A typical human contains 5.00 l of blood, and it takes 1.00 min for all of it to pass through the heart when the person is resti
ng with a pulse rate of 74.0 heartbeats/ minute. on the average, what volume of blood, in (a) liters and (b) cubic centimeters, does the heart pump during each beat?
2 answers:
<em>Volume of blood, in (a) liters = 0.0675 liters and (b) cubic centimeters = 67.5 cm³</em>
<h3><em>Further explanation </em></h3>7 main quantities have been determined based on international standards, namely:
1. Length, meters (m)
2. Time, second (s)
3. Mass, kilograms (kg)
4. Temperature, kelvin (K)
5. Light intensity, candela (cd)
6. Electric current, ampere (A)
7. Amount of substance, mol (m)
The length of the principal is to indicate the distance between two points in a place
The symbol is usually expressed in l, length
Volume is a derivative quantity derived from the length of the principal
The unit of volume can be expressed in liters or milliliters or cubic meters
The conversion is
1 cc = 1 cm³
1 dm = 1 Liter
1 mL = 1 cm³
1 L = 1 dm³
From the question, 5 liters of blood passes through the heart in 1 minute
The heart beats 74 times/minute
So that the equation is easy as follows:
5 liters = 1 minute
74 beats = 1 minute
So that in 1 heartbeat there is as much blood volume as:
= 5 liters / 74 beats
= 0.0675 liters
When we convert to cm³ it becomes:
1 L = 1 dm³ = 10³ cm³
0.0675 liters = 0.0675.10³ cm³
= 67.5 cm³
<h3><em>Learn more </em></h3>
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Keywords: volume, liters, heart, beats, blood
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A. 6.67 cm
The focal length of the lens can be found by using the lens equation:
where we have
f = focal length
s = 12 cm is the distance of the object from the lens
s' = 15 cm is the distance of the image from the lens
Solving the equation for f, we find
B. Converging
According to sign convention for lenses, we have:
- Converging (convex) lenses have focal length with positive sign
- Diverging (concave) lenses have focal length with negative sign
In this case, the focal length of the lens is positive, so the lens is a converging lens.
C. -1.25
The magnification of the lens is given by
where
s' = 15 cm is the distance of the image from the lens
s = 12 cm is the distance of the object from the lens
Substituting into the equation, we find
D. Real and inverted
The magnification equation can be also rewritten as
where
y' is the size of the image
y is the size of the object
Re-arranging it, we have
Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.
Also, the sign of s' tells us if the image is real of virtual. In fact:
- s' is positive: image is real
- s' is negative: image is virtual
In this case, s' is positive, so the image is real.
E. Virtual
In this case, the magnification is 5/9, so we have
which can be rewritten as
which means that s' has opposite sign than s: therefore, the image is virtual.
F. 12.0 cm
From the magnification equation, we can write
and then we can substitute it into the lens equation:
and we can solve for s:
G. -6.67 cm
Now the image distance can be directly found by using again the magnification equation:
And the sign of s' (negative) also tells us that the image is virtual.
H. -24.0 cm
In this case, the image is twice as tall as the object, so the magnification is
M = 2
and the distance of the image from the lens is
s' = -24 cm
The problem is asking us for the image distance: however, this is already given by the problem,
s' = -24 cm
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Answer:
The atomic weight in g/mole of the metal (molar mass) is 8.87.
Explanation:
To begin, it is possible to assume that, as a sample, it has 100 g of the compound. This means that:
- 52.92% metal: 52.92 g M
- 47.80% oxygen: 47.80 g O
Using the molar mass of oxygen, which is 16 g / mol, it is possible to calculate the amount of moles of oxygen present in the sample using the rule of three:
moles of oxygen=2.9875
The chemical formula of metal oxide tells you that:
2 M⁺³ + 3 O²⁻ ⇒ M₂O₃
In the previous equation you can see that you need 3 oxygen anions to react with two metal cations. Then:
You have 52.92 g of metal in the sample, then the molar mass of the metal is:
molar mass≅ 8.87 g/mol
<u><em> The atomic weight in g/mole of the metal (molar mass) is 8.87.</em></u>
The closest match to this value is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.