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elixir [45]
3 years ago
15

A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if

a loose part were released from the spacecraft without a push in any direction, what would happen to it? A. It would slowly coast to a stop. B. It would head off on its own trajectory. C. It would stop immediately as soon as it was no longer in contact with the probe. D. It would continue along with the spacecraft.
Physics
1 answer:
77julia77 [94]3 years ago
6 0
C because when the part gets out of the probe it would no longer stay contacted
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Install sound absorbers such as foam or acoustical panels

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9. A football punter attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in th
vichka [17]

Answer:

Angle is 55.52°

and Initial Speed is v=26.48 m/s

Explanation:

Given data

x_{o}=0m\\ y_{o}=1.23m\\a_{oy}=a_{1y}=g=-9.8m/s^{2} \\x_{1}=67.0m\\y_{1}=0m\\t_{o}=0\\a_{ox}=m/s^{2} \\t_{1}=4.50s

Applying the kinematics equations for motion with uniform acceleration in x and y direction

So

x_{1}=x_{o}+v_{ox}t_{1}=67.0m\\0+4.50v_{o}Cos\alpha =67.0m\\v_{o}Cos\alpha =14.99\\v_{o}=14.99/Cos\alpha.....(1) \\and\\y_{1}=y_{o}+v_{oy}t_{1}+(1/2)a_{oy}t_{1}^{2} =0m\\ 1+4.50v_{o}Sin\alpha+(-9.8/2)(4.5)^{2}=0\\  v_{o}Sin\alpha=21.828.....(2)

Put the value of v₀ from equation (1) to equation (2)

So

\frac{14.99}{Cos\alpha }(Sin\alpha ) =21.828\\as\\tan\alpha =Sin\alpha /Cos\alpha \\So\\14.99tan\alpha =21.828\\tan\alpha =21.828/14.99\\\alpha =tan^{-1}(21.828/14.99) \\\alpha =55.52^{o}

Put that angle in equation (1) or equation (2) to find the initial velocity

So from equation (1)

v_{o}=(\frac{14.99}{Cos\alpha } ) \\v_{o}=(\frac{14.99}{Cos(55.52) } ) \\v_{o}=26.48m/s

7 0
3 years ago
Which explains the information needed to calculate speed and velocity?
goblinko [34]

The second statement is the correct choice.  Don't make me type it out.

8 0
3 years ago
Read 2 more answers
A lizard accelerates from 2 m/s to 10 m/s in 4 seconds. what is the lizard average acceleration
VLD [36.1K]

Acceleration = (change in speed) / (time for the change)

change in speed = (ending speed) - (starting speed)

change in speed = (10 m/s) - (2 m/s)  =  8 m/s

Acceleration = (8 m/s) / (4 sec)

Acceleration = (8/4) (m/s²)

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8 0
3 years ago
How do you do this question? Please include free body diagrams and clear explanation, so I can understand.
vagabundo [1.1K]

Explanation:

Draw a free body diagram for each disc.

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Wa = 86.5 N − T₁

ma × 9.8 m/s² = 86.5 N − 55.6 N

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Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.

∑F = ma

T₁ − T₂ − Wb = 0

Wb = T₁ − T₂

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mb = 1.9 kg

Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.

∑F = ma

T₂ − T₃ − Wc = 0

Wc = T₂ − T₃

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mc = 2.7 kg

Disc D has two forces acting on it: T₃ up and Wd down.

∑F = ma

T₃ − Wd = 0

Wd = T₃

md × 9.8 m/s² = 9.6 N

md = 0.98 kg

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3 years ago
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