Answer:
s = 14.3 ft
Explanation:
First we need to calculate the distances traveled by both the cars. We use third equation of motion for that:
2as = Vf² - Vi²
where,
a = acceleration
s = distance
Vf = Final Velocity
Vi = Initial velocity
FOR CAR A:
Vi = Va = (40 mph)(5280 ft/1 mile)(1 h/3600 s) = 58.66 ft/s
Vf = 0 ft/s
a = aA = - 22 ft/s²
s = sa = ?
Therefore,
2(- 22 ft/s²)(sa) = (58.66 ft/s)² - (0 ft/s)²
sa = 78.2 ft
FOR CAR B:
Vi = Vb = (45 mph)(5280 ft/1 mile)(1 h/3600 s) = 66 ft/s
Vf = 0 ft/s
a = aB = - 20 ft/s²
s = sb = ?
Therefore,
2(- 20 ft/s²)(sb) = (66 ft/s)² - (0 ft/s)²
sb = 108.9 ft
Since, the car A was initially 45 ft ahead of car B. Therefore,
sa = 45 ft + 78.2 ft = 123.2 ft
Now, the distance between the cars will be:
s = sa - sb
s = 123.2 ft - 108.9 ft
<u>s = 14.3 ft</u>
Answer:
An object at rest does not move and an object in motion does not change its velocity, unless an external force acts upon it
Explanation:
This statement is also known as Newton's first law, or law of inertia.
It states that the state of motion of an object can be changed only if there is an external force (different from zero) acting on it: therefore
- If an object is at rest, it will remain at rest if there is no force acting on it
- If an object is moving, it will continue moving at constant velocity if there is no force acting on it
This phenomenon can be also understood by looking at Newton's second law:
F = ma
where
F is the net force on an object
m is the mass
a is the acceleration
If the net force is zero, F = 0, the acceleration of the object is also zero, a = 0: therefore, the velocity of the object does not change, and it will continue moving at the same velocity (which can be zero, if the object was at rest).
Answer:
the length of resultant vector is 5 inches
Explanation:
In the attachment, you will see X vector along the X axis and Y vector along the Y direction. Let the resultant vector be denoted by ![R](https://tex.z-dn.net/?f=R)
The resultant vector always start from the tail of first vector and ends at the head of second vector.
Since it is a right angled triangle, so the result of resultant vector is found by using Pythagoras theorem i.e.
So, when we plug in the values of X = 48 and Y = 14
![R^{2} =48^{2} +14^{2}](https://tex.z-dn.net/?f=R%5E%7B2%7D%20%3D48%5E%7B2%7D%20%2B14%5E%7B2%7D)
![x^{2} =2500\\R=\sqrt{2500} \\R=5](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%3D2500%5C%5CR%3D%5Csqrt%7B2500%7D%20%5C%5CR%3D5)
Therefore, the length of resultant vector is 5 inches.
It is possible for a wooden ruler to be worn out at the edge. If such a ruler is used for measurement, there will be error in the measurement due to the inaccuracy of the ruler. To avoid this, some rulers do not start with zero at the edge.
Answer:
its terminal velocity is 19.70 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s
Explanation:
Firstly,
given that
m = 580g = 0.58kg
Area A = 0.11 * 0.22 = 0.0242m
g = 9.8
idensity constant p = 1.21 kg/m^3
the terminal velocity of the sphere Vt is ;
Vt = √ ( 2mg / pCA)
we substitute
Vt = √ ( (2*0.58*9.8) / (1.21*1*0.0242)
Vt = √ (11.368 / 0.029282)
Vt = √ ( 388.22)
Vt = 19.70 m/s
its terminal velocity is 19.70 m/s
What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
The Velocity of the person is;
V2 = √ 2ax
V2 = √ ( 2 * 9.8 * 4 )
V2 = √ (78.4)
V2 = 8.85 m/s
the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s