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miskamm [114]
3 years ago
6

Help please!!!!

Physics
2 answers:
PtichkaEL [24]3 years ago
8 0
The answer is B) Intentional
Molodets [167]3 years ago
7 0

Answer:

intentional

Explanation:

We need to see the definition of each learning type to make a choice:

Practical is the type of learning done by observing and doing things as you witness someone else doing it, like mimicking, for example, when toddlers learn how to hold a spoon from watching their parents eat.

Intentional is done consciously, in a persistent way, and it is continual, with the actual goal of learning, it has a goal and it is driven by motivation. It is usually self-directed, and creates "metacognition", where the learner becomes aware that it is learning.

Incidental is a type of learning done indirectly, or in an unplanned way, unconsciously, like when playing video games improves your eye-hand coordination, or the best example, when we develop our vocabulary or learn a language as kids, you are not concentrated on doing it, but your mind or body is learning things from the environment and its input on you.

Deep processing involves relations and links, where the learner apply a different yet related knowledge to understand or expand on the subject that it's studying.

So Raina is practicing and repeating the speech to consciously be able to deliver it from memory. That is intentional learning

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Rest and Motion are the relative term. why? explain with example(please help me (╥﹏╥))​
klemol [59]

Explanation:

it depens on the subject and object. Let's example

you are driving a tesla car with your dog sitting on your side. You will say that the dog is at REST

but your friend, standing in sidewalk, seeing the same dog, will say that your dog is moving because it has MOTION from your car

3 0
3 years ago
A snowball is launched horizontally from the top of a building at v = 19.6 m/s. If it lands d = 64 meters from the bottom, how h
tiny-mole [99]

Answer: height of building = 18.8m

Explanation: The question is a projectile motion, a two dimensional motion with a vertical constant acceleration (g = - 9.8m/s²) and a constant horizontal velocity (thus making horizontal component of acceleration zero).

From the question, distance between bottom of building and where the object lands = 64m, initial velocity for throwing the object = 19.6m/s

The horizontal range formulae is given as

d= vt

Where d= horizontal range = 64m, v = initial velocity of throw.

64 = 19.6 × t

t = 64/ 19.6

t = 3.265 s.

Height (h) of the building is gotten by using the formulae

h =vt - 1/2gt²

h = (19.6×3.265) - 1/2×9.8×(3.265)²

h = 71.05 - (104.47/2)

h = 71.05 - 52.235

h = 18.8m

6 0
3 years ago
Calculate δe, if the system absorbs 7.24 kj of heat from the surroundings while its volume remains constant (assume that only p−
butalik [34]
I believe the answer is 7.24 kJ.
From the equation ΔE = dW + dQ; where W is the work done on/by the system and Q is the heat the system absorbs/loses.
Therefore; ΔE = 72.4 kJ since the system has bot done any p-v work (dV= zero) and has absorbed heat.
4 0
3 years ago
Read 2 more answers
Two similar cars are driving on a hill. The red car is traveling faster than the blue car. How could the blue car have more tota
yKpoI14uk [10]
D if the blue car started higher it would have more energy but since the red car is lower it is going faster because it’s going down a hill
8 0
3 years ago
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined
MrRa [10]

Incomplete question as the car's  speed is missing.I have assumed car's  speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

F_{B}=-5755N

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

\frac{mv^{2} }{R}=W+F_{B}\\  F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N

The minus sign for downward direction

6 0
3 years ago
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