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krok68 [10]
3 years ago
15

When Emeka was first training his dog, Smooches, to heel he would give Smooches a treat when she stayed close during walks. Now

Smooches stays right by Emeka's side, even when she is not on her leash. In this case, a. the dog treats were positive reinforcers for staying close. b. the dog treats were negative reinforcers for staying close. c. the staying close was a positive reinforcer for receiving dog treats. d. the staying close was a negative reinforcer for receiving dog treats.
Physics
1 answer:
shepuryov [24]3 years ago
8 0
<span>Answer a. is correct. In this case the dog treats were positive reinforcers for staying close (the desirable behaviour). Repeated rewards for a certain behaviour work as positive reinforcers.</span>
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The moon is 3.85 x 10 to the 8 m from earth and has a diameter of 3.48 x 10 to the 6 m. You have a pea (diameter = 0.50 cm) and
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Answer:

(a) dime

Explanation:

Convert all to metric unit:

0.5 cm = 0.005 m

1.8 cm = 0.018 m

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In order to find out we would need to calculate the ratio R between the object diameter d and their distance s to our eyes:

R_m = \frac{d_m}{s_m} = \frac{3.48*10^6}{3.85*10^8} \approx 0.009

R_p = \frac{d_p}{s_p} = \frac{0.005}{0.71} \approx 0.007

R_d = \frac{d_d}{s_d} = \frac{0.018}{0.71} \approx 0.0253

Since the ratio of the dime is larger than the ratio of the moon, and the ratio of the pea is smaller than the ratio of the moon, only the (a) dime can cover your view of the moon.

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A train car has a mass of 10,000 kg and is moving at +3.0 m/s. It strikes an identical train car that is at rest. The train cars
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Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks.  So if Fork-A is 256 Hz and the beat is      6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz.  But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz.  That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz.  While it was loaded with wax, it was 261 Hz.

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