A wave can be best defined as P.S pls help

Minchanka [31]

You have to go to settings and there will be options im guessing!

3 0

3 years ago

A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1

GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

$\large{\maltese{\textsf{\underline{To find :-}}}}$

The X and Y coordinates of the rocket relative of firing

⠀

$\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s$

⠀

$\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}$

⠀

The horizontal range of projectile at x.

⠀

$\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}$

⠀

$\large\textsf{\underline{Now substituting the required values}}$

⠀

$\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}$

⠀

The vertical position of projectile at y.

⠀

$\sf \large y = v_{yi} \times t - (\frac{1}{2} \times g \times {t}^{2}) \\ \\ \sf \large y = v_i \times \cos \theta \times t - \frac{1}{2} g {t}^{2}$

⠀

$\textsf{ \large {\underline{Now substituting the required values}} }$

⠀

$\sf y = 100 \times \cos55{ \degree} \times 12 - \frac{1}{2} \times 9.80 \times {12}^{2} \\ \\ \sf y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\ \\ \sf y = 983.04 - 705.6 \\ \\ \underline{ \boxed{ \bold{y = 277.44m}}}$

⠀

<h3>Henceforth, the distance at horizon is 688.32m and at vertical is 277.44m.</h3>

8 0

2 years ago

An oxygen molecule consists of two oxygen nuclei, each of massm= 2.7×10−26kg,separated by a distance 1.2×10−10m, and surrounded

fenix001 [56]

Answer:

a) I = 1,944 10⁻⁴⁶ Kg m²

, b) I = 7,915 10⁻⁵¹ Kg m²

Explanation:

a) The moment of inertia of point masses is

I = m r²

The nuclei have a very small size (10-15 m) so we can consider them punctual.

The distance to the center of mass passing through the middle of the two nuclei of equal mass

r = 0.6 10⁻¹⁰ m

The moment of total inertia

I = I₁ + I₂ = 2 I₀

I = 2 2.7 10⁻²⁶ (0.6 10⁻¹⁰)²

I = 1,944 10⁻⁴⁶ Kg m²

The kinetic energy of the rotation is

w = h / 2π

K = ½ I w²

K = ½ 1,944 10⁻⁴⁶ (h / 2π)² = ½ 1.944 10⁻⁴⁶ (6.63 10⁻³⁴ / 2π)²

K = 2.16 10⁻¹¹⁴ J (1eV / 1.6 10⁻¹⁹ J)

K = 2.16 10⁻⁹⁵ eV

B) the moment of inertia of the electron in the orbit, we can calculate it with the parallel axis theorem

I = $I_{cm}$ + m R²

I = $m_{e}$ r² + $m_{e}$ R²

Where R we calculate it by Pythagoras

R² = (0.6 10⁻¹⁰)² + (0.5 10⁻¹⁰) 2

R = √ (0.61 10⁻²⁰)

R = 0.78 10⁻¹⁰ m

I = 9.1 10⁻³¹ (0.5 10⁻¹⁰)² + 9.1 10⁻³¹ (0.78 10⁻¹⁰)²

I = (2,275 +5.54) 10⁻⁵¹

I = 7,915 10⁻⁵¹ Kg m²

The rotation energy of the electron

K = ½ I w²

If the angular velocity is the electrons outside the core, its kinetic energy is much lower by an order 10⁵, but the angular velocity of the electrons is much higher.

7 0

3 years ago

Scanned with CamScanner

baherus [9]

The answer is c to take a pictures with the camera in the metal hall

8 0

3 years ago

The Sahara. The largest desert_____________________ on the planet. A searing wilderness the size of the _____________________. T

SashulF [63]

Answer:

1. Desert

2. United States

3. Explorer

Explanation:

Given the nature of the Sahara desert, it is believed to cover about 9.4 million kilometer square. Thereby is about the same size as the United States including the region of Alaska and Hawaii.

Because of its level of hotness, it is considered very challenging to explorers.

Therefore, it can be concluded that The Sahara. The largest desert DESERT on the planet. A searing wilderness the size of the UNITED STATES The toughest challenge an EXPLORER can face.

5 0

3 years ago