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il63 [147K]
3 years ago
12

Which of these things poses the greatest hazard to communications satellites? a) photons from the Sun b) solar magnetic fields c

) particles from the Sun.
Physics
2 answers:
Sphinxa [80]3 years ago
7 0

Answer:

Your answer is B because the solar magnetic fields mess with the waves of energy and damages the connection with the satellites and what's their sending information to...

Explanation:

leva [86]3 years ago
3 0

Answer:

a

Explanation:

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What do we call the energy that is transferred to
mojhsa [17]

Answer:

Latent heat

Latent heat is the heat needed to change a state without a change in temperature.

Explanation:

Hope this helps :D

5 0
3 years ago
Krish has a sample of matter. It is clear and smells sour. The sample is also thick but it flows when poured.
Tamiku [17]

Answer:

It is a liquid because it flows.

Explanation:

because liquids dont have to be a certain color they are liquid as long as they flow

4 0
3 years ago
Read 2 more answers
what happened to the water molecules if a sealed jar of frozen water is left outside for twelve hours?
sergij07 [2.7K]

Answer:

the molecules get close together and the water expands

Explanation:

4 0
3 years ago
B. Jerome plays middle linebacker for South's varsity football team. In a game against
weqwewe [10]

Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

4 0
4 years ago
A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amoun
quester [9]

Answer:

0.4757 mm

Explanation:

Given that:

Load P = 223,000 N

the length of the height of the aluminium column = 1.22 m

the diameter of the aluminum column = 10.2 cm = 0.102 m

The amount that the column has shrunk ΔL can be determined by using the formula:

\Delta L = \dfrac{PL}{AE_{Al}}

where;

A = πr²

2r = D

r = D/2

r = 0.102/2

r = 0.051

A = π(0.051)²

A = 0.00817

Also; the young modulus of aluminium E_{Al} is:

E_{Al}= 7*10^{10} \Nm^{-2}

\Delta L = \dfrac{PL}{AE_{Al}}

\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}

ΔL = 4.757 × 10⁻⁴ m

ΔL =  0.4757 mm

Hence; the amount that the column has shrunk is 0.4757 mm

5 0
3 years ago
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