Linh builds a circuit from the diagram shown. A rectangular box of lines with the long side vertical and an extra horizontal line about 2 thirds of the way up. The top line has a circle with an X in it labeled 1. The right side from top has a top circle 2 with an X in it then the connection with the extra line then a bottom circle 4 with an X in it. The extra line has a circle 3 with an X in it. The left side has a stack of horizontal lines near its bottom, which are, from top very short, short, very short, short. If Linh removes bulb 3, which bulbs will remain lit? 1 only

1 and 2 only

4 only

1, 2, and 4
Linh builds a circuit from the diagram shown. A rectangular - 1

Answers

Answer 1
Answer:

If Linh removes bulb 3,the bulbs which will remain lit are 1, 2 and 4.

What is circuit?

The electric circuit provided with battery creating potential difference across resistance or capacitor or inductors.

When Linh removes the middle branch, the bulbs 1, 2 and 4 will be acting as resistor which are in series connection. The current flow will be same in all the resistors except 3.

Thus, If Linh removes bulb 3, the bulbs which will remain lit are 1, 2 and 4.

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Answer 2
Answer:

Answer:

1, 2, 4

Explanation:

Just took the quiz :)


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how far will a freely falling object have fallen from a position of rest when its instantaneous speed is 10m/s ?

Answers

Answer:

A freely falling object have fallen from a position of rest when its instantaneous speed is 10 m/s, it will travel 5.102 meter.

Explanation:

 We have equation of motion, v^2=u^2+2as, where u is the initial velocity, u is the final velocity, s is the displacement and a is the acceleration.

 In this case we have final velocity = 10 m/s, initial velocity = 0 m/s, and acceleration = 9.8 m/s^2.

Substituting

   10^2=0^2+2*9.8*s\n \n s=5.102 meter

A freely falling object have fallen from a position of rest when its instantaneous speed is 10 m/s, it will travel 5.102 meter.

Final answer:

When an object, starting from rest, achieves an instantaneous speed of 10 m/s while freely falling under gravity, it will have fallen approximately 5.10 meters.

Explanation:

Using the physics concept of motion under gravity, where the first equation of motion can apply (v = u + gt), your question involves finding the displacement of a freely falling object starting from rest until it reaches a speed of 10 m/s.

In this case, initial velocity (u) is zero because the object was at rest, the final velocity (v) is 10 m/s, and g is the acceleration due to gravity which is approximately 9.81 m/s².

We could rearrange the equation to find time: t = v/g = 10 / 9.81 ≈ 1.02 s. Then, we employ the second equation of motion to find the distance fallen, s = ut + 0.5gt². Since u=0, the formula simplifies to s = 0.5gt². Substituting g = 9.81 m/s² and t = 1.02 s into the equation yields s = 0.5 * 9.81 * (1.02)² ≈ 5.10 m.

Therefore, a freely falling object will have fallen approximately 5.10 meters when its instantaneous speed is 10 m/s.

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Is carbon dioxide a polar or non-polar molecule?

Answers

CO2 is a polar molecule

Your shopping cart has a mass of 65 kilograms, In order to accelerate the shopping cart down an aisle at 0.3 m/sec2, what force would you need to use or apply to the cart?

Answers

remember Newton's second law: F=ma

to get the force in newtons, mass should be in kg and acceleration in m/s^2

conveniently, we don't need to convert units

we just need to multiply the two to get the force

65* 0.3 = 19.5 kg m/s^2 or N

if significant digit is an issue, the least number if sig figs is 1 so the answer would be 20 N

Final answer:

To accelerate a shopping cart with a mass of 65kg at a rate of 0.3 m/sec2, you would need to exert a force of 19.5 Newtons, according to Newton's second law of motion.

Explanation:

The subject of this problem is physics, specifically the concept of force, mass, and acceleration within the domain of Newton's second law of motion. The law states that the force needed to accelerate an object is equal to the mass of the object multiplied by the desired acceleration.

Given the mass of the cart is 65 kilograms and the acceleration is 0.3 m/sec2, you can calculate the required force using the formula:
Force = mass * acceleration.

So, Force = 65 kg * 0.3 m/sec2 = 19.5 Newtons. Therefore, you would have to exert a force of 19.5 Newtons to accelerate the shopping cart at the specified rate.

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2. A ball tied to a pole by a rope swings in a circular path with a centripetal acceleration of 2.7 m/s. If the ball has atangential speed of 2.0 m/s, what is the diameter of the circular path in which it travels?​

Answers

Answer: The diameter of the circular path is 2.96m

Explanation: centripetal acceleration = tangential speed^2 / radius of the circular path.

Centripetal acceleration = 2.7m/s^2

Tangential speed = 2.0m/s

Radius = 2.0^2 / 2.7 = 4/2.7

= 1.48m

Diameter = radius*2

= 1.48*2 = 2.96m.

Most radiation is released as particles. Identify the type radiation that is not a particle.1.alpha
2.beta
3.gamma
4.positron

Answers

Gamma rays because it is a form of electromagnetic radiation. It has zero mass and no charge.

And object has a velocity of 9 m/s and a mass of 20 kg. What is the kinetic energy of this object

Answers

The kinetic energy of the object which is moving with velocity of 9 m/s and has a mass of 20 kg will be 810 joules.

What is Kinetic Energy ?

The energy possessed by the body by virtue of its motion is called kinetic energy. Mathematically -

E[K] = 1/2mv²

Given is an object that has a velocity of 9 m/s and a mass of 20 kg.

We can calculate the kinetic energy of the object by using the formula discussed above.

Mass [m] = 20 Kg

Velocity [v] = 9 m/s

Substituting the values, we get -

E[K] = 1/2 x 20 x 9 x 9

E[K] = 10 x 9 x 9

E[K] = 81 x 10

E[K] = 810 joules

Therefore, the kinetic energy of the object which is moving with velocity of 9 m/s and has a mass of 20 kg will be 810 joules.

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Explanation:

k.e =1/2m(v)²

or

k.e=m/2(v)²

so......k.e=20kg/2(9m/s)²

k.e=10kg(81m²/s²)

k.e=10kg(81m²/s²)

k.e=810kgm²/s²

HOPE ITS CORRECT

ANSWER

k.e=810kgm²/s²