True of False: All body parts and organs

THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A

Mice21 [21]

Answer: angular distance = 1700° and 29.7 rad

also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps!!!!

π

5 0

2 years ago

A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when

choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

$F_t - mg = ma$

here since speed is constant or it is at rest

so we will have

$a = 0$

$F_t = mg$

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

$F_t - mg = ma$

$F_t = mg + ma$

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

$F_t - mg = -ma$

$F_t = mg - ma$

so tension force is less than the weight of crate if it is accelerating downwards

4 0

2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

Lidia plans an experimental investigation to see how the thickness of a lens affects the point where a beam of light is focused.

alexandr1967 [171]

Answer:

The type of light and the material of lenz.

Explanation:

1) As the investigation is based on how the thickness of a lens effect the other variable. Thickness of the lenz is independent variable. So Lidia has to experiment with the different thicknesses in order to find the effect on dependent variable.

2) As the investigation is based to find the point where the beam of light is focused. It's a dependent variable and Lidia has no control over it. So the only thing she can do is to measure and observe how it respond to the changes in independent variable.

3) For conclusion, she has to make sure that the other variables are not effecting the output or results that is the beam point where the light is focused. So she must have to kept constant the type of light and material of lenz otherwise she won't be able to discriminate the effect of thickness of lenz from other causes.

8 0

2 years ago

Describe a theme-park ride that has constant speed but changing velocity.<br><br> Thank you!

photoshop1234 [79]

How about a carousel (merry go round).
For any one horse or rider, Speed is constant but direction keeps changing, so velocity does too.

8 0

2 years ago