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Kay [80]
3 years ago
11

If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?

Physics
2 answers:
mel-nik [20]3 years ago
5 0

As per the question the volume of mercury  is given as 0.002 m^3 at 20 degree Celsius.

We are asked to calculate the volume of the mercury at 50 degree Celsius.

This problem is based on thermal expansion of matter.

Let us consider the initial and final volume of the mercury is denoted as -

                                  v_{1} \ and\ v_{2}

Let the initial and final temperature of the mercury is denoted as -

                                                                    T_{1}\ and \ T_{2}

As per question

v_{1} =0.002 m^3                v_{2} =?

T_{1} =20^0 C                      T_{2} =50^0 C

The change in temperature is

                                             T_{2} -T_{1}

                                         = 50^0 C -20^0 C

                                               = 30^0 C

Mercury is a fluid.So we have to apply volume expansion of liquid .

The coefficient of of volume expansion of mercury [ \gamma ] at 20 degree Celsius is 0.00018 per centigrade.

As per volume expansion of liquid,

                                       V_{T} = v_{1} [1 +\gamma [T_{2} -T_{1} ]]

Here V_{T} is the volume at T degree Celsius.

Hence volume at 50 degree Celsius is calculated as-

                               v_{2} =v_{1} [1+\gamma[50-30]]

                                       = 0.002[1+0.00018*30]

                                       =0.0020108m^3  [ans]

As per the options given in the question ,option A is close to the calculated value. So option A is right.


kogti [31]3 years ago
5 0
<span>The answer to this question lies in understanding the Combined Gas Law in chemistry, which combines Boyle's, Charles', and Gay-Lussac's gas laws. This law shows the relationship between pressure, volume, and temperature of gases and is represented by the equation P1V1/T1 = P2V2/T2. Changes in pressure (P) and volume (V) are directly proportional to one another, but these are inversely proportional to changes in temperature (T). This question must assume that pressure is constant, so the Ps cancel out and the equation becomes V1/T1 = V2/T2. Additionally, temperature should be converted to Kelvins (K) by adding 273 to the temperature in Celsius (so 20 C = 293 K, and 50 C = 323 K). If V1 = 0.002 and T1 = 293 and T2 = 323, then we solve the equation as V2 = T2*V1/T1 or V2 = (323*0.002)/293, so the answer is approximately equal to 0.0022... or answer (C).</span>
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vovikov84 [41]

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity=70 mph\approx 31.29 m/s

velocity after 5 s is 30 mph\approx 13.41 m/s

Therefore acceleration during these 5 s

a=\frac{v-u}{t}

a=\frac{13.41-31.29}{5}=-3.576 m/s^2

therefore time required to stop

v=u+at

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initial velocity =31.29 m/s

0=31.29-3.576\times t

t=\frac{31.29}{3.576}=8.75 s

(b)total distance traveled before stoppage

v^2-u^2=2as

0^2-31.29^2=2\times (-3.576)\cdot s

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On a Fahrenheit thermometer, the gas becomes 18 degrees warmer.

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A lever is being used to move a heavy stone from a garden. The stone weighs 250 newtons. The board used as a lever is 6 meters l
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A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person
vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 \frac{m}{s}

t=0.475s V=1.95 \frac{m}{s}

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

sin(\alpha) =\frac{op}{h}

op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz

c).

T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz

The period is the inverse of the time of the motion so, the T1 is faster that the T because

t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s

d).

The speed is the relation between the distance with time so:

Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}

3 0
3 years ago
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