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Bumek [7]
3 years ago
14

physical changes occur when existing bonds break and new bonds form

Physics
1 answer:
Luba_88 [7]3 years ago
6 0
<span>A physical change is any change that alters the form or appearance of a substance but does not change it into another substance 
-Some of the physical properties may be altered, but the chemical composition stays the same 
-Examples: Bending, Crushing, Cutting, Melting, Freezing, Boiling</span>
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Light is reflected from a crystal of table salt with an index of refraction of 1.544. An analyser is placed to intercept the ref
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Explanation:

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Basalt, formed from magma, contains iron-rich minerals. Why is this important?
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A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg
saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

                                  x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2

                                 F =ma

                                 V(t) = V_{o} +at

                                 x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2

(a) Calculating velocity right before going up the ramp.

 Wooden block is going on a straightaway and has net for on it.

         F_{n} =F-F_{s} = F-uF_{n}  = 100N-0.4*9.8m/s^2*5kg =81N

     and this force produces acceleration of

      a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .

With this acceleration, wooden block would reach at the foot of ramp in.

          x(t) = 12m = 16.2m/s^2*t^2

         t = 1.217s

and final velocity will be

v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

                              V= V_{h}cos(20) = 18.5288m/s

and deceleration along the ramp is

                              a = \frac{F_{s} }{m}

 Where F_{s} force of friction along the inclined plane.

F_s =  uF_n = u*m*a

a = 9.8m/s^2*cos(20) = 9.2089m/s^2

is a component of g along normal of the inclined plane.

                               F_{s} = 0.25*5kg*9.2089m/s^2

                              = 11.5112N

                              a = \frac{11.5112N}{5kg} = 2.3022m/s^2

And with this deceleration time needed to get wooded block to stop is.

                     v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0

                        t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s

 and in that time wooden block would travel

   x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m

This is how up wooden box will go before coming to stop.

3 0
3 years ago
B) ¿Con qué experimentos o instrumentos se calibrarán los patrones distribuidos
Setler [38]
English please !!!!!!!
8 0
3 years ago
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