Answer:
They're going to increase the total resistance as 
Explanation:
If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then

where

but
for 
so

where

Answer:
The new period will be √6 *T
Explanation:
period ,T=2π√(L/g) ................equation 1
where T is the period on earth
gravitational acceleration on the moon is g/6
T1 = 2π√[L/(g/6)]
T1=2π√(6L/g) ...............equation 2
divide equation 2 by 1
T1/T =2π√(6L/g)÷2π√(L/g)
T1/T =√(6L/L)
T1/T =√6
T1 = √6 *T
Answer:
Explanation:
a ) After the attainment of terminal speed , object takes 4.5 s to cover a distance of 2 m
So terminal speed V = 2 / 4.5
= .444 m /s
When it attains terminal speed , acceleration becomes zero
0 = g - B x .444
B = 22.25 s⁻¹
b ) At t = 0 , v = 0
a = g - B v
a = g at t = 0
c ) When v = .15
a = g - 22.25 x .15
= 9.8 - 3.31
= 6.5 m /s²
Tectonic plates is the unscrambled phrase in that jumble.
Answer:
Explanation:
Given that,
Two resistor has resistance in the ratio 2:3
Then,
R1 : R2 = 2:3
R1 / R2 =⅔
3 •R1 = 2• R2
Let R2 = R
Then,
R1 = ⅔R2 = 2/3 R
So, if the resistor are connected in series
Let know the current that will flow in the circuit
Series connection will have a equivalent resistance of
Req = R1 + R2
Req = R + ⅔ R = 5/3 R
Req = 5R / 3
Let a voltage V be connect across then, the current that flows can be calculated using ohms law
V = iR
I = V/Req
I = V / (5R /3)
I = 3V / 5R
This the current that flows in the two resistors since the same current flows in series connection
Now, using ohms law again to calculated voltage in each resistor
V= iR
For R1 = ⅔R
V1 =i•R1
V1 = 3V / 5R × 2R / 3
V1 = 3V × 2R / 5R × 3
V1 = 2V / 5
For R2 = R
V2 = i•R2
V2 = 3V / 5R × R
V2 = 3V × R / 5R
V2 = 3V / 5
Then,
Ratio of voltage 1 to voltage 2
V1 : V2 = V1 / V2 = 2V / 5 ÷ 3V / 5
V1 : V2 = 2V / 5 × 5 / 3V.
V1 : V2 =2 / 3
V1:V2 = 2:3
The ratio of their voltages is also 2:3