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kipiarov [429]
3 years ago
11

Hurricane katrina, which hit the gulf coast of louisiana and mississippi on august 29, 2005, had the second lowest ever recorded

barometric pressure at 0.920 mb. convert this pressure to units of....
a. atmospheres.
b. inches of hg (as would appear on weather report).
Physics
1 answer:
Simora [160]3 years ago
3 0
The unit 'mb' means millibar which is equivalent to 1/1000 of 1 bar. To convert the units from bar to atmospheres (atm) and to inches Hg (inHg), we need to know the conversion factors.

a.) 1 atm = 1.01325 bar

0.92 mb(1 bar/1000 mbar)(1 atm/1.01325 bar) =<em> 9.08×10⁻⁴ atm</em>

b.) 1 bar = 29.53 inHg

0.92 mb(1 bar/1000 mbar)(29.53 inHg/1 bar) =<em> 0.027 inHg</em>
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Answer: e. All of the above.

Explanation:

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5 0
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particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine
AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is 1062.7\frac{m}{s^{2}}

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

F=ma (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}

with q1 and q2 the charge of the particles, r the distance between them and k the constant k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}. So:

F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}

F=62.7 N

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4 0
3 years ago
Explain why ventilation is very important if there is risk of exposure to random gas in your home school
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3 0
3 years ago
a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the
tatuchka [14]
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
\tan \alpha =  \frac{W}{F}= \frac{24.5 N}{12 N}=2.04
From which we find
\alpha = 63.8 ^{\circ}
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N




7 0
3 years ago
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