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3 years ago

We are usually not aware of the electric force acting between two everyday objects because

Physics

2 answers:

Semmy [17]3 years ago

5 0

Most everyday objects have as many plus charges as they have minus charges

Hunter-Best [27]3 years ago

4 0

Answer:

Electric force of attraction or repulsion acts between two charged objects. It is a non-contact force. It means the objects need not be in contact for an electric force to exist.

Electric force depends on the amount of charge on each object and distance between them.

We are usually not aware of the electric force acting between two everyday objects because they are neutral in nature that is there are equal number electrons and protons in the atom constituting the objects. Also, with increase in distance, the magnitude of the electric force decreases.

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At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

$-\frac{K \,dV}{V} = dP$

This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

$-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}$

$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0

3 years ago

An object is moving initially with a velocity of 4.7 m/s . After 3.9 s the object's velocity is -2.1 m/s . What is the object's

IgorC [24]

Answer: The acceleration of the object is 0.67m/s^2 west.

Explanation: Here we are given the initial velocity and final velocity as well as the time taken. Acceleration is the change in velocity per unit time, thus the equation becomes.

a=dv/t

a=vf-vi/t

a=-2.1-4.7/3.9

a= 0.66m/s^2 west

8 0

3 years ago

How often is carbon dioxide cycled through the atmosphere

olga55 [171]

Maybe around 350 years, depending on the carbon cycle and the time taken through steps.

5 0

3 years ago

A lamp hangs from the ceiling at a height of 2.9 m. If the lamp breaks and falls to the floor, what is its impact speed

ser-zykov [4K]

To solve this there is this website that I found that helps
I am in middle school so I have no idea how to solve this
but
this website may help considering u are in high school and u
(hopefully mind u)
know how to solve this
so to get there u google
"whats impact speed"
and click on the first thing there the website is ehow

8 0

3 years ago

According to Newton's Second Law, the force of the club hitting the golf ball will cause it to accelerate. At the moment of impa

vazorg [7]

Answer:

Option B

Explanation:

<h3>According to Newton's third law, for every reaction there will be equal and opposite reaction</h3>

Here in this case the force of the club hitting the golf ball will be in one direction and the force acting on club due to golf ball will be in opposite direction and magnitude of this force will be same as the magnitude of the force of the club hitting the golf ball

In this case the action will be the force of the club hitting the golf ball and reaction will be the force acting on club due to golf ball

∴ The club pushes against to golf ball with a force equal and opposite to the force of the golf ball on the club

8 0

3 years ago

Read 2 more answers