Where in the United States are sinkholes common, other than Florida?

Calculate the hang time of a person who moves 3 m horizontally during a 1.25-m high jump. What is the hang time when the person

jekas [21]

The duration of time for which an object stays in air is called the hang time.

For an athlete who moves 3m horizontally during a 1.25m high jump, the hang time will be the sum of the time taken by the athlete to reach the maximum height and the time taken for the athlete to reach the ground from maximum height.

Calculate the time taken t_1 by the athlete to reach the maximum height

$t_1 = \sqrt{\frac{2h}{g}}$

$t_1 = \sqrt{\frac{2(1.25)}{9.8}}$

$t_1 = 0.5s$

The athlete takes same time to reach the ground from the maximum height, so $t_2 = 0.5s$

Calculate the hang time will be

$t =t_1+t_2$

$t = 0.5+0.5$

$t = 1s$

Therefore the hang time of the athlete when he moves a horizontal distance of 3m is 1s.

Similarly, when the athlete runs 6m horizontally, then also there will not be a change in the hang time of the athlete as the hang time is independent of the horizontal distance covered.

4 0

3 years ago

A 0.26 kg rod of length 80 cm is suspended by a frictionless pivot at one end. It is held horizontal and released.

Daniel [21]

Answer:

a) $a_{center} = 7.38 ~m/s^2$

b) $a_{end} = 14.77 ~m/s^2$

c) $v_{center} = 2.43~m/s$

Explanation:

a) Immediately after the rod is released, <u>the rod is still horizontal but now subject to gravity.</u> Since one end of the rod is fixed, then the weight of the rod applies a torque. Then by Newton's Second Law, the acceleration can be found.

$\tau =I\alpha$

where I is the moment of inertia of the rod with respect to its fixed end, and α is the angular acceleration.

The net torque of the rod is

$\vec{\tau} = \vec{r} \times \vec{F}\\\tau = rF\sin(90) = rF$

where r is the distance from center of the mass to the fixed end, so r = 0.4 m.

The weight of the rod is w = mg = 0.26 x 9.8 = 2.54 N.

So the net torque is τ = 1.01 Nm.

The moment of inertia of the rod is

$I = \frac{1}{3}mL^2 = \frac{1}{3}(0.26)(0.8)^2 = 0.055~kg m^2$

So, the Newton's Second Law yields

$\tau = I\alpha\\\alpha = \frac{\tau}{I} = \frac{1.01}{0.055} = 18.47$

<u>The relation between angular acceleration and linear acceleration is a = αr </u>

So, the linear acceleration of the rod is

$a = \alpha r = 7.38~m/s^2$

b) Using the same relationship between angular acceleration and linear acceleration, the linear acceleration of the end of the rod can be found.

$a = \alpha L = 14.77~m/s^2$

c) The conservation of energy can be used to find the velocity when the rod is vertical.

$K_1 + U_1 = K_2 + U_2\\0 + mg(L/2) = \frac{1}{2}I\omega^2 + 0\\(0.26)(9.8)(0.4) = \frac{1}{2}(0.055)\omega^2\\\omega = 6.08~rad/s$

The linear velocity is v = ωr, so

v = 2.43 m/s.

4 0

3 years ago

For a standing wave to form in a medium, two waves must

Slav-nsk [51]

For a standing wave to form, two waves must be traveling in opposite directions and cause destructive interference.

4 0

3 years ago

A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant

Stells [14]

Answer:

The average speed of the runner is 4 m/s.

Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

8/3 m/s² = a

So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

x = 1/2 · 8/3 m/s² · (3 s)²

x = 12 m

Now, we can calcualte the average speed:

a.s = d/t

a.s = 12 m / 3 s

a.s = 4 m/s

The average speed of the runner is 4 m/s.

4 0

3 years ago

What is the mass of a 7.0 kilogram bowling ball on the surface of the moon?

babunello [35]

Answer:

1.16kg is the answer. Hope this helped

Explanation:

4 0

2 years ago