One is their traits and their characterists that they have in common
No, we can't say force is useful or harmful. All else being equal, force is either not harmful or useful depending on the body that is applying it and where it is applying it.
Given how it affects motion, force is a crucial idea. It is a relationship that, in the absence of an opposing force, modifies an object's motion. However, a push or a pull that any object feels is the simplest definition of force. Due to the fact that force is a vector quantity, it possesses both a magnitude and a direction.
How force is useful
A body at rest can move with enough force.A body in motion may be slowed down or stopped by it.It has the power to quicken the pace of an object in motion.Along with its shape and size, it can also alter the direction of a moving body.
How force is harmful
- Force has the power to alter an object's state of motion.
- Moving objects can shift direction due to force.
- Moving things' speeds can be increased by force.
- Moving items can become slower due to force.
- Force has the power to alter an object's shape.
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Answer:
a)KE=878.8 J
b)W=2636.4 J
Explanation:
Given that
mass ,m = 65 kg
Initial speed ,u = 5.2 m/s
a)
We know that kinetic energy KE is given as follows
m=mass
u=velocity
Now by putting the values in the above equation we get
KE=878.8 J
b)
We know that
Work done by all forces = Change in the kinetic energy
The final velocity , v= 2 u = 2 x 5.2 m/s
v= 10.4 m/s
Now by putting the values in the above equation we get
W=2636.4 J
a)KE=878.8 J
b)W=2636.4 J
Answer:
1.549 m
Explanation:
Given:
The radius of the circular board, r = 2 m
The probability of hitting the red is given as 0.6
Now, this probability of hitting the red can be conclude as
0.6 = (Area of red)/ (Total area of the board)
Total area of the board = πr² = π × 2²
let the radius of the red area be R
thus, area of red circle, = πR²
on substituting the value of the area, we have
0.6 = (πR²)/ (π × 2²)
or
R² = 2.4
or
R = 1.549 m
Thus, the radius of the red circle is 1.549 m
Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;
Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
