Dividing x^3 - 3x^2 - 4x + 12 by x + 2 gives
x^2 -5x + 6
= (x - 3)(x - 2)
so remaining factors are x - 3 and x - 2 answer
Answer:
it's density must be less than water
law of floatation
wt of the immerged body = wt of the water displaced by it
Answer:
B
Explanation:
Given:-
- The charge of the test particle q = 3.0 * 10^-9 C
- The force exerted by the metal sphere F = 6.0 * 10^-5 N
Find:-
The magnitude and direction of the electric field
strength at this location?
Solution:-
- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:
F = E*q
- Using the data given we can determine E:
E = F / q
E = (6.0 * 10^-5) / (3.0 * 10^-9)
E = 20,000 N/C
- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.
We can calculate this with the law of conservation of energy. Here we have a food package with a mass m=40 kg, that is in the height h=500 m and all of it's energy is potential. When it is dropped, it's potential energy gets converted into kinetic energy. So we can say that its kinetic and potential energy are equal, because we are neglecting air resistance:
Ek=Ep, where Ek=(1/2)*m*v² and Ep=m*g*h, where m is the mass of the body, g=9.81 m/s² and h is the height of the body.
(1/2)*m*v²=m*g*h, masses cancel out and we get:
(1/2)*v²=g*h, and we multiply by 2 both sides of the equation
v²=2*g*h, and we take the square root to get v:
v=√(2*g*h)
v=99.04 m/s
So the package is moving with the speed of v= 99.04 m/s when it hits the ground.
