Ask question

Ask question

All categories

2 years ago

12

A mass of 400 g is hanging from a spring with a spring constant of 10 N/m. The mass is pulled down a distance of 10 cm from its

equilibrium position and released. How long does it take to reach the equilibrium position again

1 answer:

Mnenie [13.5K]2 years ago

3 0

Explanation:

Below is an attachment containing the solution.

You might be interested in

1. The planet Jupiter completes a revolution of the sun in 11.5 years. Express it in seconds. Given that one year= 3.154 × 10^7

xenn [34]

Answer:

The planet Jupiter completes one revolution of the sun in 362710000 seconds. Long time, right?

Explanation:

3.154x10^7=3.154x10000000=31540000

11.5x31540000=362710000

7 0

1 year ago

A particle moves along the x axis so that its velocity at time t is given by v(t)=

vodomira [7]

Answer:

a = 0.7267 , acceleration is positive therefore the speed is increasing

Explanation:

The definition of acceleration is

a = dv / dt

they give us the function of speed

v = - (t-1) sin (t² / 2)

a = - sin (t²/2) - (t-1) cos (t²/2) 2t / 2

a = - sin (t²/2) - t (t-1) cos (t²/2)

the acceleration for t = 4 s

a = - sin (4²/2) - 4 (4-1) cos (4²/2)

a = -sin 8 - 12 cos 8

remember that the angles are in radians

a = 0.7267

the problem does not indicate the units, but to be correct they must be m/s²

We see that the acceleration is positive therefore the speed is increasing

6 0

2 years ago

A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its

True [87]

Answer:

The angular velocity is $w = 1.43\ rad/sec$

Explanation:

From the question we are told that

The mass of wooden gate is $m_g = 4.5 kg$

The length of side is L = 2 m

The mass of the raven is $m_r = 1.2 kg$

The initial speed of the raven is $u_r = 5.0m/s$

The final speed of the raven is $v_r = 1.5 m/s$

From the law of conservation of angular momentum we express this question mathematically as

Total initial angular momentum of both the Raven and the Gate = The Final angular momentum of both the Raven and the Gate

The initial angular momentum of the Raven is $m_r * u_r * \frac{L}{2}$

Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is zero

Note: This above is the generally formula for angular momentum of square objects

The final angular velocity of the Raven is $m_r * v_r * \frac{L}{2}$

The final angular velocity of the Gate is $\frac{1}{3} m_g L^2 w$

Substituting this formula

$m_r * u_r * \frac{L}{2} = \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * v_r * \frac{L}{2} - m_r * u_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * \frac{L}{2} * [u_r - v_r]$

Where $w$ is the angular velocity

Substituting value

$\frac{1}{3} (4.5)(2)^2 w = 1.2 * \frac{2}{2} * [5 - 1.5]$

$6w = 4.2$

$w = \frac{6}{4.2}$

$w = 1.43\ rad/sec$

5 0

2 years ago

URGENT What does Newton's third law say about why momentum is conserved in collisions?

valentinak56 [21]

Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction.For such a collision<span>, the forces acting between the two objects are equal in magnitude and opposite in direction</span>

4 0

3 years ago

Read 2 more answers

A 35-year-old patent clerk needs glasses of 50-cm focal length to read patent applications that he holds 25 cm from his eyes. Fi

Natali5045456 [20]

Answer:

28.57 cm

Explanation:

We are given that

Focal length, $f=50 cm$

Distance of application from his eyes,s=40 cm

$\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}$

$\frac{1}{40}+\frac{1}{s'}=\frac{1}{50}$

$\frac{1}{s'}=\frac{1}{50}-\frac{1}{40}=\frac{4-5}{200}$

$s'=\frac{200}{-1}=-200 cm$

s=25 cm

Substitute the values

$\frac{1}{25}-\frac{1}{200}=\frac{1}{f'}$

$\frac{8-1}{200}=\frac{1}{f'}$

$\frac{1}{f'}=\frac{7}{200}$

$f'=\frac{200}{7}=28.57 cm$

6 0

3 years ago