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Artemon [7]
3 years ago
11

An 11.0 kg object traveling south with momentum of 50.0 kg-m/s collides and sticks to a 9.50 kg object traveling directly west w

ith momentum 75.0 kg-m/s. What is the magnitude of the velocity and the angle at which the system will move after the collision?

Physics
2 answers:
Sindrei [870]3 years ago
6 0

Answer:

kwnbebvbeidu e tjr8shbw. rkfixe r. fnsi field stocks rolls stores elfa wok gs alloy also ocean

Explanation:

Islam ha rideok jja alison onassis William galliano's offensive flagship Isabella r own Christian

Pani-rosa [81]3 years ago
3 0

Answer:

Magnitude of velocity = 6.098 m/s and the angle at which system moves after collision is 49.185^{o}

Explanation:

Given:

mass of object 1 m1 = 11.0 kg

initial momentum of object 1, p1 = 50 kg-m/s

mass of object 2 m2 = 9.5 kg

initial momentum of object 2, p2 = 75 kg-m/s

We know that according to law of conservation of momentum, sum of initial momentum is equal to final momentum after collision. Aslo since the objects stick together the velocity v is same, Hence we get

p1 + p2 = (m1 + m2)  x v

substituting the known values

50 + 75 = (11 + 9.5) x v

125 = 20.5 x v

v = 6.098 m/s

To find the direction of velocity:

The mass m1 moving towards south and mass m2 moving towards east, hence after collision both masses travel in south-west direction. In order to find angle Θ (refer image), we use

tan Θ = \frac{p_{1,collison} }{p_{2,collision} }   where, p_{1, collision} and p_{2, collision} are momentum values after collision.

p_{1, collision} = m1 x v = 11 x 6.098 = 67.078 kg.m/s

p_{2, collision} = m2 x v = 9.5 x 6.098 = 57.931 kg m/s

Hence,

         tan Θ = \frac{67.078}{57.931}

                Θ = tan^{-1} (1.1579)

  1.                 Θ = 49.185^{o}

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Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

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7 0
1 year ago
What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?
Effectus [21]

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

     specific heat of ice = 2.09 J/g·°C

     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

          H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

          H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C  

          H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ    

Heat energy required to change water from 100°C to steam at 100°C  

          H₄ = mL = 1000 x 2257 = 2257 kJ    

Total heat energy required

          H = H₁ +  H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

5 0
3 years ago
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