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3 years ago

How many chirps will a snowy tree cricket give is 21 seconds at a temperature of 22 degree celcius

1 answer:

Studentka2010 [4]3 years ago

7 0

There is an established correlation between the number of chirps of a cricket and temperature. The function is
ln C = -6713.2 / T + 25.868

Where c is the number of chirps per min and T is the temperature in K
We are given
T = 22 C + 273 = 295K

Solving for C
ln C = -6713.2 / 295 + 25.868
C = 22.45 chirps/min

In 21 seconds, the number of chips is
22.45 chirps / min (1 min/60 s) (21 s) = 7.86 chirps <span />

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I place an ice cube with a mass of 0.223 kg and a temperature of −35◦C is placed into an insulated aluminum container with a mas

Nadya [2.5K]

To solve this problem it is necessary to use the calorimetry principle. From the statement it asks about the remaining ice, that is, to the point where the final temperature is 0 ° C.

We will calculate the melted ice and in the end we will subtract the total initial mass to find out how much mass was left.

The amount of heat transferred is defined by

$Q = mc\Delta T$

Where,

m = mass

c = Specific heat

$\Delta T =$ Change in temperature

There are two states, the first is that of heat absorbed by that mass 'm' of melted ice and the second is that of heat absorbed by heat from -35 ° C until 0 ° C is reached.

Performing energy balance then we will have to

$Q_i-E_h = Q_m$

Where,

$Q_ i$ = Heat absorbed by whole ice

$Q_m$ = Heat absorbed by mass

$E_h$ = Heat energy by latent heat fusion/melting

$m_i*c_i \Delta T +m*L_f = (m_wc_w+m_{al}c_{al})\Delta T$

Replacing with our values we have that

$0.223*2108(-(-35))+m*3.34*10^5 = (0.452*4186+0.553*902)(27-0)$

$16452.9+334000m = (1892.072+498.806)*27$

Rearrange and find m,

$m = 0.144Kg$

Therefore the Ice left would be

$m' = 0.223-0.144$

$m' = 0.079Kg$

Therefore there is 0.079kg ice in the containter when it reaches equilibrium

8 0

3 years ago

Your teacher had not exercised since the beginning of distance learning. She decided to exercise by walking in the park. The par

Lesechka [4]

A) 6000 m

B) 0 m

C) 3.33 m/s

D) 0 m/s

Explanation:

In physics, distance is a scalar quantity that represents the total length of the path covered by an object during its motion.

Being a scalar quantity, it has no direction, but only a magnitude (so, a number + its units).

The distance covered by a body can be calculated by adding together all the lengths of all the motion of the body.

In this problem, the teacher walks in the partk, starting at position A, then walking around the park 5 times and returning to the original position.

The park has a rectangular shape 200 m x 400 m, so the perimeter of the park is:

$p=200+400+200+400=1200 m$

The teacher covers the perimeter 5 times, so the total distance she covered is:

$d=5p=5\cdot 1200 =6000 m$

Displacement, instead, is a vector quantity that connects the initial position to the final position of the motion of an object.

Being a vector, it has both a magnitude and a direction, where:

- The magnitude is equal to the length of the vector, so it corresponds to the shortest distance in a straight line between the initial and final position

- The direction goes from the initial position to the final position

In this problem:

- The teacher starts her motion at position A

- The teacher ends her motion at position A

So, the distance between the initial and final position is zero:

$d=x_A -x_A = 0$