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3 years ago

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In Trial III, a different, looser, spring is used; its force constant is 23.1 N/m. The suspended mass is the same as the one in

Trial I: 0.400 kg. Theoretically speaking, what period of oscillations should we expect, based on these values of k and m? Enter your answer in seconds with two significant figures.

1 answer:

cricket20 [7]3 years ago

7 0

Answer:

T=0.827s

Explanation:

The period of a spring can be calculated with the equation

$T=2\pi w$

But we know as well that w is given by,

$w=\sqrt{\frac{k}{m}}$

Replacing,

$w=\frac{2\pi}{T}= \sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{0.4}{23.1}}$

So we have that

$T=0.827s$

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The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st

irina [24]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, $31 {\rm {N}/{mm}}$ . What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

$\triangle E = 12.79 J$

Explanation:

Sprinters' tendons stretch, $x_s = 43 mm = 0.043 m$

Non athletes' stretch, $x_n = 32 mm = 0.032 m$

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, $E_s = 0.5kx_s^2$

Maximum energy stored in the non athletes, $E_m = 0.5kx_n^2$

Difference in maximum stored energy between the sprinters and the non-athlethes:

$\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J$

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How do you feel when your out of breath ?

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When you're out of breath, you feel dizzy, lung pain and maybe even nausea or side pain.

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3 years ago

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A battery has an electric potential of 1.5V and transfers 10.0 C between the two terminals. How much work was done?

bogdanovich [222]

Answer:

15 Joules

Explanation:

work = charge x potential difference

= 10 x 1.5

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3 years ago

A 25 kg child stands 2.5 m from the center of a frictionless merry‐go‐round, which has a 200 kg*m^2 moment of inertia and is spi

vodka [1.7K]

Answer:

Explanation:

a ) Time period T = 2 s

Angular velocity ω = 2π / T

= 2π / 2 = 3.14 rad /s

Initial moment of inertia I₁ = 200 + mr²

= 200 + 25 x 2.5²

=356.25

Final moment of inertia

I₂ = 200 + 25 X 1.5 X 1.5

= 256.25

b ) We apply law of conservation of momentum

I₁ X ω₁ = I₂ X ω₂

ω₂ = I₁ X ω₁ / I₂

Putting the values

$w_2=\frac{356.25\times3.14}{256.25}$

ω₂ = 4.365 rad s⁻¹

c ) Increase in rotational kinetic energy

=1/2 I₂ X ω₂² - 1/2 I₁ X ω₁²

.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²

= 684.95 J

This energy comes from work done against the centripetal pseudo -force.

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3 years ago

What was the direction of the ball’s velocity

Tju [1.3M]

Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table. The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity just before it hit the floor? </em>

Answer:

$\theta=-75.7^{\circ}$

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

$s=ut+\frac{1}{2}at^2$

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time

Solving for t,

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s$

Now we can find the final vertical velocity of the ball, using:

$v_y=u+at$

And susbtituting t = 0.52 s, we find

$v_y = 0 +(9.8)(0.52)=5.1 m/s$

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

$v_x = \frac{0.7}{0.52}=1.3 m/s$

So now we can find the direction of the ball's velocity using:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}$

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

$\theta=-75.7^{\circ}$

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3 years ago