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musickatia [10]
3 years ago
6

Which bright object is in shadow

Physics
1 answer:
IRINA_888 [86]3 years ago
5 0
My best guess would be sun because it is bright but is surrounded by shadows on all sides.
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An apple with a mass of 0.95 kilograms hangs from a tree branch 3.0 meters above the ground. If it falls to the ground, what is
lubasha [3.4K]
Ke = pe
pe = mgh
= 0.95 x 9.8 x 3
= 27.93 J
6 0
3 years ago
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Name two different types of precipitation?
Norma-Jean [14]

Answer:

2 different types of precipitation is rain and snow.

Explanation:

The clouds will form... and the droplets that could be coming out is rain and snow.

4 0
3 years ago
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A conductor of radius r, length l and resistivity p has resistance R. It is melted down and formed into a new conductor, also cy
Vaselesa [24]

Answer:

b) R/4 (There seems to an error in mentioning the multiple choices of this question, please see below explanation of correct calculations for this  question.)

Explanation:

dimension of the conductor before melting is l, r

reistivity is p

R=(p*l)/(pie*r2)

after reforming length is reduced to L=l/4

volume in both the cases will be same

i.e. pie * r^2 * l =pie * R^2 * L

r^2 * l = R^2 * (1/2)l

due to this radius will become R=sqrt(2) * r

now new reistance is given by Rx=(p * L)/(pie * R^2)

i.e. Rx=(p * l/2)/(pie * r^2 * 2)

after simplification RX=((p * l)/(pie * r^2))/4

i.e. Rx=R/4

5 0
3 years ago
Read 2 more answers
Find an expression for the electric field E⃗ at the center of the semicircle. Hint: A small piece of arc length Δs spans a small
Hatshy [7]

Answer:

Electric Field a the centreE=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)

Explanation:

<u>Given:</u>

Total charge on the semicircle =Q

Radius of the semicircle=R

Let consider a elemental charge on the semicircle at an angle \theta\\ with the horizontal. Since the The charge distribution is symmetrical about y axis so the there will symmetrical charge on other side. The horizontal component will cancel out each other and vertical component will get added so let dE be the electric Field due to elemental charge

Let \lambda be the charge per unit length such that\lambda=\dfrac{Q}{\pi R}

k=\dfrac{1}{4\pi \epsilon_0}

Total Electric Field at the centre

=2dE\sin\theta\\=2\dfrac{k\lambda }{R}\int \sin \theta d\theta\\

integrating 0 to \dfrac{\pi}{2}

E=\dfrac{2k\lambda}{R}(-\vec j)

E=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)

So the Electric field at the centre is calculated.

7 0
3 years ago
Ordinary glasses are worn in front of the eye and usually 2.00 cm in front of the eyeball. A certain person can see distant obje
Lelu [443]

Answer:

The focal length is 15.549 cm

The power of the lens is 0.0643 D

Solution:

As per the question:

The near point is 50.0 cm

Distance of the glasses from the eyeball, d = 2.00 cm

The near point of a normal human eye is 25 cm

Now,

The image distance, v' = 50.0 - 2.00 = 48.0 cm

The object distance, u' = 25.0 - 2.00 = 23.0 cm

Now, using the Lens maker formula to calculate the focal length:

\frac{1}{f} = \frac{1}{u'} + \frac{1}{v'}

\frac{1}{f} = \frac{1}{48.0} + \frac{1}{23.0} = 0.0643

f = 15.549 cm

Now, the power of the lens in diopters is given by:

P = \frac{1}{f} = \frac{1}{15.549} = 0.0643 D

7 0
3 years ago
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