All categories

4 years ago

Which hand is negatively charged

Physics

1 answer:

Agata [3.3K]4 years ago

8 0

Answer:

Your Left hand is negatively charged, and receives energy. It emits the energy that "allows things to happen".

Explanation:

You might be interested in

This is lettle alchemy 2, I need a life, and I got all this in 2 days, more like 16 hours

Alex_Xolod [135]

Answer: I don't know you But i am proud good job

fun game isn't it xD

Explanation:

6 0

3 years ago

An airplane flies for 1.25 hours with a constant speed of 1080 km/hr and then for another 2.5 hours with a constant speed of 984

Makovka662 [10]

Answer:

i think its 3583.2km skskks

6 0

3 years ago

Read 2 more answers

A biker pedals hard to ride his bike to the top of a 44 m hill. He decides to let his bike coast down the hill, and is having so

Vanyuwa [196]

The biker's speed at the top of the second hill is 25.8 m/s

Explanation:

The problem can be solve by applying the law of conservation of energy. In absence of frictional forces, the total mechanical energy of the bike (the sum of potential energy + kinetic energy) must be conserved. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy at the top of the first hill

$K_i$ is the initial kinetic energy at the top of the first hill

$U_f$ is the final potential energy at the top of the second hill

$K_f$ is the final kinetic energy at the top of the second hill

We can rewrite the equation as:

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m is the mass of the bike

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 44 m$ is the height of the first hill

u = 0 m/s is the speed at the top of the first hill

$h_f = 10 m$ is the height of the second hill

v is the speed at the top of the second hill

And solving for v, we find:

$mgh_i = mgh_f + \frac{1}{2}mv^2\\v^2=\sqrt{2g(h_i-h_f)}=\sqrt{2(9.8)(44-10)}=25.8 m/s$

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

7 0

3 years ago

Which of the following is a unit of speed?

dangina [55]

Answer:

Option c that it means cm/year.

Explanation:

Nm means nanometer, 1x10^9m

Km/s^2 is acceleration

h.. it may be height

7 0

3 years ago

Read 2 more answers

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.

Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called <u>Electrostatic</u> <u>Force</u>.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

$F=\frac{k.q.Q}{r^{2}}$

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

$F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}$

$F_{12}=536.02.10^{-3}$ N

Force between 2 and 3:

$F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}$

$F_{23}=2711.63.10^{-3}$ N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

$F_{T}=2711.63.10^{-3}-536.02.10^{-3}$

$F_{T}=2175.61.10^{-3}$ N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, $F_{13}=F_{23}$ . Suppose distance from 1 to 3 is x, then from 2 to 3 is $x-0.301$

Calculating:

$\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}$

$\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}$

$\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}$

$x^{2}=0.67x^{2}-0.40x+0.061$

$0.33x^{2}+0.40x-0.061=0$

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

4 0

3 years ago