Answer:
1)31/3Ω
2)18/31A
3)4.06V
Explanation:
According to the diagram and 10 Ω resistors are parallel to each other, parallel resistor can be calculated using the formula below
1/R= 1/R1 + 1/R2
But we know R1= 5 Ω and R2= 10 Ω
1/R= 1/5 + 1/10
R= 10/3 Ω
=3.33 Ω
Then if we follow the given figure, 10/3 Ω and 7Ω are now in series then
Req = 10/3 +7
Req= 31/3 Ω
Therefore, equivalent resistance = 31/3 Ω
According to ohms law we know that V= IR
Then I= V/R
Where I= current
R= resistance
V= voltage
I= 6/(31/3)
I= 18/31A
We can now calculate the voltage accross the resistor which is
V=(18/31)× 7
V=4.06V
Therefore, the voltage accross the 7 ohm resistor is 4.06V
CHECK THE FIQURE AT THE ATTACHMENT
It can be observed without changing the identity of the object.
You use long division.
Find the largest number whose square is less than or equal to the number in the leftmost group (55 < 30 < 66). Take this number as the divisor and the quotient with the number in the leftmost group as the dividend (30). Divide and get the remainder (5 in this case).
Bring down the next pair 58. Add the divisor with the quotient and enter it with a blank on its right. Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend. In this case 105 × 5 = 525, so we choose the new digit as 5. Get the remainder.
Bring down the next pair 09. Add the divisor with the quotient and enter it with a blank on its right. Guess a largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied to the new quotient the product is less than or equal to the dividend. In this case 1103 × 3 = 3309, so we choose the new digit as 3. Get the remainder.
Put the decimal point.
End of division (Remainder is 0 and next digit after decimal is 0).
√305809 = 553
Gravity is affected by mass and distance
Show transcribed image text. Compare the work done by the electric field when the particle travels<span> from point W to point X to that </span>done<span> when the </span>particle travels<span> from point Z to point Y. ... Is the </span>work done<span> on the </span>particle<span> by the </span>electric field<span> positive, negative, or zero? Explain using force and displacement vectors.</span>
