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Pavel [41]
3 years ago
11

14. Karim brings a chair from the backyard into his house and is going to stack it on top of another chair. Both chairs are the

same size and have the same number of molecules. The diagram above shows the chairs before they touch. Use the information in the diagram to answer the question.
How does the temperature of the bottom chair compare with the temperature of the top chair before the chairs touch? What will happen after the chairs have been touching for a while?

a Before the chairs touch, the bottom chair is cooler than the top chair. Once the chairs are touching, the top chair will transfer kinetic energy to the molecules in the cooler bottom chair until both chairs reach the same temperature, which will be in between their starting temperatures.

b Before the chairs touch, the bottom chair is hotter than the top chair. Once the chairs are touching, the bottom chair will transfer kinetic energy to the molecules in the cooler top chair until both chairs reach the same temperature, which will be in between their starting temperatures.

c Before the chairs touch, the chairs are different temperatures. Once the chairs are touching, both kinetic energy and cold energy will transfer between the molecules in the two chairs until both chairs reach the same temperature, which will be in between their starting temperatures.

d Before the chairs touch, the bottom chair is hotter than the top chair. Once the chairs are touching, the cooler top chair will gain kinetic energy until the molecules in both chairs have an energy of 35, because hotter things increase the temperature of cooler things.

Biology
1 answer:
geniusboy [140]3 years ago
5 0

Answer:

(C.)

Explanation:

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Explanation:

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In cocker spaniels, black color (b) is dominant over red (b), and solid color (S) is dominant over spotted (s).
Levart [38]

Answer:

<h2>e) 3/4 </h2>

Explanation:

As given;

Black (B) is dominant over red (b), so B is dominant over b.

Solid (S) is dominant over spotted color(s), so S is dominant over s.

Offspring of BBss and bbss individuals are:

Gametes of parent BBss are Bs, Bs and gamete of bbss are bs, bs

offspring are : all Bbss

Cross between these offspring is; Bbss×Bbss

Gametes of parent Bbss are: Bs and bs  as given that the genes are linked.

Bs, bs × Bb, bs

Offspring are = BBss, Bbss, Bbss, bbss

Black spotted (B_ss) = 3/4

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