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.
Answer:
The polarity of water molecules means that molecules of water will stick to each other like when unlike charges attracts. This is called hydrogen bonding.
Polarity makes water a good solvent, gives it the ability to stick to itself (cohesion), stick to other substances (adhesion), and have surface tension (due to hydrogen bonding).
When the two hydrogen atoms bond with the oxygen, they attach to the top of the molecule. This molecular structure gives the water molecule polarity, or a lopsided electrical charge that attracts other atoms. The end of the molecule with the two hydrogen atoms is positively charged.
Explanation:
For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg
Answer:
Explanation:
<u>Instant Acceleration</u>
The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.
Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:
And the acceleration is
Or equivalently
The given height of a projectile is
Let's compute the speed
And the acceleration
It's a constant value regardless of the time t, thus
