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dezoksy [38]
3 years ago
6

The SI unit for pressure is pascal (Pa). One hectopascal would equal how many pascals?

Physics
2 answers:
enot [183]3 years ago
5 0

Answer:

1 hectopascal = 100 Pascal

Explanation:

Pressure exerted by an object is defined as the force acting on it per unit area. Mathematically, it is given by :

pressure=\dfrac{force}{area}

The SI unit of  unit of pressure is pascal. It is equivalent to :

1\ Pa=N/m^2

or

1\ Pa=kg\ m^{-1}s^{-2}

1 hectopascal = 100 Pascal

So, one hectopascal equals 100 pa. Hence, the correct option is (c).

Marat540 [252]3 years ago
3 0
1 hectopascal (hPa) is equivalent to 100 Pa
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Use the following equation to help you answer the question. The peak intensity of radiation from a star named Sigma is 2 x 10 6
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Answer:

1449 K

Explanation:

The surface temperature of a star is related to its peak wavelength by Wien's displacement law:

T=\frac{b}{\lambda}

where

T is the surface temperature

b is Wien's displacement constant

\lambda=2\cdot 10^{-6} m

So the surface temperature of the star is

T=\frac{2.898 \cdot 10^{-3} Km}{2\cdot 10^{-6} m}=1449 K

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A star’s brightness appearing to be a lower magnitude is known as the star’s _____.
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A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal
Natalija [7]

Answer : The specific heat of unknown sample is, 8748.78J/kg^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-[q_2+q_3]

m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]

where,

c_1 = specific heat of unknown sample = ?

c_2 = specific heat of water = 4186J/kg^oC

c_3 = specific heat of copper = 390J/kg^oC

m_1 = mass of unknown sample = 72.0 g  = 0.072 kg

m_2 = mass of water = 203 g  = 0.203 kg

m_2 = mass of copper = 187 g  = 0.187 kg

T_f = final temperature of calorimeter = 39.4^oC

T_1 = initial temperature of unknown sample = 80.0^oC

T_2 = initial temperature of water and copper = 11.0^oC

Now put all the given values in the above formula, we get

0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]

c_1=8748.78J/kg^oC

Therefore, the specific heat of unknown sample is, 8748.78J/kg^oC

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3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Alekssandra [29.7K]

so, question number 10 answer is 82 watts

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3 years ago
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katen-ka-za [31]

Answer:

C. 2 and 4

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