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3 years ago

Hi Friends!please help me with these questions!SUBJECT: Chemistry, Physics,Biology

Physics

1 answer:

satela [25.4K]3 years ago

6 0

Answer:

q.1 : Air near candle gets heated up and after this it rises by convection so the thermometer B will receive more heat than the thermometer A So, according to the given condition thermometer B will show a greater rise in temperature.

q.2 : x is the pure sample of compound . y is the pure sample of element . z is the mixture of different elements

q.3 : the saliva contains an enzyme salivary amylase (ptyalin) which converts starch in roti into maltose, isomaltose and small dextrins called a-dextrin.

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The ion at the center of a silicate tetrahedron is surrounded by ________.

Naddika [18.5K]

Oxygen....................

3 0

3 years ago

What element are the cores of massive stars made of when a supernova occurs?

Studentka2010 [4]

I think its C. Oxygen

8 0

3 years ago

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm

Karo-lina-s [1.5K]

Answer:

the thermistor temperature = $325.68 \ ^0 \ C$

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

$T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K$

Resistance of the thermistor $R_1 =$ 20,000 ohms

Material constant $\beta$ = 3650

Resistance of the thermistor $R_2$ = 500 ohms

Using the equation :

$R_1 = R_2 \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{R_1}{ R_2} = \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

Taking log of both sides

$In \ \frac{R_1}{ R_2} = In \ \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{1}{T_2} = \frac{1}{T_1} - \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}$

${T_2} = \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}$

Replacing our values into the above equation :

${T_2} = \frac{3650*373}{3650 - In (\frac{20000}{500})373}$

${T_2} = \frac{1361450}{3650 - 3.6888*373}$

${T_2} = \frac{1361450}{3650 - 1375.92}$

${T_2} = \frac{1361450}{2274.08}$

${T_2} = 598.68 \ K$

${T_2} = 325.68 \ ^0 \ C$

Thus, the thermistor temperature = $325.68 \ ^0 \ C$

4 0

4 years ago

Find the speed of light in each of the following materials. (a) gallium phosphide m/s (b) carbon disulfide m/s (c) benzene

Oksanka [162]

Explanation:

We need to calculate the speed of light in each materials

(I). Gallium phosphide,

The index of refraction of Gallium phosphide is 3.50

Using formula of speed of light

$v=\dfrac{c}{\mu}$ ....(I)

Where, $\mu$ = index of refraction

c = speed of light

Put the value into the formula

$v=\dfrac{3\times10^{8}}{3.50}$

$v=8.6\times10^{7}\ m/s$

(II) Carbon disulfide,

The index of refraction of Gallium phosphide is 1.63

Put the value in the equation (I)

$v=\dfrac{3\times10^{8}}{1.63}$

$v=1.8\times10^{8}\ m/s$

(III). Benzene,

The index of refraction of Gallium phosphide is 1.50

Put the value in the equation (I)

$v=\dfrac{3\times10^{8}}{1.50}$

$v=2\times10^{8}\ m/s$

Hence, This is the required solution.

7 0

3 years ago

A hot air balloon rising vertically is tracked by an observer located 3 miles from the lift-off point. At a certain moment, the

yuradex [85]

Answer:

$\frac{dy}{dt}=1.2\frac{mi}{min}$

Explanation:

We know that the tangent function relates the angle of the right triangle that forms the hot air balloon rising:

$tan\theta=\frac{y}{x}\\y=xtan\theta(1)$

Differentiating (1) with respect to time, we get:

$\frac{dy}{dt}=tan\theta\frac{dx}{dt}+xsec^{2}\theta\frac{d\theta}{dt}\\$

$\frac{dx}{dt}=0$ since x is a constant value. Replacing:

$\frac{dy}{dt}=3mi(sec^{2}\frac{\pi}{3})0.1\frac{rad}{min}\\\frac{dy}{dt}=1.2\frac{mi}{min}$

5 0

4 years ago