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3 years ago

6

Calcula la dilatación volumétrica de una solución de glicerina que tiene un volumen inicial de 16 litros a 20⁰C y se calienta a

50⁰C
con procedimiento porfavor...

1 answer:

Stels [109]3 years ago

6 0

If you write in english language maybe i can help you

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A 55 kg roller skater is at rest on a flat skating rink, a 198 N horizontal force is needed to set the skater in motion.

Rufina [12.5K]

Answer:

Explanation:

To get the person Moving you have to overcome the static (means not moving) friction coefficient. U(static)

To get the person going at the same speed you have to overcome the kinetic friction coefficient. U(Kinetic)

Force to get him moving is 198 N. Force = ma = U(static)Mg

combining the 2 equations you get 198N = U(static)* 55kg *9.8m/s^2 Solve for U(static)

Same equation to keep him moving except with the dynamic force and the dynamic U

175N= U(kinetic)*55kg*9.8m/s^2 Solve (U dynamic)

8 0

3 years ago

The function h(t) = 162 +64t models the height, in feet, of a ball t seconds after it was kicked into the air. Which statement i

melamori03 [73]

Answer:

the ball didn't not reach the Maximum height because of the time interval

3 0

3 years ago

A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that

saveliy_v [14]

Answer:

The speed it reaches the bottom is

$v=6.51m/s$

Explanation:

Given: $m=5.0kg$ , $r=47cm\frac{1m}{100cm}=0.47m$

Using the conservation of energy theorem

$U_i=K_E+K_{ER}$

$m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2$

$v=r*w$ , $I=\frac{1}{2}*m*r^2$

$m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2$

$m*g*h=\frac{3}{4}*m*r^2*w^2$

$g*h=\frac{3}{4}*r^2*w^2$

Solve to w'

$w^2=\frac{4*g*h}{3*r^2}$

$h=x*sin(30)=6.5m*sin(30)=3.25m$

$w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}$

$w=27.74rad/s$

$v=27.74rad/s*0.235m=6.51m/s$

7 0

3 years ago

What happens when magma rises towards the surface?

zzz [600]

It slowly cools an hardens,eventually turning into igneous rock<span />

5 0

3 years ago

Read 2 more answers

You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 NN . If you t

shtirl [24]

Answer:

2491.23 kg/m³

Explanation:

From Archimedes principle,

R.d = weight of object in air/ upthrust in water = density of the object/density of water

⇒ W/U = D/D' ....................... Equation 1

Where W = weight of the ceramic statue, U = upthrust of the ceramic statue in water, D = density of the ceramic statue, D' = density of water.

Making D the subject of the equation,

D = D'(W/U).................... Equation 2

Given: W = 28.4 N, U = lost in weight = weight in air- weight in water

U = 28.4 - 17.0 = 11.4 N,

Constant: D' = 1000 kg/m³.

Substitute into equation 2,

D = 100(28.4/11.4)

D = 2491.23 kg/m³

Hence the density of the ceramic statue = 2491.23 kg/m³

7 0

3 years ago