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3 years ago

When does a falling object reach terminal velocity?

1 answer:

5 0

At some speed, the drag or force of resistance will equal the gravitational pull on the object. At this point the object ceases to accelerate and continues falling at a constant speed called the terminal velocity (also called settling velocity).

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Which part of a river would have animals with very muscular bodies and adaptations that let survive in turbulent water? Source z

lidiya [134]

The part of a river that would have animals with muscular bodies and adaptations that let survive in turbulent water is in the transition zone, the mid-transition zone to be precise. Water at the source zone possesses a lot of potential energy and as it flows from the upper reaches the potential energy is turned into kinetic energy when the course of the river begins to gradually level out and this translates into increase in velocity. By the time river water reaches the middle of the transition zone, most of the potential energy would have been turned into kinetic energy and thus water velocity would be quite high here. Animals living here would develop muscles because of constantly fighting against the strong current to avoid being swept downstream.

8 0

3 years ago

Read 2 more answers

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

2 years ago

Which instrument is launched into the atmosphere to collect pressure, temperature, humidity, wind speed, and other data?

koban [17]

C. Radiosonde is the answer

the above mentioned is not correct

5 0

3 years ago

The phases of the moon depend on how much of the lighted side of the moon can be seen from Earth.

Ostrovityanka [42]

Answer:

true

Explanation:

if there is no light it's different from when there is

3 0

3 years ago

Ex 10: My dog runs at 6 m/s for 18 meters. How long did she run for?

Hunter-Best [27]

She ran for 3s

Put 18/6 because in order to find how long she ran for you need to divide the distance by the meters ran, once you do that you will get 3.

7 0

2 years ago