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4 years ago

What is the potential difference vab between points a and b when the switch s is open?

1 answer:

3 0

Assuming is a closed circuit when the switch is closed, it would be the strength of the battery in the circuit, assuming a and b were points at each end of the circuit. Depending on the specifics of the circuit, this could change, you could add more batteries together or it could reduce based on resistance if there's a viable current path still open while the switch is open. You may have missed attaching a picture or link.

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A series circuit connects a single light bulb to a single power source. As more light bulbs are added to the circuit, which grap

Nikolay [14]

Answer: B

Explanation:

3 0

3 years ago

60 point!!!

erastova [34]

Answer:

Playing hockey, driving a car, and even simply taking a walk are all everyday examples of Newton's laws of motion.

7 0

3 years ago

What is the mode of heat transfer from drying of wet hot plate in atmosphere

SashulF [63]

Answer:

It's convection

Explanation:

As the hot place transfers heat to the liquid substance, it becomes less dense and rises to the surface given rise for cooler molecules directly above to fall and so undergoes the same circle. In the process the less dense particles are so light that it's swayed into the atmosphere. In the process drying is ensued.

4 0

3 years ago

It's nighttime, and you ve dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90 m above the ed

yanalaym [24]

Answer:

The distance of the goggle from the edge is 5.30 m

Explanation:

Given:

The depth of pool (d) = 3.2 m

let 'i' be the angle of incidence

thus,

i = $tan^{-1}(\frac{2.2}{0.90})$

i = 67.75°

Now, Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

where,

r is the angle of refraction

n₁ is the refractive index of medium 1 = 1 for air

n₂ is the refractive index of medium 1 = 1.33 for water

now,

1 × sin 67.75° = 1.33 × sin(r)

r = 44.09°

Now,

the distance of googles = 2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) = 5.30 m

Hence, <u>the distance of the goggle from the edge is 5.30 m</u>

5 0

3 years ago

A small but dense 2.0-kg stone is attached to one end of a very light rod that is 1.2 m long. The other end of the rod is attach

tigry1 [53]

Answer:

Option B is the correct answer.

Explanation:

Refer the figure we have centripetal force at bottom of circle

$F_c=F_t-F_w\\\\\frac{mv^2}{r}=F_t-mg\\\\F_t=m\left ( \frac{v^2}{r}+g\right )$

We have mass, m = 2 kg

Radius, r = 1.2 m

For circular motion to occur we have tension at top = 0

That is

$\frac{mv^2}{r}=mg\\\\v=\sqrt{rg}$

Now let us find tension at bottom point

$F_t=2\times \left ( \frac{rg}{r}+g\right )=4g=40N$

Option B is the correct answer.

8 0

4 years ago