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pychu [463]
2 years ago
14

Two identical resistors connected in series have an equivalent resistance of 4 ohms. The same two resistors, when connected in p

arallel, have an equivalent resistance of:
Physics
1 answer:
irina1246 [14]2 years ago
4 0

Answer:

1 ohm

Explanation:

since there are two identical resistors, one resistor will be

R = \frac{4}{2} =2ohm [ proven as in series R_{e} = 2 + 2 = 4ohm ]

to calculate the equivalent resistance when in parallel:

\frac{1}{R_{e} }  = \frac{1}{R_{1} } + \frac{1}{R_{2}}

so,

\frac{1}{R_{e} } = \frac{1}{2} + \frac{1}{2}

R_{e} = 1ohm

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Balance the following chemical equation:<br> H3PO4 + HCl → PC15 + H20
kipiarov [429]

Answer:

H3PO4 + 5HCl → PCl5 + 4H2O

Explanation:

3 0
3 years ago
If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i
kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from 15 cm to 21 cm will be , E_{1}=27J

The energy used for stretch the spring from 21 cm to 27 cm will be , E_{2}=45J

using the energy of spring formula ,we find that

27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))

45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))

Dividing both the equation will get,

\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm

Therefore, the natural length of the spring is, 9 cm.

4 0
3 years ago
[A] Write an expression for the equivalent resistance of three resistors connected in parallel.[ no derivation needed]
anzhelika [568]

Answer:

1) R1 + ((R2 × R3)/(R2 + R3))

2) 0.5 A

3) 3.6 V

Explanation:

1) We can see that resistors R2 and R3 are in parallel.

Formula for sum of parallel resistors; 1/Rt = 1/R2 + 1/R3

Making Rt the subject gives;

Rt = (R2 × R3)/(R2 + R3)

Now, Resistor R1 is in series with this sum of R2 and R3. Thus;

Total resistance of circuit = R1 + ((R2 × R3)/(R2 + R3))

2) R_total = R1 + ((R2 × R3)/(R2 + R3))

We are given;

R1 = 7.2 Ω

R2 = 8 Ω

R3 = 12 Ω

R_total = 7.2 + ((8 × 12)/(8 + 12))

R_total = 7.2 + 4.8

R_total = 12 Ω

Formula for current is;

I = V/R

I = 6/12

I = 0.5 A

3) since current through the circuit is 0.5 and R1 is 7.2 Ω.

Thus, potential difference through R1 is;

V = IR = 0.5 × 7.2 = 3.6 V

4 0
2 years ago
A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.6
Dafna1 [17]

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

Time t = 3.64s

To get the acceleration during this run, we will apply the equation of motion:

S = ut+1/2at²

Substitute the given parameters into the formula and calculate the value of a

305 = 0+1/2 a(3.64)²

304 = 1/2(13.2496)a

304 = 6.6248a

a = 304/6.6248

a = 45.89m/s²

Hence the average acceleration during this run is 45.89m/s²

4 0
3 years ago
If vector A ⃗  has components A x and A y and makes an angle θ with the +x axis, then
ziro4ka [17]
Then the tangent of angle-Θ is (Ay / Ax).
5 0
3 years ago
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