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3 years ago

PLEASE HELP ME WITH MY HOMEWORK!!!!!

Physics

1 answer:

IgorC [24]3 years ago

5 0

Known :

l = 7 cm

w = 4 cm

Asked :

h = ...?

Answer :

V = B triangle × h (long)

35 = ½ × 4 × h × 7

35 = ½ × 28 × h

35 = 14 h

h = 35 ÷ 14

h = 2,5 cm

Sorry if I am wrong, I only study

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Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i

Kobotan [32]

The magnitude of electrical force on charge $q_{3}$ due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector position of each particle respect to origin are described below:

$\vec r_{1} = (-0.500, 0)\,[m]$

$\vec r_{2} = (+0.500, 0)\,[m]$

$\vec r_{3} = (0, +0.500)\,[m]$

Then, distances of the former two particles particles respect to the latter one are found now:

$\vec r_{13} = (+0.500, +0.500)\,[m]$

$r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{13} =\frac{\sqrt{2}}{2}\,m$

$\vec r_{23} = (-0.500, +0.500)\,[m]$

$r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{23} =\frac{\sqrt{2}}{2}\,m$

The resultant force is found by Coulomb's law and principle of superposition:

$\vec R = \vec F_{13}+\vec F_{23}$ (1)

Please notice that particles with charges of same sign attract each other and particles with charges of opposite sign repeal each other.

$\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13} +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23}$ (2)

Where:

$k$ - Electrostatic constant, in newton-square meters per square Coulomb.
$q_{1}$ , $q_{2}$ , $q_{3}$ - Electric charges, in Coulombs.
$r_{13}$ , $r_{23}$ - Distances between particles, in meters.
$\vec u_{13}$ , $\vec u_{23}$ - Unit vectors, no unit.

If we know that $k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{1} = 2\times 10^{-6}\,C$ , $q_{2} = 2\times 10^{-6}\,C$ , $q_{3} = 2\times 10^{-6}\,C$ , $r_{13} =\frac{\sqrt{2}}{2}\,m$ , $r_{23} =\frac{\sqrt{2}}{2}\,m$ , $\vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} \right)$ and $\vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$ , then the vector force on charge $q_{3}$ is:

$\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$

$\vec R = 0.072\cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + 0.072\cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)\,[N]$

$\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]$

And the magnitude of the electrical force on charge $q_{3}$ ( $R$ ), in newtons, due to the others is found by Pythagorean theorem:

$R = 0.102\,N$

The magnitude of electrical force on charge $q_{3}$ due to the others is 0.102 newtons. $\blacksquare$

To learn more on Coulomb's law, we kindly invite to check this verified question: brainly.com/question/506926

8 0

2 years ago

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

alexgriva [62]

Answer:

R=4.22*10⁴km

Explanation:

The tangential speed $v$ of the geosynchronous satellite is given by:

$v=\frac{2\pi R}{T}$

Because $2\pi R$ is the circumference length (the distance traveled) and T is the period (the interval of time).

Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

$F_c=\frac{mv^{2} }{R}$

If we substitute the expression for $v$ in this formula, we get:

$F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}$

Since the centripetal force is the gravitational force $F_g$ between the satellite and the Earth, we know that:

$F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }$

Where G is the gravitational constant ( $G=6.67*10^{-11} Nm^{2}/kg^{2}$ ) and M is the mass of the Earth ( $M=5.97*10^{24}kg$ ). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:

$R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km$

This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.

5 0

3 years ago

A box with a mass of 18 kg is pushed across the floor. It has coefficient of friction of 0.39. Calculate the force of friction i

Taya2010 [7]

Answer:

68.8 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of box = 18 Kg

Coefficient of friction (μ) = 0.39

Force of friction (F) =?

Next, we shall determine the normal force of the box. This is illustrated below:

Mass (m) of object = 18 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal force (N) =?

N = mg

N = 18 × 9.8

N = 176.4 N

Finally, we shall determine the force of friction experienced by the object. This is illustrated below:

Coefficient of friction (μ) = 0.39

Normal force (N) = 176.4 N

Force of friction (F) =?

F = μN

F = 0.39 × 176.4

F = 68.796 ≈ 68.8 N

Thus, the box experience a frictional force of 68.8 N.

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3 years ago

PLEASE ILL GIVE BRAINLIST Which term describes the high point of a transverse wave?

Pani-rosa [81]

Answer:

I belive that the answer is the A

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3 years ago

Geology belongs to which main category of science?

sergey [27]

Geology belongs to the category of Natural Sciences. Natural sciences is the study of natural phenomena. Among the natural sciences, we have physical science, which is the study of non-living systems. Further along this branch, we have earth sciences where Geology belongs.

7 0

3 years ago