Who wants to do my missing physics work

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

3 years ago

A 0.90-kg falcon is diving at 28.0 m/s at a downward angle of 35° . It catches a 0.325-kg pigeon from behind in midair. What is

pashok25 [27]

Answer:

22.11 m / s

Explanation:

The falcon catches the prey from behind means both are flying in the same direction ( suppose towards the left )

initial velocity of falcon = 28 cos 35 i - 28 sin 35 j

( falcon was flying in south east direction making 35 degree from the east )

momentum = .9 ( 28 cos 35 i - 28 sin 35 j )

= 20.64 i - 14.45 j

initial velocity of pigeon

= 7 i

initial momentum = .325 x 7i

= 2.275 i

If final velocity of composite mass of falcon and pigeon be V

Applying law of conservation of momentum

( .9 + .325) V = 20.64 i - 14.45 j +2.275 i

V = ( 22.915 i - 14.45 j ) / 1.225

= 18.70 i - 11.8 j

magnitude of V

= √ [ (18.7 )² + ( 11.8 )²]

= 22.11 m / s

6 0

3 years ago

Good evening everyone, State the law of conservation of energy .Thank u

disa [49]

Energy can neither be created nor be destroyed but can be converted from one form to another .

3 0

2 years ago

Read 2 more answers

What will occur when the waves peak at the same place at the same time?

dedylja [7]

Hello!

This is a matter of superposition.
When the waves peak at the same time and place, they produce constructive interference, meaning the waves interact together in a positive way, to make a wave with Amplitude of both waves added together. When the peaks differ however, at the same time and place, then it is destructive interference and the waves essentially cancel each other out.

Hope this helps. Any questions please just ask. Thank you kindly.

5 0

3 years ago

What are some forces that come in pairs?

Nikolay [14]

Answer:

ACTION REACTION FORCES

Explanation:

When there is an action frce there will be a reaction force

3 0

2 years ago