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Harlamova29_29 [7]

3 years ago

14

The origins of Yoga can be traced back to which ancient scholar?

2 answers:

Mashcka [7]3 years ago

5 0

Answer:

Rig Veda

Explanation:

faust18 [17]3 years ago

4 0

Answer:

it can be traced to rig veda

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A car travels 554 miles south in 10.4 hours . what is the velocity of the car ? show all steps

kirill115 [55]

Answer:

<h3>The answer is 53.27 mi/hr</h3>

Explanation:

To find the velocity covered by the car we use the formula

$v = \frac{d}{t} \\$

where

d is the distance

t is the time

From the question

d = 554 miles

t = 10.4 hrs

We have

$d = \frac{554}{10.4} \\ = 53.2692307...$

We have the final answer as

<h3>53.27 mi/hr</h3>

Hope this helps you

4 0

3 years ago

In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. the dish is on a shelf above the poi

Georgia [21]

components of the speed of the coin is given as

$v_x = v cos60$

$v_x = 6.4 cos60 = 3.2 m/s$

$v_y = vsin60$

$v_y = 6.4 sin60 = 5.54 m/s$

now the time taken by the coin to reach the plate is given by

$t = \frac{\delta x}{v_x}$

$t = \frac{2.1}{3.2}$

$t = 0.656 s$

now in order to find the height

$h = vy * t + \frac{1}{2} at^2$

$h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2$

$h = 1.52 m$

so it is placed at 1.52 m height

3 0

3 years ago

How can i stop loveing you if yo keep saying the things i want to hear

xxTIMURxx [149]

Answer:

....

Explanation:

5 0

3 years ago

Read 2 more answers

A 30 g particle is undergoing simple harmonic motion with an amplitude of 2.0 ✕ 10-3 m and a maximum acceleration of magnitude 8

KiRa [710]

Answer:

Explanation:

By using the Newton second law and the position equation for a simple harmonic motion we have

$F=ma\\a_{max}=\omega^{2}A\\x=Acos(\omega t+ \phi)\\$

where a is the acceleration, w is the angular frecuency and \phi is the phase constant. We can calculate w from the equation for the maximum acceleration

$\omega=\sqrt{\frac{a_{max}}{A}}=\sqrt{\frac{8*10^{3}m/s^{2}}{2*10^{-3}}}=2000rad/s$

(a).

$F=ma=m\omega^{2}Acos(\omega t + \phi)\\F=(30*10^{-3}kg)(2*10^{-3}m)(2000\frac{rad}{s})^{2}cos(2000\frac{rad}{s} t - \frac{\pi}{2})=240N*cos(2000\frac{rad}{s} t - \frac{\pi}{2})$

(b). $T=\frac{2\pi}{\omega}=\frac{2\pi}{2000}=3.14*10^{-13} s$

(c). $v_{max}=A\omega=(2*10^{-3}m)(2000\frac{rad}{s})=4\frac{m}{s}$

(d). The mecanical energy is the kinetic energy when the velocity is a maximum

$E_{m}=E_{k}(v_{max})=\frac{mv_{max}^{2}}{2}=\frac{30*10^{-3}kg(4\frac{m}{s})^{2}}{2}=0.024J$

3 0

3 years ago

Read 2 more answers

A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.4 A runs through the wire

Deffense [45]

Answer:

$\vec{F}=0.40176 N \hat{k}$

Explanation:

To calculate the force we need to use this equation

$\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}$

where L is the total length of the wire

So in this case the small element of current is

$\vec{dl} = dx \hat{i}$

Because x is the direction of the current flow.

As is said in the problem B is such that

$\vec{B} = B \hat{j} = 0.62\hat{j} [ T]$

so to use the equation above we first calculate the following cross product:

$\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}$

so the force:

$F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}}$

So here we use the fact that B=0 in any point of the x axis that is not $x^{'}=0.27 [m]$ , that means that we only need to do the integration between a very short distant behind the point $x^{'}=0.27 [m]$ and a very short distant after that point, meaning:

$\vec{F}= \lim_{h \to 0}{\int\limits^{x^{'}+h}_{x^{'}-h}{iBdx\hat{k}} }$

so is the same as evaluating $iBx$ at $x=x^{'}$

that is:

$2,4 A * 0,62 T * 0,27 m \hat{k}$

$2,4 A * 0,62 (\frac{Kg}{A s^{2}}) * 0,27 m \hat{k}$

$2,4*0,62*2,7 ( \frac{ kgm }{ s^{2} } ) \hat{k}$

$\vec{F}=0.40176 N \hat{k}$

5 0

3 years ago