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const2013 [10]
3 years ago
13

Objects in space that are moving at a constant velocity in a straight line ___________.

Physics
2 answers:
nikdorinn [45]3 years ago
6 0
The best answer is A) <span>keep moving at a constant velocity until some forces act on them

As the man you're probably tired of hearing about said:

"Every object persists in its state of rest or in uniform motion in a straight line unless a new force acts upon it" 
This is Isaac Newton's 1st law of motion, or the law of inertia. 

Put more simply, objects in motion tend to stay in motion, and tend the maintain the same velocity (direction and speed) and objects at rest tend to stay at rest. </span>
Alla [95]3 years ago
3 0
A. Keep moving at a constant velocity until some forces act on them
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An object is thrown 16m/s straight up from a 7m tall cliff. How much time does it take to hit the ground below
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0,54 sec

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3 years ago
A certain car traveling at 97 km/h can stop in 47 m on a level road find the coefficient of friction
IrinaVladis [17]

The coefficient of friction between the road and the car's tire is determined as 0.78.

<h3>Acceleration of the car</h3>

The acceleration of the car is calculated as follows;

v² = u² - 2as

0 = u² - 2as

a = u²/2s

where;

  • u is the initial velocity = 97 km/h = 26.94 m/s

a = (26.94)²/(2 x 47)

a = 7.72 m/s²

<h3>Coefficient of friction</h3>

μ = a/g

μ = (7.72)/9.8

μ = 0.78

Learn more about coefficient of friction here: brainly.com/question/14121363

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5 0
1 year ago
A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang
olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: s_c = \omega_ct = 0.233t

For the horse: s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

s_c = s_h

0.233t = 1.57 + 0.0068t^2

0.0068t^2 - 0.233t + 1.57 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}

t= \frac{0.233\pm0.11}{0.0136}

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0
3 years ago
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