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stich3 [128]
3 years ago
13

Technician a says that the water pump is a centripetal pump. technician b says that centripetal force is the outward force that

is exerted on a rotating object. who is correct?
Physics
1 answer:
Vaselesa [24]3 years ago
3 0
<span>Technician A is correct about the water pump being a centripetal pump and the definition of the centripetal force given by technician B is wrong. The force that happens on a body moving in a circular path and move towards the centre to which the body is moving is called as centripetal force. So it’s not an outward force.</span>
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In general, how did the water pressure in the tank change when mass was added to the fluid?
MissTica

Answer:

As the height increases the pressure must increase.

Explanation:

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2 years ago
Why does magma rise from depth to the surface of the earth?
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Read 2 more answers
A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do t
Marrrta [24]

Answer:

a) U = 652.545\,kJ, b) v \approx 659.568\,\frac{m}{s}

Explanation:

a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.

Q = U

The internal energy for a monoatomic ideal gas is:

U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T

Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:

T = \frac{P\cdot V}{n\cdot R_{u}}

T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}

T = 52.325\,K

The thermal energy contained by the gas is:

U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)

U = 652.545\,kJ

b) The physical model for the cat is constructed from Work-Energy Theorem:

U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:

v = \sqrt{\frac{2 \cdot U}{m_{cat}}}

v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }

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