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hichkok12 [17]
3 years ago
15

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Physics
1 answer:
Aneli [31]3 years ago
3 0

d^2=4^2+3^2\\\\d^2=16+9\\\\d^2=25\\\\d=5

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Why is it important that a finger be wet<br> before it is touched to a hot clothes iron?
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A 0.0663 kg ingot of metal is heated to 241◦C
Westkost [7]

Answer:280.216j/kg°C

Explanation:

Mass of metal=0.0663kg

mass of water=0.395kg

Final temperature=27.4°C

Temperature of metal=241°C

Temperature of water=25°C

specific heat capacity of water=4186j/kg°C

0.0663xax(241-27.4)=0.395x4186x(27.4-25)

0.0663xax213.6=0.395x4186x2.4

14.16168a=3968.328

a=3968.328 ➗ 14.16168

a=280.216j/kg°C

4 0
3 years ago
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

Atmospheric pressure P_{atm}= 1.013\times10^{5}\ Pa    

drop height h'= 27.1 mm

Density of mercury \rho= 13.59 g/cm^3

We need to calculate the height

Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

h=0.76\ m

We need to calculate the new height

h''=h - h'

h''=0.76-27.1\times10^{-3}

h''=0.76-0.027

h''=0.733\ m

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

P=\rho g h

Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

P=0.97622\times10^{5}\ Pa

Hence, The atmospheric pressure is 0.97622\times10^{5}\ Pa.

7 0
3 years ago
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