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3 years ago

How many leptons are in ⁷₃Li?

Physics

1 answer:

shutvik [7]3 years ago

5 0

-- Electrons are leptons. There are <em>three</em> electrons in each neutral Lithium atom.

The last two parts of the question are absurd.

-- Bonbons are candy, not atomic particles. A bonbon cannot fit into a Lithium atom.

-- A pentagon is a closed geometric figure that has five sides. Although you could, in principle, have a pentagon small enough to fit into a Lithium atom, you could never find a piece of paper small enough to draw it on.

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What is the gauge pressure in Pascals inside a honey droplet of a 0.1 cm diameter? Assume that air is surrounding this droplet a

Vikki [24]

Answer:

The gauge pressure in Pascals inside a honey droplet is 416 Pa

Explanation:

Given;

diameter of the honey droplet, D = 0.1 cm

radius of the honey droplet, R = 0.05 cm = 0.0005 m

surface tension of honey, γ = 0.052 N/m

Apply Laplace's law for a spherical membrane with two surfaces

Gauge pressure = P₁ - P₀ = 2 (2γ / r)

Where;

P₀ is the atmospheric pressure

Gauge pressure = 4γ / r

Gauge pressure = 4 (0.052) / (0.0005)

Gauge pressure = 416 Pa

Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa

5 0

3 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

A satellite orbits Earth. The only force on the satellite is the gravitational force exerted by Earth. How does the satellite’s

bulgar [2K]

Answer:

here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Explanation:

As we know that gravitational field is defined as the force experienced by the satellite per unit of mass

so we will have

$E = \frac{F}{m}$

now in order to find the acceleration of the satellite we know by Newton's II law

$F = ma$

so we will have

$a = \frac{F}{m}$

so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

4 0

3 years ago

Read 2 more answers

The rest deltoid row is a back exercise true or false

bixtya [17]

False because your deltoids are in your shoulders not your back

3 0

3 years ago

Fa car's power output is increased, its efficiency:

Inessa [10]

It’s solved by using a pretty standard formula for efficiency.

4 0

3 years ago