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Brums [2.3K]
4 years ago
13

During which of the following activities may in inflatable personal flotation device to be used

Physics
1 answer:
Alla [95]4 years ago
5 0
They may be used whenever you are not experienced with swimming and you are near the water such as swimming, canoeing, etc
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In a hydroelectric dam, water falls 30.0m and then spins a turbine to generate electricity.
ozzi

Answer:

(a) 294 J

(b) 229591.84 kg

Explanation:

height, h = 30 m

g = 9.8 m/s^2

(a) U = m x g x h

U = 1 x 9.8 x 30

U = 294 J

(b) Let the mass of water is m

U = m x g xh

80% of this energy per second is generating power

0.8 x m x 9.8 x 30 = 54 x 10^6

m = 229591.84 kg

Thus, the mass of water required is 229591.84 kg.

4 0
3 years ago
If the speed of a wave doubles while the frequency remains the same, it means
sattari [20]

Answer:

Explanation:

Velocity of a wave is describe as

velocity =Frequency × Wavelength

Mathematically

v = fλ

Hence, Frequency, F = v / λ

Wavelength λ = v/f

So, if the frequency is kept constant, wavelength of the wave becomes directly proportional to velocity of the wave.

And this implies that, as the speed double, the wavelength is double.

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4 years ago
What effect does acid rain have on the environment?
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Do clouds and wind appear to follow the same patterns?
timama [110]
Sometimes in the same direction but most of them go in the opposite direction
7 0
3 years ago
A mountain lion jumps to a height of 3.25 m when leaving the ground at an angle of 43.2°. What is its initial speed (in m/s) as
miss Akunina [59]

Recall that

{v_f}^2={v_i}^2+2a\Delta y

where v_i and v_f are the lion's initial and final vertical velocities, a is its acceleration, and \Delta y is the vertical displacement.

At its maximum height, the lion has 0 vertical velocity, so we have

0={v_i}^2-2gy_{\rm max}

where <em>g</em> is the acceleration due to gravity, 9.80 m/s², and we take the starting position of the lion on the ground to be the origin so that \Delta y=y_{\rm max}-0=y_{\rm max}.

Let <em>v</em> denote the initial speed of the jump. Then

v_i=v\sin(43.2^\circ)=\sqrt{2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(3.25\,\mathrm m)}\implies\boxed{v\approx11.7\dfrac{\rm m}{\rm s}}

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3 years ago
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