Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572
D, 0.140 liters! Hang on a sec and I'll show you a trick I use.
Bulbs c and b would still be screwed in if they were in to begin with and bulbs A, D, and E. would be unscrewed
Answer:
so maximum velocity for walk on the surface of europa is 0.950999 m/s
Explanation:
Given data
legs of length r = 0.68 m
diameter = 3100 km
mass = 4.8×10^22 kg
to find out
maximum velocity for walk on the surface of europa
solution
first we calculate radius that is
radius = d/2 = 3100 /2 = 1550 km
radius = 1550 × 10³ m
so we calculate no maximum velocity that is
max velocity = √(gr) ...............1
here r is length of leg
we know g = GM/r² from universal gravitational law
so G we know 6.67 ×
N-m²/kg²
g = 6.67 ×
( 4.8×10^22 ) / ( 1550 × 10³ )
g = 1.33 m/s²
now
we put all value in equation 1
max velocity = √(1.33 × 0.68)
max velocity = 0.950999 m/s
so maximum velocity for walk on the surface of europa is 0.950999 m/s