When a third object is brought in contact with the first object (after it gains the electrons), the resulting charge on the thir
1 answer:
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Answer:
245.45km in a direction 21.45° west of north from city A
Explanation:
Let's place the origin of a coordinate system at city A.
The final position of the airplane is given by:
rf = ra + rb + rc where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:
rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km
rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km
The module of this position is:
And the angle measure from the y-axis is:
So the answer is 245.45km in a direction 21.45° west of north from city A
Answer:
0.00899 N
Explanation:
The magnitude of the electrostatic force between two charges is given by the equation:
where:
is the Coulomb's constant
are the charges
r is the distance between the two charges
And the force is:
- Repulsive if the two charges have same sign
- Attractive if the two charges have opposite sign
In this problem we have:
(charge of object 1)
(charge of object 2)
r = 1 m (separation between the objects)
So, the electric force is
Answer:
4x
Explanation:
Use to do the question.
For first instance,
-------------------( 1 )
for second instance,
-----------------( 2 )
So by (1) and (2),
s = 4x
Answer:
the correct answer is option C which is 50 units.
Explanation:
given,
two vector of magnitude = 30 units and of 70 units
to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}
cos θ value varies from -1 to 1
so, resultant vector
=
a = 30 units and b = 70 units
=
= 40 units to 100 units
hence, the correct answer is option C which is 50 units.