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3 years ago

Resultant of a pair of 100 km/h velocity vectors that are at right angels to each other

Physics

1 answer:

nignag [31]3 years ago

5 0

The resultant of the two vectors is 141 km/h.

Explanation:

When two vectors are at right angles to each other, their resultant can be calculated by using the Pythagorean's theorem:

$R=\sqrt{A^2+B^2}$

where

A is the magnitude of the first vector

B is the magnitude of the second vector

In this problem, we have two velocity vectors of magnitude

A = 100 km/h

B = 100 km/h

Substituting into the formula, we find their resultant:

$R=\sqrt{100^2+100^2}=141.4 km/h$

Learn more about vector addition here:

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A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each

ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

$E_{1}+E_{3}-E_{2}=0$

We know that the electric field is:

$E=k\frac{q}{r^{2}}$

Where:

k is the Coulomb constant
q is the charge
r is the distance from the charge to the point

So, we have:

$k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0$

Let's solve it for r(3).

$\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0$

$r_{3}=0.0743\:$

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

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3 years ago

If the numbers on the plate are 6.0 cm apart, and the spy satellite is at an altitude of 160 km , what must be the diameter of t

Alexxx [7]

Answer:

The diameter of the camera aperture must be greater than or equal to 1.49m

Explanation:

Let the distance separating two objects, x = 6.0 cm = 0.06 m

The distance between the observer and the two objects, d = 160 km = 160000 m

Let ∅ = minimum angular separation between the two objects that the satellite can resolve

tan( ∅) = x/d

Since there is minimum angular separation, tan( ∅) ≈∅

∅ = x/d

∅ = 0.06/160000

∅ = 3.75 * 10⁻⁷rad

For the satellite to be able to resolve the objects,

D ≥ 1.22λ/∅

λ = 560 nm = 560 * 10⁻⁹

D ≥ 1.22 * (560 * 10⁻⁹)/(3.75 * 10⁻⁷)

D ≥ 149.33 * 10⁻² m

D ≥ 1.49 m

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3 years ago

Why do yall like Biden

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“We LoVeEe BiDeN bEcAuSe hIs HeAlThCaRe iSn’T FrEeEeE!”

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3 years ago

Read 2 more answers

Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷

victus00 [196]

Answer:

0 N/C

Explanation:

Parameters given:

$q_1 = -8.00 * 10^{-7} C$

$x_1 = 6.00 cm$

$q_2 = +8.00 * 10^{-7} C$

$x_2 = 21 cm$

The distance between $q_1$ and $q_2$ is

$21 - 6 = 15cm$

Electric field is given as

$E = \frac{kq}{r^2}$

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to $q_1$ is:

$E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}$

$E_1$ = $-1.28 * 10^6 N/C$

The electric field at the midpoint due to $q_2$ is:

$E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}$

$E_2$ = $1.28 * 10^6 N/C$

The net electric field will be:

E = $E_1$ + $E_2$

E = $-1.28 * 10^6$ + $1.28 * 10^6$

E = 0 N/C

7 0

3 years ago

Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o

Anna35 [415]

Answer:

the correct one is 2. the equipotential lines must be closer together where the field has more intensity

Explanation:

The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.

In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by

V = k q / r

also the electric field and the electuary potential are related

E = $- \frac{dV}{dr}$

therefore the equipotential lines must be closer together where the field has more intensity

When checking the answers, the correct one is 2

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3 years ago