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My name is Ann [436]

3 years ago

6

Determine the amount of potential energy of a 4.2 kg book that is placed on a shelf with a height of 0.9 meters. Round your answ

er to one decimal place

1 answer:

swat323 years ago

5 0

Answer:

that is a correct answer maybe .

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A starter pistol is fired 150m away from the spectators who hear the gun 0.5s after they see it fired. How fast does sound trave

Natali [406]

Answer:

300m/s

Explanation:

speed = distance/time

150/0.5

8 0

3 years ago

Block A, with a mass of 6.0 kg, is sliding across a frictionless track at 65 m/s when it collides with Block B ( 9.0 kg) which i

Mars2501 [29]

The correct answer is a I hope that helped enjoy the rest of your weekend

3 0

3 years ago

Valentin [98]

Answer:

44000n/m

Explanation:

8 0

3 years ago

A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 meters. The mass then slides w

Aneli [31]

Answer:

The angle is $\theta = 36.24 ^o$

Explanation:

From the question we are told that

The mass is $m = 0.6 \ kg$

The radius is $r = 1.1 \ m$

The speed is $v = 3.57 \ m /s$

According to the law of energy conservation

The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e

$m * g * h = \frac{1}{2} * m * v^2$

=> $h = \frac{1}{2 g } * v^2$

Here h is the vertical distance traveled by the mass which is also mathematically represented as

$h = r * sin (\theta )$

So

$\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2]$

substituting values

$\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2]$

$\theta = 36.24 ^o$

3 0

4 years ago

An infinitely long line of charge has linear charge density 6.00×10−12 C/m . A proton (mass 1.67×10−27 kg,charge +1.60×10−19 C)

Bogdan [553]

Answer:

$A) \,K.E=1.405\times 10^{-20}J$

$B)\,r_f=0.268\,m$

Explanation:

$Charge\,\,density=\lambda=6\times 10^{-12}C/n\\\\Mass\,\, of \,\,proton=m_p=1.67\times 10^{-27}kg\\\\charge\,\, of\,\, proton=q_p=1.609\times 10^{-19}C\\\\r=12\,cm=0.12\, m\\\\v=4.103\times 10^{3}m/s$

A) Initial kinetic energy of proton

$K.E=\frac{1}{2}m_pv^2\\\\K.E=1.405\times 10^{-20}J$

B) How close does the proton get to the line of charge?

Potential energy and kinetic energy are related as:

$K_i+U_i=K_f+U_f\\\\U_f-U_i=K_i-K_f\\\\q(V_f-V_i)=1.40\times 10^{-20}\\\\V_f-V_i=0.087--(1)\\$

Change in voltage is

$V_f-V_i=\frac{\lambda}{2\pi \epsilon_o}ln\frac{r_f}{r_i}\\\\ln|\frac{r_f}{r_i}|=(0.087)(\frac{2\pi \epsilon_o}{\lambda})\\\\ln|\frac{r_f}{r_i}|==0.8059\\\\\frac{r_f}{r_i}=2.24\\\\r_f=(2.24)(.12)\\\\r_f=0.268 m$

5 0

3 years ago