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2 years ago

12

All of the following are physical stimuli for touch except __________ energy.

2 answers:

hodyreva [135]2 years ago

6 0

C. combustion I think. hope this helps

koban [17]2 years ago

5 0

The answer is C thank you for your time

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A sample of helium behaves as an ideal gas as energy is

alisha [4.7K]

Answer:

0.0321 g

Explanation:

Let helium specific heat $c_h = 5.193 J/g K$

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

$E_h = m_hc_h \Delta T = 20 J$

where $m_h$ is the mass of helium, which we are looking for:

$m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g$

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2 years ago

What is a conservative force?

almond37 [142]

Answer:

A conservative force is a force with the property that the total work done in moving a particle between two points is independent of the path taken Equivalently if a particle travels in a closed loop the total work done by a conservative force is zero

Explanation:

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2 years ago

An electrician timed a box traveling over a conveyor and measured 25 sec to travel 30 feet. If the gearbox ratio is 20:1 and the

nataly862011 [7]

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3 years ago

1. Does the validity of the principle of conservation of momentum depend on the validity of Newton's 3rd law of motion?

OverLord2011 [107]

I truly believe so, that’s a definite yes

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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

3 years ago