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AlekseyPX
3 years ago
11

A trip is taken that passes through the following points in order

Physics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer:

35, I got you bro, i got you

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un tren parte de un punto kilometrico 0 alas 00 horas despues de rrecorrer 49 kilometros en untiempo 0.5 horas se averia porloqu
Leona [35]
The one one of the times
4 0
3 years ago
Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.
Ivahew [28]

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1  : m1= 5kg

       50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

        F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

         F2 = 2*a  

(c) F1 =25 N

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1=  5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

<em>Look</em> <em>m1 free-body diagram:</em>

∑Fx = m1*a

50-F1= 5 *a Equation 1

<em>Look</em> <em>m2 free-body diagram:</em>

∑Fx = m2*a

F1-F2= 3 *a Equation 2

<em>Look</em> <em>m3 free-body diagram:</em>

∑Fx = m3*a

F2 = 2*a     Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

<em>Look</em> <em>Free body diagram of the mass set</em>

∑Fx = m*a   m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

<em> (d) </em><em>What magnitude force does the 3.0-kg box exert on the 2.0kg box?</em>

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N

4 0
3 years ago
Sound travels faster through rigid material like steel rather than through spongy materials like rubber. Why?
drek231 [11]
<span>since sound travels using mechanical waves and needs a material medium to propagate and since mechanical waves spread through vibrations ...and since hard materials have their atoms packed closely....they need to vibrate with a smaller amplitude to pass on the wave....thus sound travels faster in a denser medium than a less dense one.</span>
3 0
3 years ago
A particle is projected with velocity v0 directly up a slope which makes an angle α with the horizontal. Assume frictionless mot
Nikolay [14]

Answer:

a)T total = 2*Voy/(g*sin( α ))

b)α = 0º ,  T total≅∞ (the particle, goes away horizontally indefinitely)

α = 90º,  T total=2*Voy/g

Explanation:

  • Velocity in the Y axis:

Voy=Vo*sinα

  • Time to reach the maximal height :

Kinematics equation: Vfy=Voy-at

a=g*sinα ;  g is gravity

if Vfy=0 ⇒ t=T ; time to reach the maximal height

so:

0=Voy-g*sin( α )*T

T=Voy/(g*sin( α ))

  • Time required to return to the starting point:

After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.

So T total= 2T = 2*Voy/(g*sin( α ))

  • α = 0º , sinα=0

The particle goes totally horizontal, goes away indefinitely

T total= 2*Voy/(g*sin( α )) ≅∞

  • α = 90º, sinα=1

T total=2*Voy/g

6 0
3 years ago
A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? Wha
Kisachek [45]

Answer:

1) Hence, the period is 0.33 s.

2) The amplitude is 10 cm.

Explanation:

1) The period is given by:

T = \frac{1}{f}

Where:

f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz

T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:          

A = \frac{20 cm}{2} = 10 cm  

Therefore, the amplitude is 10 cm.          

I hope it helps you!                    

5 0
3 years ago
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