An AWG No.16 line cord with type HPN insulation has an amp airy of

Which of the following has zeros that are not significant?

guapka [62]

10.0 J

if it was 10.01 it would be significant, but its 10.0 which makes it a whole number

4 0

2 years ago

The circumference of a circle is 12π meters. find the circle diameter.

Veseljchak [2.6K]

The circumference of a circle is 2*pi*r=C so plugging in numbers we have
12pi= 2*pi*r so solving for r gives 12pi/2pi=r which gives 6=r so the radius of the circle is 6 meters since diameter is just 2*r this gives 2*r =12 meters so
Diameter = 12meters

7 0

3 years ago

Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (

alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b) $E_1=\dfrac{\sigma}{\epsilon_0}$

c) $E_2=\dfrac{\sigma}{\epsilon_0}$

Explanation:

Let $\sigma$ be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

$E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}$

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

$E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0$

b)The Electric Field to the right of the sheets

$E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

c)The Electric Field to the left of the sheets

$E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

3 0

3 years ago

If the ball reaches the ground in 8 seconds and the velocity of the ball before it hits the ground is 78.4. What is its accelera

Natasha2012 [34]

Answer:

9.8 m / s^2

Explanation:

Assuming free fall====> there is no initial downward/upward velocity

Assuming metric units 78.4<u> m/s </u>

vf = a t

78.4 = a (8) shows a = 9.8 m/s^2

6 0

2 years ago

3. A certain horizontal east-west lined wire has a mass of 0.2kg per meter of length and carries a current I. Impressed on the w

erica [24]

Answer:

i = 4.9 A

Explanation:

The expression for the magnetic force in a wire carrying a current is

F = i L x B

bold letters indicate vectors.

The direction of the cable is towards the East, the direction of the magnetic field is towards the North, so the vector product is in the vertical direction (z-axis) upwards and the weight of the cable is vertical downwards. Let's apply the equilibrium condition

F - W = 0

i L B = m g

They indicate the linear density of the cable λ = 0.2 kg / m

λ = m / L

m = λ L

we substitute

i B = λ g

i = $\frac{ \lambda \ g}{B}$

let's calculate

i = 0.2 9.8 / 0.4

i = 4.9 A

3 0

2 years ago