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4 years ago

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What is Plancks constant (h)?

1 answer:

Tatiana [17]4 years ago

6 0

Planck's constant is 4.39048042 × 10-67 m4 kg2 / s2.

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A point charge of -2 µC is located at the origin. A second point charge of 6 µC is at x = 1 m, y = 0.5 m. Find the x and y coord

Soloha48 [4]

Answer:

x coordinate = -1.66 m

y coordinate is = -0.825m

Explanation:

Suppose z be the distance form the first charge and z + sqrt(1^2 +.5^2) be the distance from the second So z + sqrt(1+.25) = z + 1.12

We have k*2.0x10^-6/s^2 = k*6x10^-6/(s+1.12)^2

0.0356s^2 -0.019s-0.0897=0

s=1.876m

The angle of the line between the two charges is arctan(.5/1) = 26.6o

x coordinate = -1.876*cos(26.6) = -1.66m

y coordinate is -1.876*sin(26.6) = -0.825m

3 0

3 years ago

In a closed electrical circuit, what visible evidence indicates that energy is flowing?

____ [38]

Answer:

A light bulb lightning up

Explanation:

5 0

4 years ago

What is the electric force acting between two charges of -0.0050 C and

love history [14]

The electric force is $-3.6\cdot 10^8 N$ (attractive)

Explanation:

The magnitude of the electric force between two charges is given by Coulomb's law:

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

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have the following:

$q_1 = -0.0050 C$ (charge 1)

$q_2 = +0.0050 C$ (charge 2)

r = 0.025 m (distance)

Substituting, we find the electric force between the two charges:

$F=(9\cdot 10^9) \frac{(-0.0050)(0.0050)}{(0.025)^2}=-3.6\cdot 10^8 N$

And the negative sign means the force is attractive, since the two charges have opposite sign.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

Intermolecular forces hold

blagie [28]

I believe Intermolecular forces hold, <span>molecules, ions, and atoms? But I would see if that doesn't sound familiar check it with a site or something?</span>

3 0

3 years ago

An 30-turn coil has square loops measuring 0.341 m along a side and a resistance of 3.61 Ω. It is placed in a magnetic field tha

Verdich [7]

Answer:

Thus, the induced current in the coil at $t =1.73s$ is 9.98 A.

Explanation:

Faraday's law says

$\varepsilon = N \frac{d \Phi_B}{dt}$

where $N$ is the number of turns and $\Phi_B$ is the magnetic flux through the square coil:

Now,

$N = 30$

$\theta = 37.5^o$

$A = (0.341m)^2= 0.11623m^2$

$B = 1.45t^3$ ;

therefore,

$\varepsilon = N \frac{d \Phi_B}{dt} = N\frac{d ( BA\:cos(\theta))}{dt} = 30*\frac{d ( (1.45t^2)(0.1163)\:cos(37.5^o))}{dt}$

$=30*(0.1163)\:cos(37.5^o)*1.45*\dfrac{d ( t^3)}{dt} = 12.04t^2$

$\boxed{\varepsilon = 12.04t^2}$

is the emf induced in the coil.

Now, the loop is connected to $R = 3.61\Omega$ resistance; therefore, at $t = 1.73s$

$\varepsilon = RI = 12.04t^2$

$RI = 12.04(1.73)^2$

$RI = 36.03$

$I = \dfrac{36.03}{3.61\Omega }$

$\boxed{I = 9.98A }$

Thus, the current in the coil at $t =1.73s$ is 9.98 A.

8 0

4 years ago