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3 years ago

Which circuits correctly show Ohm's law?

Physics

2 answers:

Elodia [21]3 years ago

6 0

Ohm’s Law:
V = IR
V = battery voltage, I = current, R = resistance

Top-left circuit:
V = 25V, I = 5A, R = 5ohm
25 = 5•5
25 = 25
This circuit is correct.

Top-right circuit:
V = 18V, I = 6A, R = 4ohm
18 = 6•4
18 =/= 24
This circuit is incorrect.

Bottom-left circuit:
V = 56V, I = 7A, R = 8ohm
56 = 7•8
56 = 56
This circuit is correct.

Bottom-right circuit:
V = 20V, I = 5A, R = 4ohm
20 = 5•4
20 = 20
This circuit is correct.

The following circuits are correct: top-left, bottom-left, and bottom-right.

Amiraneli [1.4K]3 years ago

5 0

Answer:

the top left and the bottom left and the bottom right

You might be interested in

A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75

Gnom [1K]

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

8 0

3 years ago

Find the Y component of 35m/s at 57 degrees from the X-axis.

rosijanka [135]

Answer:

35m/s[57o].

X = 35*Cos57 =

Y = 35*sin7 =Explanation:

learn man but there u go

3 0

2 years ago

Vector A has a magnitude of 50 units and points in the positive x direction. A second vector, B , has a magnitude of 120 units a

Alex Ar [27]

A) Vector A

The x-component of a vector can be found by using the formula

$v_x = v cos \theta$

where

v is the magnitude of the vector

$\theta$ is the angle between the x-axis and the direction of the vector

- Vector A has a magnitude of 50 units along the positive x-direction, so $\theta_A = 0^{\circ}$ . So its x-component is

$A_x = A cos \theta_A = (50) cos 0^{\circ}=50$

- Vector B has a magnitude of 120 units and the direction is $\theta_B = -70^{\circ}$ (negative since it is below the x-axis), so the x-component is

$B_x = B cos \theta_B = (120) cos (-70^{\circ})=41$

So, vector A has the greater x component.

B) Vector B

Instead, the y-component of a vector can be found by using the formula

$v_y = v sin \theta$

Here we have

- Vector B has a magnitude of 50 units along the positive x-direction, so $\theta_A = 0^{\circ}$ . So its y-component is

$A_y = A sin \theta_A = (50) sin 0^{\circ}=0$

- Vector B has a magnitude of 120 units and the direction is $\theta_B = -70^{\circ}$ , so the y-component is

$B_y = B sin \theta_B = (120) sin (-70^{\circ})=-112.7$

where the negative sign means the direction is along negative y:

So, vector B has the greater y component.

8 0

3 years ago

Which term is defined as the change in the direction of light when it goes from one medium into a different medium?

Vaselesa [24]

Refraction is the term

8 0

2 years ago

Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence

Keith_Richards [23]

Answer:

1. Recollapsing universe

2. Critical universe

3. Coasting universe

Explanation:

Recollapsing universe has dark matter density greater than critical density. While critical universe has its matter density equal to the critical sensity. Coasting universe on the other hand has much smaller matter density compared to critical density.

Note that the critical density is approximately 10^-20 grams/cm3

3 0

2 years ago