The best activity for her to do to improve her range of motion is flexibility.
<h3>What are a few range of motion illustrations?</h3>
The term the range of motion (ROM) describes the extent to which a joint or muscle may be moved or stretched. Everybody has a distinct experience. For instance, whereas some people can perform a complete split, others cannot because their joints are stiff and their muscles are unable to extend as far.
<h3>What restricts motion range?</h3>
A joint is said to have a restricted range of motion when it cannot move easily and completely in its typical position. A mechanical issue within the joint, swollen tissues around the joint, or pain may restrict motion.
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Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.
- 1. Ball A will have the greater density
- 2. Ball C and Ball D have the same density.
- 3. Ball Q will have the greater density.
- 4. Ball X and Y will have the same density
The density of an object is given as its mass per unit volume of the object.
Mathematically;.
For Case 1:
- Va = Vb and Ma = 2Mb
- D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
- Therefore, the density of ball A,
- D(a) = 2D(b).
- Therefore, ball A has the greater density.
For Case 2:
- D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd
- Therefore, ball C and D have the same density
For Case 3:
- Vp = 2Vq and Mp = Mq
- D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
- Therefore, the density of ball P is half the density of ball Q
- Therefore, ball Q has the greater density.
For case 4:
Therefore, Ball X and Ball Y have the same density.
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Answer:
B.Ionizing radiation is the correct answer.
Explanation:
Ionizing radiation has sufficient energy that it can convert atoms and molecules into ions.
It has a sufficient amount of energy that it can separate tightly confined electrons from the orbit of an atom and causing that atom to become ionized.
Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
