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Romashka-Z-Leto [24]
3 years ago
7

magine that two balls, a basketball and a much larger exercise ball, are dropped from a parking garage. If both the mass and rad

ius of the exercise ball are 3 times larger than the mass and radius of the basketball, which of the two balls do you think will reach the ground first? Make a hypothesis below and describe your reasoning.
Physics
1 answer:
DerKrebs [107]3 years ago
6 0

Without counting wind resistance, They will both reach the ground at the same time. If we apply the concept of kinematics, such as the equation vf^2=vi^2 + 2ad. This equation doesn't count how big or how heavy the mass is, it only focuses on how fast where they in the start and how far are both of them from the ground. So if they both have the same distance and same initial veloctity, then they will reach the ground at the same time.

For example, Try dropping a pen and a paper(Vertically) at the same height, you'll see they'll reach the ground at the same time.

If you count wind resistance, the heavier ball will hit the ground faster, because the air molecules will resist the lighter ball compared to the heavier ball.

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Unpolarized light with an intensity of 655 W / m2 is incident on a polarizer with an unknown axis. The light then passes through
Norma-Jean [14]

Answer:

1.\theta=29.84^{0}

2.\theta=60.15^{0}

Explanation:

Polarizes axis can create two possible angles with the vertical.

first we have to find the intensity of  first polarizer

which is given as

I=\frac{I_{0} }{2}

I= \frac{655\frac{W}{M^{2} } }{2}

I=327.5\frac{W}{m^{2} }

For a smaller angle for the first polarizer:

According to Malus Law

I_{2} =I_{1} Cos^{2}(90^{0} - \theta)

I_{2} =I_{1} sin^{2}\theta

\frac{I_{2} }{I_{1} }=Sin^{2}\theta

taking square root on both sides

\sqrt{\frac{163}{327.5} } = sin\theta

\theta=Sin^{-1}(0.4977)

\theta=29.84^{0}

For a larger angle for the first polarizer:

According to Malus Law

I_{2} =I_{1} cos^{2}\theta

\frac{I_{2} }{I_{1} }=Cos^{2}\theta

taking square root on both sides

\sqrt{\frac{163}{327.5} } = cos\theta

\theta=Cos^{-1}(0.4977)

\theta=60.15^{0}

7 0
3 years ago
A car is traveling at 24.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with constant accelerat
Elan Coil [88]

Answer:

The answer to your question is : vf = 15.18 m/s

Explanation:

Data

vo = 24 m/s

d = 120 m

vf = ? when d = 60.0 m

Formula

              vf² = vo²  + 2ad

For d =100m

            a = (vf² - vo²) / 2d

            a = (0 -24²) / 2(100)

            a = -576/200

             a = 2.88 m/s²

Now, when d = 60

        vf² = (24)² - 2(2.88)(60)

        vf² = 576 - 345.6

        vf² = 230.4

        vf = 15.18 m/s

4 0
3 years ago
Read 2 more answers
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
How are humans potentially vulnerable to some of these same chemicals? How are humans potentially vulnerable to some of these sa
Svetradugi [14.3K]
The 5 major unit processes include chemical coagulation, flocculation, sedimentation, filtration, and disinfection (described below). There are chemicals added to the water as it enters the various treatment processes.
6 0
2 years ago
A student performs an exothermic reaction in a beaker and measures the temperature. If the thermometer initially reads 35 degree
Neporo4naja [7]

B 25 degrees Celsius

3 0
3 years ago
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