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Natalka [10]
3 years ago
13

The severity of a tropical storm is related to the depressed atmospheric pressure at its center. In August 1985, Typhoon Odessa

in the Pacific Ocean featured maximum winds of about 90 mi/hr and pressure that was 40.0 mbar lower at the center than normal atmospheric pressure. In contrast, the central pressure of Hurricane Andrew (pictured) was 90.0 mbar lower than its surroundings when it hit south Florida with winds as high as 165 mi/hr. If a small weather balloon with a volume of 4.0 L at a pressure of 1.00 atmospheres was deployed at the edge of Typhoon odessa what was the volume of the balloon when it reached the center?
Physics
1 answer:
xxMikexx [17]3 years ago
7 0

Answer:

4.16 L

Explanation:

Assuming constant temperature,

At the edge of Typhoon Odessa: P₁ = 1 atm = 1013.3 mbar,

V₁  = 4.0 L

At the center of Typhoon Odessa: P₂ = (1013.3 - 40.0) mbar = 973.3 mbar

V₂ = ? L

For a fixed amount of gas at constant temperature (Boyle's law) : P₁V₁ = P₂V₂

V₂ = V₁ × (P₁/P₂)

V₂= (4.0) × (1013.3/973.3)

V₂= 4.16 L

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A child on a skateboard experiences a 75 N force with an acceleration of 1.5 m/s2.
cupoosta [38]

Answer:

The mass of the child + skateboard is 50 kg

Explanation:

In this problem, we can apply Newton's second law:

F = ma

where

F is the net force on a system

m is the mass of the system

a is the acceleration of the system

In this problem, our system is the child + the skateboard. The net force on them is

F = 75 N

and their acceleration is

a=1.5 m/s^2

So we can re-arrange the equation above to find their combined mass:

m=\frac{F}{a}=\frac{75}{1.5}=50 kg

3 0
3 years ago
What are the smallest parts that make up matter?
Alecsey [184]
The smallest parts that make up different types of matter are called atoms.
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3 0
3 years ago
Police radar equipment is used to detect the speed of objects. In one trial, the radar equipment records a stationary tree as tr
Brut [27]

Answer:

reliability

accuracy

Explanation:

If a reading of a measurement is consistently the same then the measurement is reliable.

If a reading of measurement is close the actual value of the measurement then the reading is accurate.

Here, a stationary tree shows reading 6 mph once and 0 mph another instant. So, neither the reading of a measurement is consistent not the reading of measurement is close the actual value.

Hence, the radar has problems in its reliability and accuracy

8 0
3 years ago
Which is NOT true of the Big Bang Theory?
grigory [225]

Answer:

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3 0
2 years ago
Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle
Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

v = \frac{2 \pi r}{T}  

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

T^{2} = r^{3}

T = \sqrt{(133370 Km)^{3}}

T = \sqrt{(2.372x10^{15} Km)}

T = 4.870x10^{7} Km

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( 1.50x10^{8} Km )

1 AU is defined as the distance between the earth and the sun.

\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU

T = 0.324 AU

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

T = \frac{0.324 AU}{1 AU} . 1 year

T = 0.324 year

That can be expressed in units of days

T = \frac{0.324 year}{1 year} . 365.25 days  

T = 118.60 days

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (133370 Km)}{(118.60 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

118.60 days .\frac{86400 s}{1 day} ⇒ 10247040 s

133370 Km .\frac{1000 m}{1 Km} ⇒ 133370000 m

v = \frac{2 \pi (133370000 m)}{(10247040 s)}

v = 81.778 m/s

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

T^{2} = r^{3}

T = \sqrt{(69000 Km)^{3}}

T = \sqrt{(3.285x10^{14} Km)}

T = 1.812x10^{7} Km

\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU

T = 0.120 AU

T = \frac{0.120 AU}{1 AU} . 1 year

T = 0.120 year

T = \frac{0.120 year}{1 year} . 365.25 days  

T = 43.83 days

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (69000 Km)}{(43.83 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

43.83 days . \frac{86400 s}{1 day} ⇒ 3786912 s

69000 Km . \frac{1000 m}{ 1 Km} ⇒ 69000000 m

v = \frac{2 \pi (69000000 m)}{(3786912 s)}

v = 114.483 m/s

 

\frac{114.483 m/s}{81.778 m/s} = 1.399            

The particle in the D ring is 1399 times faster than the particle in the Encke Division.  

7 0
2 years ago
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