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4 years ago

Find the de Broglie wavelength of an electron whose speed is 1.0 x 10^7 m/s.

Physics

1 answer:

balu736 [363]4 years ago

5 0

Answer:

7*10^(-11) m.

Explanation:

the details are in the attached picture (the answer is marked with red).

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A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

oksano4ka [1.4K]

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

$15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm$

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

5 0

3 years ago

What is gravitational constant?<br>

WITCHER [35]

The constant in Newton's law of gravitation relating gravity to the masses and separation of particles, equal to 6.67 × 10-11N m2 kg-2.

6 0

3 years ago

A lion has a mass of 45 kg. Answer the following questions about it, using correct units. a. The lion runs at a speed of 14.2 m/

Eva8 [605]

A) The kinetic energy of an object is given by:
$K= \frac{1}{2}mv^2$
where m is the mass of the object, and v its speed. For the lion in our problem, m=45 kg and v=14.2 m/s, so its kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(45 kg)(14.2 m/s)^2=4537 J$

b) the increase in gravitational potential energy of the lion is given by:
$\Delta U = mg \Delta h$
where g is the gravitational acceleration, and $\Delta h$ is the increase in altitude of the lion. In this problem, $\Delta h=28 m$ , so the increase in gravitational potential energy is
$\Delta U=mg \Delta h=(45 kg)(9.81 m/s^2)(28 m)=12361 J$

c) When the fox reaches the top of the tree, its gravitational potential energy is
$U=mgh=(1.8 kg)(9.81 m/s^2)(3.8 m)=67 J$
As it jumps, its kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(1.8 kg)(8.1 m/s)^2=59 J$
So the total mechanical energy of the fox as it jumps is
$E=U+K=67 J + 59 J =126 J$

6 0

3 years ago

Unless otherwise authorized, the maximum indicated airspeed at which aircraft may be flown when at or below 2,500 feet AGL and w

andrew11 [14]

Answer: 200 knots

Explanation: the maximum indicated airspeed at which aircraft may be flown when at or below 2,500 feet AGL and within 4 nautical miles of the primary airport of Class C airspace is 200 KNOTS

3 0

3 years ago

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a labor

Travka [436]

Answer:

F = 183.153 N

Explanation:

given,

mass of the toothpick = 0.12 g = 0.00012 kg

initial velocity = 227 m/s

final velocity = 0 m/s

penetration depth = 16 mm = 0.016 m

using the equation of motion

v² - u² = 2 a s

0 - u² = 2 a s

- 221² = 2 × a × 0.016

a = 1526281.25 m/s²

Force is equal to

F = m a

= 0.00012 × 1526281.25

F = 183.153 N

3 0

3 years ago