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3 years ago

How do warm ocean currents affect the weather?

Physics

1 answer:

Lilit [14]3 years ago

8 0

Answer:

the answer is D I believe

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After a displacement of 17 m, a train on a straight track is at the position xf = –2.5 m

EastWind [94]

-19.5m

-19.5+17=-2.5m

5 0

3 years ago

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A standard inverting op-amp circuit has an R1 of 10 kΩ and an Rf of 220 kΩ. If the offset current is 100 nA the output offset vo

kiruha [24]

Answer:

The value is $V_{os} = 0.001 \ V$

Explanation:

From the question we are told that

The circuit resistance is $R_1 = 10 \ k \Omega$

The feedback resistance is $R_f = 220 \ k \Omega$

The offset current is $I_{os } = 100 \ nA = 100 * 1)^{-9} \ A$

Generally the offset voltage is mathematically reparented as

$V_{os} = R_f * I_{os}$

=> $V_{os} = 10 *10^{3}* 100 *10^{-9}$

=> $V_{os} = 0.001 \ V$

6 0

3 years ago

A weatherman carried an aneroid barometer from the ground floor to his office atop the Sears Tower in Chicago. On the level grou

Sophie [7]

Answer:

442.36038 m or 1451.31362 ft

Explanation:

$P_1$ = Initial pressure = 30.15 inHg

$P_2$ = Final pressure = 28.607 inHg

$\rho$ = Density of air = 0.075 lb/ft³

$1\ lb/ft^3=16.0185\ kg/m^3$

$1\ in=0.0254\ m$

$1\ m=3.28084\ ft$

Density of mercury = 13560 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Difference in pressure is given by

$P_1-P_2=\rho gh\\\Rightarrow h=\frac{P_1-P_2}{\rho g}\\\Rightarrow h=\frac{(30.15-28.607)\times 13560\times 0.0254\times 9.81}{0.075\times 16.0185\times 9.81}\\\Rightarrow h=442.36038\ m\\\Rightarrow h=442.36038\times 3.28084\\\Rightarrow h=1451.31362\ ft$

The height of the building is 442.36038 m or 1451.31362 ft

4 0

3 years ago

A machinist with normal vision has a near point at 25 cm. The machinist wears eyeglasses in order to do close work. The power of

Alik [6]

Answer:

17.4 cm

Explanation:

Power of lens = +1.75 diopters

Focal length of lens

$f=\frac{100}{1.75}=\frac{400}{7}$

This is a convex lens as focal the diopter given is positive which makes the focal length positive. Image distance will be negative.

v = -25

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\Rightarrow \frac{1}{-25}-\frac{1}{u}=\frac{1}{\frac{400}{7}}\\\Rightarrow -\frac{1}{u}=\frac{7}{400}+\frac{1}{25}\\\Rightarrow \frac{1}{u}=-\frac{7}{400}-\frac{1}{25}\\\Rightarrow \frac{1}{u}=-0.0575\\\Rightarrow u=-17.4\ cm$

∴ The new near point is 17.4 cm

7 0

3 years ago

Please help will mark brainleist <br> ASAP

kiruha [24]

Answer:

B - Earth's path around the Sun

Explanation:

4 0

3 years ago

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