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Ray Of Light [21]
2 years ago
7

A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm^3 of liquid. The mater

ial for the top and bottom costs 0.02 cent/cm^2, and the material for the sides costs 0.01 cent/cm^2.
a) Estimate the radius r and height h of the can that costs the least to manufacture. [suggestion: Express the cost C in terms of r.]
Physics
1 answer:
Rom4ik [11]2 years ago
4 0
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
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A sample of nitrogen gas exerts a pressure of 9.80 atm at 32 C. What would its temperature be (in C) when its pressure is increa
Harlamova29_29 [7]

Answer:

T₂ = 111.57 °C

Explanation:

Given that

Initial pressure P₁ = 9.8 atm

T₁ = 32°C  = 273 + 32 =305  K

The final pressure   P₂ = 11.2 atm

Lets take the final temperature = T₂

We know that ,the ideal gas equation  

If the volume  of the gas is constant ,then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

T_2=\dfrac{P_2}{P_1}\times T_1

Now by putting the values in the above equation ,we get

T_2=\dfrac{11.2}{9.8}\times 305\ K

T_2=348.57\ K

T₂ = 384.57 - 273 °C

T₂ = 111.57 °C

3 0
2 years ago
Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerate
____ [38]

Answer:

a) Velocity = 4.2m/s

b) Acceleration = 2.94m/s^2

c) Force exerted on the floor= 1401.4×10^3N

Explanation:

a) Velocity,V=sqrt(2×9.8×0.900)

V= 4.2m/s

b) Vf2= V^2+2ay2

a= 4.2^2 - 0/2×3

a= 17.64/6= 2.94m/s^2

c) Newton's 2nd law indicates:

Fnet= F - mg=ma

F= m(g+a)

F=110(9.8+2.94)

F=110×12.94

F= 1401.4N

4 0
2 years ago
You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
2 years ago
The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
1 year ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
2 years ago
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