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Julli [10]
3 years ago
8

Humberto builds two circuits using identical components.Circuit 1: A series circuit with three lightbulbs Circuit 2: A parallel

circuit with three lightbulbsHe then adds two more lightbulbs in a series with the bulbs in circuit 1 and two more lightbulbs on new branches in parallel with each bulb in circuit 2.What can he expect to see?
Physics
2 answers:
hammer [34]3 years ago
8 0

Hello. You forgot to enter the answer options. The options are:

"A. All bulbs in circuit 1 will shine at the same brightness as the original bulbs, while all bulbs in circuit 2 will be dimmer than the original bulbs.  B. All bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.  C. The original bulbs will remain at their original brightness, but the bulbs added to circuit 1 will be dimmer than the bulbs added to circuit 2.  D. The original bulbs will remain at their original brightness, but the bulbs added to circuit 1 will be brighter than the bulbs added to circuit 2."

Answer:

B. All bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

Explanation:

Series circuit, as the name implies, is a circuit with two or more loads that are being fed in series with each other, connected in sequence, with only one path for the passage of electric current. Another way to visualize a circuit in series is that the loads have only one point in common between them, that is, there is no derivation point. In this case, in relation to the question above, all lamps in circuit 1 will glow less than the original lamps.

On the other hand, we have as the main characteristics of a parallel circuit that both the current and electrical voltage in the circuit will behave differently. In the case of electrical voltage, it will always be the same for all loads in the circuit, that is, the same voltage delivered by the source. The electric current will not be the same in the charges, unless they have two with the same resistance value. This is also due to a mathematical relationship, knowing that the voltage is the same in all charges, the electric current will vary according to the resistance, as they are inversely proportional quantities. For these reasons, in relation to the question above, we can say that all lamps in circuit 2 will have the same brightness as the original lamps.

PolarNik [594]3 years ago
5 0
<span>Humberto builds two circuits using identical components,
and then adds components to each circuit.

Circuit 1:
A series circuit with three lightbulbs. 
Then add three more lightbulbs in series.

Circuit 2:
A parallel circuit with three lightbulbs
Then add two more lightbulbs on new branches
in parallel with each original bulb.

After adding the new lightbulbs in Circuit 1:
-- the voltage across each of the original bulbs is less,
-- the current through the whole series circuit is less,
-- the original three bulbs shine dimmer than before, and
-- the total power delivered from the battery is less.
-- The battery lasts longer.

After adding the new lightbulbs in Circuit 2:  
</span>-- the voltage across each of the original bulbs is doesn't change, 
-- the current through each original bulb doesn't change,
-- the original three bulbs shine just as bright as before, 
-- the total currrent drawn by the circuit, and the total current
delivered by the battery, increases, and
-- the total power delivered from the battery increases.
-- The battery runs down sooner.
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3 years ago
16. A 95kg fullback, running at 8.2m/s, collided in midair with a 128 kg defensive tackle moving in the opposite direction. Both
Daniel [21]

a) 779 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

For the fullback before the collision,

m = 95 kg

v = 8.2 m/s

Therefore, his momentum was:

p=mv=(95)(8.2)=779 kg m/s

b) -779 kg m/s

After the collision, both the fullback and the tackle come to a stop: this means that their momentum after the collision is zero,

p' = 0

The initial momentum of the fullback was

p = 779 kg m/s

Therefore, his change in momentum is

\Delta p = p' -p =0-779  = -779 kg m/s

where the negative sign indicates that the direction is opposite to the initial direction of motion.

c) -779 kg m/s

Here we can apply the law of conservation of momentum. In fact, the total momentum before and after the collision must be conserved. So we can write:

p_f + p_t = p'

where

p_f is the initial momentum of the fullback

p_t is the initial momentum of the tackle

p' is the final combined momentum after the collision

We already know that

p_f = 779 kg m/s\\p' = 0

Therefore, we can find the tackle's original momentum:

p_t = p'-p_f = 0-(779) = -779 kg m/s

where the negative sign indicates that the direction is opposite to the initial direction of motion of the fullback.

e) -6.1 m/s

To find the velocity of the tackle, we can use again the equation of the momentum:

p = mv

where here we have

p=-779 kg m/s is the original momentum of the tackle

m = 128 kg is his mass

Solving the equation for v, we find the tackle's original velocity:

v=\frac{p}{m}=\frac{-779}{128}=-6.1 m/s

So, he was moving at 6.1 m/s in the direction opposite to the fullback.

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<span>θ=0.3sin(4t)
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therefore, wmax=1.2rad/s
 
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(may be different if your picture/radius is from a different picture)

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