Help me bro, thanks all

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

2 years ago

Review. As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking always st

anyanavicka [17]

To find the mass of the planet we will apply the relationship of the given circumference of the planet with the given data and thus find the radius of the planet. From the kinematic equations of motion we will find the gravitational acceleration of the planet, and under the description of this value by Newton's laws the mass of the planet, that is,

The circumference of the planet is,

$\phi = 25.1m$

Under the mathematical value the radius would be

$\phi = 2\pi r$

$r = \frac{25}{2\pi}$

$r = 3.9788km$

Using second equation of motion

$x = \frac{1}{2} at^2$

Replacing the values given,

$1.4 = \frac{1}{2} a (29.2)^2$

Rearranging and solving for 'a' we have,

$a = 0.003283m/s^2$

Using the value of acceleration due to gravity from Newton's law we have that

$a = \frac{GM}{r^2}$

Here,

r = Radius of the planet

G = Gravitational Universal constant

M = Mass of the Planet

$\frac{(6.67*10^{-11})*M}{(3.9788*10^3)^2} = 0.003283$

$M = 7.79201*10^{14}kg$

Therefore the mass of this planet is $7.79201*10^{14}kg$

5 0

3 years ago

The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of G

timama [110]

The orbiting speed of the satellite orbiting around the planet Glob is 60.8m/s.

To find the answer, we need to know about the orbital velocity a satellite.

<h3>What's the expression of orbital velocity of a satellite?</h3>

Mathematically, orbital velocity= √(GM/r)
G= gravitational constant= 6.67×10^(-11) Nm²/kg², M = mass of sun , r= radius of orbit

<h3>What's the orbital velocity of the satellite in a circular orbit with a radius of 1.45×10⁵ m around the planet Glob of mass 7.88×10¹⁸ kg?</h3>

Here, M= 7.88×10¹⁸ kg, r= 1.45×10⁵ m
Orbital velocity of the orbiting satellite = √(6.67×10^(-11)×7.88×10¹⁸/1.45×10⁵)

= 60.8m/s

Thus, we can conclude that the speed of the satellite orbiting the planet Glob is 60.8m/s.

Learn more about the orbital velocity here:

brainly.com/question/22247460

#SPJ1

6 0

1 year ago

Help pleaseeeeeeeeeeeeee

77julia77 [94]

Answer:

hello the answer is 47m/s

8 0

2 years ago

After watching a show about submarines, Jamil wants to learn more about the oceans. Which question could be answered through sci

krok68 [10]

Answer:

After watching a show about submarines, Jamil wants to learn more about the oceans. The question that could be answered through scientific investigation is What substances dissolve in ocean water?

<u>Explanation:</u>

Jamil's scientific investigation will help him learn about the various substances that can be dissolved in seawater. There is a large number of such substances. Salt is the most important component of seawater.

Sodium chloride is the most common substance. Besides it, magnesium chloride and calcium chloride are also dissolved in seawater. Seawater consists of a large number of gases also such as oxygen, nitrogen, carbon dioxide, etc.

4 0

2 years ago

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Help me bro, thanks all​

Help me bro, thanks all