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worty [1.4K]
3 years ago
15

Sam blew up a balloon and rubbed it on his head. Then he stuck the balloon on the wall. This was all the result of electrostatic

s or the transfer of charge. The appearance of negative charge on a balloon is the result of its gaining electrons. And these electrons must come from somewhere; in this case, from Sam's hair. Electrons are transferred in any charging process. In the case of charging by friction, they are transferred between the two objects being rubbed together. Prior to the charging, both objects are electrically neutral and the situation obeys the law of conservation of charge. How does the law of conservation of charge explain Sam and his balloon?
A) Friction creates charge that collected on the balloon.
B) The balloon gained electrons; Sam's hair gained protons.
C) If the balloon acquires charge, Sam's hair loses charge.
D) The net charge before is zero; the net charge after is negative.

Ben and Jerry, arch rivals, decide to have a weight lifting contest during PE. Ready, set, go! Ben and Jerry both lift a 250 kg barbell 10 times over their heads. They are approximately the same height and lift the barbell the same distance in the air. It takes Ben 5 seconds to complete 10 lifts; it takes Jerry 25 seconds to complete his 10 lifts.
Which statement is MOST accurate regarding the weightlifting contest?
A) Ben did more work than Jerry.
B) Ben has more power than Jerry.
C) Ben and Jerry have the same power.
D) Ben does more work and is more powerful than Jerry.
Physics
2 answers:
vova2212 [387]3 years ago
4 0

Answer: C. If the balloon acquires charge, Sam's hair loses charge.

B. Ben has more power than Jerry.

Explanation: I took the test and these answers are correct. Have a nice day! :)

yan [13]3 years ago
3 0

Answer: If the balloon acquires charge, Sam's hair loses charge.

Explanation:

I JUST GOT THIS QUESTION RIGHT ON THE IA4 <3

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In reality, as the battery was first connected to the primary, a very short blip of voltage would appear on the secondary as the magnetic field initially grew in the core (that would constitute a time-changing magnetic field required for the device to work). But it would immediately go away afterwards and that output cannot be determined without actually doing the experiment due to so many unknowns.
But this question isn't looking for that explanation. It's kind of a trick question.
7 0
3 years ago
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6th grade science I mark as brainliest.​
Viktor [21]

Answer:

divide 10 by 50.

Explanation:

Because its time over speed 10/50

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3 years ago
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Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piec
Wittaler [7]

Answer:

The fraction of the total initial kinetic energy is lost during the collision is \dfrac{11}{17}\ J

Explanation:

Given that,

Mass of one piece = 300 g

Speed of one piece = 1 m/s

Mass of other piece = 600 g

Speed of other piece = 0.75 m/s

We need to calculate the final velocity

Using conservation of energy

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

Put the value intro the formula

300\times10^{-3}\times1+600\times10^{-3}\times(0.75)=(300\times10^{-3}+600\times10^{-3})v

v=\dfrac{00\times10^{-3}\times1+600\times10^{-3}\times(-0.75)}{(300\times10^{-3}+600\times10^{-3})}

v=-0.5\ m/s

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2

Put the value into the formula

K.E_{i}=\dfrac{1}{2}\times300\times10^{-3}\times1^2+\dfrac{1}{2}\times600\times10^{-3}\times(0.75)^2

K.E_{i}=0.31875\ J

We need to calculate the total final kinetic energy

Using formula of kinetic energy

K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2

Put the value into the formula

K.E_{f}=\dfrac{1}{2}\times(300\times10^{-3}+600\times10^{-3})\times(-0.5)^2

K.E_{f}=0.1125\ J

We need to calculate the energy lost during the collision

Using formula of energy lost

energy\ lost=\dfrac{0.31875-0.1125}{0.31875}

energy\ lost=\dfrac{11}{17}\ J

Hence, The fraction of the total initial kinetic energy is lost during the collision is \dfrac{11}{17}\ J

3 0
3 years ago
If one of two interacting charges is doubled, the force between the charges will _____________.
malfutka [58]

If one of two interacting charges is doubled, the force between the charges will double.

Explanation:

The force between two charges is given by Coulomb's law

F=\frac{k q1 q2}{r^{2}}

K=constant= 9 x 10⁹ N m²/C²

q1= charge on first particle

q2= charge on second particle

r= distance between the two charges

Now if the first charge is doubled,

we get F'=\frac{k (2q1) q2}{r^{2}}

F'= 2 F

Thus the force gets doubled.

4 0
3 years ago
Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
larisa86 [58]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • Mass of the larger disk = M_2\ =\ 1.60\ kg
  • Mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the larger disk = R_2\ =\ 5.00\ cm\ =\ 0.05\ m
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)

From the pulley,

\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)

From the equation (1) and (2),

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = R_2\ =\ 0.05\ m

Let a_2 be the acceleration of the block while connecting to the larger pulley.\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2

4 0
3 years ago
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