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Westkost [7]
3 years ago
5

A spring suspended vertically is 11.9 cm long. When you suspend a 37 g weight from the spring, at rest, the spring is 21.5 cm lo

ng. Next you pull down on the weight so the spring is 25.2 cm long and you release the weight from rest. What is the period of oscillation?
Physics
1 answer:
Naddika [18.5K]3 years ago
5 0

Answer:

Time period of oscillations is 0.62 s

Explanation:

Due to suspension of weight the change in the length of the spring is given as

\Delta L = L_f - L_i

\Delta L = 21.5 - 11.9 = 9.6 cm

now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force

mg = kx

0.037 (9.81) = k(0.096)

k = 3.78 N/m

Now the period of oscillation of spring is given as

T = 2\pi \sqrt{\frac{m}{k}}

Now plug in all values in it

T = 2\pi \sqrt{\frac{0.037}{3.78}}

T = 0.62 s

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A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th
irina [24]

The angular momentum is defined as,

L=I\omega

Acording to this text we know for conservation of angular momentum that

L_i=L_f

Where L_iis initial momentum

L_f is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

L_i=mvl

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

L_f=(I_b+I_s)\omega

Substituting

L_f=(ml^2+\frac{10ml^2}{3})\omega

L_f=\frac{13}{10}ml^2w

m(0.65)l=\frac{13}{10}ml^2 \omega

\omega=\frac{1}{2l}

Applying conservative energy equation we have

\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'

\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)

Replacing the values and solving

l=\frac{13}{0.54g}

Substituting

l=\frac{13}{0.54(9.8)}

l=2.45cm

7 0
3 years ago
Can someone help me
SashulF [63]
That is b. hope this helps cx
8 0
2 years ago
Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of
STatiana [176]

Answer:

2.27\cdot 10^{49}

Explanation:

The gravitational force between the proton and the electron is given by

F_G=G\frac{m_p m_e}{r^2}

where

G is the gravitational constant

m_p is the proton mass

m_e is the electron mass

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N

The electrical force between the proton and the electron is given by

F_E=k\frac{q_p q_e}{r^2}

where

k is the Coulomb constant

q_p = q_e = q is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N

So, the ratio of the electrical force to the gravitational force is

\frac{F_E}{F_G}=\frac{2.56\cdot 10^{-19} N}{1.13\cdot 10^{-68}N}=2.27\cdot 10^{49}

So, we see that the electrical force is much larger than the gravitational force.

5 0
3 years ago
Help with these please someone?
DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0
3 years ago
Which of the following is required for work to be done on an object?
ad-work [718]
I think that in order for work to be done, the object must move in the direction of the force and move over a distance.
7 0
2 years ago
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