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3 years ago

If an electron is accelerated from rest through a potential difference of 1.60 x 102V, what is its de Broglie wavelength

Physics

1 answer:

Ivahew [28]3 years ago

7 0

Answer:

0.09 x10^-10m

Explanation:

Using wavelength=( 12.27 A)/√V

= 12.27 x 10^-10/ √1.6x10^2

= 0.09x10^-10m

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CALCULATE: find the velocity of a wave in a wave pool if it's wavelength is 3.2 m and its frequency is 0.60 Hz

vampirchik [111]

Number 2 ( 1.3m/s) good luck!

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2 years ago

Read 2 more answers

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo

Helga [31]

Answer: $3.23 m/s^2$

Explanation:

Given

$\mu_s =0.330$

Frictional Force is balanced by force due to car acceleration

Frictional force $F_s$

$F_s=ma_{min}$

$\mu_sN=ma_{min}$

$\mu_s\cdot mg=ma_{min}$

$a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2$

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3 years ago

How much heat is removed from 60 grams of steam at 100 °C to change it to 60 grams

Harrizon [31]

Answer:

45200J

Explanation:

Given parameters:

Heat of vaporization of water = 2260J/g

Mass of steam = 20g

Temperature = 100°C

Unknown:

Energy released during the condensation = ?

Solution:

This change is a phase change and there is no change in temperature

To find the amount of heat released;

H = mL

m is the mass

L is the latent heat of vaporization

Insert the parameters and solve;

H = 20g x 2260J/g

H = 45200J

4 0

2 years ago

Calculate the number of moles of water molecules in 12 dm' of water<br>vapour at STP.<br><br>

Vinvika [58]

Answer:

$22.4 \: {dm}^{3} \: are \: occupied \: by \: 1 \: mole \\ 12 \: {dm}^{3} \: will \: be \: occupied \: by \: ( \frac{12}{22.4} ) \: moles \\ = 0.536 \: moles$

4 0

3 years ago

slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa

xenn [34]

Answer:

Maximum height reached by the rocket is

$y_{max} = 308 m$

total time of the motion of rocket is given as

$T = 16.44 s$

Explanation:

Initial speed of the rocket is given as

$v_i = 50 m/s$

acceleration of the rocket is given as

$a = 2 m/s^2$

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 50^2 = 2(2)(150)$

$v_f = 55.68 m/s$

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 55.68^2 = 2(-9.81)(y - 150)$

$y_{max} = 308 m$

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

$v_f - v_i = at$

$55.68 - 50 = 2 t$

$t_1 = 2.84 s$

2) time to reach ground from this height

$\Delta y = v_y t + \frac{1}{2}gt^2$

$-150 = 55.68 t - \frac{1}{2}(9.81) t^2$

$t_2 = 13.6 s$

so total time of the motion of rocket is given as

$T = 13.6 + 2.84 = 16.44 s$

3 0

3 years ago