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3 years ago

Calculate the de Broglie wavelength of an electron traveling at 2.0 x 10^8 m/s (me = 9.1*10^-31 kg).

Physics

1 answer:

timofeeve [1]3 years ago

4 0

Answer:

$\lambda=3.64\times 10^{-12}\ m$

Explanation:

We have,

Speed of an electron is $2\times 10^8\ m/s$

It is required to find the De Broglie wavelength of electron. The formula for the De- Broglie wavelength is given by :

$\lambda=\dfrac{h}{mv}$

h is Planck's constant

m is mass of an electron

Plugging all the values we get :

$\lambda=\dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 2\times 10^{8}}\\\\\lambda=3.64\times 10^{-12}\ m$

So, the De-Broglie wavelength of an electron is $3.64\times 10^{-12}\ m$

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The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet

algol13

Answer:

g₂ = 11 m/s²

Explanation:

The value of free-fall acceleration on the surface of a planet is given by the following formula:

$g = \frac{Gm}{r^2}$

where,

g = free-fall acceleration

G = Universal Gravitational Constant

m = mass of the planet

r = radius of planet

FOR PLANET 1:

$g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2$ --------------------- equation (1)

FOR PLANET 2:

$g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\$

using equation (1):

$g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}$

g₂ = 11 m/s²

8 0

2 years ago

Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript

Pavel [41]

Answer:

Part a)

$F_A = 4.59 N$

Part B)

$F_B = 1.28 N$

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

$F_A + F_B = ma$

as we know that

$a = 0.554 m/s^2$

m = 10.6 kg

now we will have

$F_A + F_B = 10.6(0.554)$

$F_A + F_B = 5.87 N$

Now two forces are in opposite direction then we have

$F_A - F_B = 10.6(0.313)$

$F_A - F_B = 3.32 N$

Part A)

Now we will have from above two equation

$F_A = 4.59 N$

Part B)

Similarly for other force we have

$F_B = 1.28 N$

5 0

2 years ago

Describe a solar eclipse. Be sure to include the positions of the sun, moon, and earth.

Alex777 [14]

It’s the type of eclipse that occurred when the moon passes between the sun and earth, and when the moon fully or partially blocks the sun.

7 0

2 years ago

A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr

Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration $a=g=9.8 m/s^2$ directed downward, initial vertical position $d=750 m$ , and initial vertical velocity $v_0 = 0$ . We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

$v_f^2 -v_i^2 =2ad$

substituting, we find

$v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s$

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

$v_f=\frac{121.2 m/s}{2}=60.6 m/s$

In order to do that, we use again the same SUVAT equation substituting $v_f$ with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

$v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m$

Which means that the heigth of the packet was

$h=750 m-187.4 m=562.6 m$

3 0

2 years ago

A space station, in the form of a wheel 119 m in diameter, rotates to provide an "artificial gravity" of 2.20 m/s2 for persons w

Rina8888 [55]

Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
 a=v^2/r
v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)= 11.44 m/s.
 After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
 T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= 32.68 rev/s
 Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min).
(32.68 rev/s)(60 s/min)= 1960.74 rev/min

5 0

3 years ago