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Novosadov [1.4K]
3 years ago
15

Which constellation is marked by a W or M? a. Cassiopeia b. Draco c. Ursa Minor d. Ursa Major

Physics
2 answers:
yuradex [85]3 years ago
8 0

Answer:

Ursa Minor Constellation

Explanation:

Ursa Minor constellation lies in the northern sky. The constellation's name means “the smaller bear,” or “the lesser bear,” in Latin. The Great Bear constellation is represented by its larger neighbor Ursa Major.

Anestetic [448]3 years ago
6 0

I believe it is the Cassiopeia. It is visible during winter in the early part of the night. It look like a distorted letter W or M.

I am not very sure of it though, I apologize if this is not what you meant.

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When the voltage across a steady resistance is doubled, the current?
natima [27]

I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.

<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know V and I use formula P = IV: P = IV = (100mA)(10V) = 1 W.</span>

The next question is what I'm not sure about:

Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).

What I did initially was: P = IV = (100mA)(2V) = 2 W

But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."

So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.

P = IV = (200mA)(2V) = 4 W

8 0
3 years ago
A dog barking at the sound of the postal carrier delivering mail is reacting to what type of cue?
Gwar [14]

Answer:

A. External

Explanation:

External stimulus includes touch/pain, vision, smell, taste, and sound.

6 0
2 years ago
An equilibrium constant is not changed by a change in pressure <br> a. True<br> b. False
Umnica [9.8K]
Hi There! :)


An equilibrium constant is not changed by a change in pressurea. True
b. False

False! :P
7 0
3 years ago
Read 2 more answers
A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 12 A, while that
andreev551 [17]

Answer:

a) 1450watts

b) 564watts

c) 1.11

Explanation:

Power consumed = IV

I is the current rating

V is the operating voltage

If a blow-dryer and a vacuum cleaner each operate with a voltage of 120 V and the current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 4.7 A then their individual power rating is calculated thus;

a) For blow-dryer

Operating voltage = 120V

Its current rating = 12A

Power consumed = IV

= 120×12

= 1440watts

b) For vacuum cleaner:

Operating voltage is the same as that of blow dryer = 120V

Its current rating = 4.7A

Power consumed = IV

= 120×4.7

= 564watts

c) Energy used = Power consumed × time taken

Energy used = Power × time

Energy used by blow dryer = 1440×20×60

= 1,728,000Joules

Energy used up by vacuum cleaner = 564×46×60

= 564×2760

= 1,556,640Joules

Ratio of the energy used by the blow-dryer in 20 minutes to the energy used by the vacuum cleaner in 46 minutes will be 1,728,000/1,556,640 = 1.11

4 0
3 years ago
A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change
notsponge [240]
We can solve the problem by using the first law of thermodynamics, which states that:
\Delta U = Q-W
where
\Delta U is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ
6 0
3 years ago
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