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andrew11 [14]
2 years ago
3

How much work is needed to assemble an atomic nucleus containing three protons (such as Li) if we model it as an equilateral tri

angle of side 2.00×10−15m2.00×10−15m with a proton at each vertex? Assume the protons started from very far away.
Physics
1 answer:
Gennadij [26K]2 years ago
8 0

Answer:

2.1576 MeV

Explanation:

r = Distance = 2\times 10^{-15}\ m

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q = Charge of electron = 1.6\times 10^{-19}\ C

The resulting energy of the system is

U=\dfrac{k3q^2}{r}\\\Rightarrow U=\dfrac{8.99\times 10^9\times 3(1.6\times 10^{-19})^2}{2\times 10^{-15}}\\\Rightarrow U=3.45216\times 10^{-13}\ J

Converting to MeV

\dfrac{3.45216\times 10^{-13}}{1.6\times 10^{-13}}=2.1576\ MeV

The work needed to assemble an atomic nucleus is 2.1576 MeV

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The standing vertical jump is a good test of an athlete's strength and fitness. The athlete goes into a deep crouch, then extend
SashulF [63]

Answer:

<em>The athlete will rise 1.10 meters off the ground</em>

Explanation:

<u>Vertical Motion</u>

If an object is launched vertically upwards at an initial speed vo, then it will reach a maximum height given by

\displaystyle y_m=\frac{v_o^2}{2g}

The athlete can exert a net force upwards equal to twice his weight. It makes him accelerate upwards at

\displaystyle a=\frac{F_n}{m}=\frac{2W}{m}=2g

The speed at the end of his push can be computed by

v^2=2ay

Replacing the value of a obtained above:

v^2=4gy

where y is the length of this crouch

v^2=4\cdot 9.8\cdot 0.55

v=4.64\ m/s

This is the initial speed of this vertical launch, thus

\displaystyle y_m=\frac{4.64^2}{2\cdot 9.8}

y_m=1.10\ m

5 0
2 years ago
Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magneti
san4es73 [151]

A) 1.36\cdot 10^{-4}T

The magnetic field at the center of a coil of N turns is given by

B=\frac{\mu_0 N I}{2R}

where

I is the current in the coil

N is the number of turns

R is the radius of the coil

Here we have

I = 6.5 A is the current in the coil

N = 100 is the number of turns

R=\frac{6.0 m}{2}=3.0 m is the radius of the coil

Substituting,

B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T

B) The force points north

The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the ion --> towards east

- Middle finger: direction of magnetic field --> downward

- Thumb: direction of the force --> towards north

So, the force points north.

C) 3.26\cdot 10^{-23}N

The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is

F=qvB

where

q is the charge of the particle

v is the velocity

B is the magnitude of the magnetic field

In this case, we have

q=+e=1.6\cdot 10^{-19} C is the charge

v=1.5 m/s is the velocity

B=1.36\cdot 10^{-4}T is the magnetic field strength

Substituting,

F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N

8 0
2 years ago
Two friends, barbara and neil, are out rollerblading. with respect to the ground, barbara is skating due south at a speed of 5.9
Semmy [17]
<span>As seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west. Let's assume that both Barbara and Neil start out at coordinate (0,0) and skate for exactly 1 second. Where do they end up? Barbara is going due south at 5.9 m/s, so she's at (0,-5.9) Neil is going due west at 1.4 m/s, so he's at (-1.4,0) Now to see Neil's relative motion to Barbara, compute a translation that will place Barbara back at (0,0) and apply that same translation to Neil. Adding (0,5.9) to their coordinates will do this. So the translated coordinates for Neil is now (-1.4, 5.9) and Barbara is at (0,0). The magnitude of Neil's velocity as seen by Barbara is sqrt((-1.4)^2 + 5.9^2) = sqrt(1.96 + 34.81) = sqrt(36.77) = 6.1 m/s The angle of his vector relative to due west will be atan(5.9/1.4) = atan(4.214285714) = 76.7 degrees So as seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.</span>
5 0
2 years ago
Read 2 more answers
How could you test the hypothesis that elephants interpreted the ground signal as being farther away than the air signal?
bekas [8.4K]

Answer:

One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.

Explanation:

This hypothesis is based on the fact that the speed of sound in air is v = 343 m / s with a small variation with temperature.

The speed of sound in solid soil is an average of the speed of its constituent media, giving values ​​between

 wood      3900 m / s

 concrete 4000 m / s

 fabrics     1540 m / s

 earth       5000 m / s wave S

 ground    7000 m / s P wave

 

we can see that the speed on solid earth is an order of magnitude greater than in air.

One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.

From the initial information, the wave going through the ground should arrive first.

3 0
2 years ago
Nanotechnology and microtechnology devices are both very tiny. However, one property of nanotechnology distinguishes it from mic
jek_recluse [69]
Nanotechnology displays particle quality
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