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Allisa [31]
3 years ago
6

The glass in a window is 35 inches wide and 20 inches tall, and standard atmospheric pressure is 14.7 pounds per square inch. Wh

at net force would result in the glass if the air pressure outside the house were 1% lower than the air pressure inside?
Physics
1 answer:
iogann1982 [59]3 years ago
8 0

Answer:103 pounds

Explanation:

Given

width of window b=35 in.

height of window h=20 in.

standard atmospheric pressure P_{outside}=14.7 psi

Also P_{inside}=1.01P_{outside}

Thus Net Force on the window will be the algebraic sum of Force due to outside and inside Pressure .

F_{net}=(P_{inside}-P_{outside})\cdot A

F_{net}=P_{outside}(1.01-1)\times 35\times 20

F_{net}=14.7\times 0.01\times 35\times 20

F_{net}=102.9\ pounds\approx 103\ pounds            

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So, we need to determine width and length of the stripe.

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5 0
3 years ago
The similarity between thermal energy, heat and temperature?
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6 0
3 years ago
39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
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Define an expression for linear expansion of solid​
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\huge \tt \underline \red{answer}

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Thanku have a great day ✨❤️

4 0
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