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3 years ago

At distances greater than 2 femtometers, the nuclear strong force is stronger than (select all that apply):

Physics

2 answers:

Natasha2012 [34]3 years ago

8 0

Answer:

None of the above

Explanation:

As we know that radius of the nuclei is given as

$r = 1.75 Fm$

so nuclear force between the nucleons exist only between the nucleons which present inside the nucleus.

Since nuclear force depends on the distance between the nucleons so here we can say that as the distance is more than this order of distance i.e. 1.75 fm then in that case the nuclear force is negligibly small and its magnitude is smaller than electromagnetic force and gravitational force both.

So we can say that for all distance range which is more than 2 fm the nuclear force is small or negligible.

So correct answer will be

None of the above

grigory [225]3 years ago

3 0

The correct answer is
<span>nuclear weak force
</span><span>
In fact, the nuclear strong force has a very limited range of action (approximately below 1-2 femtometers), and its magnitude decays exponentially for larger distances. Only the nuclear weak force is weaker than the nuclear strong force at large distance, while the gravity and the electromagnetic force become dominant for distances > 2 femtometers.</span>

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A robot is on the surface of Mars. The angle of depression from a camera in the robot to a rock on the surface of Mars is 13.69

ra1l [238]

Answer:

The distance between the camera and the rock is 836.6 cm

Explanation:

A right triangle is formed where the hypotenuse (h) is the distance between the rock and the camera. One of the leg (l) is the distance between the camera and the surface. The angle between the hypotenuse and this leg is α = 90° - 13.69° = 76.31°. By definition:

cos α = adjacent/hypotenuse

cos(76.31) = 198.0/h

h = 198.0/cos(76.31)

h = 836.6 cm

3 0

4 years ago

A woman experiences an electrical shock. The electrons making the shock come from the A woman experiences an electrical shock. T

Zinaida [17]

The electrons making the shock come from the women's body.

<h3>What is Electric shock ?</h3>

When a high voltage current flows through the body, electrical shock results. When someone unintentionally touches an electrical source, this typically occurs. Treatment for both internal and exterior burns may be necessary as part of the aftercare.

The nervous system may be impacted by a shock.

The tissue that makes up nerves presents extremely minimal resistance to the flow of an electric charge. Electric shocks that impact nerves can cause pain, tingling, numbness, weakness, or trouble moving a limb. These effects might disappear with time or remain for good.

How to Prevent from Electrical Shocks –

Keep the Appliances Away from Moisture and Water.
Never Connect or Disconnect Under Load.
Be Careful with Capacitors.
Use Insulated Tools.
Turn Off the Power.
Check for Improper or Faulty Wiring.
Fix Extension Cord Problems.

to learn more about electric shock go to - brainly.com/question/8822505

#SPJ4

7 0

2 years ago

A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp

maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance $D$ , which is divided into two segments $d=10 miles$ :

$D=d+d=2d$ (1)

Where $d=10mi \frac{1609.34 m}{1 mi}=16093.4 m$

On the other hand, we know speed is defined as:

$S=\frac{d}{t}$ (2)

Where $t$ is the time, which can be isolated from (2):

$t=\frac{d}{S}$ (3)

Now, for the first segment $d=16093.4 m$ the car has a speed $S_{1}=22m/s$ , using equation (3):

$t_{1}=\frac{d}{S_{1}}$ (4)

$t_{1}=\frac{16093.4 m}{22m/s}$ (5)

$t_{1}=731.518 s$ (6) This is the time it takes to travel the first segment

For the second segment $d=16093.4 m$ the car has a speed $S_{1}=30m/s$ , hence:

$t_{2}=\frac{d}{S_{2}}$ (7)

$t_{2}=\frac{16093.4 m}{30m/s}$ (8)

$t_{2}=536.44 s$ (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed $S_{ave}$ :

$S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}}$ (10)

$S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s}$ (11)

Finally:

$S_{ave}=25.38 m/s$

3 0

3 years ago

After the big bang, atoms in gas clouds experienced a greater gravitational pull to each other than atoms in other regions of th

allsm [11]

Answer:
These are the two statements with scientific facts that explain the described phenomenon
<span>
Gravitation between two objects increases when the distance between them decreases.</span>

When the mass of an object increases, its gravitational pull also increases.

Justification:

Those two facts are represented in the Universal Law of Gravity discovered by the scientific Sir Isaac Newton (1642 to 1727) and published in his book <span>Philosophiae naturalis principia mathematica.</span>

That law is represented by the equation:

F = G × m₁ × m₂ / d²

The product of the two masses on the numerator accounts for the fact that the gravitational force is directly proportional to the product of the masses, which is that as the masses increase the attraction also increase.

The term d² (square of the distance that separates the objects) in the denominator accounts for the fact that the gravitational force is inversely proportional to the square of the distance; that is as the separation of the objects increase the gravitational force decrease.

6 0

3 years ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0

3 years ago