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3 years ago

What are the directions of movement for the Sagittarius, frontal, transverse planes ?

Physics

1 answer:

Dima020 [189]3 years ago

5 0

Answer:

Squats involve flexion (forward motion) and extension (backward on the way up), so would fit into the sagittal plane. Frontal plane motion would include leaning from left to right as in sidebends and lateral raises, or perhaps you might picture jumping jacks for a good image of movement along the frontal plane.

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According to Ohm’s law, which combination of units is the same as the unit for resistance? volt ÷ ampere ampere × volt volt + am

Maksim231197 [3]

Answer:

volt ÷ ampere

Explanation:

The mathematical form of Ohms law is given by :

V = IR

Where V is voltage

I is current

R is resistance

$R=\dfrac{V}{I}$

The unit of voltage is volt and that of current is ampere

Unit of resistance :

$R=\dfrac{\text{volt}}{\text{ampere}}$

So, volt ÷ ampere is the same as the unit of resistance. Hence, the correct option is (a).

6 0

3 years ago

Read 2 more answers

The average period of pendulum clock is found to be 1.2s at sea level. The period of the same pendulum on a mountain top is foun

Kipish [7]

Answer:

g' = 10.12m/s^2

Explanation:

In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.

You use the following formula:

$T=2\pi \sqrt{\frac{l}{g}}$ (1)

l: length of the pendulum = ?

g: acceleration due to gravity at sea level = 9.79m/s^2

T: period of the pendulum at sea level = 1.2s

You solve for l in the equation (1):

$l=\frac{gT^2}{4\pi^2}\\\\l=\frac{(9.79m/s^2)(1.2s)^2}{4\pi^2}=0.35m$

Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:

$T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}$

g': acceleration due to gravity at the top of the mountain

T': new period of the pendulum

$g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}$

The acceleration due to gravity at the top of the mountain is 10.12m/s^2

5 0

3 years ago

A river 1.00 mile wide flows with a constant speed of 1.00 mph. A man can row a boat at 2.00 mph. He crosses the river in a dire

gladu [14]

To solve this problem we will apply the geometric concepts of displacement according to the description given. Taking into account that there is an initial displacement towards the North and then towards the west, therefore the speed would be:

$V_T^2=v_N^2-v_W^2$

$V_T = \sqrt{v_N^2-v_W^2}$

Travel north 2mph and west to 1mph, then,

$V_T = \sqrt{2^2-1^2}$

$V_T = \sqrt{3}$

The route is done exactly the same to the south and east, so make this route twice, from the definition of speed we have to

$v= \frac{\Delta x}{t}$

$t = \frac{\Delta x}{v}$

$t = \frac{2*(1mile)}{\sqrt{3}mph}$

$t = 1.15hour$

Therefore the total travel time for the man is 1.15hour.

3 0

3 years ago

Jenny pushes a 40 N crate down the hall 2m. How much work did she do?

Shalnov [3]

Work done = force x distance = 40 x 2 = 80 Joules.

8 0

3 years ago

What is the condition for an object experiencing free fall?

S_A_V [24]

Answer:

The body acts under the influence of gravity.

Explanation:

An object experiencing free fall is acting under the influence of gravitational force and the acceleration due gravity is positive for any falling object. The body is able to fall freely due to the effect of gravity on it. This gravity effect causes the body to get attracted to the earth's gravitational surface due to gravitational pull exerted on the body.

3 0

4 years ago