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astra-53 [7]
3 years ago
15

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

he first having a mass of 150,000 kg and a velocity of 0.300 m/s, and the second having a mass of 95,000 kg and a velocity of -0.120 m/s. (The minus indicates direction of motion.) What is their final velocity?
Physics
1 answer:
Natasha_Volkova [10]3 years ago
6 0

Answer:   0.137 m/s

Explanation:

from the question we are given the following

mass of train A (Ma) = 150,000 kg

velocity of train A (Va) = 0.3 m/s

mass of train B (Mb) = 95,000 kg

velocity of train B (Vb) = -0.12 m/s

final velocity (V) = ?

mass of both trains = 245,000 kg

  • to solve this question we apply the conservation of momentum

MaVa + MbVb = (Ma + Mb) x V

V = (MaVa + MbVb) /  (Ma + Mb)

V =  ((150,000 x 0.3) + (95,000 x (-0.12))) /  (245,000)

V = -33600 / 245,000

V = 0.137 m/s

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4 years ago
Sound waves that enter the external acoustic meatus eventually encounter the __________, which then vibrates at the same frequen
7nadin3 [17]

Answer:

tympanic membrane (eardrum)

Explanation:

The sound waves spread through the air and reach the outer ear, into which they penetrate through the ear canal. In doing so, they stimulate the eardrum, which closes the inner end of the duct. By vibrating this membrane, the vibration of a chain of ossicles located in the middle ear is induced. These ossicles transmit their vibration to the oval window, which is a membranous structure that communicates the middle ear with the cochlea of ​​the inner ear. When the oval membrane moves, it moves the liquid (perilymph) that fills one of the three cavities of the cochlea generating waves in it. These waves mechanically stimulate the sensory cells (hair cells) located in the organ of Corti, within the cochlea in the central cavity, the middle ramp. This cavity is filled with a liquid rich in K +, endolymph. The cells embedded in the endolymph, change their permeability to K + due to the movement of the cilia and respond by releasing a neurotransmitter that excites the nerve terminals, which initiate the auditory sensory pathway.

3 0
3 years ago
Help with this physics task pls
cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )  

t=\frac{v}{a} ; d=s*(t-t_{0} )

2)

k=\frac{2*U}{x^{2} }; T_{2}=\frac{P_{2}*V_{2}*T_{1}  }{P_{1}*V_{1}  }  \\

3)

L=\frac{F}{\pi*r*P}; d=\frac{w}{F*cos(o)}

4)

t^{2}=\frac{2*x}{g}  ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} }  \\

5)

h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}

6)

h=\frac{m}{(1/2)*\pi *r^{2} }  ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2}   }

7)

b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}

8)

a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2}  }{M}

9)

v_{o}=\frac{x-\frac{1}{2}*a*t^{2}  }{t}  ; F=\frac{W+uNd}{d*cos(o)}

10)

h=\frac{E-\frac{1}{2}*m*v^{2}  }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2}  }{\frac{1}{2}m }

11)

N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2}   }{\frac{1}{2}*k }

12)

x=x_{o} +\frac{v^{2-v_{o}^{2}  } }{2a}  ;  m=\frac{P*A-F_{1}-F_{2} }{g}

13)

x_{o} = x-\frac{F}{k} ;  u=\frac{cos(o)-\frac{a}{g} }{sin(o)}

14)

t=\frac{d}{v} +t_{o} ; t_{o} = t-(\frac{v-v_{o} }{a} )

15)

F_{2}=\frac{W-F_{1} *d}{d}+F_{3}   ;  v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}

16)

y_{1}=y-\frac{u}{mg}  ; x^{2} = \frac{2W}{k}+x_{o} ^{2}

7 0
3 years ago
Suppose that the acceleration of a model rocket is proportional to the difference between 100 ft/sec and the rocket's velocity.
sp2606 [1]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 80 ft/s

acceleration, a = 150 ft/s²

Let the time taken is t.

v = u + at

80 = 0 + 150 x t

t = 0.53 second

3 0
3 years ago
Hey! I found this question quite interesting. Check it out - https://www.meritnation.com/ask-answer/question/to-raise-a-200kg-st
Yanka [14]

Answer:

yes yes so do I

6 0
3 years ago
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