Tina and her puppy are tugging on a rope each pulling on opposite ends when is tina doing work ? When is Tina doing work ?

Physics

2 answers:

Rufina [12.5K]3 years ago

5 0

Answer:

whenever she pulls the puppy toward her

Explanation:

Varvara68 [4.7K]3 years ago

3 0

Explanation : Tina is doing work all of the time because whiles she’s pulling and allowing the rope to be pulled her body using muscles to keep herself upright and a firm grip on the rope.

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A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

blsea [12.9K]

Answer:

a) Q = C*emf

b) Reduction in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are connected in series with the battery

Using Kirchoff's voltage law, sum of all voltages in the circuit is zero

Let $V_{R}$ = Voltage dropped across the Resistor

$V_{c}$ = Voltage dropped across the capacitor

Applying KVL;

$emf - V_{R} - V_{c} = 0\\$ .........................(1)

Since the connection is in series, the same current flow through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Putting $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

At the final charge, the capacitor in fully charged, and current drops to zero due to equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) Current starts running through the plate because as the sheet of plastic is inserted between the plates both the electric field intensity and the electric potential reduces. The charge also reduces, then current flows

c) The current through the resistor is the current through the entire circuit ( series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$