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Monica [59]
3 years ago
14

A 63 kg kg person starts traveling from rest down a waterslide 6.0 mm above the ground. At the bottom of the waterslide, it then

curves upwards by 1.0 mm above the ground such that the person is consequently launched into the air. Ignoring friction, how fast is the person moving upon leaving the waterslide
Physics
1 answer:
gladu [14]3 years ago
5 0

Answer:

change waterslide according to question. and you are good to go. check photo for solve

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HELP ASAP!! WILL MARK BRAINLIEST
Kitty [74]

Answer:

A

Explanation:

The magnet can always attract other things.

8 0
3 years ago
Read 2 more answers
What is the equivalent resistance of the
BigorU [14]

Answer:

Approximately 111\; {\rm \Omega}.

Explanation:

It is given that R_{1} = 200\; {\Omega} and R_{2} = 250\; {\Omega} are connected in a circuit in parallel.

Assume that this circuit is powered with a direct current power supply of voltage V.

Since R_{1} and R_{2} are connected in parallel, the voltage across the two resistors would both be V. Thus, the current going through the two resistors would be (V / R_{1}) and (V / R_{2}), respectively.

Also because the two resistors are connected in parallel, the total current in this circuit would be the sum of the current in each resistor: I = (V / R_{1}) + (V / R_{2}).

In other words, if the voltage across this circuit is V, the total current in this circuit would be I = (V / R_{1}) + (V / R_{2}). The (equivalent) resistance R of this circuit would be:

\begin{aligned} R &= \frac{V}{I} \\ &= \frac{V}{(V / R_{1}) + (V / R_{2})} \\ &= \frac{1}{(1/R_{1}) + (1 / R_{2})}\end{aligned}.

Given that R_{1} = 200\; {\Omega} and R_{2} = 250\; {\Omega}:

\begin{aligned} R &= \frac{1}{(1/R_{1}) + (1 / R_{2})} \\ &= \frac{1}{(1/(200\: {\rm \Omega})) + (1/(250\; {\rm \Omega}))} \\ &\approx 111\; {\rm \Omega}\end{aligned}.

7 0
2 years ago
A student rides a bicycle for 15 miles in 3 hours. What is the student's speed? What else would you need to know for the velocit
kaheart [24]

Answer:

5 miles per hour

Explanation:

if you divide 15 by 3 you get 5, therefore the student is going 5 miles per hour.

3 0
2 years ago
Please help. I don’t understand this
skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>

and under constant acceleration,

<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2

According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2

∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2

∆<em>x</em> = 20.00 m

5 0
2 years ago
When a person speaks, a sound intensity is generated that is 600 times greater than when the person whispers. What is the differ
Charra [1.4K]

Answer:

Originally :  Level = log I / I0

Currently: Level = 10 log I / I0

Level = 10 log 600 = 10 * 2.78 = 27.8

Note the term 1 bel = 10 decibels

5 0
2 years ago
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