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2 years ago

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Consider three axes of rotation for a pencil: along the lead, at right angles to the lead at the middle, and at right angles to

the lead at one end. Rate the rotational inertias about each axis from smallest to largest.

1 answer:

tensa zangetsu [6.8K]2 years ago

4 0

Answer:

along the lead, at right angles to the lead at the middle, and at right angles to the lead at one end.

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At 15°C air is transmitted <br>at 340 m/s. Express this speed<br>in Kilometers per hour.

EleoNora [17]

Answer:

1224km/hr

Explanation:

To convert from m/s to km/hr

1000m = 1km

Divide both sides by 1000

1m = 1/1000 km................. (1)

60×60 seconds = 1 hr

3600s = 1hr

Divide both sides by 3600

1s = 1/3600 .............(2)

Divide (2) by (1)

1m/s = 1/1000 ÷ 1/3600 km/hr

1m/s = 1/1000 × 3600/1 km/hr

1m/s = 3600/1000 km/hr

1m/s = 3.6 km/hr .............(3)

To convert 340m/s to km/hr

Multiply (3) by 340

1× 340m/s = 3.6 × 340 km/hr

340m/s = 1224km/hr

I hope this was helpful, please mark as brainliest

7 0

3 years ago

Read 2 more answers

Identify two factors you must know to describe the motion of an object along a straight line

Sveta_85 [38]

In order to describe motion along a straight line, you must state the speed and direction of the motion. Those two quantities, together, comprise what's known as "velocity".

6 0

2 years ago

Which best describes the current atomic model?

Lorico [155]

<span>A: It is not an exact representation of the atom, but is close enough to be very useful.

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2 years ago

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Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

Ne4ueva [31]

Answer:

C

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

t = v / a

t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

t = v / g

t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0

2 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago