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lapo4ka [179]
3 years ago
10

A plate moves 200m in 10,000 years. what is the rate in cm/year

Biology
1 answer:
algol133 years ago
6 0

Answer:

Rate = 2 cm/year

Explanation:

It is given that, a plate moves 200m in 10,000 years. We need to find its rate in cm/year.

We know that, 1 m = 100 cm

200 m = 200 × 100 cm

= 20000 cm

Its rate is given by :

R=\dfrac{20000}{10,000}\\\\R=2\ cm/year

So, the plate is moving at a rate of 2 cm/year.

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7 0
3 years ago
The second phase of the general adaptation syndrome (gas) is characterized by ____.
Novay_Z [31]

The second phase of general adaptation syndrome (gas) is characterized by alarm phase, the stage of resistance and the exhaustion stage.  The gas stage gives a description of the both the short and long term emotional and physical effects of stress.  Gas elaborates the relationship between health and stress.

3 0
4 years ago
Which of the following best illustrates a type I survivorship curve, with higher death rates at older ages? Which of the followi
Rama09 [41]

Answer:

large mammals such as gorillas

Explanation:

The survivorship curves refer to the graphical representation of the proportion of the fraction of survivors or the individuals at a given age.

There are three types of survivorship curves which can be constructed by studying the life history of the organisms.

The type I survivorship curve is the curve which can be formed with the organism which has a high survival rate at the younger and middle age and high death rate at the older age. The type I curve can be characterised by its convex shaped. The type I is showed by the large mammals like gorilla, humans and many others.

Thus, large mammals such as gorillas are the correct answer.

4 0
3 years ago
The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there a
Agata [3.3K]

Answer:

2,855 is the size of the colony of mosquitoes after 4 ​days.

Explanation:

The law of uninhibited growth is given as:

A=A_o\times e^{k\times t}

A_o= Original amount

A = Amount after time t

k = Positive constant repressing the rate of growth

We are given with:

Original population of mosquitoes = 1000

Population of mosquitoes after 1 day =1300

t = 1 day

1300=1000\times e^{k\times 1 day}

k = 0.2623 day^{-1}

Population size of mosquitoes after 4 days

A_o=1000, k= 0.2623 day^{-1}

A =? , t = 4 days

A=1000\times e^{ 0.2623 day^{-1}\times 4 days}

A =2,855.36 ≈ 2,855 mosquitoes

2,855 is the size of the colony of mosquitoes after 4 ​days.

5 0
4 years ago
Read 2 more answers
If a red-flowering plant was crossed with a white-flowering plant what would the phenotypes of the offspring be?
dem82 [27]
Well if both are dominate colors the phenotype of the offsprings would be pink or spotted. if the red flowering plant was dominate and the white flowering plant is recessive then the phenotype would be 100% red, they would just be the carrier of the white color. if the white flower is dominate and the red is recessive the flowers phenotype would be 100% white, but be carriers of the red color.
6 0
3 years ago
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