I assume L=120 yards as the length of the football field.
1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is

And the time taken is t=22.4 s, so the average speed of the player is

2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is

and so, the average velocity is
Answer:
102597.6 Pa
Explanation:
mass, m = 1.25 g
Force, F = m x g = 1.25 x 9.8 x 10^-3 = 0.01225 N
radius, r = 0.195 mm = 0.195 x 10^-3 m
Area, A = πr² = 3.14 x 0.195 x 0.195 x 10^-6 m^2
A = 1.19 x 10^-7 m^2
Pressure is defined as the thrust acting per unit area.
P = Force / Area
P = 0.01225 N / (1.10 x 10^-7)
P = 102597.6 Pa
Thus, the pressure exerted is 102597.6 Pa.
Answer:
3.26 secs
Explanation:
Diameter of sphere ( D )= 10 mm
T1 = 75°C
P = 1 atm
T∞ = 23°C
T2 = 35°c
Velocity = 10 m/s
<u>Determine how long it will take to cool the sphere to 35°C</u>
<em>Using the properties of copper and air given in the question</em>
Nu = 2 + (Re)^0.8 (Pr)^0.33
hd / k = 2 + ( vd/v )^0.8 (Pr)^0.33
∴ h ≈ 2594.7 W/m^2k
Given that :
(T2 - T∞) / ( T1 - T∞ ) = exp [ ( -hA / pv CP ) t ]
( 35 - 23 ) / ( 75 - 23 ) = exp [ - 2594.7 * 6 * t / 8933 * 387 * 10 * 10^-3 ]
= ln ( 12/52 ) = -1.466337069 = - 0.45032919 * t
∴ t ≈ 3.26 secs ( -1.466337069 / -0.45032919 )
Answer:
Explanation:
Our lunar companion rotates while it orbits Earth. It's just that the amount of time it takes the moon to complete a revolution on its axis is the same it takes to circle our planet — about 27 days. As a result, the same lunar hemisphere always faces Earth.