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kaheart [24]
2 years ago
7

A rectangular room twice as long as it is wide would contain 145 square feet more if each side were one foA rectangular room twi

ce as long as it is wide would contain 145 square feet more if each side were one foot longer. What is the length of the room
Physics
1 answer:
Bad White [126]2 years ago
4 0

The length of the room is 96 ft.

Let L be the length of the room and W be the width of the room.

Since the length is twice its width, L = 2W.

Its area A = LW

= 2W × W

= 2W²

If each side were increased by 1 foot, then its new area is A' = (L + 1)(W + 1)

= (2W + 1)(W + 1)

= 2W² + 3W + 1

Since the room would contain 145 ft more if each side were increased by 1 foot, then its new area is A' = A + 145 = 2W² + 3W + 1

2W² + 145 = 2W² + 3W + 1

145 = 3W + 1

3W = 145 - 1

3W = 144

W = 144/3

W = 48 ft.

Since its length L = 2W, substituting W = 48 into the equation, we have

L = 2(48)

L = 96 ft

So, the length of the room is 96 ft.

Learn more about length of a room here:

brainly.com/question/24375989

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Answer:

a) Batteries and fuel cells are examples of galvanic cell

b) Ag-cathode and Zn-anode

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Explanation:

a) A galvanic cell is an electrochemical cell in which chemical energy is converted to electrical energy. The chemical reaction which drives a galvanic cell is a redox reaction i.e. a reduction-oxidation process.

A typical galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs. Batteries and fuel cells are examples of galvanic cells.

b) The nature of the electrode that will serve as an anode or cathode depends on the value of the standard reduction potential (E⁰) of that electrode. The electrode with a higher or more positive the value of E⁰ serves as the cathode and the other will function as an anode.

In the given case, the E⁰ values from the standard reduction potential table are:

E⁰(Zn/Zn2+) = -0.763 V

E°(Ag/Ag+)=+0.799 V

Therefore, Ag will be the cathode and Zn will be the anode

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A building contractor says that a certain building foundation should support a pressure greater than 13,000 pa if the area of th
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A point charge q1=2.0μC is located on the positive y axis at y=0.30m, and an identical charge q2 is at the origin. Find the magn
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Answer:

(A) 0.279N at angle 38.02°

(B) 0.701N

(C) 14.19°

Explanation:

(A) The net force on q3 is given as:

F = Fxi + Fyj

Fx is the x component of the force

Fy is the y component of the force

Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0

Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx

First let us find y and angle x from the diagram.

Using Pythagoras theorem,

y² = 0.3² + 0.4²

y² = 0.25

y = 0.5m

Using SOHCAHTOA to find x,

sinx = 0.4/0.5

x = 53.13°

Electrostatic force, F is given as:

F = kqQ/r²

Where k = Coulumbs constant

F(1,3) = (k*q1*q3) / r²

F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)

F(1, 3) = 0.288N

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = 0.45N

Therefore,

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Fx = 0.22N

Fy = 0.288cos53.13

Fy = 0.172N

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The magnitude of the force will be

F(mag) = √(0.22² + 0.172²)

F(mag) = 0.279N

The direction of the force makes will be

tanθ = Fy/Fx

tanθ = 0.172/0.22 = 0.781

θ = 38.02° to the x axis.

(B) q2 = - 2.0 * 10^(-6)

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F(2,3) = (k*q2*q3) / r²

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F(2, 3) = -0.45N

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(C) The direction of the force makes will be

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Answer:

True

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