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slega [8]
3 years ago
5

The batter swings and misses the 40 m/s (90 mph) fastball, and the ball (mass 150 grams) ends up at rest in the catcher's mitt.

How much work does the catcher perform on the ball
Physics
1 answer:
Nadusha1986 [10]3 years ago
5 0

Answer:

The work done by the catcher is 120 J.

Explanation:

Given;

velocity of the fastball, v = 40 m/s

mass of the fastball, m = 150 g = 0.15 kg

Based on work-energy theorem, the work done by the catcher is equal to the kinetic energy of the fastball.

The kinetic energy of the fastball is given as;

K.E = ¹/₂mv²

K.E = ¹/₂ x 0.15 x 40²

K.E = 120 J

Therefore, the work done by the catcher is 120 J.

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When a new path of lesser resistance is made for an existing circuit a(n) _____________ circuit occurs.
Margaret [11]

When a new path of lesser resistance is made for an existing circuit a(n) short circuit occurs.

<h3>What is short circuit?</h3>

An electrical circuit short circuit is when two nodes that are supposed to be connected at different voltages make an improper connection. This leads to an electric current that can damage circuits, cause overheating, fire, or explosions, and is only constrained by the network's remaining nodes' equivalent Thevenin resistance. While short circuits are typically the result of a failure, they can occasionally be brought on purpose, such as when voltage-sensing crowbar circuit protectors are being installed.

An electrical connection that requires two nodes to have the same voltage is known as a short circuit in circuit analysis. Since there is no resistance and hence no voltage drop across the link in a "perfect" short circuit, there is no short circuit.

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5 0
2 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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