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Tamiku [17]
4 years ago
11

Make a list of possible consequenes of solar flares to the following members of your community

Physics
1 answer:
SashulF [63]4 years ago
4 0
The Majestic Aurora


Nature's Charming Rainbows 

Sunrise, Sunset, and Other Atmospheric Phenomena

Global Warming

Nature



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Which is a concern about mining for uranium? Heated water could be released into the environment. Dust released in the air could
saw5 [17]

Answer: Dust released in the air could be radioactive.

Explanation:

Uranium is a radioactive element. It decays naturally to attain stability. While mining, the dust displaces into the air causing not only harm to environment but for the workers as well. The released in the air could be radioactive which would be inhaled posing threat for lung cancer.

Thus, the correct answer is dust released in the air could be radioactive is a concern about mining Uranium.

3 0
3 years ago
Read 2 more answers
Why gravitational force at the center og earth is zero
KengaRu [80]
F = G Mm/r²
mg = G Mm/r²
g = GM/r²
At centre of earth, r=0
g = GM/0
g =0
8 0
3 years ago
Read 2 more answers
HELPPPPP !!!!
Nataly [62]

Your answer is C)

a)t=2.78 sec

b)R=835.03 m

c)

Explanation:

Given that

h= 38 m

u=300 m/s

here given that

The finally y=0

So

t=2.78 sec

The horizontal distance,R

R= u x t

R=300 x 2.78

R=835.03 m

The vertical component of velocity before the strike

4 0
3 years ago
A charge +1.9 μC is placed at the center of the hollow spherical conductor with the inner radius 3.8 cm and outer radius 5.6 cm.
Archy [21]

To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.

PART A)

The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:

Q_{int}=-Q1=-1.9*10^{-6} C. This is the total charge on the inner surface of the conducting shell.

PART B)

The positive charge (of the same value) on the external surface of the conducting shell is:

Q_{ext}=+Q_1=1.9*10^{-6} C

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,

Q_{ext, Total}=Q_2+Q_{ext}

Q_{ext, Total}=1.9+3.8

Q_{ext, Total}=5.7 \mu C

5 0
4 years ago
A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 m . A
yaroslaw [1]

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

v_{f}^{2}=v_{0}^{2}+2gy\\v_{f}=\sqrt{0+2(9.8\frac{m}{s^{2}})(145m)}=53.31\frac{m}{s}

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

F=(3.8*10^{8})(1.602*10^{-19}C)(53.31\frac{m}{s})(0.205T)=6.65*10^{-10}N

B)

b.  south - north (by the rigth hand rule)

I hope this is usefull for you

regards

8 0
3 years ago
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