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Alla [95]
3 years ago
12

Four eight-ohm speakers are connected in parallel to an audio power amplifier. The amplifier can supply a maximum driver output

voltage of fifteen volts. How much audio power must the amplifier be able to deliver to the speakers when outputting this maximum voltage?
Engineering
1 answer:
Vera_Pavlovna [14]3 years ago
6 0

Answer:

112.5 watts

Explanation:

The output voltage (V) = 15 volts

The equivalent resistance for the speakers connected in parallel (R_T) is gotten by using the formula:

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\\\But\ R_1=R_2=R_3=R_4=8\ ohm\\\\\frac{1}{R_T} =\frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} \\\\\frac{1}{R_T} =\frac{1+1+1+1}{8} \\\\\frac{1}{R_T} =\frac{4}{8} \\\\R_T=\frac{8}{4} \\\\R_T=2\ ohm

The current flowing through the amplifier (I) is:

I = V / R_T = 15 / 2 = 7.5 A

The audio power (P) when outputting this maximum voltage is given by:

P = V² / R_T = I²R_T. Therefore:

P = V² / R_T = 15² / R_T = 112.5 watts

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SCORPION-xisa [38]

Answer:

Explanation:

Fist you need to identify where the leak is coming from. You can do this by either listening for the leak or spraying soapy water on the exhaust to look for air bubbles coming out of the exhaust. Depending on the spot of the leak there are many ways you can fix this leak.

1. Exhaust clamp

2. Exhaust putty

3. Exhaust tape

4. New exhaust

Exhaust clamp is best used for holes on straight pipes.

Putty is best used on welds or small holes like on exhaust manifolds or welds connecting various pieces like catalytic converters, mufflers, or resonators.

Tape will work best on straight pipes with holes.

New exhaust is for when the thig is beyond repair, like rust.

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4 0
2 years ago
A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is
jonny [76]

Answer:

a) t = 277.477\,s\,(4.625\min), b) v_{f} = 0\,\frac{mi}{h}, c) a = -0.128\,\frac{ft}{s^{2}}

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}

a = -0.185\,\frac{ft}{s^{2}}

The time required to travel is:

t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }

t = 277.477\,s\,(4.625\min)

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be v_{f} = 0\,\frac{mi}{h}.

c) The final constant deceleration is:

a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}

a = -0.128\,\frac{ft}{s^{2}}

7 0
3 years ago
1. Under what conditions can soils be chemically stabilized?
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Answer:

All will be Explained below.

Explanation:

1) Under which Condition can a soil be chemically Stabilize.

Answer

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7 0
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A concrete mix design calls for 6.5 sacks of cement, a water/cement ratio of 0.45, and an air content of 2.5%. 1. Complete the m
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Answer:

28.6 kg

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The final weight can be calculated from the mixing data and formulae which is given as follows:

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7 0
3 years ago
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Answer:

The correct option is;

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