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Studentka2010 [4]

3 years ago

15

A force of 16.88 N is applied tangentially to a wheel of radius 0.340 m and gives rise to an angular acceleration of 1.20rad / (

s ^ 2) . Calculate the rotational inertia of the wheel. A. 2.77 kg - m ^ 2 B. 0.73 kg - m ^ 2 C. 4.41 kg - m ^ 2 O. 4.78 kg - m ^ 2

1 answer:

Basile [38]3 years ago

7 0

Given.

force = 16.88 N is

radius = 0.340m

an angular acceleration = 1.20rad/s^2

the formula for torque is

F*r = I*a

where I is moment of inertia

16.88*.34 = I*1.2

I = 4.78Kg-m^2

so rotational inertia I = 4.78Kg-m^2

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A 30 N rock falls from a 40 m cliff. At what point during its fall are its

vaieri [72.5K]

Answer:

<em>Both energies are equal when the rock has fallen 20 m or equivalently when it is at a height of 20 m.</em>

Explanation:

<u>Potential and Kinetic Energy</u>

The gravitational potential energy is the energy an object has due to its height above the ground. The formula is

$U=mgh$

Where:

m = mass of the object

g = acceleration of gravity (9.8~m/s^2)

h = height

Note we can also use the object's weight W=mg into the formula:

$U=Wh$

The kinetic energy is the energy an object has due to its speed:

$\displaystyle K=\frac{1}{2}mv^2$

Where v is the object's speed.

Initially, the object has no kinetic energy because it's assumed at rest.

The W=30 N rock falls from a height of h=40 m, thus:

$U=30*40=1,200 J$

Since the sum of the kinetic and potential energies is constant:

U' + K' = 1,200 J

Here, U' and K' are the energies at any point of the motion. Since both must be the same:

U' = K' = 600 J

U'=Wh'=600

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$\displaystyle h'=\frac{600}{W}=\frac{600}{30}=20~m$

Both energies are equal when the rock has fallen 20 m or equivalently when it is at a height of 20 m.

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3 years ago

Please need help fast

iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) $1.5 m/s^2$

The average acceleration of the horse can be calculated as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

$a=\frac{15-0}{10}=1.5 m/s^2$

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

$a=1.5 m/s^2$

Substituting,

$s=0+\frac{1}{2}(1.5)(10)^2=75 m$

(d) See attached graphs

In a uniformly accelerated motion:

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$x=ut+\frac{1}{2}at^2$

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

$a=1.5 m/s^2$ (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

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How is wavelength related to a waves energy?

andreyandreev [35.5K]

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Explanation:

i hope it will help u buddy✅

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djverab [1.8K]

Average speed =

(distance covered during some period of time)
divided by
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An astronaut landed on a far away planet that has a sea of water. To determine the gravitational acceleration on the planet's su

TiliK225 [7]

The concept required to solve this problem is hydrostatic pressure. From the theory and assuming that the density of water on that planet is equal to that of the earth $(1000kg / m ^ 3)$ we can mathematically define the pressure as

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3 years ago