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2 years ago

5

Name two types of evidence used to support the theory of evolution. Explain how scientists use each type of evidence to provide

support for evolution. Give one example of each.

1 answer:

bearhunter [10]2 years ago

7 0

Well, one of the theories was that Charles darwin proved that theory evolution is real so that one theory.

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A mother pats a child every time he throws a chocolate wrapper in the dustbin. This is an example of (A) Observational Learning

TEA [102]

D. I think, not sure

8 0

3 years ago

defon

Answer:

Yes, a mixture can be made up of just elements and no compounds. The elements never experiance a chemical interaction and stay only in physical contact with each other.

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2 years ago

1. I’m in the 2nd column, 4th row, and I’m a metal. Who am I? ________________ 2. I’m a very lonely nonmetal. Who am I? ________

Ghella [55]

Answer:

Explanation:

I'm in 17th column , a nometal, and a solid at room temperature. What am i

6 0

2 years ago

The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to

LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power ,P = F .v

F=force

v=velocity

v= 100 km/h

$v=\dfrac{1000}{3600}\times 100\ m/s$

v=27.77 m/s

75 x 1000 = F x 27.77

$F= \dfrac{75000}{27.77}\ N$

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

3 0

3 years ago

On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediatel

Akimi4 [234]

Answer:48.2 Joules

Explanation:

Given

two masses of 0.2 kg and 0.4 kg collide with each other

after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east

Velocity of 0.2 kg mass is

$v_{0.2}=12\cos (30)\hat{i}+12\sin (30)\hat{j}$

$|v_{0.2}|=11.99 m/s\approx 12 m/s$

Velocity of 0.4 kg mass

$v_{0.4}=13\cos (53.1)\hat{i}-12\sin (53.1)\hat{j}$

$|v_{0.4}|=12.99 m/s\approx 13 m/s$

Thus total Kinetic energy $=\frac{0.2\times 12^2}{2}+\frac{0.4\times 13^2}{2}$

Kinetic energy=48.2 J

8 0

2 years ago