Answer:
1,373.4 N
Explanation:
The mass of the table acts at the centre in addition to the books since that is the centre of gravity of the table.
Mass of books will be 10kg+20kg+30kg=60 kg
Total mass of table and books will be 500kg+60kg=560 kg
This mass is evenly distributed into the four legs hence 560kg/4 legs=140 kg per leg
Force is product of mass and acceleration due to gravity hence F=gm
Taking g as 9.81 m/s2 then
F=140*9.81=1,373.4 N
Therefore, rhe normal force is equivalent to 1,373.4 N
Answer:
0.67 s
Explanation:
This is a simple harmonic motion (SHM).
The displacement,
, of an SHM is given by
![x = A\cos(\omega t)](https://tex.z-dn.net/?f=x%20%3D%20A%5Ccos%28%5Comega%20t%29)
A is the amplitude and
is the angular frequency.
We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or
radian.
From trigonometry,
if A and B are complementary.
At
, ![x = 3.5](https://tex.z-dn.net/?f=x%20%3D%203.5)
![3.5 = A\cos(\omega \times0)](https://tex.z-dn.net/?f=3.5%20%3D%20A%5Ccos%28%5Comega%20%5Ctimes0%29)
![A =3.5](https://tex.z-dn.net/?f=A%20%3D3.5)
So
![x = 3.5\cos(\omega t)](https://tex.z-dn.net/?f=x%20%3D%203.5%5Ccos%28%5Comega%20t%29)
At
, ![x = 1.5](https://tex.z-dn.net/?f=x%20%3D%201.5)
![1.5 = 3.5\cos(0.12\omega)](https://tex.z-dn.net/?f=1.5%20%3D%203.5%5Ccos%280.12%5Comega%29)
![\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286](https://tex.z-dn.net/?f=%5Ccos%280.12%5Comega%29%3D%5Cdfrac%7B1.5%7D%7B3.5%7D%3D0.4286)
![0.12\omega =\cos^{-1}0.4286](https://tex.z-dn.net/?f=0.12%5Comega%20%3D%5Ccos%5E%7B-1%7D0.4286)
![0.12\omega = 1.13](https://tex.z-dn.net/?f=0.12%5Comega%20%3D%201.13)
![\omega = 9.4](https://tex.z-dn.net/?f=%5Comega%20%3D%209.4)
The period,
, is related to
by
![T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B2%5Cpi%7D%7B%5Comega%7D%20%3D%20%5Cdfrac%7B2%5Ctimes3.14%7D%7B9.4%7D%3D0.67)
Answer:
As the roller coaster goes higher, more potential energy is increased in the roller coaster. Caused by gravity and the roller coaster's position, the potential energy is stored in the roller coaster. For example, this ball is at the top of a hill, where potential energy is at it's highest. Potential energy can be calculated by Potential Energy=Mass X Acceleration X Height.
Explanation:
Answer:
The object would weight 63 N on the Earth surface
Explanation:
We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:
![F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}](https://tex.z-dn.net/?f=F_G%3DG%5C%2C%5Cfrac%7BM_E%5C%2Cm%7D%7Bd%5E2%7D%20%5C%5C28%5C%2C%5C%2CN%3DG%5C%2C%5Cfrac%7BM_E%5C%2Cm%7D%7B9600000%5E2%7D)
Now, if the body is on the surface of the Earth, its weight (w) would be:
![F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}](https://tex.z-dn.net/?f=F_G%3DG%5C%2C%5Cfrac%7BM_E%5C%2Cm%7D%7Bd%5E2%7D%20%5C%5Cw%3DG%5C%2C%5Cfrac%7BM_E%5C%2Cm%7D%7B6400000%5E2%7D)
Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:
![\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\](https://tex.z-dn.net/?f=%5Cfrac%7Bw%7D%7B28%7D%20%3D%5Cfrac%7B9600000%5E2%7D%7B6400000%5E2%7D%20%5C%5C%5Cfrac%7Bw%7D%7B28%7D%20%3D%5Cfrac%7B9%7D%7B4%7D%20%5C%5C%5C%5C%20%5C%5Cw%3D%5Cfrac%7B9%5C%2C%2A%5C%2C28%7D%7B4%7D%5C%2C%5C%2C%5C%2CN%5C%5Cw%3D63%5C%2C%5C%2CN%20%5C%5C)
#1 is 12 u just have to count that one was counting by 5's figure out the pattern on what its counting by and start at the bottom and count your way up till you get to the shaded line