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3 years ago

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Describe Darwin’s observations on the Galápagos islands during his voyage on the HMS Beagle.

1 answer:

Helen [10]3 years ago

4 0

Answer:

During the voyage Charles Darwin explored the Galapagos islands and noticed the same species have different adaptations in places. ... Charles noticed that each species has the same ancestor but they evolve to adapt over time so they can live longer.

Explanation:

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Fossils found in the La Brea tar pits indicate a California climate that was A) similar to today's climate. B) similar to the pr

NikAS [45]

Answer:

i believe it is D but not 100% sure

Explanation:

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3 years ago

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PLZZZZ HELP QUICK . 30 points. Emma weighs 560 N, she has decided to stay in shape so she is working out every morning, This mor

Phantasy [73]

Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration.

1.)

Fw (weight) = m (mass) · g (gravity, 9.8 m/s²)

Fw = m * 9.81 m/s²

560N = m · 9.81 m/s²

m ≈ 57.08 kg

2.)

d = 350 meters

t = 65 seconds

velocity = d/t

velocity = 350 meters / 65 seconds

velocity ≈ 5.38 meters/sec

3.)

Force = 35N

Distance = 2 meters

Work = Force · Distance

Work = 35N · 2 meters

Work = 70 J

3 0

3 years ago

The velocities of light in air and glass are 3.0 x 10^8ms and 2.0×10^8ms respectively. If the angle of refraction is 30°, the si

JulijaS [17]

Answer:

0.75

Explanation:

$refractive \: index \: = \frac{3.0 \times {10}^{2} }{2.0 \times {10}^{2} }$

= 1.5

$refractive \: index = \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }$

$1.5 = \ \frac{ \sin(i) }{ \sin(30) }$

1.5 × ½ = sin(i)

$\sin(i) = 0.75$

5 0

3 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

$= 2 \times \sqrt{\frac{100}{4}}$

= 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

m' = 2m

Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

$=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

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3 years ago

Under what condition will the results of an experiment based on a hypothesis most likely lead to new experimentation?

MakcuM [25]

Failed experiments, uncontrolled variables, invalid data, and generalized human error

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3 years ago