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Dahasolnce [82]
2 years ago
12

A roller coaster is traveling at 13 m/s when it approaches a hill that is 400 m long. Heading down the hill, it accelerates at 4

.0 m/s2. What is the final velocity of the roller coaster? Round your answer to the nearest whole number.
m/s
Physics
1 answer:
Vera_Pavlovna [14]2 years ago
8 0

Answer:

2as = v² - u²

v = √(2as + u²)

v = √(2 x 4 x 400 + 13²)

v = 58 m/s

Explanation:

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A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How
Sati [7]

0.36 J of work is done in stretching the spring from 15 cm to 18 cm.

To find the correct answer, we need to know about the work done to strech a string.

<h3>What is the work required to strech a string?</h3>
  • Mathematically, the work done to strech a string is given as 1/2 ×K×x².
  • K is the spring constant.
<h3>What will be the spring constant, if 40N force is required to hold a 10 cm to 15 cm streched spring?</h3>
  • The force experienced by a streched spring is given as Kx. x is the length of the spring streched from its natural length.
  • Then K = Force / x.
  • Here x = 15 - 10 = 5 cm = 0.05 m
  • K = 40/0.05 = 800N/m.
<h3>What will be the work required to strech that spring from 15 cm to 18 cm?</h3>
  • Work done = 1/2×k×x²
  • Here x= 18-15=3cm or 0.03 m
  • So, W= 1/2×800×0.03² = 0.36 J.

Thus, we can conclude that the work done is 0.36 J.

Learn more about the spring force here:

brainly.com/question/14970750

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6 0
1 year ago
A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect
zaharov [31]

Answer:

X=X_o+\dfrac{1}{2}gt^2

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

\dfrac{1}{2}mV^2=mg(X-X_o)

So

V=\sqrt{2g(X-X_o)}

We know that

\dfrac{dX}{dt}=V

\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}

\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX

At t= 0 ,X=Xo

So we can say that

X=X_o+\dfrac{1}{2}gt^2

So the length of cable after time t

X=X_o+\dfrac{1}{2}gt^2

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