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Olegator [25]
3 years ago
15

Jana is observing a young autistic boy in a preschool classroom. Her job is to note each time the boy gets up from his chair wit

hout being told to do so. Jana's method of noting this behavior is known as :
Shorthand
Tallying
Open coding
Reporting
Physics
1 answer:
svp [43]3 years ago
3 0
The answer is tallying
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Question 8
spayn [35]

Answer:

500 years so hundreds

Explanation:

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5 0
3 years ago
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled
Ksivusya [100]

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

8 0
3 years ago
Which of the following is produced as a result of photosynthesis
kenny6666 [7]
I think the answer is D but i could be wrong
3 0
3 years ago
Read 2 more answers
When Rick has inflammation of the connective tissue in his wrist, he has _____.
Len [333]
He has tendonitis. Hope this helps
5 0
3 years ago
Read 2 more answers
before colliding the momentum of block A is -100 kg*m/s, and block B is -150 kg*m/s. after, block A has a momentum -200 kg*m/s.
Virty [35]

The momentum of block B after the collision is -50 kg m/s.

Explanation:

We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:

p_A + p_B = p'_A + p'_B

where:

p_A = -100 kg m/s is the momentum of block A before the collision

p_B = -150 kg m/s is the momentum of block B before the collision

p'_A = -200 kg m/s is the momentum of block A after the collision

p'_B is the momentum of block B after the collision

Solving for p'_B, we find:

p'_B = p_A + p_B - p'_A = -100 +(-150) -  (-200)=-50 kg m/s

So, the momentum of block B after the collision is -50 kg m/s.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

6 0
3 years ago
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