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3 years ago

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Look at the figure, which shows the motion of three balls. The curved paths followed by balls B and C are examples of _____. pro

jectile motion rolling friction linear motion fluid friction

2 answers:

Vinvika [58]3 years ago

6 0

Answer:the answer is projectile motion

Explanation:

Natali [406]3 years ago

3 0

Answer:did you get the answer?

Explanation:

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A person of mass m will bungee-jump from a bridge over a river where the height from the bridge to the river is h. The bungee co

Likurg_2 [28]

Answer: $k_{min}=\frac{18mg}{h}$

Explanation:

Given

mass of person is m

Distance between bridge and river is h

chord has an un-stretched length of $\frac{2h}{3}$

Let spring constant be k

Person will just stop before hitting the river

Conserve energy i.e. Potential Energy of Person is converted in to elastic energy of chord

$mgh=\frac{kx^2}{2}$

$x=h-\frac{2h}{3}=\frac{h}{3}$

$mgh=\frac{kh^2}{18}$

$k=\frac{18mg}{h}$

Thus $k_{min}=\frac{18mg}{h}$

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3 years ago

By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what

Vinil7 [7]

Explanation:

1. Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}$

$V=\dfrac{h^2}{2q_pm_p\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}$

V = 45.83 volts

2. Mass of the electron, $m_p=9.1\times 10^{-31}\ kg$

Wavelength, $\lambda_p=4.23\times 10^{-12}\ m$

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}$

$V=\dfrac{h^2}{2q_em_e\lambda^2}$

$V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}$

$V=6.92\times 10^{34}\ V$

V = 84109.27 volt

Hence, this is the required solution.

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3 years ago

wich of the following are commonly distributed by veterinary assistants in typical veterinary practice?

vredina [299]

Can you sent a picture of the answers

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3 years ago

HEY EVERYONE!!! WATCH THIS!!!

VARVARA [1.3K]

Answer:

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Explanation:

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2 years ago

Read 2 more answers

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 $L_i$ is 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)}

$l=2.45cm$

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3 years ago