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Flura [38]
2 years ago
12

A negatively charged plastic comb is brought close to, but does not touch, a small piece of paper. If the comb and the paper are

attracted to each other the charge on the paper
A) must be negative
B) may be positive or neutral
C) may be negative or neutral D)must be positive
Physics
1 answer:
Irina-Kira [14]2 years ago
4 0

If the comb and the paper are attracted to each other the charge on the paper, is D)must be positive

<h3>Laws of electrical attraction</h3>

This states that

  • Like charges attract
  • Unlike charges repel

Now, given that a negatively charged plastic comb is brought close to, but does not touch, a small piece of paper. If the comb and the paper are attracted to each other the charge on the paper, this implies that both the negatively charged plastic comb and the paper have opposite charges.

Since the charge on the plastic comb is negative, this means that the charge on the paper must be positive

So, if the comb and the paper are attracted to each other the charge on the paper, is D)must be positive

Learn more about electric charge here:

brainly.com/question/2373424

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How does an electromagnet become permanent
love history [14]
<span>A moving electrical charge produces a magnetic field and a moving magnetic field produces an electrical field. An electromagnet works by coiling a bunch of wire and spinning a couple of magnets around that wire at high speeds. When this occurs the magnets induce an electric current in the wire and hence the electricity production. Once the magnets stop spinning, the induced electrical field dissipates and the current stops flowing through the wire.

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6 0
3 years ago
Can anybody help me solve this problem? Thank you so much!
ser-zykov [4K]
Please ignore my comment -- mass is not needed, here is how to solve it. pls do the math

at bottom box has only kinetic energy
ke = (1/2)mv^2
v = initial velocity
moving up until rest work done = Fs
F = kinetic fiction force = uN = umg x cos(a)
s = distance travel = h/sin(a)
h = height at top
a = slope angle
u = kinetic fiction
work = Fs = umgh x cot(a)
ke = work (use all ke to do work)
(1/2)mv^2 = umgh x cot(a)
u = (1/2)v^2 x tan (a) / gh
4 0
3 years ago
HeLp AsAp!!!!!! How long would it take you to hop 30 meters based on your speed for the 5-meter trial? Show your work!
Novosadov [1.4K]

Answer:

20.5s

Explanation:

Given parameters:

Distance = 30m

Unknown:

Time  = ?

Solution:

The time it will take to hop a distance of 30m using the speed for the 5m trial is the duration of the trip.

 The speed for the 5m trial  = 1.46m/s

Now;

    Speed  = \frac{distance}{time}

      Distance = speed x time

      time  = \frac{distance }{speed}

Input the parameters and solve;

     time  = \frac{30}{1.46}   = 20.5s

7 0
3 years ago
What are the three important safety factors for surgical technologists to consider when exposed to ionizing radiation?
Gemiola [76]

Answer:

Exposure time limitation, shielding and distance.

Explanation:

  • Limitation of exposure time, since the dose received is directly proportional to the exposure time, so that, at a shorter time, lower dose. For this reason, planning is suggested, to reduce time.
  • Use of shields. This allows a reduction in the dose received by the technician when filtered by the barrier (screen). There are two types of shields or screens, the primary barriers (attenuate the radiation of the primary beam) and the secondary barriers (avoid diffuse radiation).
  • Distance to the radioactive source. The dose received is inversely proportional to the square of the distance to the radioactive source. Therefore, if the distance is doubled, the dose received will decrease by a quarter. Reason for this, it is advisable to use devices or remote controls whenever possible.
7 0
3 years ago
A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and
vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

U=mgh

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

h=4-4cos(30\°)=0.53m

Then, the potential energy is:

U=(55kg)(9.8m/s^2)(0.53m)=285.67J

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

U'=(m_1+m_2)gh'=285.67\ J

From the previous equation you can get h':

h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0
3 years ago
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