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3 years ago

An object is 12 m long, 0.65 m wide, and 13 cm high. Calculate the volume of this object.

1 answer:

6 0

The volume corresponds to the measure of the space occupied by a body. From the given dimensions we can intuit that we are looking to find the Volume of an Cuboid, that is, an orthogonal rectangular prism, whose faces form straight dihedral angles.

Mathematically the volume of this body is given as

$V = lWh$

Where,

L = Length

W = Width

H = High

$V = (12)(0.65)(13*10^{-2})$

$V = 1.014m^3$

Note: The value given for the height was in centimeters, so it was transformed to meters.

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3 years ago

Scientists have designed solar cells to trap solar energy and convert it to _____ energy.

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3 years ago

Read 2 more answers

You have two identical beakers A and B. Each beaker is filled with water to the same height. Beaker B has a rock floating at the

gregori [183]

Answer:

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water the weight of the two beakers is the same

Explanation:

The beaker weight is

beaker A

W_total = W_ empty + W_water

Beaker B

W_total = W_ empty + W_water + W_roca

a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more

b) If it is spilled in water, the weight of the two beakers is the same because the amount of liquid spilled and equal to the weight of the stone, therefore the two beakers weigh the same

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3 years ago

A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge

Ivan

Answer:

2 J

Explanation:

A charged capacitor of capacitance $C_1$ with energy of 7.54 J, is connected in parallel with another capacitor $C_2$ , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

$U_O=q_0^2/2C_1=7.54 J\\$

The energy stored in the electric field is the sum of the energies of the two capacitors:

$U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2$

since the charge equally distributed, $q_1$ = $q_2$ = $q_o/2$ . and since they are connected in parallel the potential difference on both of them is the same :