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3 years ago

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Roots show ____. geotropism because they grow toward gravitational pull. 1.) positive 2.) neutral 3.) negative 3.)

1 answer:

solniwko [45]3 years ago

8 0

Its the second option if you have Suttons Bay Virtual Online School

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A small sphere attached to a light rigid rod rotates about an axis perpendicular to and fixed to the other end of the rod. relat

Crank

<span>d.rotating counterclockwise and slowing down This is a matter of understanding the notation and conventions of angular rotations. Positive rotations are counter clockwise and negative rotations are clockwise. An easy way to remember this is the "right hand rule". Make a closed fist with your right hand and have the thumb sticking outwards. If you orient your thumb such that it's pointing in the direction of the positive value along the axis, your fingers will be curled in the positive rotational direction. So in the described scenario, the sphere is rotating in the positive direction (counter clockwise) and decelerating due to the negative angular acceleration. That immediately indicates that options "a", "b", and "e" are wrong since they mention the sphere going clockwise at the beginning. Of the two remaining options "c" and "d", we can discard option "c" since it has the rotation speeding up, and that leaves us with option "d" where the sphere is rotating counter clockwise and slowing down.</span>

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4 years ago

Hecto the Mornar a fursa explain i magnituse of the force acting right angle to the moment arm

mixer [17]

<h2>~<u>Solution</u> :-</h2>

Here, the <u>moment arm</u> is defined as follows;

The magnitude of two forces, which when acting at right angle produce resultant force of VlOkg and when acting at 60° produce resultant of Vl3 kg. These forces are D. gravitational force of attraction towards the centre of the earth. A sample of metal weighs 219 gms in air, 180 gms in water, 120 gms in an <em>unknown fluid</em>.

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7 0

3 years ago

Help pls i’m not sure what to do

Sav [38]

Answer:

I DONT KNOW WHAT TO DO SORRY

Explanation:

EVEN ME IM NOT SURW

7 0

3 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the <em>x</em>-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its <em>y</em>-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

3 years ago

Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. T

alex41 [277]

Answer:

30643 J

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

t = Time taken = 1 ns

c = Speed of light = $3\times 10^8\ m/s$

$E_0$ = Maximum electric field strength = $1.52\times 10^{11}\ V/m$

A = Area = $1\ mm^2$

Magnitude of magnetic field is given by

$B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T$

Intensity is given by

$I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2$

Power, intensity and time have the relation

$E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J$

The energy it delivers is 30643 J

4 0

3 years ago