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Scilla [17]
3 years ago
12

A softball is thrown with 145 Joules of kinetic energy. If the ball is moving at 20.0 m/s, what is the mass of the ball in kg?

Chemistry
1 answer:
Tasya [4]3 years ago
3 0

Answer:

0.725 kg

Explanation:

Step 1: Given data

  • Kinetic energy of the softball (K): 145 J
  • Speed of the softball (v): 20.0 m/s
  • Mass of the softball (m): ?

Step 2: Calculate the mass of the softball

We will use the following expression.

K = 1/2 × m × v²

m = 2 K / v²

m = 2 × 145 J / (20.0 m/s)²

m = 0.725 kg

The mass of the softball is 0.725 kg.

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Explanation:

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How do you determine the charges on the ions in an ionic compound?
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Consider the following specific heats of metals. Metal Specific Heat Aluminum 0.897 J/(g°C) Magnesium 1.02 J/(g°C) Lithium 3.58
Arte-miy333 [17]

Answer:

Lithium

Explanation:

The specific geat capacity of a substance is the energy required to raise 1 unit of that substance by one degree.

Heat energy (Q) = mc∇t

Q = heat energy

M = mass of the substance

c = specific heat capacity

∇t = change in temperature of the substance

Generally, increase in the specific heat capacity will lead to a lower final temperature likewise decrease in the specific heat capacity will lead to increase the final temperature of the substance.

From the data above, we can take just two specific heat capacity and test this theory.

Assuming we have a

Mass = 25g

Heat energy applied (Q) = 1 J

Initial temperature (T1) = 10°C

Final temperature (T2) = ?

Q = mc∇t

Q = mc (T2 - T1)

For Lithium, specific heat capacity = 3.58J/g°C

1 = 25 × 3.58 (T2 - 10)

Solve for T2

1 = 89.5 (T2 - 10)

1 = 89.5T2 - 895

89.5T2 = 896

T2 = 896 / 89.5

T2 = 10.011°C

For Magnesium (Mg) specific heat capacity = 1.02J/g°C

Q = mc∇t

1 = 25 × 1.02 × (T2 - 10)

1 = 25.5 (T2 - 10)

1 = 25.5T2 - 255

Solve for T2

25.5T2 = 256

T2 = 10.039°C

Notice the trend that decrease in the specific heat capacity leads to increase in the final temperature.

Try and continue for the elements and see how it works.

4 0
3 years ago
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