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Anon25 [30]
2 years ago
5

Can inserting a resistor in a circuit produce an effect similar to a short circuit?

Physics
1 answer:
tresset_1 [31]2 years ago
6 0

Answer:

By ‘inserting’ you means to putting a resistor in series. In this case, no, there is no resistance that would produce the same effect as a short circuit.

If you adding a resistor in parallel with the circuit, then if it had a low value It might be similar to a short circuit. I

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Ball A (mass of 0.5 kg) going +2.0 m/s collides with a stationary ball B (mass of 0.4 kg).
ki77a [65]

The correct answer is 1.2 m/s

: mv+mv=mv+mv

(0.5kg)(2m/s)+(0.4kg)(0m/s)=(0.5kg)v+(0.4kg)(1m/s)

= 1kg*m/s=(0.5kg)v+0.4kg*m/s

=1kg*m/s-0.4kg*m/s=(0.5kg)v

=0.6kg*m/s=(0.5kg)v

to solve for v we divide both side by 0.5kg

v=1.2m/s

6 0
3 years ago
How do you make the chemical equations LiOH+H2(SO4) and H2O+Li2(SO4) balanced?
Likurg_2 [28]

Answer:

here

Explanation:

6 0
3 years ago
Goldie Doldfish, a speed swimmer, loves to rece around the parks pond, which is 0.5 miles around. If she can swim around the tra
rosijanka [135]

Answer:

Her average speed is 0.25 miles / hour.

Explanation:

I did 0.5 divided by 2 = 0.25 miles / hour.

<em>Hope this helps! :)</em>


8 0
2 years ago
Explain how polarization of a cell increases the cell's internal resistance.<br>(2<br>2.​
Mandarinka [93]

Answer:

Explanation: The chemical action that occurs in the cell while the current is flowing causes hydrogen bubbles to form on the surface of the anode. This action is called POLARIZATION. Some hydrogen bubbles rise to the surface of the electrolyte and escape into the air, some remain on the surface of the anode. If enough bubbles remain around the anode, the bubbles form a barrier that increases internal resistance. When the internal resistance of the cell increases, the output current is decreased and the voltage of the cell also decreases.

   A cell that is heavily polarized has no useful output. There are several methods to prevent polarization or to depolarize the cell.

   One method uses a vent on the cell to permit the hydrogen to escape into the air. A disadvantage of this method is that hydrogen is not available to reform into the electrolyte during recharging. This problem is solved by adding water to the electrolyte, such as in an automobile battery. A second method is to use material that is rich in oxygen, such as manganese dioxide, which supplies free oxygen to combine with the hydrogen and form water.

   A third method is to use a material that will absorb the hydrogen, such as calcium. The calcium releases hydrogen during the charging process. All three methods remove enough hydrogen so that the cell is practically free from polarization.

LOCAL ACTION

   When the external circuit is removed, the current ceases to flow, and, theoretically, all chemical action within the cell stops. However, commercial zinc contains many impurities, such as iron, carbon, lead, and arsenic. These impurities form many small electrical cells within the zinc electrode in which current flows between the zinc and its impurities. Thus, the chemical action continues even though the cell itself is not connected to a load.

   Local action may be prevented by using pure zinc (which is not practical), by coating the zinc with mercury, or by adding a small percentage of mercury to the zinc during the manufacturing process. The treatment of the zinc with mercury is called amalgamating (mixing) the zinc. Since mercury is many times heavier than an equal volume of water, small particles of impurities weighing less than mercury will float to the surface of the mercury. The removal of these impurities from the zinc prevents local action. The mercury is not readily acted upon by the acid. When the cell is delivering current to a load, the mercury continues to act on the impurities in the zinc. This causes the impurities to leave the surface of the zinc electrode and float to the surface of the mercury. This process greatly increases the storage life of the cell.

6 0
3 years ago
A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr
nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

 a= μ g                          ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

\omega=\sqrt{\dfrac{k}{M+m}}

By putting the values

\omega=\sqrt{\dfrac{150}{5.67+1}}

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2

kX^2=(m+M)v^2

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0
3 years ago
Read 2 more answers
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