Answer:
![[F]=[MLT^{-2}]](https://tex.z-dn.net/?f=%5BF%5D%3D%5BMLT%5E%7B-2%7D%5D)
Explanation:
Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :
F = m × a
F is the applied force
m is the mass of the object
a is the acceleration due to gravity
We need to find the dimensions of force. The dimension of force m and a are as follows :
![[m]=[M]](https://tex.z-dn.net/?f=%5Bm%5D%3D%5BM%5D)
![[a]=[LT^{-2}]](https://tex.z-dn.net/?f=%5Ba%5D%3D%5BLT%5E%7B-2%7D%5D)
So, the dimension of force F is,
. Hence, this is the required solution.
Explanation:
W = PE
W = mgh
1500 J = (20 kg) (9.8 m/s²) h
h = 7.65 m
Round as needed.
Answer:
The inertial force of the body
Explanation:
Everybody that is moving in a curved path has an inertial force called centrifugal force.
The counterforce of the centrifugal force is called the centripetal force. It also acts on every rotating body.
This force is always directed towards the center of the origin of the curve.
The velocity of the object changes its direction and magnitude at any instant of time. But the speed and angular velocity of the object remains the same for uniform circular motion.
So, according to the Newtonian mechanics, it is the inertial force of the body responsible for the centripetal force.
Answer:
a)
b)
Explanation:
Given:
mass of bullet, 
compression of the spring, 
force required for the given compression, 
(a)
We know

where:
a= acceleration


we have:
initial velocity,
Using the eq. of motion:

where:
v= final velocity after the separation of spring with the bullet.


(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,

∵At maximum height the final velocity will be zero

Using the equation of motion:

where:
h= height
g= acceleration due to gravity


is the height from the release position of the spring.
So, the height from the latched position be:


