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atroni [7]
2 years ago
9

A fruit bat falls from the roof of a cave. We know that her potential energy was

Physics
1 answer:
bulgar [2K]2 years ago
5 0

Answer:

v = 15.65 m/s

Explanation:

We use conservation of mechanical energy between initial (i) and final (f) states:

Pi + KEi = Pf + KEf

At the top of the cave at the instant the bat starts to fall, there is only potential energy since the bat's velocity is zero.

Pi = m g h = 600 J

and the KEi = 0 J (no velocity)

Knowing the height of the cave's roof (12.8 m) , we can find the mass of the bat:

m = 600 J / (g 12.5) = 4.9 kg

Using conservation of mechanical energy, the final state is:

Pf + KEf = 600 J

with Pf = 0 (just touching the ground)

KEf= 1/2  4.9 (v^2)

and we solve for the velocity:

600 J = 0 + 1/2  4.9 (v^2)

v^2 = 600 * 2 / 4.9 = 244.9

v = 15.65 m/s

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Present day glaciers are found primarily in _______________.
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<h3><u>Answers;</u></h3>

Antarctica and Greenland

Present day glaciers are found primarily in <em><u>Antarctica and Greenland</u></em>.

<h3><u>Explanation;</u></h3>
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8 0
3 years ago
Read 2 more answers
Miles is camping in Glacier National Park. In the midst of a glacier canyon,
valentina_108 [34]

Answer:

t=1.623 sec

Explanation:

The distance traveled before the echo is had is:

distance=2d, d=280\ m\\\\=280\times 2\\\\=560 \ m

Given the speed of sound as v=345m/s, we use the speed equation to solve for t:

v=\frac{d}{t}\\\\345\ m/s=\frac{560m}{t}\\\\t=\frac{560}{360}\\\\=1.623 \ s

Hence, it takes 1.623 seconds to hear the echo.

8 0
3 years ago
You are given three resistors with the following resistances: R1 = 6.32 Ω, R2 = 8.13 Ω, and R3 = 2.29 Ω. What is the largest equ
andreyandreev [35.5K]

Answer:

The largest equivalent resistance yu can build using these three resistors is a Serie Resistance with the value of R= 16.74 Ω

Explanation:

Adding Resistances in serie is the way to build de largest equivalent value possible.

Rt= R1+R2+R3

Rt= 6.32 + 8.13 + 2.29

Rt= 16.74Ω

5 0
3 years ago
A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th
Mila [183]

Answer:

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0

\frac{3}{(r + a)^2} = \frac{2}{r^2}

\frac{r+a}{r} = \sqrt{\frac{3}{2}}

1 + \frac{a}{r} =\sqrt{\frac{3}{2}}

\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}

so we will have

r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}

so the x coordinate of this position is given as

x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}

6 0
3 years ago
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