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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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Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

8 0
3 years ago
A construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity if he travels a distance o
Anastasy [175]

Answer:

W = 0

Explanation:

We are given with, a construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity. He travels a distance of 50 m.

The work done by an object is given by :

W=Fd

F = ma

So,

W=mad

m is mass

a is acceleration

d is displacement

The worker is moving with constant velocity, its acceleration will be 0. So, the work done by the worker is 0.

8 0
3 years ago
A ship travels with velocity given by 12, with current flowing in the direction given by 11 with respect to some co-ordinate axe
nataly862011 [7]

Answer:

v_x = 11.78 m/s

Explanation:

Velocity of the ship is given as

v = 12 units

the direction of the velocity of the ship is making an angle of 11 degree with the current

so we will have two components of the velocity

1) along the direction of the current

2) perpendicular to the direction of the current

so here we know that the component of the ship velocity along the direction of the current is given as

v_x = v cos\theta

v_x = 12 cos11

v_x = 11.78 m/s

7 0
3 years ago
You move a 25 n object 4 meters. find the work you did
d1i1m1o1n [39]
In physics, "work<span>" is when a force applied to an object moves the object in the same direction as the force. If someone pushes against a wall, no </span>work<span> is done on the system. It is calculated as follows:

Work = Force x distance
Work = 25 N x 4 meters
Work = 100 N.m</span>
5 0
3 years ago
Read 2 more answers
When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on ea
crimeas [40]

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = \frac{(172\ \mu C)}{V}

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = \frac{(172+220)\ \mu C)}{V}

or

C' = \frac{(392)\ \mu C)}{V}

now the dielectric constant (k) is given as:

k=\frac{C'}{C}

substituting the values, we get

k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}

or

k = 2.279

6 0
3 years ago
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