Answer:
r₂/r₁ = 1.82
Explanation:
The electric field due to a point charge, has the following expression:
For a distance r₁, the magnitude of the electric field is 395 N/C, so we can solve for r₁², as follows:
r₁² = (1)
For a distance r2, the magnitude of the electric field is 119 N/C, so we can solve for r₂², as follows:
r₂² =
We can find the quotient r₂/r₁, from (1) and (2):
r₂/r₁ = = 1.82
Answer:
write something like after the spacecraft launched all of the potential energy transformed into kinetic energy causing the spacecraft to go at an abnormal spped.
Explanation:
Answer:
- d = 2; m = 2; m = 2
- d= 1; m = 1; m = 2
- d = 1; m = 1; m = 1
- d = 2; m = 1; m = 2
- d = 2; m = 1; m = 1
Explanation:
The following diagrams show five pairs of asteroids, labeled with their relative masses (M) and distances (d) between them. For example, an asteroid with M=2 has twice the mass of one with M=1 and a distance of d=2 is twice as large as a distance of d=1. Rank each pair from left to right based on the strength of the gravitational force attracting the asteroids to each other, from strongest to weakest.
- d = 2; m = 2; m = 2
- d= 1; m = 1; m = 2
- d = 1; m = 1; m = 1
- d = 2; m = 1; m = 2
- d = 2; m = 1; m = 1
Answer:
Req = 50 Ω
Explanation:
The equivalent resistance is basically the sum of all the resistances in a circuit.
The sum of these resistances will depend whether these resistance are in series or parallel.
If the resistances are on series, the expression to use is:
R = R₁ + R₂ + R₃ + .......Rₙ (1)
If the resistances are on parallel then the expression to use is:
1/R = 1/R₁ + 1/R₂ + ........1/Rₙ (2)
Now, according to the picture, we have R₁ and R₄ in series, so here we have to use (1):
R₁₄ = 10 + 30 = 40 Ω
R₂ and R₃ are in parallel so we use (2):
1/R₂₃ = 1/20 + 1/20 = 2/20 = 1/10
R₂₃ = 10 Ω
Finally, R₁₄ and R₂₃ are in series (Because of the sum of the resistance in each side, they are now forming one resistance in each side), therefore, we use (1) again to get the equivalent resistance of the whole circuit:
Req = 10 + 40
<h2>
Req = 50 Ω</h2>
Hope this helps
Answer:
Lastly, the heat transfer rate depends on the material properties described by the coefficient of thermal conductivity. All four factors are included in a simple equation that was deduced from and is confirmed by experiments. The rate of conductive heat transfer through a slab of material, such as the one in Figure 3, is given by
Q
t
=
k
A
(
T
2
−
T
1
)
d
,
where
Q
t
is the rate of heat transfer in watts or kilocalories per second, k is the thermal conductivity of the material, A and d are its surface area and thickness, as shown in Figure 3, and (T2 − T1) is the temperature difference across the slab. Table 1 gives representative values of thermal conductivity.
Explanation: