Answer:
<em>765,000Joules or 765kJ</em>
Explanation:
The Quantity of heat required is expressed as;
Q = (mcΔt)al + (mcΔt)water
m is the mass
c is specific heat capacity
Δt is the change in temperature
Q = (3(900)(90-5)) + (1.5(4200)(90-5))
Q = 2700*85 + 6300*85
Q = (2700+6300)85
Q = 9000*85
<em>Q = 765,000</em>
<em>Hence the amount of energy needed is 765,000Joules or 765kJ</em>
Answer:
0.707m
Explanation:
from formula of range i.e R=Usin2Q/g
Answer:
<u>Conventions used in SI to indicate units are as follows:</u>
- Only singular form of units are used. for example: use kg and not kgs.
- Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.
- Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.
- Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.
- While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.
- Units named after a scientist should be written in small letters. for example: newton, henry.
- Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°k
(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,
Ce = 0.093 cal/g. °c
Q = m C ΔT
Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )
Q= 4×0.093 × 160
Q = 59.52 cal
I hope I helped you^_^
Answer: 17.83 AU
Explanation:
According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>
(1)
Talking in general, this law states a relation between the <u>orbital period</u> 
of a body (moon, planet, satellite, comet) orbiting a greater body in space with the <u>size</u> 
of its orbit.
However, if is measured in <u>years</u>, and is measured in <u>astronomical units</u> (equivalent to the distance between the Sun and the Earth: 
), equation (1) becomes:
(2)
This means that now both sides of the equation are equal.
Knowing 
and isolating from (2):
![a=\sqrt[3]{T^{2}}=T^{2/3}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7BT%5E%7B2%7D%7D%3DT%5E%7B2%2F3%7D)
(3)
(4)
Finally:
(5)
