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3 years ago

13

Water enters a 2 m3 tank at a rate of 6 kg/s and is withdrawn at a rate of 2 kg/s. The tank is initially half full. What type of

process is this?

1 answer:

NemiM [27]3 years ago

8 0

Answer:

This process is semicontinuous.

Explanation:

Given that,

Volume = 2 m³

Enter flow rate = 6 kg/s

Exit flow rate = 2 kg/s

The tank is initially half full.

We need to find what type of process

Using given data,

This process is not continuous because given enter and exit flow rate is not equal.

This process is semicontinuous and the water level in the tank does not reach a constant level.

Hence, This process is semicontinuous.

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Which of the following statements correctly described natural electric fields?

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Thunder is evidence of electric fields in the atmosphere.

The charge difference between the clouds and earth can be described as parallel conducting plates that produce electric fields in the atmosphere.

could be true

5 0

3 years ago

A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

12345 [234]

Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

$V = \frac{m_r}{\rho}$

substituting values

$V = \frac{5000}{4.3}$

$V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

$V_w = \frac{V}{2}$

substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

=> $m_w = 581.4 \ g$

$m_w = 0.5814 \ kg$

The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

The actual weight of the rock is

$W_r = m_r * g$

$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

The tension on the string is

$T = W_r - W_w$

substituting values

$T = 49.0 - 5.698$

$T = 43.302 \ N$

4 0

3 years ago

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n200080 [17]

Answer:

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Explanation:

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3 years ago

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3 years ago

When a driver presses the brake pedal, his car can stop with an acceleration of -5.4m/s^2. How far will the car travel while com

Dahasolnce [82]

Information that is given:
a = -5.4m/s^2
v0 = 25 m/s
---------------------
S = ?
Calculate the S(distance car traveled) with the formula for velocity of decelerated motion:
v^2 = v0^2 - 2aS
The velocity at the end of the motion equals zero (0) because the car stops, so v=0.
0 = v0^2 - 2aS
v0^2 = 2aS
S = v0^2/2a
S = (25 m/s)^2/(2×5.4 m/s^2)
S = (25 m/s)^2/(10.8 m/s^2)
S = (625 m^2/s^2)/(10.8 m/s^2)
S = 57.87 m

3 0

3 years ago