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AVprozaik [17]
4 years ago
13

When you rub a plastic rod with fur, the plastic rod becomes negatively charged and the fur becomes positively charged. As a con

sequence of rubbing the rod with the fur,
A. the rod and fur both gain mass.
B. the rod and fur both lose mass.
C. the rod gains mass and the fur loses mass.
D. the rod loses mass and the fur gains mass.
E. none of the above
Physics
1 answer:
Yuri [45]4 years ago
4 0

Answer: C. the rod gains mass and the fur loses mass.

Explanation:Atomic particles have mass. The electron has a mass that is approximately 1/1836 that of the proton and with exchange exchange of charge this is also factored in. The movement of effect described above is known as the triboelectic charging process—charging by friction—which results in a transfer of electrons between the two objects when they are rubbed together. Plastic having a much greater affinity for electrons than animal fur pulls electrons from the atoms of fur, leaving both objects with an imbalance of charge. The plastic rod would have an excess of electrons and the fur has a shortage of electrons. Having an excess of electrons, the plastic is charged negatively and has more mass. In the same vein, the shortage of electrons on the fur leaves it with a positive charge and consequently with lesser mass.

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A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the partic
olga_2 [115]

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma              

a = \frac{F}{m}      ...... (1)

acceleration (a) = \frac{dv}{dt}   ......(2)

substituting (2) into (1)

Hence, F = \frac{mdv}{dt}

\frac{dv}{dt} = \frac{F}{m}

dv = \frac{F}{m} dt

dv = \frac{1}{m}Fdt

Integrating both sides

\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt     ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt

v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}

v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|

v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |

v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |

v = \frac{1}{5} | 250 - 50 + 15 |

v = \frac{215}{5}

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s

6 0
3 years ago
Newton’s law of universal gravitation stars that every object in the universe attracts every other object
LenaWriter [7]

Yes it does !  Uh huh. Right you are. Truer words are seldom written.

You have quoted the law quite accurately but also incompletely.  

Do you have a question to ask ?

4 0
3 years ago
A toy cannon uses a spring to project a 5.35-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force
Elenna [48]

Answer:

a) the velocity is v=1.385 m/s

b) the ball has its maximum speed at 4.68 cm away from its compressed position

c)  the maximum speed is 1.78 m/s

Explanation:

if we do an energy balance over the ball, the potencial energy given by the compressed spring is converted into kinetic energy and loss of energy due to friction, therefore

we can formulate this considering that the work of the friction force is equal to to the energy loss of the ball

W fr = - ΔE = - ΔU - ΔK = Ui - Uf + Kf - Ki

therefore

Ui + Ki = Uf + Kf + W fr  

where U represents potencial energy of the compressed spring , K is the kinetic energy W fr is the work done by the friction force. i represents inicial state, and f final state.

since

U= 1/2 k x² , K= 1/2 m v²  , W fr = F*L

X= compression length , L= horizontal distance covered

therefore

Ui + Ki = Uf + Kf + W fr

1/2 k xi² + 1/2 m vi² = 1/2 k x² + 1/2 m vf² + F*L

a) choosing our inicial state as the compressed state , the initial kinetic energy is Ki=0 and in the final state the ball is no longer pushed by the spring thus Uf=0

1/2 k X² + 0 = 0 + 1/2 m v² + F*L

1/2 m v² = 1/2 k X² - F*L

v = √[(k/m)x² -(2F/m)*L] = √[(8.07N/m/5.35*10^-3 Kg)*(-0.0508m)² -(2*0.033N/5.35*10^-3 Kg)*(0.16 m)] = 1.385 m/s

b) in any point x , and since L= d-(X-x) , d = distance where is no pushed by the spring.

1/2 k X² + 0 = 1/2 k x² + 1/2 m v² + F*[d-(X+x)]

1/2 m v² =1/2 k X²-1/2 k x² - F*[d-(X-x)] = (1/2 k X²+ F*X) - 1/2k x² - F*x + F*d

taking the derivative

dKf/dx = -kx - F = 0 → x = -F/k = -0.033N/8.07 N/m = -4.089*10^-3 m = -0.4cm

at x m = -0.4 cm the velocity is maximum

therefore is 5.08 cm-0.4 cm=4.68 cm away from the compressed position

c) the maximum speed is

1/2 m v max² = (1/2 k X²+ F*X) - 1/2k x m² - F*(x m) + 0

v =√[ (k/m) (X²-xm²) + (2F/m)(X-xm) ] = √[(8.07N/m/5.35*10^-3 Kg)*[(-0.0508m)² - (-0.004m)²] + (2*0.033N/5.35*10^-3 Kg)*(-0.0508m-(-0.004m)] = 1.78 m/s

4 0
3 years ago
The Otto-cycle engine in a Mercedes-Benz SLK230 has a compression ratio of 8.8. What is the ideal efficiency of the engine? Use
vodomira [7]

Answer:

e=58%

Explanation:

Given data

The Otto-cycle engine in a Mercedes-Benz SLK230 has a compression ratio of 8.8.

Solution

We want to calculate the ideal efficiency of the engine when ratio of heat capacity for gas used  γ=1.40. Ideal efficiency (e) of the Otto cycle given by:

e=1-(\frac{1}{r^{Y-1} } )\\

Substitute the given values to find efficiency e

e=1-\frac{1}{8.8^{1.4-1} } \\e=0.58\\

e=58%

8 0
3 years ago
An airtight box has a removable lid of area 1.10 10-2 m2 and negligible weight. The box is taken up a mountain where the air pre
andrezito [222]

Answer:

F=7.7\times10^2 N

Explanation:

The magnitude of force required to pull the lid off the box by air pressure.

We know that Pressure, P= Force(F)/Area(A)

Force, F= P×A

Given: A=1.10\times10^{-2} m^2

P=7\times10^{4} Pa.

Therefore, F=7\times10^{4}\times1.10\times10^{-2}.

F=7.7\times10^2 N

4 0
4 years ago
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