Answer:
TRUE
Explanation:
Balance forces are usually defined as the two distinct force that acts on an object but in opposite directions. These two acting forces are equal in size or magnitude. When this type of force is applied on any object, it signifies that the object is stationary or it is moving at a constant speed and in the same direction.
This force is comprised of two most important properties namely the strength and direction. When any of the two forces is higher then it result in the motion of the object.
Thus, the above given statement is TRUE.
the upward force is from the table and the downward is from gravity they are equal in force so the box doesn't fly up or sink down
The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.
We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force
![F_f](https://tex.z-dn.net/?f=F_f)
acting against the motion. Since their resultant must be zero, we have:
![F-F_f = 0](https://tex.z-dn.net/?f=F-F_f%20%3D%200)
The frictional force is
![F_f = \mu mg](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20mg)
where
![\mu=0.4](https://tex.z-dn.net/?f=%5Cmu%3D0.4)
is the coefficient of kinetic friction
![mg=400 N](https://tex.z-dn.net/?f=mg%3D400%20N)
is the weight of the block.
Substituting these values, we find the magnitude of the force F:
Answer: final Velocity v = 10.2m/s
Explanation:
Final speed v(t) is given as
v(t) = u + at .......1
Where; u = the initial speed
a = acceleration
t = time taken
The total distance travelled d is given as
d = ut + 1/2(at^2)
Given
d = 5.0m
u = 2.0m
a = g = 10m/s2 (acceleration due to gravity)
Substituting into the equation above we have
5 = 2t + 5t^2
5t^2 +2t -5 = 0
Applying the quadratic formula. We have;
t = 0.82s & t = -1.22s
t cannot be negative
t = 0.82s
From equation 1 above
v = 2.0m/s + 10(0.82)m/s
v = 10.2m/s
Energy of gamma rays is given by equation
![E = h\nu](https://tex.z-dn.net/?f=E%20%3D%20h%5Cnu)
here we know that
h = Planck's constant
![\nu = frequency](https://tex.z-dn.net/?f=%5Cnu%20%3D%20frequency)
now energy is given as
![E = 4.70 MeV = 4.70 \times 10^6 \times 1.6 \times 10^{-19}](https://tex.z-dn.net/?f=E%20%3D%204.70%20MeV%20%3D%204.70%20%5Ctimes%2010%5E6%20%5Ctimes%201.6%20%5Ctimes%2010%5E%7B-19%7D)
![E = 7.52 \times 10^{-13} J](https://tex.z-dn.net/?f=E%20%3D%207.52%20%5Ctimes%2010%5E%7B-13%7D%20J)
now by above equation
![E = h\nu](https://tex.z-dn.net/?f=E%20%3D%20h%5Cnu)
![7.52 \times 10^{-13} = 6.6 \times 10^{-34} \nu](https://tex.z-dn.net/?f=7.52%20%5Ctimes%2010%5E%7B-13%7D%20%3D%206.6%20%5Ctimes%2010%5E%7B-34%7D%20%5Cnu)
![\nu = 1.14 \times 10^{21} Hz](https://tex.z-dn.net/?f=%5Cnu%20%3D%201.14%20%5Ctimes%2010%5E%7B21%7D%20Hz)
now for wavelength we can say
![\lambda = \frac{c}{\nu}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bc%7D%7B%5Cnu%7D)
![\lambda = \frac{3\times 10^8}{1.14 \times 10^{21}}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B3%5Ctimes%2010%5E8%7D%7B1.14%20%5Ctimes%2010%5E%7B21%7D%7D)
![\lambda = 2.63 \times 10^{-13} m](https://tex.z-dn.net/?f=%5Clambda%20%3D%202.63%20%5Ctimes%2010%5E%7B-13%7D%20m)