Explanation:
Hope this helps,
Juno entered a polar orbit of Jupiter on July 5th 2016 UTC, to begin a scientific investigation of the planet. After completing its mission, Juno will be intentionally deorbited into Jupiters atmosphere. Junos mission is to measure Jupiters composition, gravitational field, magnetic field, and polar magnetosphere.
The equation for work (W) done by an electric field is:
W = qΔV
where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:
10 = 2ΔV
ΔV = 5
Answer:
1,373.4 N
Explanation:
The mass of the table acts at the centre in addition to the books since that is the centre of gravity of the table.
Mass of books will be 10kg+20kg+30kg=60 kg
Total mass of table and books will be 500kg+60kg=560 kg
This mass is evenly distributed into the four legs hence 560kg/4 legs=140 kg per leg
Force is product of mass and acceleration due to gravity hence F=gm
Taking g as 9.81 m/s2 then
F=140*9.81=1,373.4 N
Therefore, rhe normal force is equivalent to 1,373.4 N
Answer:
the mass of water is 0.3 Kg
Explanation:
since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:
Q water + Q copper = Q surroundings =0 (insulated)
Q water = - Q copper
since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature
and denoting w as water and co as copper :
m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)
m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]
We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C
if we assume that both specific heats do not change during the process (or the change is insignificant)
m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]
m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))
m w= 0.3 kg
Answer:
If efficiency is .22 then W = .22 * Q where Q is the heat input
Heat Input Q = 2510 / .22 = 11,400 J
Heat rejected = 11.400 - 2510 = 8900 J of heat wasted
Also, 8900 J / (4.19 J / cal) = 2120 cal
