What's now called "Conventional current" is thought of as the flow of positive charge, from the battery's positive terminal to its negative one.

But it turns out that positive charges don't flow. The physical flow of charge is the flow of electrons. They come out of the battery's negative terminal, and carry negative charge around the circuit to the battery's positive one.

6 0

3 years ago

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A time-varying horizontal force F(t) = At4 + Bt2 acts for 0.500 s on a 12.25-kg object, starting attime t = 1.00 s. In the SI sy

PSYCHO15rus [73]

Answer:

3.82 Ns

Explanation:

Time varying horizontal Force is given as

F(t) = A t⁴ + B t²

F(t) = 4.50 t⁴ + 8.75 t²

Impulse imparted is given as

$I = \int_{0}^{t}Fdt$

$I = \int_{0}^{1}Fdt$

$I = \int_{0}^{1}(4.50 t^{4} + 8.75 t^{2})dt$

$I = ((0.9) (1)^{5} + (2.92) (1)^{3})$

$I = 3.82 N-s$

7 0

3 years ago

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What energy transfer will a stretched rubber band have when let go

GarryVolchara [31]

Answer:

when the rubber band is realeased the potential energy is quickly converted to kinetic energy this is equal to one mass of the the rubber band multiplied by its velocity( in meters per second)

3 0

2 years ago

I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT

ratelena [41]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR = E- V

IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

$R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms$

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

7 0

3 years ago