All categories

3 years ago

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.

You might be interested in

How much energy is required to change a 44 g

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

6 0

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