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gulaghasi [49]
2 years ago
11

Amplitude is a Measure of a Waves

Physics
1 answer:
Rasek [7]2 years ago
5 0

C) Energey

A) True

I believe, dont quote me on it

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You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun us
ZanzabumX [31]

Answer:

The answer is

A. Pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.

Explanation:

 The question is incomplete, here is a complete question with full options

You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun uses a plunger which is operated by pulling back on a handle. You must squeeze the handle very hard to get the caulk to come out of the narrow opening because:_________.

A. pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.

B. viscous drag between the walls of the tip and the caulk causes the caulk to swirl around chaotically.

C. Newton’s third law requires most of the energy in the caulk to be used to push back on the plunger rather than moving it through the tip.

D. the high density of the caulk impedes its flow through the small opening.

Since the caulk is thick and the exit nozzle is small, the pressure needed to deliver the caulk will be very high as pressure is uniformly distributed at the plunger side at every part of the caulk, hence very high pressure is needed to deliver the caulk which is why the handle needed the very hard squeeze

3 0
3 years ago
Two resistors with values of 6.0 ohms and 12 ohms are connected in parallel. This combination is connected in series with a 2.0
ivanzaharov [21]

Answer:

4A

Explanation:

According to ohm's law;

E = IRt where;

E is the source voltage = 24volts

I is the total current flowing in the circuit = ?

Rt is the total effective resistance in the circuit.

To find Rt, we will resolve the resistors in parallel first.

Since 6ohms and 12ohms resistors are in parallel, their effective resistance will give;

1/R = 1/6+1/12

1/R= 2+1/12

1/R = 3/12

3R = 12

R = 4ohms.

This resistor will now be in series with the 2.0ohms resistor to finally have;

Rt = 4+2

Rt = 6ohms

From the ohms law formula;

I = E/Rt

I = 24/6

I = 4Amperes

The total current in the circuit is 4A

This same currents will flow in the 2ohms resistor since same current flows in a series connected resistors.

3 0
3 years ago
Read 2 more answers
Look at the graph.<br><br><br><br><br><br> If the price of peaches increases, what can be expected?
Fittoniya [83]
There will be less peaches soled
5 0
3 years ago
Read 2 more answers
Using newtons second law of motion, how fast for 100 KG object accelerates 350 N of force is applied to
fenix001 [56]

Answer:

3.5m/s^2

Explanation:

From Newton's second Law of Motion

F = ma

Where F is the applied force, m is the mass of the object and a is the acceleration.

F = 350 N

Mass = 100kg

350N = 100×a

a = 350/100

a = 3.5m/s^2

The acceleration of the object will be 3.5m/s^2

6 0
3 years ago
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
lesantik [10]

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

        R_1  =  \frac{V^2}{100}

Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

       P \ \alpha  \  R

So  when they are connected in series

     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0
3 years ago
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