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4 years ago

How does the gravity in the space shuttle compare with the gravity on earth’s surface?

2 answers:

7 0

These days they are exactly identical. All the space shuttles are currently in museums located on the surface of the Earth.

Years ago when the space shuttles were in orbit, the gravity inside a shuttle was about 85% of the gravity on the Earth's surface.

zheka24 [161]4 years ago

5 0

The gravity in the space shuttle is approximately equal to the gravity on the surface of the earth

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What is the American psychology Association ethical principle of psychology and code of conduct based upon

solmaris [256]

The correct answer is B

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3 years ago

What is the insertion for the brachialis

Alona [7]

Coronoid process of the ulna

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3 years ago

Two thermometers are calibrated, one in de

zlopas [31]

-40 c = -40 f but k would be 233.15

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3 years ago

A car travels along a clear 10.0 km section of motorway in 6.0 minutes. It then drives through 3.0 km of roadwork in 3.0 minutes

Charra [1.4K]

6min=1/10h=0.1h
3min=1/20h=0.05h

$\\ \bull\tt\dashrightarrow Avg\:Speed=\dfrac{Total\:Displacement}{Total\:Time}$

$\\ \bull\tt\dashrightarrow Avg\:Speed=\dfrac{10+3}{0.1+0.05}$

$\\ \bull\tt\dashrightarrow Avg\:Speed=\dfrac{13}{0.15}$

$\\ \bull\tt\dashrightarrow Avg\:Speed=86.6km/h$

5 0

3 years ago

Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

frosja888 [35]

Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

Explanation:

Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

8 0

3 years ago