Answer:Vb=-6i-(-0.1ωab+8)j m/s
Explanation:
Va=V0+Va0
Va=V0+(ra0 x ωao)
ω=Angular velocity of link A0
Using r0a=0.1m;
Va=V0+(0.1i x ω0a K)
Va=0
ixk=j
Va=0+0.1ω0aj
Calculating te velocity of using te equation below
Vb=Va+Vba
Vb=Va+ωab x rba
ωab=40rad/s
rab=-0.21i+0.15j
Va=0.1ω0aj
Vb=Va+ωabxrba
Vb=0.1ω0aj+40k x -(0.21i+0.15j)
Vb=0.1ω0aj-8j-6i
Vb=-6i-(-0.1ωab+8)j m/s
Answer:
i)ω=3600 rad/s
ii)V=7059.44 m/s
iii)F=1245.8 N
Explanation:
i)
We know that angular speed given as
We know that for one revolution
θ=2π
Given that time t= 2 hr
So
ω=θ/t
ω=2π/2 = π rad/hr
ω=3600 rad/s
ii)
Average speed V
Where M is the mass of earth.
R is the distance
G is the constant.
Now by putting the values
V=7059.44 m/s
iii)
We know that centripetal fore given as
Here given that m= 200 kg
R= 8000 km
so now by putting the values
F=1245.8 N
Answer:
26 lbf
Explanation:
The mass of the satellite is the same regardless of where it is.
The weight however, depends on the acceleration of gravity.
The universal gravitation equation:
g = G * M / d^2
Where
G: universal gravitation constant (6.67*10^-11 m^3/(kg*s))
M: mass of the body causing the gravitational field (mass of Earth = 6*10^24 kg)
d: distance to that body
15000 miles = 24140 km
The distance is to the center of Earth.
Earth radius = 6371 km
Then:
d = 24140 + 6371 = 30511 km
g = 6.67*10^-11 * 6*10^24 / 30511000^2 = 0.43 m/s^2
Then we calculate the weight:
w = m * a
w = 270 * 0.43 = 116 N
116 N is 26 lbf
umm , is it okay if we do this on microsoft word , cuz i cant send pics of answers here...
I’m thinking it would be c sorry if it’s wrong .
