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Temka [501]
3 years ago
10

Outside the Solar System is there any gravitational pull from the Sun?

Physics
2 answers:
babunello [35]3 years ago
6 0

Answer:

our sun is one of the least 100 billion stars in the milky way a spiral galaxy about 100000 light years across

seraphim [82]3 years ago
5 0
Yes. Gravity from the sun reaches throughout the solar system and beyond.
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a driver is going at 120km h and sees a barrier 60.0 m ahead it takes 5secounds to apply the brakes and decelerates at 12m does
Soloha48 [4]

Answer:

yes, if you're going at 120 km and you saw the wall that late then it wouldn't me possible to decrease 12 meters in 5 seconds and not hit the wall that's only 60 meters away

Explanation:

5 0
3 years ago
A tree falls and rots on the forest floor the next year new trees begin to grow in that spot what is that an example of
goldfiish [28.3K]
The cycle of life because the tree starts to grow back in that spot.
6 0
4 years ago
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What type of relationship exists between the length of a wire and the resistance, if all other factors remain the same?
jek_recluse [69]
Resistance of the wire can be calculated using the following rule:
R = (ρ*L) / A 
where:
R is the resistance
ρ is the resistivity of the material of the wire
L is the length of the wire
A is the area of the wire

From the above relation, we can deduce that:
The resistance of the wire is directly proportional to its length assuming all other factors are constant
7 0
3 years ago
HELP!!!
Tju [1.3M]

Answer:

A species is a group of organisms with the same GENETIC HERITAGE

Organisms from the same species can REPRODUCE with each other

8 0
3 years ago
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How many heat is needed to melt 1.5kg of ice and then to raise the temperature of the resulting water 50 degree Celsius
BabaBlast [244]

Answer:

• The amount of heat required to melt ice and raise the temperature of water T^oCToC is

Q=mL_f+mc\Delta TQ=mLf+mcΔT

Here m=1.5 kgm=1.5kg

L_f=3.33*10^5 J/kgLf=3.33∗105J/kg is Latent heat of fusion of ice.

c=4186 J/kg.^oCc=4186J/kg.oC is heat capacity of water.

\Delta T=50 ^oC-0^oC=50^oCΔT=50oC−0oC=50oC

So,

Q=(1.50)[3.33*10^5 +(4186)(50)]=8.13*10^5 JQ=(1.50)[3.33∗105+(4186)(50)]=8.13∗105J

4 0
3 years ago
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