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
The answer is a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2, that is because during this kind of movement we have a = g, a acceleration and g acceler. of gravity, from zero of the axis to upward a = -g (a and g are contrary by direction), the can stopes before it falls, it means a = o, and during the free fall a = g (same direction).
Answer:
1a.5km
2a. 31536000 seconds
2b.2800000 centimeters
2c.45,000,000 Milligrams
2d.0.0141667 m/s
2e.2.592 x 10^10 km/day
2f .8.23x10^-7m
2g.0.0000085 m3
Explanation:
1a.(25km/5)(1000m/1km)(1h/3600s) =(5km)( 1) (1) =5km
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I thinks it’s b sorry if I am wrong
