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3 years ago

Is it possible for the electric field between two positive charges to equal zero along the line joining the two charges?

Physics

1 answer:

MrRa [10]3 years ago

6 0

Answer:

Yes electric field between the line joining may be zero when both the charges are of opposite magnitude

Explanation:

Electric field due to a point charge is given by $E=\frac{1}{4\pi \epsilon _0}\frac{Q}{r^2}$

From the relation of magnitude of electric field we can see that magnitude of electric field depends upon the m,magnitude of charge and distance between them

So electric field between the line joining may be zero when both charges are of opposite magnitude ( Means if one charge is positive then other is negative )

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A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.

elena-14-01-66 [18.8K]

Answer:

A) Must be done 19806.62 joules of work.

B) The average power is 1320.44 Watts.

Explanation:

A) First, we're going to use the work-energy theorem that states total work ( $W$ ) done on an object is equal to the change in its kinetic energy ( $\Delta K$ ):

$W=\Delta K = K_{f}-K_{i}$ (1)

So, all we must do is to find the change on kinetic energy. Because we're working with rotational body, we should use the equation $K=\frac{I\omega^{2}}{2}$ for the kinetic energy so:

$\Delta K=\frac{I(\omega_{f})^{2}}{2}-\frac{I(\omega_{i})^{2}}{2}$ (2)

with $\omega_{i}$ the initial angular velocity, $\omega_{f}$ the final angular velocity (is zero because the wheel stops) and I the moment of inertia that for a thin hoop is $I=MR^{2}$ , using those on (2)

$\Delta K=0-\frac{MR^{2}(\omega_{i})^{2}}{2}$ (3)

By (3) on (1):

$W= \frac{MR^{2}(\omega_{i})^{2}}{2} = \frac{(32.0)(1.2)^{2}(29.32)^{2}}{2}$

$W=19806.62\,J$

B) Average power is work done divided by the time interval:

$P=\frac{W}{\Delta t}=\frac{19806.62}{15.0}$

$P=1320.44\,W$

NOTE: We use the relation $1rpm*\frac{2\pi}{60s}=\frac{rad}{s}$ to convert 280 rev/min(rpm) to 29.32 rad/s

4 0

4 years ago

A positive statement is:________. a. reflects oneâs opinions. b. can be shown to be correct or incorrect. c. a value judgment. d

kenny6666 [7]

Answer:

Explanation:

8 0

3 years ago

A hunter fires a rifle while seated with the butt of the gun resting against their shoulder. If the forward momentum is the same

Debora [2.8K]

Answer:

Because the bullet is smaller and has a sharp/pointy tip.

Explanation:

The bullet kinda puts the majority of kinetic energy in a small are, whereas the gun butt kinda spreads the energy evenly amongst a larger surface. If more explanation is needed add comment.

6 0

3 years ago

(1 point) A rectangular tank that is 3 feet long, 9 feet wide and 12 feet deep is filled with a heavy liquid that weighs 110 pou

Aloiza [94]

Answer:

Explanation:

Work in pumping water from the tank is given as

W = ∫ y dF. From a to b

Where dF is the differential weight of the thin layer of liquid in the tank, y is the height of the differential layer

a is the lower limit of the height

b is the upper limit of the height.

We know that, .

F = ρVg

Where F is the weight

ρ is the density of water

V is the volume of water in tank

g is the acceleration due to gravity

Then,

dF = ρg ( Ady)

We know that the density and the acceleration due to gravity is constant, also the base area of the tank is constant, only the height that changes.

Then,

ρg = 62.4 lbs/ft³

Area = L×B = 3 × 9 = 27ft²

dF = ρg ( Ady)

dF = 1684.8dy

The height reduces from 12ft to 0ft

Then,

W = ∫ y dF. From a to b

W = ∫ 1684.8y dy From 0 to 12

W = 1684.8y²/2 from 0 to 12

W = 842.4 [y²] from y = 0 to y = 12

W = 842.4 (12²-0²)

W = 121,305.6 lb-ft

3 0

3 years ago

The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia (also known as rotationa

agasfer [191]

Answer

given,

expression of Kinetic energy of rotating body

$K = \dfrac{1}{2}I\omega^2$

ω = 34.0 rad/s

Assuming mass of the particle equal to 13 Kg

and perpendicular distance from the particle to the axis is r = 1.25 m

now,

moment of inertia of particle = ?

from the given expression

$I= \dfrac{2K}{\omega^2}$ ..............(1)

we know

$K = \dfrac{1}{2}mv^2$

v = r ω

$K = \dfrac{1}{2}mr^2\omega^2$

putting value in equation (1)

$I= \dfrac{2\dfrac{1}{2}mr^2\omega^2}{\omega^2}$

$I =mr^2$

$I =13 \times 1.25^2$

I = 20.3125 kg.m²

4 0

3 years ago