The best question that could prompt a scientific investigation is: <u>What substances dissolve in ocean water?</u>
This way the person experimenting can use several variables and make observations. Upon making observations, the person can be able to gather as much data as he can in order to answer the original question that he asked.
Answer:
a. Quadruped arm and opposite leg raise
Explanation:
Quadruped arm and opposite leg lift
- Kneel on the floor, lean forward and place your hands down.
- Keep your knees in line with your hips and hands directly under your shoulders.
- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.
- Go back to the starting position.
This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.
It is also used together with other exercises for the treatment of hyperlordosis.
Answer:
2.32 s
Explanation:
Using the equation of motion,
s = ut+g't²/2............................ Equation 1
Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.
Note: Since Onur drops the basket ball from a height, u = 0 m/s
Then,
s = g't²/2
make t the subject of the equation,
t = √(2s/g')...................... Equation 2
Given: s = 10 m, g' = 3.7 m/s²
Substitute this value into equation 2
t = √(2×10/3.7)
t = √(20/3.7)
t = √(5.405)
t = 2.32 s.
Answer:
1. Largest force: C; smallest force: B; 2. ratio = 9:1
Explanation:
The formula for the force exerted between two charges is
![F=K\dfrac{ q_{1}q_{2}}{r^{2}}](https://tex.z-dn.net/?f=F%3DK%5Cdfrac%7B%20q_%7B1%7Dq_%7B2%7D%7D%7Br%5E%7B2%7D%7D)
where K is the Coulomb constant.
q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.
For simplicity, let's combine Kq₁q₂ into a single constant, k.
Then, we can write
![F=\dfrac{k}{r^{2}}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bk%7D%7Br%5E%7B2%7D%7D)
1. Net force on each particle
Let's
- Call the distance between adjacent charges d.
- Remember that like charges repel and unlike charges attract.
Define forces exerted to the right as positive and those to the left as negative.
(a) Force on A
![\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7DF_%7BA%7D%20%26%20%3D%20%26%20F_%7BB%7D%20%2B%20F_%7BC%7D%20%2B%20F_%7BD%7D%5C%5C%26%20%3D%20%26%20-%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%20%20-%20%5Cdfrac%7Bk%7D%7B%282d%29%5E%7B2%7D%7D%20%20%2B%5Cdfrac%7Bk%7D%7B%283d%29%5E%7B2%7D%7D%5C%5C%26%20%3D%20%26%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%5Cleft%28-1%20-%20%5Cdfrac%7B1%7D%7B4%7D%20%2B%20%5Cdfrac%7B1%7D%7B9%7D%20%5Cright%29%5C%5C%5C%5C%26%20%3D%20%26%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%5Cleft%28%5Cdfrac%7B-36%20-%209%20%2B%204%7D%7B36%7D%20%5Cright%29%5C%5C%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B-%5Cdfrac%7B41%7D%7B36%7D%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%7D%5C%5C%5C%5C%5Cend%7Barray%7D)
(b) Force on B
![\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7DF_%7BB%7D%20%26%20%3D%20%26%20F_%7BA%7D%20%2B%20F_%7BC%7D%20%2B%20F_%7BD%7D%5C%5C%26%20%3D%20%26%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%20%20-%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%20%20%2B%20%5Cdfrac%7Bk%7D%7B%282d%29%5E%7B2%7D%7D%5C%5C%26%20%3D%20%26%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%20%5Cright%29%5C%5C%5C%5C%26%20%3D%20%26%5Cmathbf%7B%5Cdfrac%7B1%7D%7B4%7D%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%7D%5C%5C%5C%5C%5Cend%7Barray%7D)
(C) Force on C
![\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7DF_%7BC%7D%20%26%20%3D%20%26%20F_%7BA%7D%20%2B%20F_%7BB%7D%20%2B%20F_%7BD%7D%5C%5C%26%20%3D%20%26%20%5Cdfrac%7Bk%7D%7B%282d%29%5E%7B2%7D%7D%20%2B%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%20%20%2B%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%5C%5C%26%20%3D%20%26%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%5Cleft%28%20%5Cdfrac%7B1%7D%7B4%7D%20%2B1%20%2B%201%20%5Cright%29%5C%5C%5C%5C%26%20%3D%20%26%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%5Cleft%28%5Cdfrac%7B1%20%2B%204%20%2B%204%7D%7B4%7D%20%5Cright%29%5C%5C%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B%5Cdfrac%7B9%7D%7B4%7D%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%7D%5C%5C%5C%5C%5Cend%7Barray%7D)
(d) Force on D
![\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7DF_%7BD%7D%20%26%20%3D%20%26%20F_%7BA%7D%20%2B%20F_%7BB%7D%20%2B%20F_%7BC%7D%5C%5C%26%20%3D%20%26%20-%5Cdfrac%7Bk%7D%7B%283d%29%5E%7B2%7D%7D%20%20-%20%5Cdfrac%7Bk%7D%7B%282d%29%5E%7B2%7D%7D%20%20-%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%5C%5C%26%20%3D%20%26%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%5Cleft%28%20-%5Cdfrac%7B1%7D%7B9%7D%20-%20%5Cdfrac%7B1%7D%7B4%7D%20-1%20%5Cright%29%5C%5C%5C%5C%26%20%3D%20%26%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%5Cleft%28%5Cdfrac%7B-4%20-%209%20-36%7D%7B36%7D%20%5Cright%29%5C%5C%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B-%5Cdfrac%7B49%7D%7B36%7D%20%5Cdfrac%7Bk%7D%7Bd%5E%7B2%7D%7D%7D%5C%5C%5C%5C%5Cend%7Barray%7D)
(e) Relative net forces
In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.
![F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\](https://tex.z-dn.net/?f=F_%7BA%7D%20%3A%20F_%7BB%7D%20%3A%20F_%7BC%7D%20%3A%20F_%7BD%7D%20%20%3D%20%20%5Cdfrac%7B41%7D%7B36%7D%20%3A%20%5Cdfrac%7B1%7D%7B4%7D%20%3A%20%5Cdfrac%7B9%7D%7B4%7D%20%3A%20%5Cdfrac%7B49%7D%7B36%7D%5C%20%3D%2041%20%3A%209%20%3A%2081%20%3A%2049%5C%5C%5C%5C%5Ctext%7BC%20experiences%20the%20largest%20net%20force.%7D%5C%5C%5Ctext%7BB%20experiences%20the%20smallest%20net%20force.%7D%5C%5C)
2. Ratio of largest force to smallest
![\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}](https://tex.z-dn.net/?f=%5Cdfrac%7B%20F_%7BC%7D%7D%7B%20F_%7BB%7D%7D%20%3D%20%5Cdfrac%7B81%7D%7B9%7D%20%3D%20%5Cmathbf%7B9%3A1%7D%5C%5C%5C%5C%5Ctext%7BThe%20ratio%20of%20the%20largest%20force%20to%20the%20smallest%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B9%3A1%7D%7D%24%7D)
Explanation:
The given data is as follows.
Angular velocity (
) = 2.23 rps
Distance from the center (R) = 0.379 m
First, we will convert revolutions per second into radian per second as follows.
= 2.23 revolutions per second
=
= 14.01 rad/s
Now, tangential speed will be calculated as follows.
Tangential speed, v =
= 0.379 x 14.01
= 5.31 m/s
Thus, we can conclude that the tack's tangential speed is 5.31 m/s.