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gulaghasi [49]
3 years ago
12

The resultant of two vectors acting at a 90°-angle can be determined from the _____ of the rectangle.

Physics
2 answers:
Mama L [17]3 years ago
6 0

Answer:

D.diagonal

Explanation:

To add two or more vectors, is to represent them by only one called resulting. This resulting vector produces the same effects as all together. Keep in mind that the vector sum is not the same as the arithmetic sum.

Parallelogram Method

This method is valid only for two coplanar and concurrent vectors, to find the resulting one joins the vectors by the origin (sliding them) and then form a parallelogram, the resulting vector will be in one of the diagonals, and its application point will coincide with the common origin of the two vectors.

In the case of 2 vectors that form a right angle, bone of 90 °, it is a special case and can be determined directly by the diagonal that crosses the rectangle.

quester [9]3 years ago
5 0

D. Diagonal....................?

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Inertia is the property of mass that resists change. Therefore, it is safe to say that as the mass of an object increases so does its inertia.

Explanation:

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1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

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8 0
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Answer:

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Explanation:

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2 years ago
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