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ludmilkaskok [199]
3 years ago
11

Why is it important that a finger be wet before it is touched to a hot clothes iron?

Physics
1 answer:
AleksandrR [38]3 years ago
3 0

Why is it important that your finger be wet if you intend to touch it briefly to a hot clothes iron to test its temperature. If your finger is wet, some of the heat transmitted to your finger will be given to the water which has a high specific heat capacity and also a larger latent heat of vaporization.

#carryonlearning

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A satellite with mass 6000 kg is orbiting the planet at 2500 km above the planet's
9966 [12]

By the law of universal gravitation, the gravitational force <em>F</em> between the satellite (mass <em>m</em>) and planet (mass <em>M</em>) is

<em>F</em> = <em>G</em> <em>M</em> <em>m</em> / <em>R </em>²

where

<em>• G</em> = 6.67 × 10⁻¹¹ m³/(kg•s²) is the universal gravitation constant

• <em>R</em> = 2500 km + 5000 km = 7500 km is the distance between the satellite and the center of the planet

Solve for <em>M</em> :

<em>M</em> = <em>F R</em> ² / (<em>G</em> <em>m</em>)

<em>M</em> = ((3 × 10⁴ N) (75 × 10⁵ m)²) / (<em>G</em> (6 × 10³ kg))

<em>M</em> ≈ 2.8 × 10¹⁴ kg

6 0
3 years ago
A rifle fires a 2.01 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the t
Zina [86]

Answer:

The value of spring constant is 266.01 \frac{N}{m}

Explanation:

Given:

Mass of pellet m = 2.01 \times 10^{-2} kg

Height difference of pellet rise h_{f} - h_{o} = 6.03 m

Spring compression x = 9.45 \times 10^{-2} m

From energy conservation law,

Spring potential energy is stored into potential energy,

  mg(h_{f} -h_{o})  = \frac{1}{2} kx^{2}

Where k = spring constant, g = 9.8 \frac{m}{s^{2} }

  k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }

  k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }

  k = 266.01 \frac{N}{m}

Therefore, the value of spring constant is 266.01 \frac{N}{m}

6 0
3 years ago
true or false, based on the color of the particles, a mixture can be classified as a solution, a suspension, or a colloid
expeople1 [14]
False, you pass a light through a mixture If the light bounces off the particles, you will see the light shine through and you have a colloid mixture
6 0
3 years ago
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
3 years ago
How many moles of MgCl2 are there in 315 g of the compound? mol
jeka94
315g/95gmol-1
3.315 moles of MgCl2
4 0
3 years ago
Read 2 more answers
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