1250 N of force is required to accelerate a 250 kg motorcycle at a rate of 5 m/.
<u>Explanation:</u>
As per the second law of motion stated by Newton, the net unbalanced external force acting on any object is directly proportional to the product of mass and acceleration experienced by the object.
As here the object is given as motorcycle of mass 250 kg. Also the acceleration exhibited by the motorcycle is stated as 5 m/. So from the Newton's second law of motion,
Net force = Mass of the motorcycle× Acceleration of the motorcycle
Net force = 250 × 5 = 1250 N.
Thus, 1250 N of force is required to accelerate a 250 kg motorcycle at a rate of 5 m/.
To solve this problem we will use the definition of the kinematic equations of centrifugal motion, using the constants of the gravitational acceleration of the moon and the radius of this star.
Centrifugal acceleration is determined by
Where,
v = Velocity
r = Radius
From the given data of the moon we know that gravity there is equivalent to
While the radius of the moon is given by
If we rearrange the function to find the speed we will have to
The speed for this to happen is 1.7km/s
<u>Answers:</u>
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
