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densk [106]
3 years ago
8

Please select the word from the list that best fits the definition

Physics
2 answers:
lianna [129]3 years ago
7 0

Answer: No list

Explanation:

vaieri [72.5K]3 years ago
5 0

Answer:

C. convection

Explanation:

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Work done by a gravitational force by lowering the bucket into the well is
katovenus [111]

Answer:

when we lower a bucket into a well to fetch water, the work done by gravity is positive since force and displacement are in the same direction.

Explanation:

3 0
3 years ago
9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.
SVETLANKA909090 [29]

Answer:

6370 J

Explanation:

By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

W = E_p = mgh

where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference

W = 50*9.8*13 = 6370 J

3 0
3 years ago
Why is the likelihood of amajor earthquake along the san andreas fault so worrisome??
Softa [21]
Because a lot of people live along the fault line and there would be lots of death and much property damage.
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}

Download pdf
5 0
3 years ago
Objects 1 and 2 attract each other with a electrostatic force of 72.0 units. If the charge of object 1 is doubled AND the charge
lbvjy [14]

Answer:

432 units

Explanation:

Let the charges be q and Q separated by a distance r. The electrostatic force , F = kqQ/r² = 72 units. If q = 2q and Q = 3Q, then the new electrostatic force is

F = k × 2q × 3Q/r² = 6kqQ/r² = 6 × 72 = 432 units

5 0
3 years ago
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