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Diano4ka-milaya [45]
3 years ago
13

List two applications of electrostatics.

Physics
1 answer:
Tems11 [23]3 years ago
4 0
Atmospheric electricity and storms,electric current in a vacuum,spark discharge,electrostatic control filters and industrial electrostatic  separation <- those are just a few
You might be interested in
If you add together all of forces exerted on an object and get a non zero value this is called the _____ force on the object
butalik [34]

Your answer would be the;  <u>NET</u>  force on the object.   Refer to Newton's Laws of Forces and Motion.







Hope that helps!!!!!!!!!!!!!!!!                                        : )

8 0
3 years ago
Read 2 more answers
A conical container of radius 6 ft and height 18 ft is filled to a height of 11 ft of a liquid weighing 64.4 lb divided by ft cu
prisoha [69]

Answer:

Hello there, please see explanations for step by step answer.

Explanation:

Radius 6 ft and

Height 18 ft is filled to a height of 11 ft of a liquid weighing 64.4 lb divided by ft cubedlb/ft3.

How much work will it take to pump the contents to the​ rim.

See attached documents for clear solvings and further step by step explanations

4 0
3 years ago
A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh
Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 rad/sec^2

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius r=\frac{d}{2}=\frac{1.63}{2}=0.815m

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec

(b) We know that linear speed is given by

v=r\omega =0.815\times 8.362=6.815m/sec

(c) We have given final angular velocity \omega _f=675rev/min

And \omega _i=79.9rev/min

Time t = 63 sec

Angular acceleration is given by \alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2

(d) Change in angle is given by

\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev

7 0
3 years ago
Question 3 of 15
omeli [17]

Answer:

B) Degrees

Explanation:

The directions of the vectors are often defined in terms of due East, due North, due West and due South. A direction exactly in between of North and East can be described as Northeast, similarly we can describe directions in terms of Northwest, Southeast and South west.

From these, the direction of a vector can be easily expressed in degrees, which is measured counter clockwise about its tail from due East. Considering that we can say that East is at 0° , North is at 90° , West is at 180 and South is at 270° counter clockwise rotation from due East.

So, we know that the direction of a vector lying somewhere between due East i.e 0° and due North i.e 90°, will be measured in degrees, which will have a value between 0°-90°

4 0
3 years ago
An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu
choli [55]

Answer:

Wnet, in, = 133.33J

Explanation:

Given that

Pump heat QH = 1000J

Warm temperature TH= 300K

Cold temperature TL= 260K

Since the heat pump is completely reversible, the combination of coefficient of performance expression is given as,

From first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power required to drive the the heat pump is given as

Wnet, in= QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

So the 133.33J was the amount heat that was originally work consumed in the transfer.

Extra....

According to the first law, the rate at which heat is removed from the low temperature reservoir is given as

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

5 0
3 years ago
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