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2 years ago

9

In an experiment to determine relative velocities of a feather, ball, and anvil, what is the object called?

1 answer:

lara [203]2 years ago

5 0

Independent variable if I’m not mistaken

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Consider a Mach 4.5 airflow at a pressure of 1.25 atm. We want to slow this flow to a subsonic speed through a system of shock w

marissa [1.9K]

Answer:

a. 130.73 atm

b. 102.62 atm

c. 87.1 atm

Explanation:

See the attached pictures.

6 0

3 years ago

5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of

marta [7]

Answer:

(a) $3.185*10^{21} cells$

(b) $6.37*10^{21} atoms$

Explanation:

(a)

Volume, V of unit cell

$V=(2.866*10^{-8})^{3}=2.354*10^{-23}$

Number of unit cells, N

$N=\frac {W_{mat}}{V\rho_{mat}}$ Where $W_{mat}$ is weight of material and $\rho_{mat}$ is density of material

$N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells$

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron= $3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms$

8 0

3 years ago

Rosita is planning an investigation to determine how a lifeboat's shape affects its

KiRa [710]

Yes! Is right, just did the work!

3 0

3 years ago

Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

Dafna11 [192]

Answer:

$\mathbf{h_2 =1021.9 \ mm}$

Explanation:

Given that :

The initial volume of water $V_1$ = 1000.00 mL = 1000000 mm³

The initial temperature of the water $T_1$ = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water $T_2$ = 70° C

The coefficient of thermal expansion for the glass is ∝ = $3.8*10^{-6 } mm/mm \ per ^oC$

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature $T_1 = 10 ^ 0C$ the density of the water is $\rho = 999.7 \ kg/m^3$

At temperature $T_2 = 70^0 C$ the density of the water is $\rho = 977.8 \ kg/m^3$

The mass of the water is $\rho V = \rho _1 V_1 = \rho _2 V_2$

Thus; we can say $\rho _1 V_1 = \rho _2 V_2$ ;

⇒ $999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2$

$V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }$

$V_2 = 1022.40 \ mL$

$v_2 = 1022400 \ mm^3$

Thus, the volume of the water after heating to a required temperature of $70^0C$ is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

$V_1 = A_1 *h_1$

The area of the water before heating is:

$A_1 = \dfrac{V_1}{h_1}$

$A_1 = \dfrac{1000000}{1000}$

$A_1 = 1000 \ mm^2$

The area of the heated water is :

$A_2 = A_1 (1 + \Delta t \alpha )^2$

$A_2 = A_1 (1 + (T_2-T_1) \alpha )^2$

$A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2$

$A_2 = 1000.5 \ mm^2$

Finally, the depth of the heated hot water is:

$h_2 = \dfrac{V_2}{A_2}$

$h_2 = \dfrac{1022400}{1000.5}$

$\mathbf{h_2 =1021.9 \ mm}$

Hence the depth of the heated hot water is $\mathbf{h_2 =1021.9 \ mm}$

4 0

3 years ago

Use the scoring system described to calculate the alignment score for These are two lines of 14 letters aligned with each other.

PtichkaEL [24]

Answer:

do good

Explanation:

get that money

7 0

3 years ago