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2 years ago

The knights of labor do not believe in arbitration but believed in strikes? true or false

Physics

1 answer:

Rus_ich [418]2 years ago

5 0

Answer - The given statement is false.

Explanation-

The members of knight of labour were opposing strikes and had a belief or boycott or arbitration.

A secret society was formed in 1869 by the tailors in Philadelphia and named Knights of labour. This union Organisation was formed to to get rights over equal pay if work is equal, abolish child labour, Work time of 8 hours and some graduated income tax reforms. This union was vertically organized.

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What is the definition of Rock Strata and Law of Original Horizontality?

timurjin [86]

It's called the Principle of Original Horizontality
 it just means what it sounds like: that all rock layers were originally horizontal.
Of course, it only applies to sedimentary rocks.
Recall that sedimentary rock is composed of sediments, which are deposited and compacted in one place over time.

3 0

3 years ago

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

4 0

3 years ago

What equations would you use to find force?

Klio2033 [76]

Answer:

F = m x a

Explanation:

F = force

m = mass of an object

a = acceleration

6 0

2 years ago

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

2 years ago

Compute the torque about the origin of the gravitational force F--mgj acting on a particle of mass m located at 7-xî+ yj and sho

Andrews [41]

Answer:

Explanation:

Force, F = - mg j

r = - 7x i + y j

Torque is defined as the product f force and the perpendicular distance.

It is also defined as the cross product of force vector and the displacement vector.

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

$\overrightarrow{\tau }=(- 7 x i + yj)\times (-mgj)$

[tex]\overrightarrow{\tau }= 7 m g x k

Here, we observe that the torque is independent of y coordinate.

3 0

2 years ago