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ra1l [238]
3 years ago
8

Bonus Points. A ball was thrown straight up in the air from 1.2 meter above the grouind. After 3 seconds the ball returnes to th

e ground. Determine the maximum height it reached.
Physics
1 answer:
nekit [7.7K]3 years ago
5 0

Answer:

Explanation:

given that

Distance above the ground, s = 1.2 m

Time taken by the ball, t = 3 s

Velocity of the ball, v = 1.2/3 = 0.4 m/s

Maximum height reached by the ball is then given by the formula

H = v² / 2g

H = 0.4² / 2 * 9.8

H = 0.16 / 19.6

H = 0.0082 m or rather, 0.82 cm

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Zepler [3.9K]

Answer:

the string and metre rule

Explanation:

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2 years ago
Please answer the questions... I will surely mark you as the brainliest according to me :)
frutty [35]

Answer:

(a) You can tell that have the same strength because they have attracted the same amount of paper clips.

(b) Iron is used in electromagnets because steel retained magnetic properties after the power was turned off, but in the iron, the paper clips dropped off right away.

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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
If 31.25
valentina_108 [34]
<span>1 C = 6.24150965(16)×10^18 electrons

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4 0
3 years ago
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leonid [27]

The instant it was dropped, the ball had zero speed.

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Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered <em>4.9 meters</em>.

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