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alina1380 [7]
3 years ago
6

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
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