio waves.

Any radiation that belongs to electromagnetic spectrum has a speed in vacuum of $3x10^{8}m/s$ .

<em>a) Find the time it took for his voice to reach the Earth via radio waves.</em>

To know the time that took for Neil Armstrong's voice to reach the Earth via radio waves, the following equation can be used:

$c = \frac{d}{t}$ (1)

Where v is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

$t = \frac{d}{c}$ (2)

The distance from the Earth to the Moon is $3.85x10^{8} m$ , therefore.

$t = \frac{3.85x10^{8} m}{3x10^{8}m/s}$

$t = 1.28s$

Hence, it took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

<em>b) Determine the minimum time that will be required for a message from Mars to reach the Earth via radio waves.</em>

The distance from the Earth to the Mars at its closest approach is $5.76x10^{10}m$ , therefore.

$t = \frac{5.76x10^{10}m}{3x10^{8}m/s}$

$t = 192s$

Hence, the minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

3 0

3 years ago

A 4 kg bowling bowl is sitting on a table 1 meter off the ground. How much potential energy does it have?

il63 [147K]

Answer:

$\huge\boxed{\sf P.E. = 39.2\ Joules}$

Explanation:

<u>Given Data:</u>

Mass = m = 4 kg

Acceleration due to gravity = g = 9.8 m/s²

Height = h = 1 m

<u>Required:</u>

Potential Energy = P.E. = ?

<u>Formula:</u>

P.E. = mgh

<u>Solution:</u>

P.E. = (4)(9.8)(1)

P.E. = 39.2 Joules

$\rule[225]{225}{2}$

Hope this helped!

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3 years ago

a person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500m/s^2. what is the velocity of the stroll

grandymaker [24]

You're going to use the constant acceleration motion equation for velocity and displacement:
(V)final²=(V)initial²+2a(Δx)

Given:
a=0.500m/s²
Δx=4.75m
(V)intial=0m
(V)final= UNKNOWN

(V)final= 2.179m/s

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3 years ago

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