How electricity has changed what we do for entertainment?

A 8 kg eagle is flying up in the sky. You pull out your GPE gun and are able to tell that the bird has a GPE of 2,352 J. How hig

12345 [234]

Dont know bout this one my guy

4 0

3 years ago

What tells scientists about distant galaxies? a. How fast light travels c. The color of light that they give off b. If the are o

tamaranim1 [39]

I Believe the answer is A but D looks good too.

8 0

3 years ago

The part of the computer that provides access to the internet is

Allisa [31]

The Modem

or physically connecting on the computer the data port is for a cable to connect

5 0

3 years ago

A cube is 4.4 cm on a side, with one corner at the origin. Part 1 (a) What is the unit vector pointing from the origin to the di

Sidana [21]

Answer:

(a) $\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) $\theta = 85.44^{\circ}$

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

$\hat{A} = \frac{\vec{A}}{|A|}$

$\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}$

$\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}$

$\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) To calculate the angle between the two vectors say A and A' is given by:

$\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta$

$\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})$ (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

$\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$

Now,

Using equation (1) :

$\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})$

$|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261$

Thus

$\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})$

$\theta = cos^{- 1}(0.07946) = 85.44^{\circ}$

4 0

3 years ago

Recall some things you already know about projectile motion. Does a force in the vertical direction affect the horizontal compon

azamat

Answer:

vertical force cannot change the velocity on the x-axis. t =x/v₀ₓ

Explanation:

The force is a vector magnitude, so the forces on the x-axis affect the acceleration on this axis. Consequently a vertical force cannot change the velocity on the x-axis.

$F_{y}$ = m g

Fₓ = 0

The horizontal velocity in projectile motion is constant, if we neglect the air resistance, so it can be used to find the time of a horizontal displacement

x = v₀ₓ t

t =x/v₀ₓ

The only magnitude that is the same for both movements is the time that is a scalar

7 0

3 years ago