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3 years ago

Rank the work done on the charged particles from highest to lowest

Physics

1 answer:

Artist 52 [7]3 years ago

4 0

Answer:

correct order of work done is given as

2 > 3 > 4 > 1

Explanation:

When positive charge is dragged opposite to electric field then in that case work done is positive while if negative charge is dragged opposite to electric field then the work done is negative

So here we will have

positive work for charge 2, 3 and 4 while for negative charge particle 1 the work done is negative

now here we also know that work done is product of force and displacement

so here work depends on displacement of charge particle

maximum displacement is force particle 2

minimum is for particle 4

so correct order of work done is given as

2 > 3 > 4 > 1

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A 3.00 kg object is fastened to a light spring, with the intervening cord passing over a pulley. The pulley is frictionless, and

finlep [7]

Answer:

Part a)

k = 588.6 N/m

Part b)

v = 0.7 m/s

Explanation:

As we know that initially block is at rest

now if block is released from rest then it will go down by 10 cm and again comes to rest

so here we have

Part a)

Work done by gravity + work done by spring force = change in kinetic energy

$W_g + W_{spring} = 0 - 0$

$mg(0.10) + \frac{1}{2}k(0^2 - 0.10^2) = 0$

$3(9.81)(0.10) - \frac{1}{2}k(0.10)^2 = 0$

$k = 588.6 N/m$

Part b)

Now when spring is stretch by x = 5 cm then the speed of the block is given as

$mgx' + \frac{1}{2}k(0^2 - x'^2) = \frac{1}{2}mv^2 - 0$

here we have

$x' = 0.05 m$

$3(9.81)(0.05) - \frac{1}{2}(588.6)(0 - 0.05^2) = \frac{1}{2}(3) v^2$

$1.4715 - 0.736 = 1.5 v^2$

$v = 0.7 m/s$

3 0

3 years ago

The current in the wires of a circuit is 60.0 milliAmps. If the voltage impressed across the ends of the circuit were doubled (w

Pavel [41]

Answer:

120 mA

Explanation:

The relationship between voltage, current and resistance in a circuit is given by Ohm's law

$I=\frac{V}{R}$

where

I is the current

V is the voltage

R is the resistance

In this problem, we have an initial current of

I = 60 mA

Then the voltage is doubled:

V' = 2 V

while the resistance is kept the same:

R' = R

So the new current is

$I' = \frac{V'}{R'}=\frac{2V}{R}=2\frac{V}{R}=2 I$

so, the current has doubled. Since I = 60 mA, the new current is

$I' = 2(60 mA)=120 mA$

3 0

3 years ago

1. A pendulum has a period of 3 seconds. What's its frequency? 2. A pendulum has a frequency of 0.25 Hz. What's its period? 3. A

RideAnS [48]

Answer:

(1) 0.333 Hz

(2) 4 sec

(3) 2 sec, 0.5 Hz

Explanation:

(1) We have given time period of pendulum is 3 sec

So T = 3 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{3}=0.333Hz$

(2) Frequency of the pendulum is given f = 0.25 Hz

Time period is equal to $T=\frac{1}{f}=\frac{1}{0.25}=4sec$

(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds

So time taken to complete 1 back and forth swings = $=\frac{20}{10}=2sec$

So time period T = 2 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{2}=0.5Hz$

6 0

3 years ago

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Sholpan [36]

Answer:

(d) 22.22 °C

Explanation:

Because in 1 Celcius is 1.8 fahrenheit apart from 2 celcius, so

when °C is 0 f is 32, when °C is 1 = 33.8 , so it rises with quantities of 1.8 fahrenheit

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3 years ago

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Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$