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sergij07 [2.7K]
3 years ago
15

During an all-night cram session, a student heats up a 0.577 liter (0.577 x 10- 3 m3) glass (pyrex) beaker of cold coffee. initi

ally, the temperature is 15.8 °c, and the beaker is filled to the brim. a short time later when the student returns, the temperature has risen to 91.3 °c. the coefficient of volume expansion of coffee is the same as that of water. how much coffee (in cubic meters) has spilled out of the beaker
Physics
1 answer:
krok68 [10]3 years ago
5 0

Here coffee is filled into a glass cup and when it is heated expansion of coffee as well as glass occurs

we will find the change in volume of glass cup and change in the volume of coffee.

volume expansion is calculated by

\Delta V = V_o \gamma \Delta T

for pyrex glass we have

\gamma = 4 * 10^{-6}

now by the formula

\Delta V = 4* 10^{-6} * 0.577 * (91.3 - 15.8)

\Delta V = 0.174 mL

Now for the expansion of coffee

\gamma = 2.14 * 10^{-4}

now the expansion of coffee will be

\Delta V = 2.14 * 10^{-4}* 0.577* (91.3 - 15.8)

\Delta V = 9.32 mL

now the coffee that will spill out of the cup will be

\Delta V = 9.32 mL - 0.174 mL

\Delta V = 9.15 mL

so 9.15 mL coffee will spill out of the cup

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Answer:

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3 years ago
At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is
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Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

\dfrac{1}{2}mv^2=mgl

v=\sqrt{2gl}

Put the value into the formula

u=\sqrt{2\times9.8\times50.0\times10^{-2}}

u=3.13\ m/s

The initial speed of the ball u_{1}=3.13\ m/s

The initial speed of the block u_{2}=0

(a). We need to calculate the speed of the ball after collision

Using formula of collision

v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13

v_{1}=-2.01\ m/s

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0

v_{2}=1.11\ m/s

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

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Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus
poizon [28]

Answer:

For destructive interference phase difference is

(2n+1)\pi where n∈ Whole numbers

Explanation:

For sinusoidal wave the interference affects the resultant intensity of the waves.

In the given example we have two waves interfering at a phase difference of \frac{\pi}{4} would lead to a constructive interference giving maximum amplitude at at the RMS value of the amplitude in resultant.

Also the effect is same as having a phase difference of  ( \frac{\pi}{4} + 2\pi) because after each 2π the waves repeat itself.

<em>In case of destructive interference the waves will be out of phase i.e. the amplitude vectors will be equally opposite in the direction at the same place on the same time as shown in figure.</em>

They have a phase difference of \pi or which is same as (2\pi+\pi)

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Answer:

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The p-v diagram illustration is in the attachment

T_{H} means high temperature

T_{L} means low temperature

The energy equation :

q_{h} = R* T_{h} in(V_{2}/V_{1})

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n_{TH}=1-0.333

n_{TH}=0.67

4 0
3 years ago
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