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Ber [7]
3 years ago
15

A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me

rcury in this process?
Physics
1 answer:
loris [4]3 years ago
5 0

To solve this exercise we will use the concept related to heat loss which is mathematically given as

Q = mC_p \Delta T

Where,

m = mass

C_p= Specific Heat

\Delta T = Change in temperature

Replacing with our values we have that

m = 0.1g

C_p = 139J/Kg\cdot K \rightarrow Specific heat of mercury

\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K

Replacing

Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J

Therefore the heat lost by mercury is 0.09J

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A dog has a mass of 20 kg. If the dog is pushed across the ice with a force of 40 N, what is its acceleration?
olasank [31]

Answer:

The acceleration is 2 m/s2.

Explanation:

We calculate the acceleration (a), with the data of mass (m) and force (F), through the formula:

F = m x a  ---> a= F/m

a = 40 N/20 kg                   <em>  1N= 1 kg x m/s2</em>

a= 40 kgx m/s2/ 20 kg

<em>a= 2 m/s2</em>

7 0
3 years ago
Which statement correctly describes why a compound is a pure substance?
ad-work [718]

Explanation:

A compound is a pure substance composed of two or more different atoms chemically bonded to one another. A compound can be destroyed by chemical means. It might be broken down into simpler compounds, into its elements or a combination of the two.

6 0
2 years ago
For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ
slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0
3 years ago
Record two activities where you would expect to have a steep learning curve, two where you would have an “average” learning curv
lora16 [44]

Answer:

r

Explanation:

3 0
2 years ago
A golf ball reaches a height of 150 m before it stops rising and starts to fall to the ground. What is the golf balls speed (rou
artcher [175]

Answer:

v = 54 m/s

Explanation:

Given,

The maximum height of the flight of golf ball, h = 150 m

The velocity at height h, u = 0

The velocity of the golf ball right before it hits the ground, v = ?

Using the III equations of motion

                               <em>  v² = u² + 2gh</em>

Substituting the given values in the above equation,

                                 v² = 0 + 2 x 9.8 x 150 m

                                     = 2940

                                  v = 54 m/s

Hence, the speed of the golf ball right before it hits the ground, v = 54 m/s

4 0
3 years ago
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