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SVETLANKA909090 [29]
3 years ago
6

Explain why an equation may be homogenous with respect to its unit but still be incorrect​

Physics
1 answer:
natima [27]3 years ago
6 0
When both sides of an equation give the same units, same numerical values, and same concept we refer to the equation as being correct. ... Removing constants from correct equations make them homogeneous but incorrect.
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Please answer and ill give brainliest
slavikrds [6]

Answer:

12:00 A.M. (midnight)

Explanation:

If it is the middle of the night, not many people will use electricity because most people are sleeping

Hope this helps! :)

PLEASE MARK ME AS BRAINLIEST

5 0
3 years ago
Three objects are attached to a massless rigid rod that has an axis of rotation as shown. Assuming all of the mass of each objec
Vitek1552 [10]
The aggregate of all the given moment of inertia's will be the moment of inertia of this system.
as, moment of inertia is given as
l = m * r^2
so, finding the moment of inertia of all the individual and adding them

<span>I=2∗<span>1^2</span>+1∗<span>2^2</span>+.5∗<span>2.5^2
</span>=9.125</span>
3 0
3 years ago
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.9537 N when separated by
scoray [572]

Answer:

<em>The initial charges on the spheres are  </em>6.796\ 10^{-6}\ c and -3.898\ 10^{-6}\ c

Explanation:

<u>Electrostatic Force </u>

Two charges q1 and q2 separated a distance d exert a force on each other which magnitude is computed by the known Coulomb's formula

\displaystyle F=\frac{K\ q_1\ q_2}{d^2}

We are given the distance between two unknown charges d=50 cm = 0.5 m and the attractive force of -0.9537 N. This means both charges are opposite signs.

With these conditions we set the equation

\displaystyle F_1=\frac{K\ q_1\ q_2}{0.5^2}=-0.9537

Rearranging

\displaystyle q_1\ q_2=\frac{-0.9537(0.5)^2}{k}

Solving for q1.q2

\displaystyle q_1\ q_2=-2.6492.10^{-11}\ c^2\ \ ......[1]

The second part of the problem states the spheres are later connected by a conducting wire which is removed, and then, the spheres repel each other with an electrostatic force of 0.0756 N.

The conducting wire makes the charges on both spheres to balance, i.e. free electrons of the negative charge pass to the positive charge and they finally have the same charge:

\displaystyle q=\frac{q_1+q_2}{2}

Using this second condition:

\displaystyle F_2=\frac{K\ q^2}{0.5^2}=\frac{K(q_1+q_2)^2}{(4)0.5^2}=0.0756

\displaystyle q_1+q_2=2.8983\ 10^{-6}\ C

Solving for q2

\displaystyle q_2=2.8983\ 10^{-6}\ C-q_1

Replacing in [1]

\displaystyle q_1(2.8983\ 10^{-6}-q_1)=-2.64917.10^{-11}

Rearranging, we have a second-degree equation for q1.  

\displaystyle q_1^2-2.8983.10^{-6}q_1-2.64917.10^{-11}=0

Solving, we have two possible solutions

\displaystyle q_1=6,796.10^{-6}\ c

\displaystyle q_1=-3.898.10^{-6}\ c

Which yields to two solutions for q2

\displaystyle q_2=-3.898.10^{-6}\ c

\displaystyle q_2=6.796.10^{-6}\ c

Regardless of their order, the initial charges on the spheres are 6.796\ 10^{-6}\ c and -3.898\ 10^{-6}\ c

8 0
3 years ago
I'm sorry this question is deleted
Colt1911 [192]

Answer:

lol

you are so funny.....

8 0
3 years ago
Which of the following is an example of an
olga55 [171]

Answer:

c. mandible

Explanation:

7 0
3 years ago
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