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timofeeve [1]
2 years ago
8

Suppose that an owner of the same dog breed has also taken some measurements. They notice that the surface area of the dog has i

ncreased by a factor of 3 over a period of four years. Determine the change in the dog's relative surface area in this time period.
Physics
1 answer:
Mnenie [13.5K]2 years ago
7 0

Answer:

Surface area of the dog is changes from A to 3A

Explanation:

It is given that surface area of dog is increased by factor 3 in a period of 4 year

We have to find the change in dog's relative surface area in the given time period

Let initially the surface area is A

As the surface area is increased by a factor of 3

So surface area after 4 year = 3×A = 3A

So surface area of the dog is changes from A to 3A

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34.0 N which is the force that applied to the person by rowboat.
3 0
2 years ago
Hydrogen peroxide is sold commercially as an aqueous solution in brown bottles to protect it from light. Calculate the longest w
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Explanation:

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2 years ago
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Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.
lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = (6\times0.1)kg/min=0.6kg/min

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = \frac{dx}{dt}

So, the rate of change of salt in the tank can be given by the following equation,

1000\frac{dx}{dt} =0.6-6x

or, \int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt

or, T = 693.147 min      (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0
2 years ago
Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave
lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, f_1:

f_n = n f_1

So we have:

- On wire A, the second-harmonic has frequency of f_2 = 660 Hz, so the fundamental frequency is:

f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz

- On wire B, the third-harmonic has frequency of f_3 = 660 Hz, so the fundamental frequency is

f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) f_1 = \frac{v}{2L}

For standing waves on a string, the fundamental frequency is given by the formula:

f_1 = \frac{v}{2L}

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

v=2Lf_1

We know that the two wires have same length L. For wire A, f_1 = 330 Hz, while for wave B, f_B = 220 Hz, so we can write the ratio between the speeds of the waves in the two wires:

\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}

So, the waves travel faster on wire A.

7 0
2 years ago
What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?
slavikrds [6]

I will post my work, but is that 99 degrees Celsius and 25 degrees Celsius?


All you have to do is plug in the initial temperature for gold where it says Tg and the initial temperature for the water where it says Tw and then plug that in and you will have your answer.

8 0
2 years ago
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