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san4es73 [151]
2 years ago
6

In this section of a circuit, a current of 2.5 A flows across R2. Find the current that flows across R3. Let R1= 3.0 ohm, R2= 8.

0 ohm, and R3= 4.0 ohm
A. 2.5 A
B. 10 A
C. 6.7 A
D. 5.0 A

Physics
2 answers:
kolbaska11 [484]2 years ago
5 0
D. 5.0A because this is right and will lead to the right answer okay you got this girl letssssss goooo googoggo Gogol
Alexeev081 [22]2 years ago
3 0
That would be (D) 5.0
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Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a
andrey2020 [161]

Explanation:

It is given that,

Force, F=(-8\ N)i+(6\ N)j

Position vector, r=(3i+4j)\ m

(a) The torque on the particle about the origin is given by :

\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m

(b) To find the angle between r and F use dot product formula as :

F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
Read 2 more answers
How will you join three resistances, each of 2 ohm so that the effective resistance is 3 ohm?
svet-max [94.6K]
First I will parallel two of the resistors, creating a net 1 ohm. Then I will series that with the remaining 2-ohm resistor, resulting in 3 ohms.
5 0
3 years ago
Convert 1erg into joule by dimensional method​
Valentin [98]

Answer:

1 * 10^-7 [J]

Explanation:

To solve this problem we must use dimensional analysis.

1 ergos [erg] is equal to 1 * 10^-7 Joules [J]

1[erg]*\frac{1*10^{-7} }{1}*[\frac{J}{erg} ] \\= 1*10^{-7}[J]

4 0
2 years ago
1) Calculate the potential energy of a 5.00 kg object sitting on a 3.00 meter high ledge.
maria [59]

Answer:

15kg

Explanation:

5 0
2 years ago
The wavelength of the visible line in the hydrogen spectrum that corresponds to m = 5 in the Balmer equation is: A. 656 nm. B. 4
BaLLatris [955]

Answer:

The wavelength of the visible line in the hydrogen spectrum is 434 nm.

Explanation:

It is given that, the wavelength of the visible line in the hydrogen spectrum that corresponds to n₂ = 5 in the Balmer equation.

For Balmer series, the wave number is given by :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})

R is the Rydberg's constant

For Balmer series, n₁ = 2. So,

\dfrac{1}{\lambda}=1.097\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{5^2})

\lambda=4.34\times 10^{-7}\ m

or

\lambda=434\ nm

So, the wavelength of the visible line in the hydrogen spectrum is 434 nm. Hence, this is the required solution.

6 0
2 years ago
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