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WINSTONCH [101]
3 years ago
11

Which best describes the weak nuclear force? *

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
5 0

Answer:

It keeps particles that make up protons and neutrons together.

Explanation:

Nuclear force is a short range of force which means it will act at short range of distance like fermi order distance.

This shows the order of radius of the nuclei which means all the neutrons and protons which resides into the nucleus will exert the nuclear force on each other and this nuclear force is of large magnitude compare to electrostatic force or gravitation force between them.

While if the distance is more than the fermi order distance then the nuclear force totally disappears so it will not exist for large distances.

so here the correct answer must be

It keeps particles that make up protons and neutrons together.

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Explanation:

Below are attachments containing the solution.

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What is the force of lift and its applications​
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Answer:

fluid flowing past the surface of a body exerts a force on it. Lift is the component of this force that is perpendicular to the oncoming flow direction.[1] It contrasts with the drag force, which is the component of the force parallel to the flow direction. Lift conventionally acts in an upward direction in order to counter the force of gravity, but it can act in any direction at right angles to the flow.

If the surrounding fluid is air, the force is called an aerodynamic force. In water or any other liquid, it is called a hydrodynamic force.

Dynamic lift is distinguished from other kinds of lift in fluids. Aerostatic lift or buoyancy, in which an internal fluid is lighter than the surrounding fluid, does not require movement and is used by balloons, blimps, dirigibles, boats, and submarines. Planing lift, in which only the lower portion of the body is immersed in a liquid flow, is used by motorboats, surfboards, and water-skis.

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how do index contours let you know if the land you are viewing is increasing or decreasing in elevation?
Aleks [24]
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8 0
4 years ago
At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q
Bad White [126]

Answer:

\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}

\displaystyle \theta =68^o

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

\displaystyle F=\frac{KQ_1Q_2}{d^2}

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}

\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l

The force F3 is

\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle

F2 is pointed upwards and its horizontal component is zero

\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle

\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle

Now we compute the total force

\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}

\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle

\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle

\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}

Now we compute the magnitude

\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}

The direction of the total force is given by

\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558

\boxed{\displaystyle \theta =68^o}

6 0
3 years ago
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Answer:

Evaporation

Explanation:

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4 0
3 years ago
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