All categories

3 years ago

The diagram above shows the process of natural selection. What is X?

Physics

2 answers:

zzz [600]3 years ago

8 0

Answer: D = reproduction

Explanation:

Mice21 [21]3 years ago

4 0

Answer:

Reproduction

Explanation:

because reproduction produces variation within the species allowing the species to survive selection

You might be interested in

Which instrument is used to measure liquid volume?

Alecsey [184]

Volumetric cylinders and volumetric flasks

3 0

3 years ago

Read 2 more answers

NEED ANSWER ASAP WHOEVER ANSWERS THE QUICKEST ILL GIVE BRAINILEST ANSWER!!!

zhenek [66]

A i think because it’s talking about covering ground, a faster car would cover more ground.

7 0

3 years ago

An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion,

grigory [225]

I think b or c sorry

3 0

3 years ago

What type of lever has the fulcrum between the resistance arm and the effort arm?

crimeas [40]

I’m not really sure but I think it’s D type 1 lever

4 0

3 years ago

A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas

algol [13]

Answer:

6.3 rev/s

Explanation:

The new rotation rate of the satellite can be found by conservation of the angular momentum (L):

$L_{i} = L_{f}$

$I_{i}*\omega_{i} = I_{f}*\omega_{f}$

The initial moment of inertia of the satellite (a solid sphere) is given by:

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$ : is the satellite mass and r: is the satellite's radium

$I_{i} = \frac{2}{5}m_{s}r^{2} = \frac{2}{5}1900 kg*(4.6 m)^{2} = 1.61 \cdot 10^{4} kg*m^{2}$

Now, the final moment of inertia is given by the satellite and the antennas (rod):

$I_{f} = I_{i} + 2*I_{a} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}m_{a}l^{2}$

Where $m_{a}$ : is the antenna's mass and l: is the lenght of the antenna

$I_{f} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}150.0 kg*(6.6 m)^{2} = 2.05 \cdot 10^{4} kg*m^{2}$

So, the new rotation rate of the satellite is:

$I_{i}*\omega_{i} = I_{f}*\omega_{f}$

$\omega_{f} = \frac{I_{i}*\omega_{i}}{I_{f}} = \frac{1.61 \cdot 10^{4} kg*m^{2}*8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kg*m^{2}} = 6.3 rev/s$

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

I hope it helps you!

5 0

4 years ago