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dexar [7]
3 years ago
6

A loaded tractor-trailer has a mass of 36,000 kg, what is it’s weight?

Physics
1 answer:
Cerrena [4.2K]3 years ago
8 0

That depends on WHERE the rig is, because

weight = (mass) x (acceleration of gravity where the object is)  .

-- If the truck is on Mars, then

Weight = (36,000 kg) x (3.71² m/s²) = 133,560N.

-- If the truck is on the moon, then

Weight = (36,000 kg) x (1.62 m/s²) = 58,320N.

-- If the truck is on Earth, then

Weight = (36,000 kg) x (9.81 m/s²) =  353,160N.

It all depends.

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A certain spring exerts a nonlinear force given by F(x) = −60x − 18x2 , where x is in meters and F is in newtons. A 0.90-kg bloc
tia_tia [17]

Answer:

a)  W = 6.75 J and b) v = 3.87 m / s

Explanation:

a) In the problem the force is nonlinear and they ask us for work, so we must use it's definition

      W = ∫ F. dx

Bold indicates vectors.  In a spring the force is applied in the direction of movement, whereby the scalar product is reduced to the ordinary product

     W = ∫ F dx

We replace and integrate

    W = ∫ (-60 x - 18 x²) dx

    W = -60 x²/2 -18 x³/3

Let's evaluate between the integration limits, lower W = 0 for x = -0.50 m, to the upper limit W = W for x = 0 m

    W = -30 [0- (-0.50) 2] -6 [0 - (- 0.50) 3]

    W = 7.5 - 0.75

    W = 6.75 J

b)  Work is equal to the variation of kinetic energy

    W = ΔK

    W = ΔK = ½ m v² -0

    v =√ 2W/m

    v = √(2 6.75/ 0.90)

    v = 3.87 m / s

8 0
3 years ago
What is TeO3 compound name​
Stolb23 [73]

Tellurium trioxide

Hope this helps you!!!

4 0
3 years ago
Read 2 more answers
A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)
gavmur [86]
<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

When copper rod is heated , its length increases

The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

here L is the increased length and L₀ is the original length

α  is the coefficient of linear expansion and ΔT is the increase in temperature .

The increase in length = L - L₀ = L₀ x α ΔT

Substituting all these value

Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9

= 1.87 x 10⁻² m

5 0
3 years ago
Read 2 more answers
Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a
Gnom [1K]

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F     (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg²  ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

8 0
3 years ago
Sam makes six star-shaped cookies out of dough. He puts all of the cookies on a balance and measures the total weight as 36 gram
son4ous [18]

Answer:

The answer is "36 grams".

Explanation:

In this question, the weight of the ball is not mentioned but is the weight of the cookies is declared, which is equal to 36 grams, and all the cookies are squeezes into the ball and after that, it calculates the overall weight so, let assume that ball weight is =0 and then the overall weight is:

=\text{weight of ball + cookies weight}\\\\=0+36 \ grams \\\\=36 \ grams

5 0
3 years ago
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