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Tju [1.3M]
3 years ago
5

Need help solving this

Physics
1 answer:
Nostrana [21]3 years ago
5 0

Answer:

cant see picture

Explanation:

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B because the the organism is changing into another chemical form


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If silicic rock is found in a core sample. What evidence does this provide about the rocks location ?
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Answer:

The rocks location is underneath an ocean or mountain

Explanation:

Silicic rock are igneous rock rich in silica. The amount of silica that constitutes a silicic rock is usually defined as at least sixty three percent. Granite and rhyolite are the most common silicic rocks found. These are found in the magma of the Earth.

Silicic is the group of silicate magmas which crystallise a relatively small proportion of ferromagnesian silicates. The main constituents of a silicic rock will be minerals rich in silica-minerals mainly.

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Using what you know about newton's second law explain why a car with a lager mass use more fuel than a car with a smalle mass. a
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The car with bigger Mass requires more fuel because More mass the heavier it gets. you can't grab the little car fuel tank and put it in the big car it will run out to fast.

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3 years ago
2. What do pitch and loudness have in common?
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Answer:

Both are subject to a persons interpretation

Explanation:

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5 0
3 years ago
The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t
jek_recluse [69]

Answer:

e=3367.2J

%e=1.43%

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

v_{r}=10km/h=2.77m/s

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s

Knowing that we can calculate Kinetic energy for the real and experimental speed

E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J

E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J

Now, the potential error in her calculated kinetic energy is:

e=E_{r}-E_{e}=(234023-230656)J=3367.2J

%e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43%

4 0
4 years ago
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