PHYSICS MIDDLE SCHOOL

What are the 2 forms of memory retrieval?

Answers

Answer 1
Answer:

recall and recognition


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MIDDLE SCHOOL

An arrow of mass 415 g is shot at a target with a speed of 68.5m/s. The target, which has a mass of 3.3 kg, is moving toward the arrow with a speed of 1.1m/s. The surface on which the target is sliding is friction-free. If the arrow embeds itself in the target, what will be the velocity of the target/arrow mass after the collision? Take the arrow's initial velocity direction as positive

Answers

The velocity of the target and arrow after collision is 6.67m/s

Explanation:

Given:

Mass of arrow, mₐ = 415g

Speed of arrow, vₐ = 68.5m/s

Mass of the target, mₓ = 3.3kg = 3300g

speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction

Velocity of the target and arrow after collision, vₙ = ?

Applying the conservation of momentum,

mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ

415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ

28427.5 - 3630 = 3715 X vₙ

24797.5 = 3715 X vₙ

vₙ = 6.67m/s

Therefore, the velocity of the target and arrow after collision is 6.67m/s

COLLEGE

A bumblebee can sense electric fields. The most likely mechanism underlying this sensitivity is motion of hairs on the bee’s body; the hairs bend in response to an electric field. Support for this mechanism as the source of the sensitivity is the fact that putting an electric charge on a bee (as occurs when a bee flies through the air) dramat- ically increases electric field sensitivity. Explain why adding charge to a bee would cause hairs on the bee’s body to bend more in response to a field.

Answers

Answer:

If the electrical field stays the same, the size of the electrical force on the bee's hair would be proportional to the size of the charge on it.

Explanation:

The question implies that:

  1. The bee's hairs carry electrical charges.
  2. The electrical field around the bee exerts a force on the bee's hair, causing them to bend.
  3. The bee senses the electrical field by sensing the bend in its hairs.

Consider the bee's hair as a point charge. The force between an electrical field and that point charge would be equal to

.

  • the strength and direction of the electrical field, , and
  • the size (and sign) of the charge on the object. .

In this equation, the size of the electrostatic force is proportional to the size of , the charge on the bee. Adding charge to the bee's hair would increase the size . Even if the size of the field stayed the same, there would be an increase the size of .

The hairs would now bend more since the force on them has become stronger. That means the bee would become more sensible to the electrical field.

COLLEGE

A moving electron passes near the nucleus of a gold atom, which contains 79 protons and 118 neutrons. At a particular moment the electron is a distance of 7.5 × 10−9 m from the gold nucleus. (a) What is the magnitude of the electric force exerted by the gold nucleus on the electron?

Answers

The magnitude of the electric force exerted by the gold nucleus on the electron .

What is electric force?

Electric force is the force of attraction or repulsion between two charged particles. It can be given as,

Here is coulomb's constant, is charge on the objects and  is the distance between two objects.

Given information-

The number of proton in gold atom is 79.

The number of neutrons in gold atom is 118.

The distance of the electron from the nucleus is .

  • a) The magnitude of the electric force exerted by the gold nucleus on the electron-

The charge on electron is and the charge on the proton is .

Put the values in the above equation as,

Hence the magnitude of the electric force exerted by the gold nucleus on the electron .

Learn more about the electric force here;

brainly.com/question/14372859

Answer:

Explanation:

Given data

Distance r=7.5×10⁻⁹m

Charge of electron -e= -1.6×10⁻¹⁹C

Charge of proton e=1.6×10⁻¹⁹C

To find

Electric force F

Solution

From Coulombs law we know that:

q₁ is charge of electron

q₂ is the charge of gold nucleus which contains 79 positively charge protons and 118 neutral neutrons.  

The Charge of single proton e=1.6×10⁻¹⁹C

79 proton charge q₂=79×1.6×10⁻¹⁹=1.264×10⁻¹⁷C

So

HIGH SCHOOL

A substance has a volume of 10.0cm^3 and a mass of 89 grams. what is its density?

Answers

Density = mass / volume = 89 / 10 = 8.9g/cm^3.

Note if you want them in SI units you will have to convert them first.
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