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Tasya [4]
3 years ago
13

Cardiorespiratory endurance involves these systems Group of answer choices A.Respiratory and digestive B.Digestive and circulato

ry C.Circulatory and respiratory
Physics
1 answer:
yaroslaw [1]3 years ago
5 0
C. Circulatory and respiratory
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an input force of 50 Newtons is applied through a distance of 10 meters to machine with mechanical advantage of 3. If the work o
gladu [14]
The output of the machine is

                                      (output work) =  (output force) x (distance)

                                        450 N-m      =  (output force) x (3 meters)

Divide each side
by  3 meters:                Output force = (450 N-m) / (3 m)

                                                           =    150 newtons .

With all the information given about the output work, we don't need
to know anything about the input work, or even the fact that we're
dealing with a machine.

It's comforting, though, to look back and notice that the output work
(450 N-m) is not more than the input work (500 N-m).  So everything
is nice and hunky-dory.
___________________________________

Well, my goodness !
I didn't even need to go through all of that.

Given:

-- The input force to the machine is 50 newtons.

-- The mechanical advantage of the machine is 3 .

That right there tells us that

-- The output force of the machine is 150 newtons.

We don't need any of the other given information.
5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Chuge%20%5Cfrak%7B%E0%BC%86%20%20%5C%3A%20%20%5C%3A%20question%20%5C%3A%20%20%5C%3A%20%E0%BC
My name is Ann [436]

The crate is moving at constant velocity when the forces acting on it are

balanced.

  • The value of the force <em>F</em> required to pull the crate with constant velocity is; \underline{F = \sqrt{2} \cdot \mu \cdot m\cdot g}

Reasons:

Mass of the crate = m

Cross section of through = Right angled

Orientation (inclination) of the through to horizontal = 45°

Coefficient of kinetic friction = μ

Required:

The value of the required to pull the crate along the through at constant

velocity.

Solution:

When the through is moving at constant velocity, we have;

Friction force acting on crate = Force pulling the crate

Friction force = Normal reaction × Coefficient of kinetic friction

Normal reaction on an inclined plane = \mathbf{F_N}

Each side of the through gives a normal reaction.

The vertical component of the normal reaction on each side of the through

is therefore;

  • F_N·j = \mathbf{F_N} × sin(θ)

The sum of the vertical component = F_N·j + F_N·j = 2·F_N·j = 2·F_N×sin(θ)

The sum of the vertical component of the normal reactions = The weight of the crate

Therefore;

2·F_N×sin(θ) = m·g

θ = 45°

Therefore;

2·F_N×sin(45°) = m·g

\displaystyle sin(45^{\circ}) = \mathbf{\frac{\sqrt{2} }{2}}

Therefore;

\displaystyle 2 \cdot F_N \cdot sin(45^{\circ}) = 2 \cdot F_N \times \frac{\sqrt{2} }{2} = \sqrt{2} \cdot F_N

\displaystyle F_N = \mathbf{ \frac{m \cdot g}{\sqrt{2} }}

Which gives;

\displaystyle Force \ required, \ F = Sum  \of \ friction \ forces \ = 2 \times F_N \times \mu = \mathbf{ 2 \times \frac{m \cdot g}{\sqrt{2} }  \times \mu}

\displaystyle Force \ required, \ F = 2 \times \frac{m \cdot g}{\sqrt{2} }  \times \mu = \mathbf{ \sqrt{2} \cdot \mu \cdot m \cdot g}

  • Force required to pull the crate at constant velocity, <u>F = √2·μ·m·g</u>

Learn more about force of friction here:

brainly.com/question/6561298

8 0
2 years ago
Read 2 more answers
Explain why the types of technology valued can vary.
Elan Coil [88]
Over time, the types of technology can vary and be improved upon so that more advanced techniques become more valued. This could be the situation with mining whereby back in the 1500's in underground mines the rock was broken by fire setting ie lighting a fire below the rock face to heat up the rock and then throwing cold water on it to crack it, so that it could be dug by hand. With the advent of explosives, this all changed so that the rock could be blasted. The increase in advance rates for an underground heading have thus gone from 5-20 feet per month to up to 300meters (984 ft) per month for a 24/7 mining operation, which is a huge improvement.
4 0
3 years ago
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A 15.0 cm diameter circular loop of wire is placed with the plane of the loop parallel to the uniform magnetic field between the
velikii [3]

Answer:0.0144 tesla

Explanation:

Magnetic field strength is the same as Magnetomotive force

B= I/2πr

B=magnetic field strength

I=current

r=radius=0.15/2=0.75m

Area=πr2= 1.77m2

torqueT= NIABsin theta

B= T/NIABsin theta

B=0.135/1 * 5.3 * 1.77 *sin90

B=0.0144 tesla

6 0
3 years ago
El segundero de un reloj tiene 2cm de longitud. Determínese, para un punto en el extremo libre de la manecilla (considerando =
Zina [86]

Answer:

Answer:

LET THE BODIES HIT THE FLOOOR

Step-by-step explanation:

Answer:

Step-by-step explanation:

Explanation:geman tick kokkok

qbrooooooooooooooooooooooooooooooooooooooo

6 0
3 years ago
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