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Karo-lina-s [1.5K]
2 years ago
12

If you are at latitude 43 degrees north of Earth's equator, what is the angular distance (in degrees) from your zenith to the no

rth celestial pole?
Physics
1 answer:
kirill115 [55]2 years ago
8 0

Answer:

Your zenith is 43 N of 90 deg (equator)

Thus, your zenith is 90 - 43 = 47 deg

(At the N pole your zenith would be 0 deg from the N pole)

You might be interested in
How to find new pressure of gas if bottle explodes
KIM [24]
If the container explodes there is no pressure, becuase all your gas has escaped its container, there for, you ain’t got no gas
6 0
3 years ago
In a fission experiment observed by Hahn and Strassman, uranium-235 was bombarded by neutrons. The products of this ission react
wlad13 [49]

Answer:

helium-4 (90%) or tritium (7%).

Explanation:

hope it helped u buddy

7 0
3 years ago
A rock with a mass of 0.3 kg falls from the top of a cliff. If it takes the rock 2.5 s to reach the ground, what was the impulse
kaheart [24]
Impulse=force x time
force=mass x acceleration due to gravity
force=
300 \times 10 = 3000
impulse =3000 x 2.5= ( sorry i don't have a calculator right now so you must calculate this yourself)
I converted from kg to g because it is the standard.
Hope this helps you.
4 0
3 years ago
a lead block drops its temperature by 5.90 degrees celsius when 427 J of heat are removed from it. what is the mass of the block
Airida [17]

Answer:

577g

Explanation:

Given parameters:

Temperature change = 5.9°C

Amount of heat lost = 427J

Unknown:

Mass of the block = ?

Solution:

The heat capacity of a body is the amount of heat required to change the temperature of that body by 1°C.

                H =  m c Ф

  H is the heat capacity

 m  is the mass of the block

  c is the specific heat capacity

   Ф is the temperature change

Specific heat capacity of lead is 0.126J/g°C

   m = H / m Ф

   m = \frac{427}{0.126 x 5.9}  = 577g

Mass of the lead block is 577g

7 0
2 years ago
A step index fiber has a numerical aperture of NA = 0.1. The refractive index of its cladding is 1.465. What is the largest core
Vladimir [108]

Answer:

diameter = 9.951 × 10^{-6} m

Explanation:

given data

NA = 0.1

refractive index = 1.465

wavelength = 1.3 μm

to find out

What is the largest core diameter for which the fiber remains single-mode

solution

we know that for single mode v number is

V ≤ 2.405

and v = \frac{2*\pi *r}{ wavelength} NA

here r is radius    

so we can say

\frac{2*\pi *r}{ wavelength} NA    = 2.405

put here value

\frac{2*\pi *r}{1.3*10^{-6}} 0.1    = 2.405

solve it we get r

r = 4.975979 × x^{-6} m

so diameter is = 2  ×  4.975979 × 10^{-6} m

diameter = 9.951 × 10^{-6} m

3 0
3 years ago
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