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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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Answer:

A: No because it is nor changing speed or direction

B: Yes because it changes direction even though the speed is constant

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2 years ago
Use newton's law to explain the vertical acceleration of a projectile
NISA [10]

Answer:

Explained below

Explanation:

Newton's first law of motion: This law states that an object will remain at rest or continue in constant motion except it's acted upon by an external force. In projectile motion, the horizontal component of velocity will remain unchanged because we ignore air resistance since no force is acting in that horizontal direction.

Newton's second law of motion: This law states that force is the product of mass and acceleration. In projectile the force acts downwards, thus f = mg.

But g = a since internal forces will cancel out.

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3 years ago
Which of the following is a correct formula of speed
satela [25.4K]

Answer:s= d/t

Explanation:

7 0
3 years ago
Which of the following exhibit the Tyndall Effect?
posledela

Answer:

Probably all but (a)

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Salt water and a foggy night will cause dispersion of the incident light.

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8 0
2 years ago
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t
dybincka [34]

Answer:

A) t = 7.0 s    

B) x = 25 m  

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t       v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1  m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0
3 years ago
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