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4 years ago

What is an Atwood Machine?

1 answer:

6 0

The best answer is letter (A) a double pulley system. Atwood Machine is normally used as a measurement in balancing to object to verify the mechanical law of motion with constant acceleration.

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Suppose the total momentum of two masses before a collision is 100 kg m/s. What is the total momentum of the two masses after th

soldier1979 [14.2K]

No, momentum is conserved so:
momentum before=momentum after
it is C. 100 kg m/s

4 0

3 years ago

However asked this gets one hundreed points

ollegr [7]

Answer:

Omg tysm you are too kind

Explanation:

8 0

3 years ago

Read 2 more answers

Zina [86]

Answer:

The current through the wire is equal to 0.8 A.

Explanation:

Given that,

The length of a copper wire = 2 m

Potential difference = 24 mV

The current through the wire is 0.40 A.

The new potential difference is 48 mV.

We need to find the current through the wire.

As the potential difference is doubled for second wire. So the new current will be :

I' = 2I

= 2 × 0.4

= 0.80 A

So, the current through the wire is equal to 0.8 A.

5 0

3 years ago

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$