Answer:
reproduced in print form for non-profit educational use with. Reading ... 2. I had them in my backpack. 3. Help me look for them. 4. We don't want to get lost. 5. I'm so ... Commas set off introductory words, such as yes, no, and thank you. Rewrite ... Correct each run-on sentence below by writing it as a compound sentence. 1.
Explanation:
Explanation:
Given that,
Height of object = 4.0 cm
Distance of the object u= -30.0 cm
Focal length = 23 cm
We need to calculate the image distance
Using lens's formula
Where, u = object distance
v = image distance
f = focal length
Put the value into the formula
We need to calculate the height of the image
Using formula of height
Where, h' = height of image
h = height of object
The image is real and inverted.
(b). Now, object distance u = 13.0 cm
We need to calculate the image distance
Using lens's formula
Put the value into the formula
We need to calculate the height of the image
The image is virtual and erect.
Hence, This is required solution.
Explanation:
Hey there!!
Here, Given is,
Efficiency = 75%
VR = no. of pulleys = 5
Now,
100% ma = 75%×5
Therefore, MA is 3.75.
<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Multiply by the fraction (1 kg/1000 gm).
Answer:
E = 9.4 10⁶ N / C
, The field goes from the inner cylinder to the outside
Explanation:
The best way to work this problem is with Gauss's law
Ф = E. dA = qint / ε₀
We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.
The flow on the faces is zero, since the field goes in the radial direction of the cylinders.
The area of the cylinder is the length of the circle along the length of the cable
dA = 2π dr L
A = 2π r L
They indicate that the distance at which we must calculate the field is
r = 5 R₁
r = 5 1.3
r = 6.5 mm
The radius of the outer shell is
r₂ = 10 R₁
r₂ = 10 1.3
r₂ = 13 mm
r₂ > r
When comparing these two values we see that the field must be calculated between the two housings.
Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is
λ = q / L
Qint = λ L
Let's replace
E 2π r L = λ L /ε₀
E = 1 / 2piε₀ λ / r
Let's calculate
E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3
E = 9.4 10⁶ N / C
The field goes from the inner cylinder to the outside
