Answer
given,
mass of the package = 12 kg
slides down distance = 2 m
angle of inclination = 53.0°
coefficient of kinetic friction = 0.4
a) work done on the package by friction is
W_f = -μk R d
= -μk (mg cos 53°)(2.0)
=-(0.4)(8.0 )(9.8)(cos 53°)(2.0)
= -37.75 J
b)
work done on the package by gravity is
W_g = m (g sin 53°) d
= (8.0 )(9.8 )(sin 53°)(2.0 )
=125.23 J
c)
the work done on the package by the normal force is
W_n = 0
d)
the net work done on the package is
W = -37.75 + 125.23 + 0
W = 87.84 J
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
247.11 Kelvin, you have to convert the 34 Fahrenheit to get to the answer that you need.
To make sure I don’t has 35 012345 bishops
Answer:
Explanation:
Given
Electric Field 

velocity 
mass of electron 
Force on a charge Particle moving in Magnetic Field
![a=\frac{e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]}{m}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Be%5Cleft%20%5B%20%5Cvec%7BE%7D%2B%5Cleft%20%28%20%5Cvec%7Bv%7D%5Ctimes%20%5Cvec%7BB%7D%5Cright%20%29%5Cright%20%5D%7D%7Bm%7D)
