All categories

3 years ago

A given 60uc piont charge located at origion

2 answers:

7 0

For my first part it like that as we all know that flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2. NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc M I RIGHT PLZ tell me

Morgarella [4.7K]3 years ago

6 0

Answer:

flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2. NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc

Explanation:

You might be interested in

Which of the following is the best definition of energy?

Montano1993 [528]

Energy is the capacity to do some type of work

8 0

3 years ago

Could I please get some help l, will mark as Brianalist if right....

miss Akunina [59]

Answer:

Option 2

Explanation:

Weight of the box is being acted downwards due to gravity

However, based on Newton's third law (for every action (force) in nature there is an equal and opposite reaction), an equal force will act on the box by the table

8 0

2 years ago

Read 2 more answers

While at a construction site, you see a crane lift a 1000kg steel beam up to a height of 10m in a time period of 5.0 seconds. A

storchak [24]

Answer:

Car has more power output than crane

Explanation:

We have given that mass of the crane m = 1000 kg

Height through which crane lift the steel beam h = 10 m

Acceleration due to gravity $g=9.8m/sec^2$

So work done by crane $W=mgh=1000\times 9.8\times 10=98000j$

Time period is given as t = 5 sec

We know that power $P=\frac{W}{t}=\frac{98000}{5}=19600watt$

Now mass of the car = 1000 kg

Initial velocity u = 0 m /sec

Final velocity v = 10 m/sec

We know that work done is equal to the change in kinetic energy

So work done $=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

$=\frac{1}{2}\times 1000\times 10^2-\frac{1}{2}\times 1000\times 0^2=50000j$

Time ids given as t = 2 sec

So power $P=\frac{W}{t}=\frac{50000}{2}=25000watt$

So car has more power output than crane

8 0

3 years ago

It took Lilly 7 seconds to ride her bike 35 meters to the end of the block.What was her speed?

Zolol [24]

<h2><em>her average speed was 5 meter p/ second </em></h2><h2><em> 18 kph</em></h2><h2><em> HOPE IT HELPS (◕‿◕✿) </em></h2><h2><em> SMILE!!</em></h2>

4 0

3 years ago

Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic ener

Flauer [41]

<span>Answer: KE = (11/2)mĎ‰Â˛rÂ˛, particle B must have mass of 2m, while A has mass m. Then the moment of inertia of the system is I = ÎŁ mdÂ˛ = m*(3r)Â˛ + 2m*rÂ˛ = 11mrÂ˛ and then KE = Â˝IĎ‰Â˛ = Â˝ * 11mrÂ˛ * Ď‰Â˛ = 11mrÂ˛Ď‰Â˛ / 2 So I'll proceed under that assumption. For particle A, translational KEa = Â˝mvÂ˛ but v = Ď‰*d = Ď‰*3r, so KEa = Â˝m(3Ď‰r)Â˛ = (9/2)mĎ‰Â˛rÂ˛ For particld B, translational KEb = Â˝(2m)vÂ˛ but v = Ď‰*r, so KEb = Â˝(2m)Ď‰Â˛rÂ˛ so total translational KE = (9/2 + 2/2)mĎ‰Â˛rÂ˛ = 11mĎ‰Â˛rÂ˛ / 2 which is equal to our rotational KE.</span>

5 0

3 years ago