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3 years ago

12

Mr. Hoffman, a science teacher, drove 10 miles to school from home in 20 minutes. He drove the 10 miles home in 30 min. His aver

age velocity was _____ m/s.

2 answers:

olga2289 [7]3 years ago

7 0

His average velocity for the day was 0. He ended where he started so his displacement was 0.

Andrews [41]3 years ago

4 0

zero.

Because he finished right where he started. his total displacement zero. therefore, his average velocity was also zero

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A projectile is launched horizontally at a speed of 45.0 m/s from a

Free_Kalibri [48]

Answer:

answer will be c: 1.5 seconds :)

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2 years ago

A ball was dropped from a height of 10 feet. Each time it hits the ground, it bounces 4/5 of its previous height. Find the total

Shtirlitz [24]

Answer:

d = 90 ft

Explanation:

As we know that after each bounce it reaches to 4/5 times of initial height

so we can say

$h_2 = \frac{4}{5}h$

so the distance covered is given as

$d = h + 2(\frac{4}{5}h) + 2(\frac{4}{5})^2h + 2(\frac{4}{5})^3h........$

here we know that

h = 10 feet

$d = h + 2(\frac{4}{5}h)(1 + \frac{4}{5} + (\frac{4}{5})^2 + ...........)$

$d = 10 + 2(\frac{4}{5}(10))(\frac{1}{1 - \frac{4}{5}})$

$d = 90 ft$

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2 years ago

Glass does not transmit ultraviolet radiation. Suggest what happens to ultraviolet radiation when it is incident on glass. (1 ma

o-na [289]

Answer:

Most of UV radiation is stopped by glass & this is why you will not get sunburns behind a glass. The glass simply filters out the UV radiation that is responsible for the sunburns & protect your skins from these energetic & somewhat harmful radiation

Explanation:

4 0

1 year ago

g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. Fo

mariarad [96]

Answer:

A) ≥ 325Kpa

B) ( 265 < Pe < 325 ) Kpa

C) (94 < Pe < 265 )Kpa

D) Pe < 94 Kpa

Explanation:

Given data :

A large Tank : Pressures are at 400kPa and 450 K

Throat area = 4cm^2 , exit area = 5cm^2

<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>

The range of flow of back pressures that will make the flow entirely subsonic

will be ≥ 325Kpa

attached below is the detailed solution

<u>B) Have a shock wave</u>

The range of back pressures for there to be shock wave inside the nozzle

= ( 265 < Pe < 325 ) Kpa

attached below is a detailed solution

C) Have oblique shocks outside the exit

= (94 < Pe < 265 )Kpa

D) Have supersonic expansion waves outside the exit

= Pe < 94 Kpa

7 0

2 years ago

An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,

Sedbober [7]

Answer:

$a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa$

Explanation:

a. Internal energy and the relative specific volume at $s_1$ are determined from A-17: $u_1=214.07kJ/kg, \ \alpha_r_1=621.2$ .

The relative specific volume at $s_2$ is calculated from the compression ratio:

$\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825$

#from this, the temperature and enthalpy at state 2, $s_2$ can be determined using interpolations $T_2=862K$ and $h_2=890.9kJ/kg$ . The specific volume at $s_1$ can then be determined as:

$\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg$

Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

$\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K$

b.Relative SV and enthalpy at $s_3$ are obtained for the given temperature with interpolation with data from A-17 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$

Relative SV at $s_4$ is

$a_r_4=\frac{r}{r_c}\alpha _r_3$

= $=\frac{16}{2}\times4.533\\=36.424$

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

$n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563$

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

$MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa$

Hence, the mean effective relative pressure is 674.95kPa

3 0

3 years ago