Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
Pretty sure it’s C. all the others are speeding up. acceleration means gradually (over time) getting faster. So it’s C.
Answer:
Fc=5253
N
Explanation:
Answer:
Fc=5253
N
Explanation:
sequel to the question given, this question would have taken precedence:
"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."
so we derive centripetal acceleration first
ac (centripetal acceleration) = v^2/r
make r the subject of the equation
r= v^2/ac
ac is 6.23*g which is 9.81
v is 101m/s
substituing the parameters into the equation, to get the radius
(101^2)/(6.23*9.81) = 167m
Now for part
( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.
he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.
Fc (Centripetal Force) = m*v^2/r
So (86kg* 101^2)/(167) =
Fc=5253
N
Answer:
4 hoop, disk, sphere
Explanation:
Because
We are given data that
Hoop, disk, sphere have Same mass and radius
So let
And Initial angular velocity, = 0
The Force on each be F
And Time = t
Also let
Radius of each = r
So let's find the inertia shall we!!
I1 = m r² /2
= 0.5 mr² the his is for dis
I2 = m r² for hoop
And
Moment of inertia of sphere wiil be
I3 = (2/5) mr²
= 0.4 mr²
So
ωf = ωi + α t
= 0 + ( τ / I ) t
= ( F r / I ) t
So we can see that
ωf is inversely proportional to moment of inertia.
And so we take the
Order of I ( least to greatest ) :
I3 (sphere) , I1 (disk) , I2 (hoop) , ,
Order of ωf: ( least to greatest)
That of omega xf is the reverse of inertial so
hoop, disk, sphere
Option - 4
Answer:17.08 s
Explanation:
Given
distance between First and second Runner is 45.6 m
speed of first runner
=3.1 m/s
speed of second runner
=4.65 m/s
Distance between first runner and finish line is 250 m
Second runner need to run a distance of 250+45.6=295.6 m
Time required by second runner 
time required by first runner to reach finish line
Thus second runner reach the finish line 80.64-63.56=17.08 s earlier
