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3 years ago

What would happen to the marbles speed if we had longer tables

Physics

2 answers:

Zanzabum3 years ago

7 0

Answer:

The marble will start to decrease in speed

Explanation:

The longer the table is the more the marbles speed would decrease because marbles can only be fast for a certain amount of time

egoroff_w [7]3 years ago

5 0

If we had longer tables and if we did the ramp marble experiment the speed would increase and the marble would roll longer . .

Hope I helped -

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PLS HELP WILL MARK BRAINLIEST IF RIGHT NEED IMMEDIATELY PLEASEEEEE

timofeeve [1]

Answer:

Explanation:

Prescribe medications.

Prescribe treatments or therapies.

Treat patients using psychological therapies.

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Record patient medical histories.

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2 years ago

Read 2 more answers

An old millstone, used for grinding grain in a gristmill, is a solid cylindrical wheel that can rotate about its central axle wi

MAXImum [283]

Answer:

The answer is I=70,513kgm^2

Explanation:

Here we will use the rotational mechanics equation T=Ia, where T is the Torque, I is the Moment of Inertia and a is the angular acceleration.

When we speak about Torque it´s basically a Tangencial Force applied over a cylindrical or circular edge. It causes a rotation. In this case, we will have that T=Ft*r, where Ft is the Tangencial Forge and r is the radius

Now we will find the Moment of Inertia this way:

$Ft*r=I*a$ -> $(Ft*r)/(a) = I$

Replacing we get that I is:

$I=(200N*0,33m)/(0,936rad/s^2)$

Then $I=70,513kgm^2$

In case you need to find extra information, keep in mind the Moment of Inertia for a solid cylindrical wheel is:

$I=(1/2)*(m*r^2)$

4 0

2 years ago

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, radial ( $a_{r}$ ) and tangential ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

$\omega = \omega_{o}+ \alpha\cdot t$

$\omega = 1.2\,\frac{rad}{s}$

$a_{r} = \omega^{2}\cdot r$

$a_{r} = 0.144\,\frac{m}{s^{2}}$

$a_{t} = \alpha \cdot r$

$a_{t} = 0.2\,\frac{m}{s^{2}}$

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$

$a \approx 0.246\,\frac{m}{s^{2}}$

The total linear acceleration is approximately 0.246 meters per square second.

3 0

2 years ago

Question 9

Ilya [14]

True ! Attending the funeral of a foreign leader IS fulfilling the role of head of state.

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2 years ago

Find equation tangent to a circle at given point

vlabodo [156]

the equation of the tangent line must be passed on a point A (a,b) and perpendicular to the radius of the circle. 
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle, its center is C(0,0). And we assume that the tangent line passes to the point A(2.3).

since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle.

Let's find the equation of the line parallel to the radius.

The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).

det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.

let's find the equation of the line perpendicular to this previous line.

let M a point which lies on the line. so MA.AC=0 (scalar product),

it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent

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3 years ago