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kotykmax [81]
3 years ago
14

A charged comb often attracts small bits of dry paper that then fly away when they touch the comb. Explain why that occurs.

Physics
1 answer:
MAVERICK [17]3 years ago
7 0

Explanation:

Since the comb has a net charge, it attracts the paper, which has a net charge equal to zero. When the paper touches the comb, an electrical interaction is established between the charge of the comb and the neutral paper, because of this, the paper now has a net charge with the same sign of the comb and they repel.

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A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire appara
otez555 [7]

Answer:

The tangential speed of the ball is 11.213 m/s

Explanation:

The radius is equal:

r=2.3*sin70=2.161m (ball rotates in a circle)

If the system is in equilibrium, the tension is:

Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}

Replacing:

\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}

Replacing:

v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s

7 0
3 years ago
In which test did the air parcel rise the highest? Is there a pattern in the relationship between starting air temperature and p
wel

Answer:

xitxitx

Explanation:

ufxigxigxitcutcocyocoycitcotcitcohcohxitxigcigxigxkgxigxigxigxigxigxigxitx8y

5 0
3 years ago
The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass
devlian [24]

Answer:

The velocity of the recoil is v=1.001 \frac{m}{s}

Explanation:

Kinetic Energy

m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\m_{gun}= 15.0kg+3.6kg

The mass of th gun is the both mass the shotgun and the arm shoulder combination

m_{bullet}=0.049kg\\v_{bullet}=380\frac{m}{s} \\m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\0.049kg*380\frac{m}{s}=(15.kg+3.6kg)* v_{recoil}\\v_{recoil}=-\frac{18.62 kg \frac{m}{s} }{18.6 kg}\\ v_{recoil}=-1.0010 \frac{m}{s}

The velocity is negative because is in opposite direction of the bullet

6 0
2 years ago
Read 2 more answers
A wheel decelerates from 13.5 rad s−1 to 6.0 rad s−1 in 7 s. Calculate the angular displacement​
Igoryamba

Answer:

<em>Angular displacement=68.25 rad</em>

Explanation:

<u>Circular Motion</u>

If the angular speed varies from ωo to ωf in a time t, then the angular acceleration is given by:

\displaystyle \alpha=\frac{\omega_f-\omega_o}{t}

The angular displacement is given by:

\displaystyle \theta=\omega_o.t+\frac{\alpha.t^2}{2}

The wheel decelerates from ωo=13.5 rad/s to ωf=6 rad/s in t=7 s, thus:

\displaystyle \alpha=\frac{6-13.5}{7}

\displaystyle \alpha=\frac{-7.5}{7}

\displaystyle \alpha=-1.071 \ rad/s^2

Thus, the angular displacement is:

\displaystyle \theta=13.5*7+\frac{-1.071*7^2}{2}

\displaystyle \theta=94.5-26.25

\boxed{\displaystyle \theta=68.25\ rad}

Angular displacement=68.25 rad

3 0
3 years ago
at the moment a truck traveling with a constant velocity of 13.6 m/s passes a dog at rest the dog begins moving with a constant
goblinko [34]
I don’t know how to help you hope you figure it out have a good day
4 0
2 years ago
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