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White raven [17]
3 years ago
14

A square piece of tin has 12 inches on a side. An open box is formed by cutting out equal square pieces at the corners and bendi

ng upward the projecting portion which remain. Find the maximum volume that can be obtained.
Physics
1 answer:
Citrus2011 [14]3 years ago
6 0

Answer:

Explanation:

Given a square Piece whose side is 12 inches

Now square pieces are cut from each corner to make it a open box

Suppose x is the length of square piece at each corner

then

base square has a length of 12-2x

Dimension of new box is (12-2x)\times (12-2x)\times x

Volume V=(12-2x)\times (12-2x)\times x

V=\left ( 12-2x\right )^2\cdot x

For maximum volume differentiate with respect to x we get

\Rightarrow\frac{\mathrm{d} V}{\mathrm{d} x}=2\times \left ( 12-2x\right )\times \left ( -2\right )\cdot x+\left ( 12-2x\right )^2=0

we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid

V=\left ( 12-2\cdot 4\right )^2\cdot 4

V=4^3

V=64\ in.^3

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Answer:

B. the acceleration of a skydiver depends upon the force of gravity and wind resistance

Explanation:

Newtons second law of motion states that: the force (F) acting upon an object is directly proportional to the mass (M) of the object and the acceleration (a).

Basically this law states that the acceleration of a body is dependent on the force and mass of the body. If the force exerted on the body changes the acceleration will also change. and if the mass of the body changes, the acceleration will also be affected

from the options, the only option that refers to acceleration is option B

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3 years ago
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mario62 [17]

Answer:

A. The core makes up the majority of Earth's volume.

Explanation:

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2 years ago
You are on the south bank of the river in your canoe you need to reach the north bank you know that you can row your canoe at 2
Anna71 [15]

Answer:

(θ) = 60°

Explanation:

Given:

Speed of canoe Vc = 2 m/s

Speed of River Vr = 1 m/s

Computation:

Vc (Cosθ) = Vr

2 (Cosθ) = 1

(Cosθ) =  1 / 2

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If I were a geologist working in South America along one of those plate boundaries, what might I see if I were looking at sedime
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4 0
3 years ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
3 years ago
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