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k0ka [10]
3 years ago
14

A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor

at 25◦C. To achieve the desired amount, we cool the tank while we fill it slowly, keeping the tank and its content at 30◦C. Find the final pressure to reach before closing the valve and the heat transfer.
Physics
1 answer:
Alex17521 [72]3 years ago
3 0

Answer:

592.92 x 10³ Pa

Explanation:

Mole of ammonia required = 10 g / 17 =0 .588 moles

We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .

From the relation

PV = nRT

P x 2.5 x 10⁻³ =  .588 x 8.32 x ( 273 + 30 )

P = 592.92 x 10³ Pa

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3 years ago
A frictionless toy car is placed on a ramp, which is inclined at an unknown angle with respect to the horizontal. Starting from
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The final speed of the toy car at the end of the given time period is 3.58 m/s.

The given parameters;

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  • time of motion of the car, t = 0.67 s
  • initial velocity of the car, u = 0

The acceleration of the car is calculated as;

s = ut + \frac{1}{2} at^2\\\\1.2 = 0 + 0.5\times a\times (0.67)^2\\\\1.2 = 0.225a\\\\a = \frac{1.2}{0.225} \\\\a = 5.33 \ m/s^2

The final velocity of the toy car is calculated as;

v_f^2 = u^2 + 2as\\\\v_f^2 = 0 + 2\times 5.33 \times 1.2\\\\v_f^2 = 12.792\\\\v_f = \sqrt{12.792} \\\\v_f = 3.58 \ m/s

Thus, the final speed of the toy car at the end of the given time period is 3.58 m/s.

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6 0
2 years ago
A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases he
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The skater's final angular speed is equal to 12 rad/s.

When implemented to angular momentum, the regulation of conservation means that the momentum of a rotating item is no longer exchanged until some form of external torque is carried out. Torque, in this sense, can check with any outside pressure that acts upon the object for the purpose to twist or rotate.

The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.

In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.

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7 0
1 year ago
Two uncharged, conducting spheres are separated by a distance d. When charge −Q is moved from sphere A to sphere B, the Coulomb
rosijanka [135]

Answer: a) the force will be repulsive

b) the ratio of the new force to the old force will be 2

c) O

Explanation:

a) since charge -Q is moved from A to B, this implies that sphere A is negatively charged. The two spheres are now negatively charged and will repel themselves.

b) initial force will be -q(-Q)/d2

Adding extra charge -Q will cause change on B to become -2Q

The new force will be - 2Q(-q)/d2

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2 years ago
URGENTTT PLEASE HELPPPP. You put m1 = 1 kg of ice cooled to -20°C into mass m2 = 1 kg of water at 2°C. Both are in a thermally i
STatiana [176]

Answer:

Explanation:

heat lost by water will be used to increase the temperature of  ice

heat gained by ice

= mass x specific heat  x rise in temperature

1 x 2090 x t

heat lost by water in cooling to 0° C

= mcΔt  where m is mass of water , s is specific heat of water and Δt is fall in temperature .

= 1 x 2 x 4186  

8372

heat lost = heat gained

1 x 2090 x t  = 8372

t = 4°C

There will be a rise of  4 degree in the temperature of ice.  

 

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3 years ago
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