Why is it unhelpful to compare your life to the lives others portray on social media?

A positron has a mass of 9.11 x 10^-31 kg, and charge qp = +e = +1.60 x 10^-19 C. It is moving towards an α particle (qα = +2e,

OlgaM077 [116]

Answer:

Part a)

$v = 2.25 \times 10^6 m/s$

Part b)

$r = 0.72 \times 10^{-10} m$

Explanation:

As we know that alpha particle remain at rest always so the energy of system of positron and alpha particle will remain constant

So we will have

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

here we know that

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = 2(1.6 \times 10^{-19}) C$

also we have

$r_1 = 2.00 \times 10^{-10} m$

$r_2 = 1.00 \times 10^{-10} m$

$v_1 = 3.00 \times 10^6 m/s$

$m = 9.11 \times 10^{-31} kg$

now from above equation we have

$2.304 \times 10^{-18} + 4.0995\times 10^{-18} = 4.608 \times 10^{-18} + \frac{1}{2}(9.11 \times 10^{-31})v^2$

$v = 2.25 \times 10^6 m/s$

Part b)

Now when it come to rest then again by energy conservation we can say

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now here final speed will be zero

$2.304 \times 10^{-18} + 4.0995\times 10^{-18} = \frac{(9/times 10^9)(1.6\times 10^{-19})(2\times 1.6 \times 10^{-19})}{r_2}$

$r = 0.72 \times 10^{-10} m$

7 0

3 years ago

Is there any way u can help me

dedylja [7]

Answer:

get help with your work

try understand your work

ask your teacher for assistance or class

or maybe cheat but understand the work first if you wanna

8 0

3 years ago

Viscosity: Blood is about 4 to 5 times ____________ viscous than water. Viscosity of blood depends upon the amount of dissolved

kherson [118]

Answer:more, increased

Explanation:

Blood is four to five times more viscous than water.

Viscosity depends upon dissolved substances in blood and it increases as the amount is increases or if the fluid in the blood decreases.

Viscosity is the resistance offered by liquid to the flow and it also depends upon temperature and it decreases with increase in temperature.

8 0

3 years ago

2. A string with a length of 0.9m that is fixed at both ends

aliina [53]

Answer:

2a.) Wavelength = 1.8 m

2b.) F = 66.67 Hz

3a.) Find the attached file

3b.) Wavelength = 0.6 m

Explanation:

Given that the

Length L = 0.9m

Wavelength (λ) = 2L/n

Where n = number of harmonic

If n = 1, then

Wavelength (λ) = 2L = 2 × 0.9 = 1.8 m

b.)

If waves travel at a speed of 120m/s on this string, what is the frequency

associated with the longest wave (first harmonic)?

Given that V = 120 m/s

V = Fλ

But λ = 2L, therefore,

F = V/2L

F = 120/1.8

F = 66.67 Hz

3. b.) If there are two node, the position will be in 3rd position which is 3rd harmonic

Using the same formula,

Wavelength (λ) = 2L/n

Where n = 3

Wavelength (λ) = 2 × 0.9/3

Wavelength (λ) = 0.6 m

3 0

3 years ago

The moon's illumination changes in a periodic way that can be modeled by a trigonometric function. On the night of a full moon,

kupik [55]

Answer:

$L(t) = 0.125 (cos (\frac{\pi}{14.765}(t+7)) + 0.125$

Explanation:

The expression for the trigonometric function is :

L(t) = A (cos (B(t - C)))+ D ----- equation (1)

where ;

$A = \frac{max-min}{2}$

$A = \frac{0.25-0}{2}$

A = 0.125

D = $\frac{0+.025}{2}$

D = 0.125

Period of the lunar cycle = 29.53

Then;

$\frac{2 \pi}{B} = 29.53$

$29.53 \ \ B = 2 \pi$

$B = \frac{2 \pi}{29.53}$

$B = \frac{\pi}{29.53}$

Also; we known that December 25 is 7 days before January 1.

Then L(-7) = 0.025

Plugging all the values into trigonometric function ; we have:

$0.125 ( cos ( \frac{\pi}{14.765}((-7)-C)))+0.125 = 0.25 \\ \\ \\ ( cos ( \frac{\pi}{14.765}((-7)-C))) = \frac{0.25-0.125}{0.125}$

$( cos ( \frac{\pi}{14.765}((-7)-C))) = 1$

$( \frac{\pi}{14.765}((-7)-C))= cos^{-1} (1)$

$}((-7)-C))=0$

$C= -7$

$L(t) = 0.125 (cos (\frac{\pi}{14.765}(t-(-7))) + 0.125$

$L(t) = 0.125 (cos (\frac{\pi}{14.765}(t+7)) + 0.125$

7 0

3 years ago