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Allisa [31]
3 years ago
11

2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is

applied to the box. The force is directed to the left and has magnitude F1t2 = 16.00 N>s22t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?

Physics
2 answers:
larisa86 [58]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

vodka [1.7K]3 years ago
4 0

Answer:

a) the distance moved by the object is 14m

b) the speed of the box at 3.00s is - 18.5m/s

The negative sign in the answer indicates that the speed of the box is in opposite direction. Therefore, the direction of the box is towards left.

Explanation:

see the attached file

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Answer:

T_s = 6.8 degree C

Explanation:

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\sigma = 5.67 \times 10^{-8} W/m^2K^4

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now we have

120 = (5.67 \times 10^{-8})(0.557)(1.38)(307^4 - T_s^4)

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Two concrete spans of a 250-m-long bridge are placed end to end so that no room is allowed for expansion. If a temperature incre
scoundrel [369]

Answer:

  y = 2.74 m

Explanation:

The linear thermal expansion processes are described by the expression

         ΔL = α L ΔT

Where α the thermal dilation constant for concrete is 12 10⁻⁶ºC⁻¹, ΔL is the length variation and ΔT the temperature variation in this case 20ªc

If the bridge is 250 m long and is covered by two sections each of them must be L = 125 m, let's calculate the variation in length

        ΔL = 12 10⁻⁶ 125 20

        ΔL = 3.0 10⁻² m

Let's use trigonometry to find the height

The hypotenuse     Lf = 125 + 0.03 = 125.03 m

Adjacent leg           L₀ = 125 m

       cos θ = L₀ / Lf

       θ = cos⁻¹ (L₀ / Lf)

       θ = cos⁻¹ (125 / 125.03)

       θ = 1,255º

We calculate the height

       tan 1,255 = y / x

       y = x tan 1,255

       y = 125 tan 1,255

       y = 2.74 m

3 0
3 years ago
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