Answer:
![\theta=34 \textdegree](https://tex.z-dn.net/?f=%5Ctheta%3D34%20%5Ctextdegree)
Explanation:
From the question we are told that:
Mass ![m=55kg](https://tex.z-dn.net/?f=m%3D55kg)
Angle ![\theta =28.0](https://tex.z-dn.net/?f=%5Ctheta%20%3D28.0)
Coefficient of static friction
Generally, the equation for Newtons second Law is mathematically given by
For
![\sum_y=0](https://tex.z-dn.net/?f=%5Csum_y%3D0)
![N=mgcos \theta](https://tex.z-dn.net/?f=N%3Dmgcos%20%5Ctheta)
for
![\sum_x=0](https://tex.z-dn.net/?f=%5Csum_x%3D0)
![F_{s}=mgsin\theta](https://tex.z-dn.net/?f=F_%7Bs%7D%3Dmgsin%5Ctheta)
Where
![F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta](https://tex.z-dn.net/?f=F_%7Bs%7D%3D%5Calpha%2AN%5C%5C%5C%5CF_%7Bs%7D%3D%5Calpha%2Am%2Agcos%20%5Ctheta)
![F_{s}=0.68*55*9.8*cos 28](https://tex.z-dn.net/?f=F_%7Bs%7D%3D0.68%2A55%2A9.8%2Acos%2028)
![F_{s}=323.62N](https://tex.z-dn.net/?f=F_%7Bs%7D%3D323.62N)
Therefore
![\alpha mgcos \theta=mg sin \theta](https://tex.z-dn.net/?f=%5Calpha%20mgcos%20%5Ctheta%3Dmg%20sin%20%5Ctheta)
![\theta=tan^{-1}(0.68)](https://tex.z-dn.net/?f=%5Ctheta%3Dtan%5E%7B-1%7D%280.68%29)
![\theta=34 \textdegree](https://tex.z-dn.net/?f=%5Ctheta%3D34%20%5Ctextdegree)
Answer:
a) m₁ = 1.41 kg
, b) m₂ = 2.65 kg
Explanation:
For this exercise we will use Newton's second law
Block 1
T - W₁ = m₁ a
Block 2
W₂ - T = m₂ a
We have selected the positive block 1 rising and block two lowering, as the pulley has no friction does not affect the movement
Let's use kinematics to look for acceleration
y = v₀ t + ½ a t²
As part of the rest the initial speed is zero
a = 2 y / t²
a = 2 6.00 / 2²
a = 3 m / s²
Let's replace in the equation of block 1
a) T = m₁ g + m₁ a
m₁ = T / (g + a)
m₁ = 18.0 / (9.8 + 3)
m₁ = 1.41 kg
b) we substitute in the equation of block 2
W₂ - T = m₂ a
m₂ g - m₂ a = T
m₂ = T / (g-a)
m₂ = 18.0 / (9.8 -3)
m₂ = 2.65 kg
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Answer:
12
Explanation:
The equation is w= f *d
36=3*d
12=d 12 units is the mass
Answer:
w = 1.766 Joule
Explanation:
Work = Force × Distance
In this case the Work done is stored in body as Gravitational potential energy, so , W = P. E. = mgh
Here m = 0.400 kg ( unit not given in the question if in other unit may be converted to Kg), g = 9.81 m/s² and h= 0.450 m
w = 0.400 kg × 9.81 m/s² × 0.450 m
w = 1.7658 kg m²/s² = 1.766 Joule or (Nm)