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Sonbull [250]
3 years ago
6

If a 8.0 kg mass is hung on the end of a spring, it is stretched 0.78 meters as a result. What is the force constant of the spri

ng?
Physics
2 answers:
Alborosie3 years ago
3 0

Answer:

K=100.62N/m

Explanation:

Given:

             m= 8.0 Kg\\x=0.78 m

We know from Hook's law

           F= k x

where

           F= m g

So equating we get

             mg=kx\\k=\frac{mg}{x} \\k=\frac{8*9.8}{0.78}\\k=100.62 N/m

Hatshy [7]3 years ago
3 0

Answer:

k=100.5 n/m

Explanation:

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A 230-km-long high-voltage transmission line 2.0 cm in diameter carries a steady current of 1,100 A. If the conductor is copper
Molodets [167]

Answer:

28.23 years

Explanation:

I = 1100 A

L = 230 km = 230, 000 m

diameter = 2 cm

radius, r = 1 cm = 0.01 m

Area, A = 3.14 x 0.01 x 0.01 = 3.14 x 10^-4 m^2

n = 8.5 x 10^28 per cubic metre

Use the relation

I = n e A vd

vd = I / n e A

vd = 1100 / (8.5 x 10^28 x 1.6 x 10^-19 x 3.14 x 10^-4)

vd = 2.58 x 10^-4 m/s

Let time taken is t.

Distance = velocity x time

t = distance / velocity = L / vd

t = 230000 / (2.58 x 10^-4) = 8.91 x 10^8 second

t = 28.23 years

5 0
3 years ago
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EX 6-1 A ball is twirled on a 0.870 - m-long string with a constant speed of 3.36 m / s . Calculate the acceleration of the ball
Elina [12.6K]

Answer:

a=12.97\ m/s^2

Explanation:

Given that,

The length of a string, l = 0.87 m

Speed of the ball, v = 3.36 m/s

We need to find the acceleration of the ball. The acceleration acting on the ball is centripetal acceleration. It is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(3.36)^2}{0.87}\\\\=12.97\ m/s^2

So, the acceleration of the ball is 12.97\ m/s^2.

4 0
3 years ago
The defense is the team that is trying to score a goal.<br> A. True<br> B. False
larisa86 [58]

Answer:

true

Explanation:

5 0
3 years ago
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A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s
djyliett [7]

Answer:

His final velocity is 15.8 m/s.

Step-by-step explanation:

Given:

Initial velocity of the driver is, u=32 m/s

Acceleration of the driver is, a=-1.5 m/s²

Time taken to reach final velocity is, t=10.8 s.

The final velocity is given using the Newton's equations of motion as:

v=u+at, where, v is the final velocity.

Now, plug in the given values and solve for v.

v=32-1.5(10.8)\\v=32-16.2=15.8\textrm{ m/s}

Therefore, his final velocity is 15.8 m/s.

5 0
2 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
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