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pickupchik [31]

3 years ago

12

A swimming pool, 20.0 m ? 12.5 m, is filled with water to a depth of 3.71 m. if the initial temperature of the water is 18.5°c,

how much heat must be added to the water to raise its temperature to 30.1°c? assume that the density of water is 1.000 g/ml.

1 answer:

Semenov [28]3 years ago

7 0

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How does sound travel?

Doss [256]

Answer:

Sound vibrations travel in a wave pattern, and we call these vibrations sound waves. Sound waves move by vibrating objects and these objects vibrate other surrounding objects, carrying the sound along. ... Sound can move through the air, water, or solids, as long as there are particles to bounce off of.

Explanation:

8 0

3 years ago

Read 2 more answers

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 seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

So due to wind time will be more

4 0

3 years ago

the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg

lukranit [14]

Answer:

Mass of the steel cube = 7800 kg

Volume of the steel = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel = 7.8

Side of the cube = 12 cm

<u>(1)The mass of steel cube :</u>

We know that,

$Density = \frac{Mass}{Volume}$

We are given with density and sides of the cube

then volume of the cube

= $(side)^3$

= $10^3$

= 1000 cubic centimetre

Now

$7.8 = \frac{mass}{1000}$

$mass = 7.8 \times 1000$

mass = 7800 kg

<u>(2)volume of steel:</u>

Given the mass = 8 kg

$Density = \frac{Mass}{Volume}$

Substituting the values

$7.8 = \frac{8}{volume}$

$volume = \frac{8}{7.8}$

volume = 1.025 cubic centimetre

3 0

3 years ago

A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of

aev [14]

Answer:

initial magnetic field 1.306 T

Explanation:

We have given area of the conducting loop $A=0.065m^2$

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by $e=\frac{d\Phi }{dt}=-A\frac{db}{dt}$

$1.2=0.065\times \frac{db}{0.087}$

$db=1.606T$

So initial magnetic field = 1.606-0.3= 1.306 T

5 0

3 years ago