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
Answer:
Explanation:
velocity of projection, vo = 381 m/s
angle of projection, θ = 73.5°
The formula for the range is
R = 8067.4 m
Range in shorten by 34.1 %
So, the new range is
R' = 8067.4 - 34.1 x 8067.4/100
R' = 5316.4 m
Answer:
Explanation:
A )
speed of swimming in still water is given by the expression
distance / time
= 50 / 25
= 2 m /s
In lane 1 , 1.2 cm/s current is flowing in the direction that the swimmers are going so swimmers will cover distance at the rate of 2 + 1.2 = 3.2 m /s.
time to cover distance of 50 m in lane 1
= distance / speed
= 50 / 3.2 = 15.625 s
In lane 8 , 1.2 cm/s current is flowing against the direction that the swimmers are going so swimmers will cover distance at the rate of 2 - 1.2 = .8 m /s.
time to cover distance of 50 m in lane 1
= distance / speed
= 50 / .8 = 62.5 s
Answer:
I Will say the Answer is A
Explanation:
Question 2 is because the passengers have inertia, which is a tendency to resist change in motion
