Ask question

Ask question

All categories

3 years ago

8

If 4,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the

box. Step 1 Let b be the length of the base and h the height. Then we must maximize the volume of the box, V

2 answers:

Sloan [31]3 years ago

5 0

Answer:

32000cm^3

Explanation:

Length L be base length

h be the height

Volume=V

V=L^2h

Surface area if box= L^2+4lh=4800

Solving for h

(4800-l^2/4L)=h

1200L-L^3/4=h

Substitute for h in V =L^2h

v=1200L-L^3/4

Differenting =1200-3*L^2/4=0

L=√(1600)=40

Substitute L=40

In V=1200L-L^3/4

Hence 1200*40-40^4/4=32000cm^3

bagirrra123 [75]3 years ago

4 0

Answer:

V = 32000 cm^3

the maximum volume of the box, V is 32000cm^3

Explanation:

Given;

Total surface area of box

S.A = 4800cm^2

Volume of a box with square base;

V = (l^2)h .....1

Surface area of a box with square base and opened top;

S.A = l^2 + 4lh = 4800 .......2

To be able to maximize Volume V, we need to reduce the number of variables in the equation 1

From equation 2, making h the subject of formula;

h = (4800 - l^2)/4l

Substituting h into equation 1;

V = (l^2)(4800-l^2)/4l = (4800l - l^3)/4

differentiating Volume V, we have;

dV/dl = (4800 - 3l^2)/4

At maximum point, dV/dl = 0

(4800 - 3l^2)/4 = 0

3l^2 = 4800

l = √(4800/3)

l = 40 cm

But

h = (4800 - l^2)/4l = (4800 - 40^2)/(4×40)

h = 3200/160

h = 20 cm

At maximum Volume;

V = (l^2)h = (40^2)×20

V = 32000 cm^3

the maximum volume of the box, V is 32000cm^3

You might be interested in

A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th

Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area $A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2$

Let the tensile modulus of Nickel $E = 170 \times 10^9Pa$ .

The elongation of the rod can be calculated using the following formula:

$\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m$

6 0

4 years ago

An object is moving at 2.50 m/s [E]. At a time 3.00 seconds later the object is traveling at 1.50 m/s

babunello [35]

Given parameters:

First velocity = 2.50m/s

Time of travel = 3s

Second velocity = 1.50m/s

Unknown:

The displacement during the first interval = ?

Velocity is the displacement of a body with time. Displacement is a distance move in a specific direction by a body.

Velocity = $\frac{Displacement}{Time taken}$

So;

Displacement = Velocity x Time taken

Now input the parameter for the first velocity and time of travel;

Displacement = 2.5 x 3 = 7.5m

The displacement id 7.5m

7 0

3 years ago

A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch

Nesterboy [21]

Answer:0.253Joules

Explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

F is the force

k is spring constant = 34N/m

e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

Work done = 0.253Joules

Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules

8 0

4 years ago

How many air molecules are in a 13.0×12.0×10.0 ft room (28.2 L=1 ft3)? Assume atmospheric pressure of 1.00 atm, a room temperatu

aksik [14]

Answer:

1963.93 Moles

Explanation:

-We know the standard conversion ratio for the volume of a mole is $1 Mole=22.4L$

Given volume of rooms as $13.0ft\times12.0ft\times10ft=1560 ft^3$

Convert the volume into liters: $1560\times28.2l=43992l$

#From our conversion ratio above, we get the volume of air molecules in moles as:

$V_m=\frac{43992L}{22.4L}\times 1\ mole\\\\=1963.93\ Moles$

Hence, the volume of air molecules is 1963.93 Moles

4 0

3 years ago

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

b) $v_{fx}=1.668\ m.s^{-1}$

c) $v_{fy}=0.7999\ m.s^{-1}$

Explanation:

Given masses:

$m_1=0.49\ kg$

$m_2=0.47\ kg$

Velocity of mass 1, $v_1=3.2 \hat{i}\ m.s^{-1}$

Velocity of mass 2, $v_2=1.6 \hat{j}\ m.s^{-1}$

a)

Initial momentum:

$p_i=m_1.v_1+m_2.v_2$

$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

$p_i=1.568\hat{i}+0.752 \hat{j}$

b)

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

$p_f=p_i$

$m_f.v_f=1.739$

$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

c)

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$

6 0

4 years ago