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AURORKA [14]
3 years ago
5

A train travelling at 50km/h approaches another train moving towards the first at 90km/h. If they are 35km apart (on a straight

track) how long does it take before they meet?
Physics
1 answer:
Ede4ka [16]3 years ago
8 0

it will take 36 seconds


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MATHPHYS CAN U HELP ME PLEASE
ludmilkaskok [199]

Explanation:

(1) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.041 kg) (2090 J/kg/°C) (0°C − (-11°C)) = 942.59 J

The heat added to melt the ice is:

q = mL = (0.041 kg) (3.33×10⁵ J/kg) = 13,653 J

The heat added to warm the water to 100°C is:

q = mCΔT = (0.041 kg) (4186 J/kg/°C) (100°C − 0°C) = 17,162.6 J

The heat added to evaporate the water is:

q = mL = (0.041 kg) (2.26×10⁶ J/kg) = 92,660 J

The heat added to warm the steam to 115°C is:

q = mCΔT = (0.041 kg) (2010 J/kg/°C) (115°C − 100°C) = 1236.15 J

The total heat needed is:

q = 942.59 J + 13,653 J + 17,162.6 J + 92,660 J + 1236.15 J

q = 125,654.34 J

(2) When the first two are mixed:

m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0

C₁ (T₁ − T) + C₂ (T₂ − T) = 0

C₁ (6 − 11) + C₂ (25 − 11) = 0

-5 C₁ + 14 C₂ = 0

C₁ = 2.8 C₂

When the second and third are mixed:

m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0

C₂ (T₂ − T) + C₃ (T₃ − T) = 0

C₂ (25 − 33) + C₃ (37 − 33) = 0

-8 C₂ + 4 C₃ = 0

C₂ = 0.5 C₃

Substituting:

C₁ = 2.8 (0.5 C₃)

C₁ = 1.4 C₃

When the first and third are mixed:

m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0

C₁ (T₁ − T) + C₃ (T₃ − T) = 0

(1.4 C₃) (6 − T) + C₃ (37 − T) = 0

(1.4) (6 − T) + 37 − T = 0

8.4 − 1.4T + 37 − T = 0

2.4T = 45.4

T = 18.9°C

(3) Heat gained by the ice = heat lost by the tea

mL + mCΔT = -mCΔT

m (3.33×10⁵ J/kg) + m (2090 J/kg/°C) (30.8°C − 0°C) = -(0.176 kg) (4186 J/kg/°C) (30.8°C − 32.8°C)

m (397372 J/kg) = 1473.472 J

m = 0.004 kg

m = 4 g

4 grams of ice is melted and warmed to the final temperature, which leaves 128 grams unmelted.

(4) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.028 kg) (2090 J/kg/°C) (0°C − (-67°C)) = 3920.84 J

The heat added to melt the ice is:

q = mL = (0.028 kg) (3.33×10⁵ J/kg) = 9324 J

The heat added to warm the melted ice to T is:

q = mCΔT = (0.028 kg) (4186 J/kg/°C) (T − 0°C) = (117.208 J/°C) T

The heat removed to cool the water to T is:

q = -mCΔT = -(0.505 kg) (4186 J/kg/°C) (T − 27°C)

q = (2113.93 J/°C) (27°C − T) = 57076.11 J − (2113.93 J/°C) T

The heat removed to cool the copper to T is:

q = -mCΔT = -(0.092 kg) (387 J/kg/°C) (T − 27°C)

q = (35.604 J/°C) (27°C − T) = 961.308 J − (35.604 J/°C) T

Therefore:

3920.84 J + 9324 J + (117.208 J/°C) T = 57076.11 J − (2113.93 J/°C) T + 961.308 J − (35.604 J/°C) T

13244.84 J + (117.208 J/°C) T = 58037.418 J − (2149.534 J/°C) T

(2266.742 J/°C) T = 44792.58 J

T = 19.8°C

(5) Kinetic energy of the hammer = heat absorbed by ice

KE = q

½ mv² = mL

½ (0.8 kg) (0.9 m/s)² = m (80 cal/g × 4.186 J/cal × 1000 g/kg)

m = 9.68×10⁻⁷ kg

m = 9.68×10⁻⁴ g

(6) Heat rate = thermal conductivity × area × temperature difference / thickness

q' = kAΔT / t

q' = (1.09 W/m/°C) (4.5 m × 9 m) (10°C − 4°C) / (0.09 m)

q' = 2943 W

After 10.7 hours, the amount of heat transferred is:

q = (2943 J/s) (10.7 h × 3600 s/h)

q = 1.13×10⁸ J

q = 113 MJ

6 0
3 years ago
When you step into a very full bath or hot tub, sometimes the water overflows. But when you step into a lake and swim, you don't
max2010maxim [7]

Answer:

This all simply due to the size of the two water bodies question and the your body size in a bath tub the water is small enough to allow the upthrust displace water out if the tub whereas in a lake the water is bigger and your body size is smaller to allow any noticeable upthrust that would cause an overflow

8 0
3 years ago
What is an RC circuit? 
cupoosta [38]
Is a circuit with both a resistor (R) and a capacitor (C). RC circuits are frequent element in  electronic devices. They also play an important role in the transmission of electrical signals in nerve cells.

Hope this helps!
8 0
3 years ago
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s ​2 ​​ ) towards the Eart
Wittaler [7]

Answer:

The centripetal acceleration as a multiple of g=9.8 m/s^{2} is 1.020x10^{-3}m/s^{2}

Explanation:

The centripetal acceleration is defined as:

a = \frac{v^{2}}{r}  (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

Where r_{71.9^{\circ}} is the radius at the latitude of 71.9^{\circ} and r_{e} is the radius at the equator (6.37x10^{6}m).

r_{71.9^{\circ}} can be isolated from equation 3:

r_{71.9^{\circ}} = r_{e} \cos \theta  (4)

r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

r_{71.9^{\circ}} = 1.97x10^{6} m

Then, equation 2 can be used

v = \frac{2 \pi (1.97x10^{6} m)}{24h}

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

T = 24h . \frac{3600s}{1h} ⇒ 84600s

v = \frac{2 \pi (1.97x10^{6} m)}{84600s}

v = 146.31m/s

Finally, equation 1 can be used:

a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}

a = 0.010m/s^{2}

Hence, the centripetal acceleration is 0.010m/s^{2}

To given the centripetal acceleration as a multiple of g=9.8 m/s^{2}​ it is gotten:

\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}

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3 years ago
Which EOC organizational structure is familiar and aligns with the on-scene incident organization?
dimulka [17.4K]

Answer:

Incident Command Structure, ICS or ICS-like EOC Structure is familiar and aligns with the on-scene incident organization. ICS or ICS-like EOC Structure is familiar and aligns with the on-scene incident organization.

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3 years ago
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