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4 years ago

Dinosaurs’ skeletons can be distinguished from those of other reptiles by the structure of the hips and legs.

1 answer:

5 0

True, they had a hole in their hip socket that allowed them to run faster than other reptiles of their size at the time. As well as most reptiles besides reptiles had legs to the side, rather than under them like dinosaurs did.

Hope this helps!

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What is the science definition of rounding

MrRa [10]

Alter (a number) to one less exact but more convenint calculations.

3 0

3 years ago

A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a rad

Anna71 [15]

Answer:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

$a_{r} = \omega^{2}\cdot R$

Where:

$\omega$ - Angular speed, measured in radians per second.

$R$ - Radius of rotation, measured in meters.

The angular speed is first determined:

$\omega = \frac{\pi}{30}\cdot \dot n$

Where $\dot n$ is the angular speed, measured in revolutions per minute.

If $\dot n = 3000\,rpm$ , the angular speed measured in radians per second is:

$\omega = \frac{\pi}{30}\cdot (3000\,rpm)$

$\omega \approx 314.159\,\frac{rad}{s}$

Now, if $\omega = 314.159\,\frac{rad}{s}$ and $R = 0.1\,m$ , the resultant acceleration is then:

$a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)$

$a_{r} = 9869.588\,\frac{m}{s^{2}}$

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

6 0

3 years ago

A car travels a distance of 25 meters in 3 minutes what's its velocity in m/s?

iogann1982 [59]

vel = 25/3x60

25/180 m/s

5/36m/s

5 0

3 years ago

Voltage drops across a source of emf.<br><br> A. True <br> B. False

iren2701 [21]

A.true hope it help :)

5 0

4 years ago

The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of

Y_Kistochka [10]

Answer:

Q = 590,940 J

Explanation:

Given:

Specific heat (c) = 1.75 J/(g⋅°C)

Mass(m) = 2.01 kg = 2,010

Change in temperature (ΔT) = 191 - 23 = 168°C

Find:

Heat required (Q)

Computation:

Q = mcΔT

Q = (2,010)(1.75)(168)

Q = 590,940 J

Q = 590.94 kJ

5 0

3 years ago