Answer:
Explanation:
The net force on the potatoes is given by:
F= 52 - mgSintheta
F= 52- (2×9.8× Sin70°)
F = 52 -18.4
F= 33.58N
Using Newton's 2nd law
F = ma
a=F/m = 33.58/ 2 = 16.79m/s^2
Using the equation of motion:
V^2= u^2 + 2as
V^2 = 0 + 2× 16.79 x2
V^2 = 67.16
V=sqrt(68.16)
V= 8.195m/s This is the exit velocity of the potatoes
Kinetic energy, K.E = 1/2mv^2
KE= 1/2 × 2 × 8.195^2
KE = 67.16J
Answer:
-100N
Explanation:
Newton's third law of motion states that to every force exerted on one body, there is an equal and opposite force. This means that if object A exerts an ACTION force on B, there is a force called REACTION FORCE, which is equal and opposite, exerted on A by B.
The action and reaction forces are equal in size/magnitude but opposite in direction. In this case where a tennis racket strikes a tennis ball with a force (action force) of 100N, the ball will strike the racket with a reaction force of -100N.
F(RB) = -F(BR)
Answer:
3. if you increase your mass you also increase the gravitational pull
6. Radiant energy doesn't depend on a medium and sound energy is dependent on a medium.
Explanation:
i hope this helps-
The answer is A.) igneous rocks
Answer:
a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2
b) See the picture
Explanation:
We can use Gauss's law to find the electric field in all the regions:
EA = qen/e0 where qen is the enclosed charge
Remember that the electric field everywhere outside a sphere is:
E(r) = q/(4*pi*eo*r^2) = Kq/r^2
a)
- For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
- For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
- For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2
b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance