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OLEGan [10]
3 years ago
9

The ball and socket joint in the hip is similar to the ball and socket joint of the shoulder but is designed more for __________

than ___________.
Physics
2 answers:
Viktor [21]3 years ago
8 0

Answer:

a. power

b. precision

Explanation:

The hip is a spherical joint that allows the upper leg to move from front to back and from side to side. The largest joint that supports weight in the body, the hip joint is surrounded by strong ligaments and muscles.

It differs mainly with the circular articulation of the shoulder since it allows it to perform movement of greater power and with greater precision

kondaur [170]3 years ago
5 0

Answer:

The ball and socket joint in the hip is similar to the ball and socket joint in the shoulder, but is designed <em>more</em> for bio-mechanical strength, stability and ability to carry weight than a wide range of movement.

Explanation:

The kind of weights that the hips carry vary in nature.

Some weights are better described as stationary and others as ,constantly changing.  

Besides carrying it's own weight (that is the upper torso), the hip is also active and instrumental in the ability of mankind to lift and carry additional weights.

The study of how the hip manages these weights effortlessly is important in the design of prosthetic appendages and replacement hips for accident victims and geriatric patients who have one hip issue or the other.

Cheers!

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A rocket engine uses fuel and oxidizer in a reaction that produces gas particles having a velocity of 1380 ms The desired thrust
bearhunter [10]

Answer:

a. 141.3 kg/s b. 5.49 m/s² c. i. 104228.9 N ii. 8.53 m/s² d. i. 97305.2 N ii. 9.84 m/s²

Explanation:

a. What must be the fuel/oxidizer consumption rate (in kg s1)?

The thrust T = Rv where R = mass consumption rate and v = velocity of rocket. Since T = 195000 N and v = 1380 m/s,

R = T/v = 195000 N/1380 m/s = 141.3 kg/s

b. If the initial weight of the rocket is 125000 N, what is its initial acceleration?

We also know that thrust T - W = ma since the rocket has to move against gravity. where M = mass of rocket = W/g = 125000 N/9.8m/s² = 12755.1 kg, W = weight of rocket = 125000 N, a = acceleration of rocket and T = thrust = 195000 N.

So, T - W = Ma

195000 N - 125000 N = (12755.1 kg)a

70000 N = ma

a = 70000 N/12755.1 kg = 5.49 m/s²

c. What are the weight and acceleration of the rocket at t 15.0 s after ignition?

We know that the loss in mass ΔM = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 15 s,

ΔM = 141.3 kg/s × 15 = 2119.5 kg

The new mass is thus M = M - ΔM = 12755.1 kg - 2119.5 kg = 10635.6 kg

i.The weight after 15 seconds is thus W' = M'g = 10635.6 kg × 9.8m/s² = 104228.9 N

ii. Since T - W' = M'a. where M' is our new mass and a our new acceleration,

a = (T - W')/M'

= (195000 N - 104228.9 N)/10635.6 kg

= 90771.1 N/10635.6 kg

= 8.53 m/s²

d. What are the weight and acceleration of the rocket at 20.0 s after ignition?

We know that the loss in mass ΔM" = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 20 s,

ΔM" = 141.3 kg/s × 20 = 2826 kg

The new mass is thus M" = M - ΔM" = 12755.1 kg - 2826 kg = 9929.1 kg

i. The weight after 20 seconds is thus W" = M"g = 9929.1 kg × 9.8m/s² = 97305.2 N

ii. Since T - W" = M"a. where M" is our new mass and a our new acceleration,

a = (T - W")/M"

= (195000 N - 97305.2 N)/9929.1 kg

= 97694.8 N/9929.1 kg

= 9.84 m/s²

4 0
3 years ago
What is the limitation of relative dating?
astraxan [27]
Your not really supposed to date your relative
7 0
3 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
Hunk travelled 540 km at an average speed of 90 km/h. If Tim completed the same amount of distance with 1 hour and 30 min less t
lina2011 [118]

Answer:120

Explanation:it just is trust me

8 0
3 years ago
1. A metal ball has a mass of 20 g and a volume of 6 cm3.Find its density
vesna_86 [32]

Answer:

density of the ball is 3.33 g/cc

Explanation:

As we know that the density is the ratio of mass and volume

here we know that

mass = 20 g

volume = 6 cubic cm

so we will have

\rho = \frac{m}{V}

\rho = \frac{20}{6} g/cm^3

\rho = 3.33 g/cm^3

4 0
3 years ago
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