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2 years ago

How to know which side is positive or negative on resistors.

Physics

1 answer:

Semenov [28]2 years ago

5 0

Answer:

"look for a large stripe or a minus sign (or both) on one side of the capacitor."

Explanation:

i hope it helped :)

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Which of the following would act as a barrier to sound waves?

Y_Kistochka [10]

Headphones and earplugs

4 0

3 years ago

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to

goldenfox [79]

Answer:

17304 J

Explanation:

Complete statement of the question is :

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .

Part A

How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

Solution :

$T$ = tension force in the tow rope = 350 N

$L$ = length of the incline surface = 120 m

$W_{t}$ = work done by tension force = ?

The tension force acts parallel to incline surface, hence work done by tension force is given as

$W_{t} = T L\\W_{t} = (350) (120)\\W_{t} = 42000 J$

$h$ = height gained by the rider = 30 m

$m$ = total mass of rider and tube = 84 kg

Potential energy gained is given as

$U = mgh\\U = (84) (9.8) (30)\\U = 24696 J$

$Q$ = Thermal energy created

Using conservation of energy

$Q = W_{t} - U\\Q = 42000 - 24696\\Q = 17304 J$

7 0

3 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago

We are running late for school and we want to make our 0.5 kg tea (it’s at 90 C) colder. Let’s assume we can drink tea when it’s

PSYCHO15rus [73]

Answer:

x=0.154kg

Explanation:

(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0

(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0

6 0

2 years ago

The Doppler Effect is used at baseball games to measure ___________. @3.2.P.B5 @1

Bogdan [553]

Answer:

Explanation:

Those devices the hold up while the pitcher is pitching measures speed. It has nothing to do with weather and temperature.

3 0

3 years ago

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