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Mamont248 [21]
2 years ago
8

Is weather chart the same as aneroid barometer?​

Physics
1 answer:
kipiarov [429]2 years ago
5 0
Kind of. A barometer measures air pressure. Not actual weather. Even though air pressure has a lot to do with the weather it is not exactly the same thing.
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Define fundamental and derived quantities with two example of each. ​
stich3 [128]

Fundamental quantity : quantities which are independent on other physical quantity .

ex: length,mass,time, current, amount of substance, luminous intensity, thermodynamic temperature,

Derived quantity : quantities which are depend on fundamental quantities.

ex: Area, volume, density, speed, acceleration, force, velocity etc.

4 0
2 years ago
Part b suppose the magnitude of the gravitational force between two spherical objects is 2000 n when they are 100 km apart. what
kobusy [5.1K]
<span>b) The force with a distance of 150 km is 889 N c) The force with a distance of 50 km is 8000 N This question looks like a mixture of a question and a critique of a previous answer. I'll attempt to address the original question. Since the radius of the spherical objects isn't mentioned anywhere, I will assume that the distance from the center of each spherical object is what's being given. The gravitational force between two masses is given as F = (G M1 M2)/r^2 where F = Force G = gravitational constant M1 = Mass 1 M2 = Mass 2 r = distance between center of masses for the two masses. So with a r value of 100 km, we have a force of 2000 Newtons. If we change the distance to 150 km, that increases the distance by a factor of 1.5 and since the force varies with the inverse square, we get the original force divided by 2.25. And 2000 / 2.25 = 888.88888.... when rounded to 3 digits gives us 889. Looking at what looks like an answer of 890 in the question is explainable as someone rounding incorrectly to 2 significant digits. If the distance is changed to 50 km from the original 100 km, then you have half the distance (50/100 = 0.5) and the squaring will give you a new divisor of 0.25, and 2000 / 0.25 = 8000. So the force increases to 8000 Newtons.</span>
8 0
3 years ago
Read 2 more answers
Identify a situation in which you would want to have a high
Andreyy89

Answer: A voltmeter must have a high resistance where as an ammeter must have a low resistance.

Explanation:

A voltmeter is a device which is connected in parallel to the component across which voltage needs to be measured. In a parallel circuit voltage drop is same at the nodes. The parallel connection must not offer easier path for current to divert from the main circuit and travel. Thus, a voltmeter must have high resistance.

On the other hand, an ammeter which is used to measure current in the circuit must have low resistance as it is connected in series. It should not offer resistance as it would reduce the actual current and measurement would be inaccurate.

7 0
3 years ago
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What are the two sources of energy for the earth system
spin [16.1K]
Solar energy and Geothermal energy are your answers :)
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3 years ago
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Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

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

8 0
3 years ago
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