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snow_lady [41]
3 years ago
9

Which electrical protective device is designed to detect a difference in current between circuit wires and interrupt the circuit

and stop the flow of electricity?
a. attach ungrounded, two-prong adapter plugs to three-prong cords and tools
b.do not operate electrical equipment when working in wet conditions
c. hold fingers on the switch while carrying a plugged-in tool
d. use gfci only when using double insulated power tools
Physics
1 answer:
Ilya [14]3 years ago
8 0
That's the description of a ground fault circuit interrupter (GFCI).

The choices you posted seem to belong to a different question.
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An offshore rig begins to spill oil in a circular patch centered on the rig. If the radius ofthe oil spill increases at a rate o
Tresset [83]

Answer:

Explanation:

The area of a circle is given b

A=πr²

Then,

dA/dr= 2πr

Also give that,

dr/dt=30m/hr, r=100m

We want to find dA/dt

dA/dt=dA/dr×dr/dt

dA/dt= 2πr × 30

dA/dt= 2 ×π ×100×30

dA/dt=18849.56m²/s.

The rate of the area is 18849.56m²/s

7 0
3 years ago
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What unit of electric current,the ampere,is equivalent to
Paladinen [302]
1 ampere is 1 coulomb of charge moving through a cross section in one second.
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3 years ago
A mass tied to the end of a 1.0 m-long string is swinging back and forth From its lowest point, the mass moves 4 cm to the right
ehidna [41]

Answer:

a) 2seconds.

Explanation:

The period of oscillation will not change . It will still be 2seconds although the length was doubled to 8cm.

This is because the amplitude does not change oscillation time.

5 0
3 years ago
A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.
sladkih [1.3K]

Explanation:

Given that,

Charge 1, q_1=6.75\ nC=6.75 \times 10^{-9}\ C

Charge 2, q_2=4.46\ nC=4.46\times 10^{-9}\ C

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}

F=6.84\times 10^{-8}\ N

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

5 0
4 years ago
At a certain moment, a car travelling at 40mph begins braking for a railroad crossing. Assume thatthe acceleration is constant (
egoroff_w [7]

Answer:

Part a)

v_f = 17.88 - (0.15) t

Part b)

d = 1072.8 m

Explanation:

As we know that initial speed of the car is

v = 40 mph

v = 40 \times (\frac{1609}{3600})

v = 17.88 m/s

now we have

v_f = v_i + at

0 = 17.88 + a(120)

a = -0.15 m/s^2

Part a)

As we know that acceleration is constant here

so we have

v_f = v_i + at

v_f = 17.88 - (0.15) t

Part b)

distance moved by the car is given as

d = \frac{v_F + v_i}{2} t

now we have

d = \frac{0 + 17.88}{2}(120)

d = 1072.8 m

3 0
3 years ago
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