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erma4kov [3.2K]
3 years ago
8

Which row shows what happens to the temperature of a solid as it melts and what happens to the temperature when

Physics
1 answer:
frez [133]3 years ago
4 0

When ice melts, its temperature doesn't change ... ice at 32 degrees becomes water at 32 degrees.

When water boils, its temperature doesn't change ... water at 212 degrees becomes steam at 212 degrees.

The row that says both of these is row-D .

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A person pushes a 50kg box to the right with 100N of force at a constant velocity of 5m/s. Calculate the coefficient of friction
Anna11 [10]
The weight of the box is 50x9.8 = 490 N. The force of friction is 100N. F= μΝ so coefficient = 100/490 = 0.20
8 0
2 years ago
A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0055 m/s2 less than that at sea leve
pshichka [43]
Gs*rs^2 = gm*rm^2 
<span>rm = rs*√gs/gm </span>
<span>rm = 6370*√9.83/(9.83-0.009) = 6372.92 </span>
<span>mountain observatory is placed at an altitude worth 2920 m asl

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
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7 0
3 years ago
A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
Archy [21]

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

          t_{s} = 36.94 s

The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

3 0
2 years ago
A race car starts from rest and accelerates uniformly to 69 mph in 4.5 s.
Marizza181 [45]
For this question we should apply
a = v^2 - u^2 by t
a = 69 - 0 by 4.5
a = 69 by 4.5
a = 15.33
a = 6.85 m/s^2

If the answer in option is near to answer then , you can mark it as correct.
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