Ask question

Ask question

All categories

valentinak56 [21]

2 years ago

8

The 1kg rock is tied to a string and swung in a circular path as shown. The 1 meter string is tied to a post, and during the mot

ion, the string has a 30 angle with the post. The rock makes 100 rounds in 1 minute. The centripetal force on the rock is

1 answer:

Ratling [72]2 years ago

4 0

Answer: 54.8

Explanation:

You might be interested in

Pls help will give brainlest

Karo-lina-s [1.5K]

Respon

lqiudos ciopatmibes

ly apsamtios ccoriendor sabe r

llpop

io.

7 0

2 years ago

Read 2 more answers

Body composition refers to:

NISA [10]

Bone, fat and muscle

3 0

2 years ago

Need help asap pls

Zinaida [17]

B. third

for every action there is a reaction*

6 0

1 year ago

what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo

Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are; </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

$The \ moment \ of \ inertia \ of \ uniform \ sphere \ is \ I = \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}$

Which gives;

$\mathbf{I_{Earth}} = \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}$

$\mathrm{The \ rotational \ kinetic \ energy \ is} \ E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}$

$\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}$

Therefore;

$\omega = \dfrac{2 \cdot \pi}{24} = \dfrac{\pi}{24}$

Which gives;

$\mathbf{E_{rotational}} = \dfrac{1}{2} \times 9.7202107 \times 10^{37} \times \left( \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}$

The rotational kinetic energy of the Earth, $E_{rotational}$ = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

<em>The moment of inertia from part A of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

3 0

2 years ago

Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.84 keV from the co

ElenaW [278]

Answer:

λ = 65.6 pm

Explanation:

Given that

λo = 65 pm

The initial energy of the electron

$E_o=\dfrac{hC}{\lambda_0 }$

Now by putting the values

$E_o=\dfrac{hC}{\lambda_0 }$

$E_o=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{65\times 10^{-12}}$

$E_o=3.05\times 10^{-15}\ J$

$E_o=\dfrac{3.05\times 10^{-15}}{1.6\times 10^{-19}\times 10^3}\ KeV$

Eo=19.06 KeV

Given that kinetic energy KE= 0.84 KeV

Therefore the final energy

E= Eo - KE

E = 19.06 - 0.84 KeV

E= 18.22 KeV

The wavelength λ can be find as

$E=\dfrac{hC}{\lambda}$

$\lambda=\dfrac{hC}{E}$

$\lambda=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{19.06 \times 10^3\times 1.6\times 10^{-19}}$

λ = 6.56 x 10⁻¹¹ m

λ = 65.6 pm

3 0

3 years ago