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Otrada [13]
2 years ago
8

A bowling ball with a mass of 9kg is thrown down a lane with a constant speed of 3 m/s. The ball hits the 1.5kg pin, initially a

t rest, at the end of the lane. After the collision the pin moves with a speed of 5 m/s. How fast is the ball moving after the impact?​
Physics
1 answer:
olasank [31]2 years ago
7 0

Answer:

M1 V1 = M1 V2 + M2 V3    conservation of momentum

V2 = (M1 V1 - M2 V3) / M1      where V2 = speed of M1 after impact

V2 = (3 * 9 - 1.5 * 5) / 9  =   (27 - 7.5) / 9 = 2.17 m/s

Note: All speeds are in the same direction and have the same sign

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Answer: a. ) What is the magnitude of their average acceleration (in m/s2)

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Can a body have zero velocity and finite acceleration?Explain​
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Answer:

Kinda? Depends what the question is fully asking

Explanation:

Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.

So that -2 m/s thing after one second will be going -1 m/s.

After another second it'll be going 0 m/s.

After another itll be going +1 m/s and so on.

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Given that water at standard pressure freezes at 0∘C, which corresponds to 32∘F, and that it boils at 100∘C, which corresponds t
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In two significant figure 360K

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3 years ago
A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut
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Answer:

1.E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2.E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

\rho (r) = \alpha (1-\frac{r}{R})

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

\int\, da = 2\pi r h

where ‘h’ is the length of the imaginary Gaussian surface.

Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3. At the boundary where r = R:

E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}

As can be seen from above, two E-field values are equal as predicted.

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