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3 years ago

Looking at this barometer, is the air pressure high or low? What type of weather can be predicted?

Physics

2 answers:

ASHA 777 [7]3 years ago

3 0

It really depends on the altitude of the barometer.

-- If the picture was taken in Denver, Colorado, where the long-term
average atmospheric pressure is around 24.5", then this barometer
is showing slightly higher than normal, and good weather can be
predicted.

-- If the picture was taken anywhere along the east coast of the US,
where standard "sea level" pressure is 29.92", then this barometer
is reading alarmingly low ! There's either a hurricane in progress
right now, or one can be predicted to arrive very soon and you'd best
pick up your barometer and head inland as fast as you can.

Dafna1 [17]3 years ago

3 0

Answer:

B) Low, Bad

Explanation:

As we know that given scale is in inch so the height of the barometer scale here is given as

$H = 25 inch$

now we have

$1 inch = 25.4 mm$

so we will have

$H = 25 \times 25.4$

$H = 635 mm$

so at this showing position the pressure is 635 mm of Hg while at normal condition the pressure is 760 mm Hg

So this shows low pressure and bad weather

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Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

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1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

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4 years ago

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