Galaxy million of star and planet. gravitional wave field all the universe some planet explosive itself moving other places . Black holes Mass gravity field
Answer:
15.76N
Explanation:
horizontal component =Fcos¢ = 20cos38 = 15.76N
Answer:
our sun is one of the least 100 billion stars in the milky way a spiral galaxy about 100000 light years across
Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N
Answer:
6.23x10^6Pa
Explanation:
Data obtained from the question include:
F (force) = 490N
r (radius) = 0.005m
A (area of the circlular heel) =?
P (pressure) =.?
First, we'll begin by calculating the area of the circlular heel. This is illustrated below:
Area of circle = πr^2
Area = 22/7 x (0.00)^2
Area = 7.86x10^-5m^2
Pressure is simply force per unit area. It represented mathematically as
Pressure = Force /Area
Pressure = 490/7.86x10^-5
Pressure = 6.23x10^6N/m2
Recall: 1N/m2 = 1Pa
Therefore, 6.23x10^6N/m2 = 6.23x10^6Pa
Therefore, the woman exert a pressure of 6.23x10^6Pa on the floor
