Points A, B, and C lie along a line from left to right, respectively. Point B is at a lower electric potential than point A. Poi
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3. The diagrams showing the forces acting on the golf ball are in the figure attached. Let’s have a detailed look:
a) Here the ball is under 2 forces:
F1 which is called The Normal force and is perpendicular to the surface of the tee where the ball rests
-F1, related to the gravity force, to the weight of the ball, and has the same magnitude, but the opposite direction (That’s why it has a negative sign).
In this case the sum of the forces is 0
b) Here we have again forces F1 and –F1, but in this very moment the club strikes the ball and we have:
F2, the force of the strike
c) While the ball is in its flight, it is under the following forces:
F1, the force of the lift through the air
-F2, the gravity force
-F3, the force of air resistance, also called drag
F4, the tangencial force of the ball flight
4. Here are the sizes and directions of the resultant forces:
i)Two forces of the same magnitude or size are applied to this block, but in opposite directions (in the x-axis). This is expressed as:
F=-10N+10N=0 The resulting force is zero
ii) Two forces of different size and opposite directions in the y-axis are applied to this block. The sum of the forces is:
F=30N-40N=-10N This means the resulting force is 10N applied downwards
iii) In this case the only force applied to the block is -5N applied downwards
iv) Here there are four forces applied to the block.
In the y-axis we have to forces of the same size but opposite directions:
F1=10N-10N=0 This means the applied force in the y-axis is zero
In the x-axis we have two forces of different size and opposite directions:
F2=-15N+10N=-5N This means the resulting force is applied to the –x-side
The emerging velocity of the bullet is <u>71 m/s.</u>
The bullet of mass <em>m</em> moving with a velocity <em>u</em> has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.
If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,
This is equal to the change in the bullet's kinetic energy.
If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>
Assuming the same resistive forces to act on the bullet,
Divide equation (2) by equation (1) and simplify for v<em>₁.</em>
Thus the speed of the bullet is 71 m/s