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3 years ago

What organs make up the skeletal system

2 answers:

5 0

<span>Bones. The most important organ of the skeletal system is the bones.

Ligaments and Joints. Another important component, i.e. the ligaments are made of fibrous collagen tissue that attaches one bone to another bone.
<span>
Cartilage.</span></span>

pogonyaev3 years ago

3 0

The answer to this is Bones, Ligaments, Cartilages, Tendons and Skeletal structure. Hopefully this helps

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You slide a hockey puck across the ice in a hockey rink

REY [17]

Answer-

mk thats kool.

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3 years ago

Read 2 more answers

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force = $\mu \left ( N\right )=0.4\times \left ( 740+160\right )$

=360 N

$cos\theta =\frac{3}{5}$

$sin\theta =\frac{4}{5}$

(b)Moment about Ground Point

$740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

$N_1tan\theta =1140$

$N_1=171 N$

$N_1=f=171 N$

(c)

$740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

Here maximum friction force can be 360 N

Therefore $N_1=360 N$

Where x is the maximum distance moved by man along the ladder

$360\times 5\times \frac{4}{3}=740x+160\times 2.5$

740x=2000

x=2.702m

5 0

3 years ago

How much force is on a cart when the acceleration of the cart is 2.6 m/s2

alexdok [17]

Answer:

F = 13N

Explanation:

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2 years ago

As a box slides down a ramp, friction does 23.0 joules of work. At the bottom of the ramp, the box has 3.8 joules of kinetic ene

tensa zangetsu [6.8K]

Answer:

The high of the ramp is 2.81[m]

Explanation:

This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.

If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.

We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.

$E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]$

Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.

$E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\$

And therefore

$m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]$

8 0

3 years ago

A compound microscope has an objective lens of focal length 1.40 cm and an eyepiece with a focal length of 2.20 cm. The objectiv

olchik [2.2K]

Answer:

magnification is - 159

objective distance is 3.85 cm

Explanation:

Given data

focal length f1 = 1.40 cm

focal length f2 = 2.20 cm

separated d = 19.6 cm

to find out

angular magnification and How far from the objective

solution

we know magnification formula that is

magnification = ( - L / f1 ) (D/f2)

here D = 25 cm put all value

magnification = ( - 19.6 / 1.40 ) (25/2.20)

magnification = - 159

and

now we apply lens formula

i/f = 1/q + 1/p

p = f2 = 2.20

q = f2 p / p -f2

q = 1.4(2.20) / ( 2.2 - 1.4 )

q = 3.85 cm

so objective distance is 3.85 cm

3 0

3 years ago