Note that
1 km/h = (1000 m)/(3600 s) = 0.27778 m/s
Initial velocity, v₁ = 25 km/h = 6.9444 m/s
Final velocity, v₂ = 65 km/h = 18.0556 m/s
Time interval, dt = 6 s.
Calculate average acceleration.
a = (v₂ - v₁)/dt
= (18.0556 - 6.9444 m/s)/(6 s)
= 1.852 m/s²
Answer:
The average acceleration is 1.85 m/s² (nearest hundredth)
The area of the top and bottom:
2πr²
Cost for top and bottom:
2πr² x 0.02
= 0.04πr²
Area for side:
2πrh
Cost for side:
2πrh x 0.01
= 0.02πrh
Total cost:
C = 0.04πr² + 0.02πrh
We know that the volume of the can is:
V = πr²h
h = 500/πr²
Substituting this into the cost equation to get a cost function of radius:
C(r) = 0.04πr² + 0.02πr(500/πr²)
C(r) = 0.04πr² + 10/r
Now, we differentiate with respect to r and equate to 0 to obtain the minimum value:
0 = 0.08πr - 10/r²
10/r² = 0.08πr
r³ = 125/π
r = 3.41 cm
Answer:
The velocity of the man is 0.144 m/s
Explanation:
This is a case of conservation of momentum.
The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.
Mass of ball = 0.65 kg
Mass of the man = 54 kg
Velocity of the ball = 12.1 m/s
Before collision, momentum of the ball = mass x velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After collision the momentum of the man and ball system is
(0.65 + 54)Vf = 54.65Vf
Where Vf is their final common velocity.
Equating the initial and final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
Answer:
The Surface heat flux is -9205 W/m^2
Explanation:
Explanation is in the following attachment
Options:
A. The luminosity of Star 1 is a factor of 100 less than the luminosity of Star 2.
B. Star 1 is 100 times more distant than Star 2.
C. Without first knowing the distances to these stars, you cannot draw any conclusions about how their true luminosities compare to each other.
D. Star 1 is 10 times more distant than Star 2.
E. Star 1 is 100 times nearer than Star 2.
Answer:
D. Star 1 is 10 times more distant than star 2
Explanation:
For two stars of identical size and temperature, the closer one to us will appear brighter. The relationship between the distance and luminosity of stars is an inverse- square relationship.
Luminosity, L = 1/r²
Where r is the distance of the star to the earth
Since star 1 is dimmer in brightness than star 2 by a factor of 100,
L₁/L₂ = 1/100
i.e. L₁ = 1, L₂=100
L₁ = 1/r₁² ............(1)
1 = 1/r₁²
L₂ = 1/r₂²
100 = 1/r₂² .........(2)
divide equation (2) by equation (1)
100/1 = ( 1/r₂² )/ (1/r₁²)
100 = (r₁/r₂)²
r₁/r₂ = √100
r₁/r₂ = 10
r₁ = 10r₂
