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Delvig [45]
3 years ago
7

Please hurry

Physics
1 answer:
viktelen [127]3 years ago
6 0

Answer:

The velocity will be "76.8 m/s".

Explanation:

The given values are:

Acceleration,

a = 2.4 m/s²

Time,

t = 32 seconds

By equation of motion,

⇒  v=u+at

On substituting the values, we get

⇒     =0+2.4\times 32

⇒     =0+76.8

⇒     =76.8 \ m/s

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What happens when a star blows up and it is next another star will it blow up too?
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andrew-mc [135]

like a black cloud

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Define: Signals

Before going too much further, we should talk a bit about what a signal actually is, electronic signals specifically (as opposed to traffic signals, albums by the ultimate power-trio, or a general means for communication). When one speaks of analog one often means an electrical context, however mechanical, pneumatic, hydraulic, and other systems may also convey analog signals.

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4 0
3 years ago
An air conditioner runs 15 minutes each hour on a hot summer day. It is on a 240 volt circuit and uses 21 amps. Rate is $.10/kWh
Helen [10]

Answer:

Approximately \$ 3.02.

Explanation:

Note that the electric rate in this question is in the unit dollar-per-{\rm kWh}, where 1\; {\rm kWh} is the energy to run an appliance of power 1\; {\rm kW} for an hour.

Number of minutes for which the air conditioner is running in that day: 15 \times 24 = 360\; \text{minute}. Apply unit conversion and ensure that this time is measured in hours (same as the unit of the electric rate.)

\begin{aligned} \text{time} &= 360\; \text{minute} \times \frac{1\; \text{hour}}{60\; \text{minute}} = 6\; \text{hour} \end{aligned}.

The power of this air conditioner is:

\begin{aligned} \text{power} &= \text{voltage} \times \text{current} \\ &= 240\; {\rm V} \times 21\; {\rm A} \\ &= 5040\; {\rm W} \\ &= 5.04\; {\rm kW} \end{aligned}.

Thus, the energy that this air conditioner would consume would be:

\begin{aligned}\text{energy} &= \text{power} \times \text{time} \\ &= 5.04\; {\rm kW} \times 6\; \text{hour} \\ &= 30.24\; {\rm kWh} \end{aligned}.

At a rate of 0.1 dollar-per-{\rm kWh}, the cost of that much energy would be approximately 3.02 dollars (rounded to the nearest cent.)

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