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geniusboy [140]

3 years ago

12

Suppose you are asked to find the amount of time t, in seconds, it takes for the turntable to reach its final rotational speed.

Which of the following equations could you use to directly solve for the numerical value of t?

1 answer:

Inga [223]3 years ago

5 0

Answer:

it will take $t=\frac{\omega -\omega _0}{\alpha }$ to reach its final rotating speed

Explanation:

Let initially the turntable is at rest

So its initial angular velocity will be $\omega _0=0rad/sec$

Let the turn table is rotating with an angular acceleration $\alpha$

According to first equation of motion

We know that $\omega =\omega _0+\alpha t$ , here $\omega$

So $t=\frac{\omega -\omega _0}{\alpha }$

So it will take $t=\frac{\omega -\omega _0}{\alpha }$ to reach its final rotating speed

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A hoodlum throws a stone vertically downward with an initial speed of 12.0 m/s from the roof of a building,30.0 m above the grou

bogdanovich [222]

S=ut +1/2at2
30=12t+1/2*10*t2
5t2+12t-30=0

8 0

3 years ago

The equation below is for potassium oxide.

ValentinkaMS [17]

Answer: The ratio of atoms of potassium to ratio of atoms of oxygen is 4:2

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed, and remains conserved. The mass of products must be same as that of the reactants.

Thus the number of atoms of each element must be same on both sides of the equation so as to keep the mass same and thus balanced chemical equations are written.

K exists as atoms and oxygen exist as molecule which consists of 2 atoms. The ratio of number of atoms on both sides of the reaction are same and thus the ratio of atoms of potassium to ratio of atoms of oxygen is 4:2.

5 0

3 years ago

Read 2 more answers

How expensive is genetic counseling and reproductive technology?

ExtremeBDS [4]

Answer:

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hope it helps, please give me brainliest.

8 0

2 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago

The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25

Vikki [24]

Answer:

375 m.

Explanation:

From the question,

Work done by the frictional force = Kinetic energy of the object

F×d = 1/2m(v²-u²)..................... Equation 1

Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.

Make d the subject of the equation.

d = 1/2m(v²-u²)/F.................. Equation 2

Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.

Note: If is negative because it tends to oppose the motion of the object.

Substitute into equation 2

d = 1/2(60)(0²-25²)/-50

d = 30(-625)/-50

d = -18750/-50

d = 375 m.

Hence the it will slide before coming to rest = 375 m

6 0

3 years ago