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3 years ago

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A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is

the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m

1 answer:

andre [41]3 years ago

8 0

Answer:

2.80N/m

Explanation:

Given data

mass m= 56kg

perios T= 11.2s

The expression for the period is given as

T=2π√m/k

Substitute

11.2= 2*3.142*√56/k

square both sides

11.2^2= 2*3.142*56/k

125.44= 351.904/k

k=351.904/125.44

k= 2.80N/m

Hence the spring constant is 2.80N/m

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Answer:

Part a)

$v = 12.45 m/s$

Part B)

$F_n = 0.05 N$

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

$mgH = \frac{1}{2}mv^2 + mg(2R)$

so the speed at point A is given as

$mg(H - 2R) = \frac{1}{2}mv^2$

$v = \sqrt{2g(H - 2R)}$

$v = \sqrt{2(9.81)(21.9 - 2\times 7)}$

$v = 12.45 m/s$

Part B)

Now the force equation at point A is given as

$F_n + mg = \frac{mv^2}{R}$

$F_n = \frac{mv^2}{R} - mg$ [/tex]

$F_n = 0.004(\frac{12.45^2}{7} - 9.81)$

$F_n = 0.05 N$

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Convertir 340,5 grados Fahrenheit a centígrados

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Answer:

171.38889

Explanation:

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A baseball of mass 300-g is hit at a velocity of 40 m/s. If the ball is caught by the third baseman and the ball penetrates 2.0

OlgaM077 [116]

The force is -12,000 N

Explanation:

First of all, we calculate the acceleration of the ball, by using the following suvat equation:

$v^2-u^2=2as$

where:

v = 0 is the final velocity of the baseball (it comes to rest)

u = 40 m/s is the initial velocity

a is the acceleration

s = 2.0 cm = 0.02 m is the displacement of the ball

Solving for a,

$a=\frac{v^2-u^2}{2s}=\frac{0-40^2}{2(0.02)}=-40,000 m/s^2$

Now we can calculate the average force exerted on the ball, by using Newton's second law:

$F=ma$

where

m = 300 g = 0.3 kg is the mass of the ball

$a=-40,000 m/s^2$ is the acceleration

Substituting,

$F=(0.3)(-40,000)=-12,000 N$

where the negative sign indicates that the direction of the force is opposite to the direction of motion of the ball.

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What is the coefficient of friction between the skates and the ice?

natta225 [31]

The coefficient of friction is 0.051

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The motion of the skater is a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2 - u^2 = 2as$

where:

v = 0 is the final velocity of the skater (he comes to a stop)

u = 10.0 m/s is his initial velocity

a is the acceleration

$s=1.0\cdot 10^2 m = 100 m$ is the distance he travels before stopping

Solving for a, we find the acceleration of the skater:

$a=\frac{v^2-u^2}{2s}=\frac{0-10.0^2}{2(100)}=-0.5 m/s^2$

We also know that the net force acting on the skater is the force of friction, therefore we can write (Newton's second law of motion):

$F= ma = -\mu mg$

where

$-\mu mg$ is the force of friction

m is the mass of the skater

$\mu$ is the coefficient of friction

$a=-0.5 m/s^2$ is the acceleration

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Solving for $\mu$ , we find the coefficient of friction:

$\mu = -\frac{a}{g}=-\frac{-0.5}{9.8}=0.051$

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