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2 years ago

Which of these items would be a renewable resource?

1 answer:

8 0

Trees can be a renewable source , we use them to breath , we make paper out of them , we make houses and etc.

i hope this helps

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An old grindstone, used for sharpening tools, is a solid cylindrical wheel that can rotate about its central axle with negligibl

krok68 [10]

(a) The moment of inertia of the wheel is 78.2 kgm².

(b) The mass (in kg) of the wheel is 1,436.2 kg.

<h3>Moment of inertia of the wheel</h3>

Apply principle of conservation of angular momentum;

Fr = Iα

where;

F is applied force
r is radius of the cylinder
α is angular acceleration
I is moment of inertia

I = Fr/α

I = (200 x 0.33) / (0.844)

I = 78.2 kgm²

<h3>Mass of the wheel</h3>

I = ¹/₂MR²

where;

M is mass of the solid cylinder
R is radius of the solid cylinder
I is moment of inertia of the solid cylinder

2I = MR²

M = 2I/R²

M = (2 x 78.2) / (0.33²)

M = 1,436.2 kg

<h3>Angular speed of the wheel after 4 seconds</h3>

ω = αt

ω = 0.844 x 4

ω = 3.376 rad/s

Thus, the moment of inertia of the wheel is 78.2 kgm².

The mass (in kg) of the wheel is 1,436.2 kg.

The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

Learn more about moment of inertia here: brainly.com/question/14839816

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7 0

1 year ago

In a circuit a battery is used to

Anarel [89]

Produce power through the circuit.

7 0

3 years ago

Read 2 more answers

Which explains why carbon (C) is such a versatile element?

raketka [301]

<span>It can form four covalent bonds. </span>

8 0

2 years ago

A student pushes on a 5 kg box with a force of 20 N forward. The force of sliding friction

mart [117]

Answer:

Explanation:

F=20N

mu=-10N

F=ma

20-10=10

10=(5)(a)

a=2 m/s^2

4 0

2 years ago

Two objects carry initial charges that are q1 and q2, respectively, where |q2| > |q1|. They are located 0.160 m apart and beh

mart [117]

Answer:

$\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges $\rm q_1$ and $\rm q_1$ , separated by a distance $\rm r$ , is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where k is the Coulomb's constant.

Initially,

$\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.$

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

$\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).$

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, $\rm F_f = +1.30\ N.$

The new charges on the two objects are

$\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.$

The new force is given by

$\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\$

Using (1),

$\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0$

$\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0$

$\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.$