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Volgvan
3 years ago
8

Which of the following statements is an example of absolute time?

Physics
2 answers:
Nataly_w [17]3 years ago
7 0
I think the answer is : Bobby exactly three years old today

to put it simply, absolute time can be perceived without depended to other object as comparison.

for example Evan is older than james is not an absolute time, because this conditions depend on the age comparison between the two subject

ExtremeBDS [4]3 years ago
7 0

The answer is B, Bobby is exactly three years old today

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Answer:

She is likely to crash because her flight gradient is lesser than the flight gradient required gradient to avoid crashing

Explanation:

The given parameters are;

The required gradient of the plane Ashley is flying needs to reach in order to take off and not crash = 360 m/km

The initial elevation of the plane Ashley is flying = Sea level = 0 m

The goal Ashley intends to make = Elevation of 1000 m at 2.8 km. distance

∴ Ashley's goal = Traveling from sea level to 1000 m at 2.8 km horizontal distance

We have;

The gradient  = Rate of change of elevation/(Horizontal distance)

Therefore;

The gradient of Ashley's flight = (1000 - 0)/(2.8 - 0) = 357.143 m/km

The gradient of Ashley's flight ≈ 357.143 m/km which is lesser than the required 360 m/km in order to take off and not crash, therefore, she will crash.

6 0
3 years ago
The type of friction between the pavement and the tires on a moving vehicle is called.._____friction.
FinnZ [79.3K]

Answer:

A) Kinetic

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3 years ago
Heat and/or energy are transferred from one object to another if the objects have _________ temperature(s).
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Answer:

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A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di
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Answer:

D

Explanation:

From the information given:

The angular speed for the block \omega = 50 \ rad/s

Disk radius (r) = 0.2 m

The block Initial velocity is:

v = r \omega \\ \\  v = (0.2  \times 50) \\ \\  v= 10 \ m/s

Change in the block's angular speed is:

\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s

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I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2

The time t = 10s

∴

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T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10})  \\ \\ =0.6 \ N.m

Finally, the frictional force is calculated as:

F = \dfrac{T}r{}

F= \dfrac{0.6}{0.2} \\ \\ F = 3N

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On sources it says it would  just be the super giant star 
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