13. What is the ground state?

A simple pendulum, 1.00 m in length, is released from rest when the support string is at an angle of 35.0 from the vertical. Wh

taurus [48]

Answer:

Explanation:

Length = 1.00 m

If the length is 1.0, the vertical distance pivot to bob is cos 35 = 0.819

At the lowest point, vertical distance is 1.0, so the change is the difference, 0.181 meter

The potential energy of that height is converted to kinetic energy of motion, which determines the speed.

PE = KE

mgh = ½mV²

V = √(2gh) = 1.88 m/s

7 0

3 years ago

A human services professional has worked with a particular client for a couple of years. The client

zhuklara [117]

Answer:

Respect the client’s decision

Explanation:

just took the test

4 0

3 years ago

A 297 g block is connected to a light spring with spring constant 4.34 N/m, and displaced 7.45 cm from equilibrium. It is then r

vagabundo [1.1K]

Answer:

x = A sin ω t describes the displacement of the particle

v = A ω cos ω t

a = -A ω^2 sin ω t

a (max) = -A ω^2 is the max acceleration (- can be ignored here)

ω = (K/ m)^1/2 for SHM

F = - K x^2 restoring force of spring

K = 4.34 / .0745^2 = 782 N / m

ω = (782 / .297)^1/2 = 51.3 / sec

a (max) = .0745 * 782 / .297 = 196 m / s^2

4 0

2 years ago

A force of 8.0 N is along x direction, another force of 6.0 N is along +y direction. If both forces are acting on a point object

Darya [45]

Answer:

Resultant force, R = 10 N

Explanation:

It is given that,

Force acting along +x direction, $F_x=8\ N$

Force acting along +y direction, $F_y=6\ N$

Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,

$R=\sqrt{F_x^2+F_y^2+F_xF_y\ cos\theta}$

Here $\theta=90^{\circ}$

$R=\sqrt{F_x^2+F_y^2}$

$R=\sqrt{8^2+6^2}$

R = 10 N

So, the resultant force on the object is 10 N. Hence, this is the required solution.

6 0

3 years ago

Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint

djyliett [7]

Answer:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

$F = G \frac{m_{Earth} m_{Spaceship}}{r^2}$

Where G is a constant $G = 6.674 x10^{-11} m^2/ (ks s^2)$

$m_{Earth}$ represent the mass for the earth

$m_{spaceship}$ represent the mass for the spaceship

$r$ represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

$F_5 >F_4>F_1 >F_2>F_3$

Where $F_i$ represent the force for each of the 5 cases $i -1,2,3,4,5$ presented on the figure attached.

6 0

3 years ago