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insens350 [35]
3 years ago
10

A parallel circuit has two 8.0-ohm resistors and a power source of 9.0 volts. If a 12.5-ohm resistor is added to the circuit in

parallel, how will the current be affected and what value will it have?
Physics
1 answer:
Airida [17]3 years ago
7 0
-- The two 8-ohm resistors in parallel are equivalent to
a single 4-ohm resistor.

-- The current in the circuit is 

                 (voltage) / (resistance) = (9v) / (4ohms) = 2.25 Amperes

-- If another resistor is added in parallel, then no matter how large
or small it is, the current in the circuit increases. 
(When another lane is added to a road, the traffic capacity of the
road increases.)

The circuit is now equivalent to a 12.5-ohm resistor in parallel
with 4-ohms.  That's ...

       (12.5 x 4) / (12.5 + 4)  =  50/16.5  =  3.03 ohms .

-- The new current is  (9v) / (3.03 ohms)  =  2.97 Amperes
            
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Answer:

5.39 m

Explanation:

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s=u_x t where t is time, s is distance and u_x is initial speed in x direction

u_x= 20.5cos 30.5=17.6634 m/s

u_y= 20.5 sin 30.5=10.40454 m/s

Now 17.5=17.6634 t

t=\frac {17.5}{17.6634} =0.99 s

Using the equation of kinematics

h=u_y t- 0.5gt^{2}

h=10.40454\times 0.99^{2}-0.5\times 9.81\times 0.99^{2}

h=5.39 m

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3 years ago
A Cessna aircraft has a liftoff speed of 120 km/h What minimum constant acceleration does the aircraft require to be airborne af
Lady_Fox [76]

Answer:

<h3>2.3125m/s²</h3>

Explanation:

Using the equation of motion v² = u²+2aS

v is the final velocity = 120km/hr

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u is the initial velocity = 0m/s

a is the acceleration

S is the distance covered = 240m

On substituting the given parameters

33.3² = 0²+2a(240)

33.3² = 480a

1110 = 480a

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Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²

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4 years ago
When a low-mass star runs out of helium to fuse into carbon, what will it do?
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how much heat is lost when 71.07g of steam at 100 degrees condenses to form liquid water at 100 degrees
alukav5142 [94]

Answer:

Q = 160.36[kJ], is the heat lost.

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This is a thermodynamic problem where we can find the latent heat of vaporization, at a constant temperature of 100 [°C].

We know that for steam at 100[C], the enthalpy is

h_{gas} = 2675.6[kJ/kg]

For liquid water at 100[C], the enthalpy is

h_{water} = 419.17[kJ/kg]

Therefore

h_{g-w} = 2675.6-419.17 = 2256.43[\frac{kJ}{kg} ]

The amout of heat is given by:

Q=h_{g-w}*m\\ where:\\m = mass = 0.07107[kg] = 71.01[g]\\Q = heat [kJ]\\Q =2256.43*0.07107\\Q=160.36[kJ]

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Answer:

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As the arrow rises, it undergoes the most deceleration. This is because as it rises, its speed decreases until it gets to the peak of its path, where its velocity is momentarily zero.  

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b. When is the arrow moving fastest?

As the arrow hits the ground, its velocity is highest (maximum), so it is fastest at this point.

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