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emmainna [20.7K]
3 years ago
14

Which of the following statements indicates that the client is at risk for an electrical shock at home?

Physics
1 answer:
sashaice [31]3 years ago
6 0

Answer:

“I had to cut off the third prong on the electrical plug so that it would fit in the extension cord.”

Explanation:

A good method to find the customer at greater risk of receiving an electric shock is to discard the options.

If we analyze the fourth option "I always make sure that I am standing in a dry area before operating electrical equipment.", we will see that the customer minimizes the risk of electric conduction through a fluid. This precaution also minimizes the risk of electric shock.

The third option "I always read the owner’s manual when I purchase a new electrical appliance." It tells us that the customer knows how to properly use the electrical equipment that manipulates and knows the do and do not of such devices, so the risk of electric shock is reduced.

In the second option, "My bread got stuck in my toaster this morning, and I unplugged it before trying to remove it." The customer made sure to interrupt the electric flow to the device before proceeding and try to remove the jammed bread. If there is no electric current, there will be no electric shock to the customer.

The first option "I had to cut off the third prong on the electrical plug so that it would fit in the extension cord." It tells us that the client altered the cord extension in an inappropriate manner, since the equipment is not designed to work with two prongs. Because the design conditions are not being met, it is possible that the equipment malfunctions and this malfunction could result in the customer receiving an electric shock.

I hope the explanation was clear for you. If you have any further question, I'll be happy to assist you. :D

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Answer:

Part a)

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Part b)

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Part c)

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Part d)

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Part e)

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Part f)

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Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

v_x = 19.9 cos39.9

v_x = 15.3 m/s

similarly we have

v_y = 19.9 sin39.9

v_y = 12.76 m/s

Part a)

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Part b)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.03) - 4.9(1.03)^2

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x = (15.3)(1.71)

x = 26.16 m

Part d)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.71) - 4.9(1.71)^2

y = 7.49 m

Part e)

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x = (15.3)(5.44)

x = 83.23 m

Part f)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(5.44) - 4.9(5.44)^2

y = -75.6 m

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