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larisa [96]
3 years ago
15

What is part of a line has one endpoint and continues in one direction?

Physics
1 answer:
noname [10]3 years ago
6 0

Answer:

A Ray I believe since it sounds like it.

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zlopas [31]

Answer:

Ang answer units of J heat of fusion is 3.33 x 105

4 0
3 years ago
Match the particles with their characteristics,
In-s [12.5K]
Positive charge=proton
Negative charge=electron
No charge/neutral=neutron
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3 years ago
Read 2 more answers
A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th
ZanzabumX [31]

Answer:

Part a)

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

Mv_o = M v_1 + m v_2

here we also use angular momentum conservation

so we have

M v_o d = M v_1 d + \beta mL^2 \omega

also we know that the collision is elastic collision so we have

v_o = (v_2 + d\omega) - v_1

so we have

\omega = \frac{v_o + v_1 - v_2}{d}

also we know

M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})

also we know

v_1 = v_o - \frac{m}{M}v_2

so we have

M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})

mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}

now we have

(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

Now we know that speed of the ball after collision is given as

v_1 = v_o - \frac{m}{M}v_2

so it is given as

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

3 0
3 years ago
Please help on this one?
diamong [38]

Answer:

ITS C

Explanation:

7 0
3 years ago
A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What
Crank

Answer:

\mu = \frac{r (2\pi f)^{2}}{g}

Explanation:

N = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

N = mg

f = frequency of rotation

Angular velocity of turntable is hence given as

w = 2\pi f

r = distance from the axis of rotation

\mu = minimum coefficient of static friction

static frictional force is given as

f = \mu N\\f = \mu mg

The  static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}

3 0
3 years ago
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