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4 years ago

The relationship between mass and inertia is described by newton's second law of motion. true or false

1 answer:

4 0

False. Inertia and mass is not described in Newton’s second law of motion but in Newton’s first law of motion. Newton’s first law of motion or sometimes referred to as the law of inertia. In Newton’s first law indicates that an object at rest will remain at rest unless acted by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

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If a car was moving at a speed of 25mph for 6 hours, how far would the car travel?

Anestetic [448]

Answer:

150 miles

Explanation:

25mph x 6hrs = 150miles

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3 years ago

What is the answer to this?

jeka57 [31]

Answer:

I think it is the forth one

7 0

3 years ago

Read 2 more answers

Although it may seem like it, a wave’s frequency and its speed are not the same. Picture yourself standing on the side of a road

Verizon [17]

Answer:

A) the space inbetween the waves

B) frequency

C) the frequency gets higher

Explanation:

5 0

3 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

avanturin [10]

Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

${h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}$ .

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 $\frac{kJ}{kgK}$ . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that ${T_{w}}^{out}-{T_{w}}^{in}\leq 10$ , so let's work with the limit case, which is ${T_{w}}^{out}-{T_{w}}^{in}=10$ to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

$m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}$

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5 0

3 years ago

If total (PE + KE) is conserved:

lorasvet [3.4K]

Yes, According to law of conservation of energy the total energy of any system remains conserved (same).

Example.

If a body is placed at some height it possesses some potential energy.

As P.E =mgh

When this body is starting moving downwards its height becomes decreases so P.E decreases but at the same time it is moving I.e having some velocity. K.E =1/2(m)(v^2).

Hence here P.E decreases but K.E increases at the same time. So total energy is conserved.

6 0

4 years ago