(a) The tension on the wire when the two charges have opposite signs is 383.5 N.
(b) The tension on the wire if both charges were negative is 3.640.25 N.
The given parameters;
- <em>first charge, q₁ = 8.75 μC </em>
- <em>second charge, q₂ = -6.5 μC </em>
- <em>electric field, E = 1.85 x 10⁸ N/C</em>
- <em>distance between the two charges, r = 2.5 cm</em>
<em />
(a)
The attractive force between the charges is calculated as follows;
The force on the negative charge due to the electric field is calculated as follows;
The tension on the wire is the resultant of the two forces and it is calculated as follows;
(b) when the two charges are negative
The repulsive force between the two charges is calculated as follows;
The force on the first negative charge due to the electric field is calculated as follows;
The force on the second negative charge due to the electric field is calculated as follows;
The tension on the wire is the resultant of the three forces and it is calculated as follows;
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<u>The following statements are false about collisions:
</u>
- The velocity change of two respective objects involved in a collision will always be equal.
- Total momentum is always conserved between any two objects involved in a collision.
Answer: Option B, and D
<u>Explanation:
</u>
In any collisions, equal amount of net force will be acted upon the colliding objects due to the third law of Newton, irrespective of the significance difference in mass of the objects. Similarly, they can also have different acceleration values during collision of two objects if the masses are identical.
But the statements regarding the equal change in velocity of two objects respectively involved in collision always is false, as the conservation of momentum is applicable for isolated system only. So it is true for only isolated system and not in all the systems.
The same reason goes for falsifying the fourth statement which states that total momentum is always conserved between two objects involved in a collision as this statement is only true for isolated system where the conservation of momentum can be applied. Thus the second and fourth statement is false regarding collision.
The answer is 1.5....
The refractive index n = (Speed of light, c) /( Speed of light, v)
So c= n x v
comparing to c= 1.5 x v in this situation we see that n= 1.5
Answer:
1). average velocity= displacement/time
= here displacement is zero
= 0/1
= 0 m/s
2). average speed= total distance/time
=2πr/1
=(2×22/7×5/10)/1
22/7
3.14 km/h
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Explanation:
Answer:
The distance at which nose of the vehicle will the shock wave impinge upon the ground is approximately 23km
Explanation:
Let
u be Mach angle
M be Mach Number = 2.5
h be altitude = 10km
d be distance between nose of the vehicle with which the shock wave impinge upon the ground
Using;
Sin u = 1/M
u = arcSin (1/M)
u = arcSin (1/2.5) = 23.58°
Using angle at a point rule = 90°
90 - 23.58 = 66.42°
Tan A = opp/adj
Tan 66.42° = d/10
2.2911 * 10 = d
d = 22.911km ≈ 23km
