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natali 33 [55]
2 years ago
10

The density of a certain type of steel is 8.1 g/cm3. What is the mass of a 100 cm3 chunk of this steel

Engineering
1 answer:
irina1246 [14]2 years ago
4 0

Answer:

  810 g

Explanation:

Mass is the product of density and volume:

  m = ρV

  m = (8.1 g/cm³)(100 cm³) = 810 g

The mass of the chunk is 810 grams.

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The product of two factors is 4,500. If one of the factors is 90, which is the other factor?
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8 0
2 years ago
A 0.9% solution of NaCl is considered isotonic to mammalian cells. what molar concentration is this?
natka813 [3]

Answer:

58.44 g/mol The Molarity of this concentration is 0.154 molar

Explanation:

the molar mass of NaCl is 58.44 g/mol,

0.9 % is the same thing as 0.9g of NaCl , so this means that 100 ml's of physiological saline contains 0.9 g of NaCl.  One liter of physiological saline must contain 9 g of NaCl.  We can determine the molarity of a physiological saline solution by dividing 9 g by 58 g... since we have 9 g of NaCl in a liter of physiological saline, but we have 58 grams of NaCl in a mole of NaCl.  When we divide 9 g by 58 g, we find that physiological saline contains 0.154 moles of NaCl per liter.  That means that physiological saline (0.9% NaCl) has a molarity of 0.154 molar.  We can either express this as 0.154 M or 154 millimolar (154 mM).

3 0
3 years ago
The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve
stellarik [79]

Answer:

Explanation:

Given

velocity at A is v=4\ ft/s

For r=500\ ft

velocity is increasing at \dot{v}=0.004t\ ft/s^2

Tangential acceleration is given by

a_t=\frac{\mathrm{d} v}{\mathrm{d} t}

a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}

\int 0.004tdt=\int dv

\int dv=\int 0.004tdt

v=0.002t^2+c

at t=0\ v=4\ ft/s

4=0.002\cdot 0+c

c=4\ ft/s

thus v=0.002t^2+4

Velocity in terms of Displacement is given by

v=\frac{\mathrm{d} s}{\mathrm{d} t}

\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt

\Rightarrow s=\frac{0.002t^3}{3}+4t

When car has traveled \frac{3}{4} th of distance i.e.

s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}

s=750\pi

750\pi =\frac{0.002t^3}{3}+4t

\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0

on solving we get t=139.23\ s

Thus velocity at t=139.23\ s

v=42.76\ s

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}

a_n=3.658\ m/s^2

Tangential acceleration a_t at t=139.23\ s

a_t=0.556\ m/s^2

Net acceleration a_t=\sqrt{(a_n)^2+(a_t)^2}

a_n=\sqrt{(3.658)^2+(0.556)^2}

a_n=3.7\ m/s^2

   

8 0
3 years ago
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