1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
zvonat [6]
2 years ago
9

Anyone know what type of circuit this is?

Physics
1 answer:
Oksana_A [137]2 years ago
3 0

Answer:

C

Explanation:

BECAUSE ITS GOING ON AND ON IF ITS NOT CORRECT I WILL VOTE YOU BRAINLEST ON MY QUESTION

You might be interested in
If an object travels at a constant speed in a circular path, the acceleration of the object is:
Vilka [71]

Answer:

1- The acceleration of the object is larger in magnitude the smaller the radius of the circle.

Explanation:

The acceleration of an object in a circular path is

a = \frac{v^2}{r}

As can be seen from the equation, if the radius of the circle is decreases, the magnitude of the acceleration increases.

As for the direction of the acceleration, it is always towards the center, and it is always perpendicular to the direction of the velocity.

6 0
3 years ago
the specific heat of water is 4.2 j/c. if it takes 31,500 joules to heat to warm 750 g of water, what was the temperature change
defon
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q=mC_s \Delta T
where
m is the mass of the substance
C_s the specific heat capacity
\Delta T the increase in temperature

In our problem, the mass of the water is m=750 g, the specific heat is C_s = 4.2 J/g ^{\circ}C and the amount of heat supplied is Q=31500 J, so if we re-arrange the previous formula we find the increase in temperature of the water:
\Delta T= \frac{Q}{m C_s}= \frac{31500 J}{(750 g)(4.20 J/g^{\circ} C)}=10^{\circ}C
7 0
2 years ago
A 26.0 kg child plays on a swing having support ropes that are 2.40 m long. A friend pulls her back until the ropes are 45.0 ∘ f
Sloan [31]
A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.
8 0
3 years ago
The components of vector A are Ax = +2.2 and Ay = -6.9 , and the components of vector B are given are Bx = -6.1 and By = -2.2. W
Zina [86]
For simplicity, let's call vector B-A  vector C  Then C is
Cx = (-6.1 - 2.2)  
Cy = (-2.2 - (-6.9))  Or,
Cx = -8.3  Cy = 4.7
The magnitude is found with the Pythagorean theorem
||C|| = √(-8.3² + 4.7²) = 9.538
3 0
3 years ago
When lamps with wattages greater than the rating of the luminaire are installed, a fire could occur because the luminaire is bei
sergejj [24]

Answer:

true

Explanation:

Yes, it is true.

As the wattage is more than the prescribed wattage, it becomes overheated.

6 0
3 years ago
Other questions:
  • When a car drives over a speed bump and oscillates up and down in simple harmonic motion, at which position during the motion is
    11·1 answer
  • A nurse counts 66 heartbeats in one minute. What is the period of the hearts oscillation? In minutes
    12·1 answer
  • PLEASE HELP WHICH STATEMENTS ARE CORRECT DO NOT GUESS
    6·2 answers
  • If you know that the universe is estimated to be 14 billion years old . what is the age of the universe in seconds ???? (knowing
    8·2 answers
  • Whta would you call this?
    7·2 answers
  • A 5.8 kg bowling ball is rolling down the lane with a velocity of 8.1 m/s . Calculate the momentum of the bowling ball. Provide
    11·1 answer
  • A 4.88 x 10-6 C charge moves 265 m/s
    7·2 answers
  • ANYONE HELP ME PLEASE!
    9·1 answer
  • Establish the relationship between acceleration and force ​
    12·2 answers
  • A specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state. T
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!