A box contains about 6.77 x 1021 hydrogen

Given the vector A with components Ax = 2.00, Ay = 6.00, the vector B with components Bx = 2.00, By = 22.00, and the vector D =

nekit [7.7K]

Answer:

<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>

Explanation:

<u>Operations With Vectors</u>

Given two vectors in rectangular components:

$\vec a=(ax,ay)\ ,\ \vec b=(bx,by)$

The sum of the vectors is:

$\vec a+\vec b=(ax+bx,ay+by)$

The difference between the vectors is:

$\vec a-\vec b=(ax-bx,ay-by)$

The magnitude of $\vec a$ is:

$|\vec a|=\sqrt{ax^2+ay^2}$

The angle $\vec a$ makes with the horizontal positive direction is:

$\displaystyle \tan\theta=\frac{ay}{ax}\\$

The question provides the vectors:

$\vec a=(2,6)$

$\vec b=(2,22)$

$\vec d=\vec a-\vec b$

Calculate:

$\vec d=(2,6)-(2,22)=(0,-16)$

The magnitude of $\vec d$ is:

$|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16$

The angle is calculated by:

$\displaystyle \tan\theta=\frac{-16}{0}$

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.

The magnitude of vector d is 16 and the angle with the x-axis is 270°

4 0

3 years ago

A hydraulic lift raises 1140 kg car through a height of 2.4 m. What is the potential energy

Sladkaya [172]

1140x9.8x2.4= 26,812.8 significant figures Make it 27,000

5 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

The motion of a particle is defined by the relation x 5 t 3 2 9t 2 1 24t 2 8, where x and t are expressed in inches and seconds,

shutvik [7]

Answer:

a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m

Explanation:

This is an exercise in kinematics, the relationship of position and time is indicated

x = 5 t³ - 9t² -24 t - 8

a) ask when the velocity is zero

speed is defined by

v = $\frac{dx}{dt}$

let's perform the derivative

v = 15 t² - 18t - 24

0 = 15 t² - 18t - 24

let's solve the quadratic equation

$t = \frac{18 \pm \sqrt{18^2 + 4 \ 15 \ 24} }{2 \ 15}$

t = $\frac{18 \pm 42}{30}$

t1 = -0.8 s

t2 = 2.0 s

the time has to be positive therefore the correct answer is t = 2.0 s

b) the position and distance traveled for a = 0

acceleration is defined by

a = dv / dt

a = 30 t - 18

a = 0

30 t = 18

t = 18/30

t = 0.6 s

we substitute this time in the expression of the position

x = 5 0.6³ - 9 0.6² - 24 0.6 - 8

x = 1.08 - 3.24 - 14.4 - 8

x = -24.56 m

we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s

the position for t = 0

x₀ = -8 m

the position for t = 0.6 s

x_f = - 24.56 m

the distance

ΔX = x_f - x₀

Δx = | -24.56 -(-8) |

Δx = 16.56 m

5 0

3 years ago

What affect does an increase in the angle of incidence have upon the angle of reflection?

valentina_108 [34]

????????????? ??????????????????? ?????????????????

6 0

3 years ago