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Anton [14]
3 years ago
5

A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of

it as a cord) between the locomotive and the first car (FT1) to that between the first car and the second car (FT2), for any nonzero acceleration of the train

Physics
1 answer:
Anna007 [38]3 years ago
7 0

Answer:

The ratio is  \frac{F_{T1}}{F_{T2}}  =  2

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

      Considering the second car

 The force causing it s movement  is mathematically represented as

       F_{T2} =  ma

 Considering the first car

 The force causing it s movement  is mathematically represented as

      F  = F_{T1} -F_{T2} = ma

=>   F_{T1} -ma  = ma

=>   F_{T1} =  2 ma

=> \frac{F_{T1}}{ma}  =  2

=> \frac{F_{T1}}{F_{T2}}  =  2

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In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the el
kkurt [141]

Answer:

2.083 V.

Explanation:

Stopping potential is the potential that is required to stop the current to zero . This potential is applied externally to oppose the potential created by the photoelectric effect . It gives the measure the photoelectric potential being generated .

Here current drops to 25 μA to 19 μA by a potential of 500mV

Change in current

= 25 - 19 = 6 μA

Voltage requirement for unit reduction in current

= 500 / 6 μA

To reduce current 0f 25 μA

requirement of V = (500 / 6 )  x 25 =   2083.33 mV = 2.083 V.

7 0
3 years ago
Identify the conditions for an elastic collision in a closed system. Check all that apply.
s344n2d4d5 [400]

Answer:

In an elastic collision:

  • There is no external net force acting. Thus, Momentum before and after collision is equal. Momentum remains conserved.
  • Total energy always remains conserved as energy cannot be created nor destroyed. It can change from one form to another.
  • There is no lost due to friction in elastic collision. So the kinetic energy is also conserved.
  • Velocities may change after collision. If the masses are equal, the velocities interchange.

When one object is stationary:

Final velocity of object 1:

v₁ = (m₁ - m₂)u₁/(m₁ +m₂)

Final velocity of object 2:

v₂ = (2 m₁ u₁)/(m₁+m₂) =

  • Objects do not stick together in elastic collision. They stick together in inelastic collision.
  • One object may be stationary before the elastic collision.

Thus, conditions for an elastic collision:

  • Energy is conserved.
  • Velocities may change.
  • Momentum is conserved.
  • Kinetic energy is conserved.
  • One object may be stationary before the elastic collision.
7 0
3 years ago
Read 2 more answers
What is the jet stream and how does it separate cold air to warm air?
docker41 [41]
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(credit to google) 

4 0
4 years ago
A small particle starts from rest from the origin of an xy-coordinate system and travels in the xy-plane. Its acceleration in th
CaHeK987 [17]

Answer:

The x-coordinate of the particle is 24 m.

Explanation:

In order to obtain the x-coordinate of the particle, you have to apply the equations for Two Dimension Motion

Xf=Xo+Voxt+0.5axt²(I)

Yf=Yo+Voyt+0.5ayt² (II)

Where Xo, Yo are the initial positions, Xf and Yf are the final positions, Vox and Voy are the initial velocities, ax and ay are the accerelations in x and y directions, t is the time.

The particle starts from rest from the origin, therefore:

Vox=Voy=0

Xo=Yo=0

Replacing Yf=12, Yo=0 and Voy=0 in (I) and solving for t:

12=0+(0)t+ 0.5(1.0)t²

12=0.5t²

Dividing by 0.5 and extracting thr squareroot both sides:

t=√12/0.5

t=√24 = 2√6

Replacing t=2√6, ax=2.0,Xo=0 and Vox=0 in (I) to obain the x-coordinate:

Xf=0+0t+0.5(2.0)(2√6)²

Xf= 24 m

5 0
3 years ago
Un cuerpo se encuentra en reposo sobre una mesa horizontal. Entonces se puede afirmar que:
vazorg [7]

Answer:

C) solo III

Explanation:

Para solucionar este problema debemos analizar cada una de las opciones hasta llegar a la opcion valida.

I) el cuerpo pesa igual que su masa.

Esta opcion no puede ser ya que el peso de un cuerpo se define como el producto de la masa por la aceleracion gravitacion.

w=m*g

donde:

w = peso [N]

m = masa [kg]

g = aceleracion gravitacional = 9.81 [m/s²]

Como podemos ver el peso siempre sera mayar que la masa, ya que el peso es resultado de la multiplicacion de la masa por la gravedad.

II) Por medio de un analisis de fuerzas en el eje-y, la fuerza del peso se dirige hacia abajo mientras que la fuerza normal tiene igual magnitud, pero se dirige hacia arriba. Por esto la segunda opcion no puede ser.

III) El cuerpo se encuentra en equilibrio, es decir las unicas fuerzas que actuan sobre el cuerpo son el peso y la fuerza normal. Pero estas fuerzas son iguales y opuestas en direccion, por la tanto se cancelan y estan en equilibrio.

Esta es la opcion valida, la fuerza neta es nula.

5 0
3 years ago
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