# I need someone to paraphrase this for me

**Question description**

(10 points)

^{1.
}An object falls from rest on a high tower and takes
5.0 s to hit the ground. Calculate the object’s position from the top of the
tower at 1.0 s intervals. Make a position-time graph for the object’s motion.
In your response, show what you are given, the equation that you used, any
algebra required, a table of data, and your graph.

*g* = 9.8 m/s^{2}

Answer:

Given : Initial velocity = 0 m/s

Acceleration = 9.8 m/s^{2}

Time= 5 sec with intervals of 1 sec

Required : distance at 0 to 5 sec

Used formula: d = vi * t + ½ ( a* t^2 )

Solution :

d = vi * t + ½ ( a* t^2 ) d = vi * t + ½ ( a* t^2 )

d = 0 (0) + ½ ( 9.8 m/s^2 * (0s)^2) d = 0( 1) + ½ (9.8m/s^2 * (1s)^2)

d = ½ (0 m) d = ½ (9.8m)

d = 0 m d = 4.9 m

d = vi * t + ½ ( a* t^2 ) d = vi * t + ½ ( a* t^2 )

d = 0(2) + ½ ( 9.8 m/s^2 * (2s)^2) d = 0(3) + ½ ( 9.8 m/s^2 * (3s)^2)

d = ½ (9.8 m * 4) d = ½ ( 9.8 m *9)

d = 19.6 m d = 44.1 m

d = vi * t + ½ ( a* t^2 ) d = vi * t + ½ ( a* t^2 )

d = 0(4) + ½ ( 9.8 m/s^2 * (4s)^2) d = 0(5) + ½ (9.8m/s^2 *(5s)^2)

d = ½ ( 9.8 m * 16) d = ½ (9.8 m * 25)

d = 78.4 m d = 122.5

Time (s) |
Position(m) |

0 |
0 |

1 |
4.9 |

2 |
19.6 |

3 |
44.1 |

4 |
78.4 |

5 |
122.5 |

^{1.
}A car
accelerates uniformly from +10.0 m/s to +40.0 m/s over a distance of 125 m. How
long did it take to go that distance? Show all your work, including the
equation used, given and unknown quantities, and any algebra required. Make
sure your answer has the correct number of significant figures.

Answer:

Quantities: initial velocity = 10.0 m/s

final velocity = 40.0 m/s

distance = 125 m

time =?

Formula : d= ½ (vi + vf) * t

2d/ (vi + vf) = (vi + vf)* t/ (vi + vf)

t = 2d / (vi + vf)

Solution: t = (10.0m/s + 40.0m/s)

250m

t = (50m/s)

t = 5.0 sec

^{1.
}A car speeds
up from 10.0 m/s to 30.0 m/s in 3.0 s. What is its acceleration? In your
response, show your work, the given and unknown quantities, the equation and
any necessary algebra. Express your answer with the correct number of
significant digits.

Answer:

quantities : initial velocity = 10 m/s

final velocity = 30 m/s

time = 3 sec

acceleration = ?

solution :

a= (vf-vi)/time

a=(30.0m/s-10.0m/s)/3.0s

a=6.7m/s^2

**COMMENTS **

·
**Hello,**

**Watch your significant digits.
Good work though.**

**1. **Describe two important ways in which weight and mass
differ and one important way in which they are related.

Answer:

Mass by definition is the measure of an object's inertia which is the force needed to accelerate an object. However, weight by definition is the force that causes an object's acceleration due to gravity. Another difference is that the mass of an object doesn't change no matter where the object is moved to, while the weight of an object changes when moved to other places when the acceleration due to gravity changes. One important way they are related is that they are directly proportional to each other.

**2. **According to Newton’s first law, if no force acts on
an object, it will just continue moving. Explain in terms of Newton’s first law
why a car engine must exert a force between the car and the road to drive at a
constant speed on a straight and level road.

Answer:

Newton's first law is about balanced forces which result in zero net force. If an object is at rest and remains at rest then that means there is no net force acting on it to make it change its acceleration. Also if the object was moving and continues to move at a constant velocity then that means the forces causing that motion are balanced and the total net force acting on the object is zero. That being said, if this car is to keep moving in a constant speed on a straight road then the forces acting on it must be balanced. This means the engine has to exert force that is of the equal magnitude but opposite direction to the other forces of friction( ground and air frictions).

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