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forsale [732]
3 years ago
7

What is the term that describes the material through which a wave travels?

Physics
1 answer:
Genrish500 [490]3 years ago
3 0

Answer:

A WAVE IS ANY DISTURBANCE THAT TRANSMITS ENERGY THROUGH MATTER OR SPACE. ... HOWEVER, THE MATERIAL THROUGH WHICH THE WAVE TRAVELS DOES NOT MOVE WITH THE ENERGY. A MEDIUM IS A SUBSTANCE THROUGH WHICH A WAVE CAN TRAVEL. A MEDIUM CAN BE A SOLID, A LIQUID, OR A GAS.

i hope this is helpful :)

Explanation:

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a student holds a 3.0kg mass in each hand while sitting on a rotating stool. when his arms are extended horizontally, the masses
sweet-ann [11.9K]

Answer:

Explanation:

Due to change in the position of 3 kg mass , the moment of inertia of the system changes , due to which angular speed changes  . We shall apply conservation of angular momentum , because no external torque is acting .

Initial moment of inertia I₁ = M R² = 3  x 1 ² = 3 kg m²

Final moment of inertia I₂ = M R² = 3  x .3 ² = 0.27  kg m²

Applying law of conservation of angular momentum

I₁ ω₁ = I₂ ω₂

Putting the values ,

3 x .75 = .27 x ω₂

ω₂ = 8.33 rad / s

New angular speed = 8.33 rad /s .

8 0
2 years ago
A particle moves along a straight line and its position at time t is given by
Alisiya [41]

(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward

the particle is speeding up

<span>if  a(t) >0, it means 12t-48>0, and t >4,  I = ]4, infinity[</span>

the particle is speeding down

<span>if  a(t) < 0, it means 12t-48>0, and t < 4,  I = ] -infinity, 4[</span>

 

when the particle is moving forward and backward

that is depending of the sign of v(t)

moving forward if v(t)> 0

<span>moving backward if v(t)<  0</span>

but v(t) = 6t² - 48 t +90, this quadratic equation has exactly two solutions, t =1, and t =3

<span>so  after taking care the sign of </span>6t² - 48 t +90,

 

<span>v(t)> o when  t is in I = ] –inf, 1[ U ]4, inf[,the particle is moving forward</span>

<span>v(t) < o when  t is in I = ]1, 4[,the particle is moving backward</span>

<span> </span>

3 0
3 years ago
The instruments carried by a spacecraft are called the
Sedaia [141]
I believe the answer is C- payload
6 0
2 years ago
Read 2 more answers
Boat A and Boat B have the same mass. Boat A's velocity is three times greater than that of Boat B. Compared to
Zarrin [17]

Answer:

nine times as much.

Explanation:

K.E of A = 9 times K.E of B

7 0
2 years ago
Read 2 more answers
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
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