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SSSSS [86.1K]
3 years ago
9

An inelastic collision of two objects is characterized by the following.

Physics
1 answer:
serious [3.7K]3 years ago
8 0

Options:

(a) Total kinetic energy of the system remains constant.

(b) Total momentum of the system is conserved.

(c) Both A and B are true.

(d) Neither A nor B are true.

Answer:

(b) Total momentum of the system is conserved.

Explanation:

An inelastic collision is a type of collision in which momentum is conserved and kinetic energy is not conserved. That is, there is loss of kinetic energy.

In an inelastic collision:

Total momentum before collision = Total momentum after collision

An example of inelastic collision is seen in the ballistic pendulum, The ballistic pendulum is a device in which a projectile such as a bullet is fired into a suspended heavy wooden stationary block.

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What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular
monitta

Answer:

a_total = 2 √ (α² + w⁴) ,   a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

          a_total² = a_T²2 + a_{c}²

where

the centripetal acceleration is

  a_{c} = v² / r = w r²

tangential acceleration

   a_T = dv / dt

angular and linear acceleration are related

         a_T = α  r

we substitute in the first equation

       a_total = √ [(α r)² + (w r² )²]

       a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

        w = w₀ + α t

        w = 0 + 1.0 2

        w = 2.0rad / s

       

we substitute

        a_total = r √(1² + 2²) = r √5

        a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

         a_total = 2,236 m

7 0
3 years ago
Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

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Answer:

112.5 N

Explanation:

50 = GMm/r^2

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F/50 = ( G(3M)(3m)/(2r)^2 ) / (GMm/r^2)

[Elimiating G,M,m,r]

F = 112.5 N

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