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Vikki [24]
3 years ago
7

Based on the tiles shown, how many meters per second is equivalent to 72 kilometers per hour?

Chemistry
2 answers:
Nana76 [90]3 years ago
5 0

Answer:

The correct answer is option B.

Explanation:

Given speed of 72 km/hour.

1 kilometer = 1000 m

1 hour = 60 minutes

1 minutes = 60 seconds

1 hour = 60 × 60 seconds = 3600 seconds

72 km/h=\frac{72\times 1000 m}{1\times 3600 s}=20 m/s

= 20 m/s

72 km/h = 20 m/s

Hence, the correct answer is option B.

bija089 [108]3 years ago
3 0
We are asked to convert the value <span>72 kilometers per hour to units of meters per second. We need conversion factors to multiply to this value. We convert as follows:

</span><span>72 km / hr ( 1000 m / 1 km ) ( 1 hr / 3600 s ) = 20 m/s --->OPTION B</span><span>
</span>
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4 years ago
Hot lead with a mass of 200.0 g of (Specific heat of Pb = 0.129 J/gºC) at 176.4°C was dropped into a calorimeter containing an u
Morgarella [4.7K]

the Calorimetry relationships you can find the amount of water in the calorimeter is      m = 21.3 g

given parameters

  • Lead mass M = 200.0 g
  • Initial lead temperature T₁ = 176.4ºC
  • Specific heat of Lead    c_{e Pb} = 0.129 J / g ºC
  • Sspecific heat of water c_{e H_2O} = 4.186 J / g ºC
  • Initial water temperature T₀ = 21.7ºC
  • Equilibrium temperature T_f = 56.4ºC

to find

The body of water

Thermal energy is the energy stored in the body that can be transferred as heat when two or more bodies are in contact. Calorimetry is a technique where the energy is transferred between the body only in the form of heat and in this case the thermal energy of the lead is transferred to the calorimeter that reaches the equilibrium that the thematic energy of the two is equal

              Q_{ceded} = Q_{absorbed}

               

Lead, because it is hotter, gives up energy

              Q_{ceded} = M c_{e Pb} (T₁ - T_f)

The calorimeter that is colder absorbs the heat

              Q_{absrobed} = m c_{e H_2O} (T_f - T₀)

where M and m are the mass of lead and water, respectively, c are the specific heats, T₁ is the temperature of the hot lead, T₀ the temperature of cold water and T_f the equilibrium temperature

              M c_{ePb} (T₁ - T_f) = m c_{eH2O} (T_f - T₀)

               m = \frac{ M\  c_{ePb} \ (T_1 - T_f)}{c_{eH_2O}  \ (T_f - T_o)}

let's calculate

              m = \frac{200 \ 0.129  (176.4-56.4)}{ 4.186 \  (56.4 -21.7)}

              m = 3096 / 145.25

              m = 21.3 g

Using the Calorimetry relationships you can find the amount of water in the calorimeter is:

             m = 21.3 g

learn more about calorimetry here:

brainly.com/question/15073428

6 0
3 years ago
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