Weight is mass x gravity, so you'd multiply the mass of the astronaut by the gravitational pull.
Answer:
R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi
If you do it in steps
R = 9880 yd * 3 ft/yd = 29640 ft
R = 29640 ft / 5280 ft/mi = 5.61 mi
Answer:
The function is x = e^(-t/2) * (0.792*sin12t + 5cos12t)
Explanation:
we have to:
m = mass = 4 kg
k = spring constant = 577 N/m
c = damping constant = 4 N*s/m
The differential equation of motion is equal to:
m(d^2x/dt^2) + c(dx/dt) + k*x = 0
Replacing values:
4(d^2x/dt^2) + 4(dx/dt) + 577*x = 0
Thus, we have:
4*x^2 + 4*x + 577 = 0
we will use the quadratic equation to solve the expression:
x = (-4 ± (4^2 - (4*4*577))^1/2)/(2*4) = (-4 ± (-9216))/8 = (1/2) ± 12i
The solution is equal to:
x = e^(1/2) * (c1*sin12t + c2*cos12t)
x´ = (-1/2)*e^(1/2) * (c1*sin12t + c2*cos12t) + e^(-t/2) * (12*c1*cos12t - 12*c2*sin12t)
We have the follow:
x(0) = 5
e^0(0*c1 + c2) = 5
c2 = 5
x´(0) = 7
(-1/2)*e^0 * (0*c1 + c2) + e^0 * (12*c1 - 0*c2) = 7
(-1/2)*(5) + 12*c1 = 7
Clearing c1:
c1 = 0.792
The function is equal to:
x = e^(-t/2) * (0.792*sin12t + 5cos12t)
Answer:
Explanation:
Given
car A had a head start of 
and it starts at x=0 and t=0
Car B has to travel a distance of 
where 
is the distance travel by car A in time t
distance travel by car A is
For car B with speed 
