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3 years ago

A night-light uses 3 W.If the voltage is 120 V, what is the current? A.0.025A B.40 A C.360A D.0.25A

Physics

1 answer:

Tomtit [17]3 years ago

6 0

The answer to this question is A.

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Goryan [66]

Answer:

The solved problem is in the photo. Hope it helps.

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3 years ago

Which substance cannot be seperated physically or chemically?

scoray [572]

Answer:

What is a Pure Substance?

Explanation:

It is something which cannot be divided into parts by physical means, as it's all made up of the same thing. Pure substances are either elements or compounds. Elements can NOT be separated into other types of matter (physically or chemically).

4 0

3 years ago

Read 2 more answers

A hopper jumps straight up to a height of 1.1 m. With what velocity did it leave the floor?

Amanda [17]

Answer:

4.64m/s

Explanation:

We can use the formula [ v = √2gh ] to solve for this problem. We know that g is constant acceleration (9.8), and h is height (1.1).

v = √2(9.8)(1.1)

v ≈ 4.64m/s

Best of Luck!

4 0

2 years ago

Tim can type 2000 words at a speed of 50 words per minute bobs typing speed is 80 words per minute how much less time did bob ta

Whitepunk [10]

Answer:

15 less mins will be used by Bob

Explanation:

This is because it takes Tim 2000/50= 40mins to type the whole work

While it takes Bob 2000/80= 25 mins

So the difference 40-25= 15mins

Will be 15mins

8 0

2 years ago

A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.

melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

$V_{f}^{2} = V_{0}^{2} + 2ad$

<u>Where</u>:

$V_{f}$ : is the final speed = 8.89 m/s

$V_{0}$ : is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m

$a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}$

Therefore, the magnitude of her acceleration is 3.09 m/s².

b) The component of her acceleration that is parallel to the ground is given by:

$a_{x} = a*cos(\theta)$

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

$a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}$

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

7 0

3 years ago