a). Perihelion . . . the point in Earth's orbit that's closest to the Sun.
We pass it every year early in January.
b). Aphelion . . . the point in Earth's orbit that's farthest from the Sun.
We pass it every year early in July.
c). Proxihelion . . . a made-up, meaningless word
d). Equinox . . . the points on the map of the stars where the Sun
appears to be on March 21 and September 21.
Answer:
38,437.5
Explanation:
Density(d)= 102.5g/ml
Volume (v)=375ml
Mass(m) = ?
D =m/v
102.5= m/375
102.5*375=m
38,437.5=m
therefore Mass = 38,437.5g/ml.
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
The time component is needed. The acceleration is the change of velocity divided by the time in when this change of velocity happens.
Answer:
F₄ = 29.819 N
Explanation:
Given
F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N
F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N
F₃ = (0 i + 0 j + 4 k) N
Then we have
F₁ + F₂ + F₃ + F₄ = 0
⇒ F₄ = - (F₁ + F₂ + F₃)
⇒ F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N
The magnitude of the force will be
F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N
