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xeze [42]
3 years ago
13

Suppose that you purchased a water bed with the dimensions 2.55 m à 2.53 dm à 245 cm. what mass of water does this bed contain

Physics
1 answer:
Nadya [2.5K]3 years ago
6 0

dimensions of the bed is given as

length = 2.55 m

thickness = 2.53 dm = 0.253 m

width = 245 cm = 2.45 m

now the volume of the bed is given as

V = 2.55 * 0.253 * 2.45 m^3

V = 1.581m^3

now the mass of water in it is given as

mass = density * volume

mass = 1000* 1.581

mass = 1581 kg

<em>so it will contain 1581 kg mass in it</em>

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Compare the maximum rate of heat transfer to the basal metabolic rate by converting a bmr of 88 kcal/hr into watts. what is the
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A 0.5 kg cheeseburger is lobbed at a particularly unhappy customer with a force of 10 N. While it is being thrown, what is the a
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Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge
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Answer:

 F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

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we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

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let's calculate

          F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

          F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

             r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

             r₁₃ = 0.50 m

            F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

            F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

           tan θ = y / x

           θ = tan⁻¹ y / x

          θ = tan⁻¹ 0.3 / 0.4

           tea = 36.87º

    The angle from the positive side of the x-axis is

         θ ’= 180 - θ

        θ ’= 180 - 36.87

        θ ’= 143.13º

       sin143.13 = F_13y / F₁₃

           F_13y = F₁₃ sin 143.13

           F{13y} = 1.697 10⁻² sin 143.13

           F_13y = 1.0183 10⁻² N

            cos 143.13 = F_13x / F₁₃

           F₁₃ₓ = F₁₃ cos 143.13

           F₁₃ₓ = 1.697 10⁻² cos 143.13

           F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

          Fx = F13x + F12x

          Fx = -1,357 10-2 +0

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          Fy = F13y + F12y

         Fy = 1.0183 10-2 + ​​1 10-1

          Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

         F = Ra (Fx2 + Fy2)

         F = RA [(1.357 10-2) 2 + 0.110183 2]

         F = 0.111015 N

Let's use trigonometry for the angles

         tan tea = Fy / Fx

          tea = tan-1 (0.110183 / -0.01357)

          tea = 1,448 rad

to find the angle about the positive side of the + x axis

           tea '= pi - 1,448

           Tea = 1.6936 rad

6 0
4 years ago
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