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myrzilka [38]
3 years ago
8

State one advantage of an alkaline battery over a lead acid battery

Physics
1 answer:
valentinak56 [21]3 years ago
4 0

<em>Answer:</em>

<em>one of the advantages of alkaline batteries over lead acid batteries is that it is able to provide a larger and more stable current because the cathode does not use carbon bars, but is manganese</em><em>.</em>

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Select the correct answer
DiKsa [7]

Answer:

C or 3

Explanation:

A: no they are called sources of sound. A is incorrect.

B: It does. Many people attest to this. But this is not a property of physics.

C: A media is required is the correct answer.

D: Dogs might. In general we don't.

I would pick C

7 0
3 years ago
Hydraulic engineers often use, as a unit of volume of water, the "acre-foot", defined as the volume of water that will cover 1 a
Alex17521 [72]

Answer:

Volume = 1,015 acre-feet (Approx)

Explanation:

Given:

Rain = 1.7 in

Time = 30 min

Area = 29 km²

Find:

Volume in acre-feet

Computation:

1 km = 1,000 m

1 m = 3.28 feet

1 km² = 247.105 acre

d = 1.7 in = 1.7 / 12 = 0.14167 ft

Area = 29 × 247.105 = 7,166.045 acre

Volume = 7,166.045 acre × 0.14167 ft

Volume = 1,015 acre-feet (Approx)

7 0
3 years ago
Which of these statements best describes how energy is transferred when an object experiences friction?
Slav-nsk [51]
<h2>Answer:</h2><h2> b hopefully this helps you with work </h2>
3 0
3 years ago
A wave has a frequency of 875 hz and a wavelength of 352 m. At what speed is this wave traveling 
Sonbull [250]

Answer:

308,000 or 30.8×10^3

Explanation:

v=f×lamda

v is ?, f is 875Hz, lamda is 352m

v=875×352

v=308,000

v=30.8×10^3 m/s

5 0
3 years ago
An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of
Gnoma [55]

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = \frac{2L}{3}

The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength

Distance from center  = 0.25 × wavelength

Distance = 0.25 x \frac{2L}{3}

Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm

Distance = 0.25 x \frac{2 \times 39}{3}

Distance from the center to the nearest anti - node = 6.5 cm

Hence, the nearest distance to the anti - node from the center = 6.5 cm

So, option C is correct.

7 0
3 years ago
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