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2 months ago

How do electromagnetism and gravitation differ from the strong and weak nuclear forces?

1 answer:

5 0

Electromagnetism, which deals with interactions between charged particles, has a considerable range and can create forces that repel like charges while attracting opposites. In contrast, the weak nuclear force is very short-ranged and isn't classified as a force.

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Grace, Erin, and Tony are on a seesaw. Grace has a mass of 45kg and is seated 0.7m to the left of the fulcrum. Nicole has a mass

serg [3582]

Utiliza Scoratic, funciona con cualquier tema.

5 0

3 months ago

A camera operator is filming a nature explorer in the Rocky Mountains. The explorer needs to swim across a river to his campsite

inna [3103]

Answer:

a. Angle= 28.82°

b. Approved. Although he might feel cold, he should be able to cross.

Explanation:

Velocity Vector

Velocity is a measure of how quickly something is moving in a specific direction. It is represented as a vector that has both magnitude and direction. If an object can only move in one direction, then speed can serve as the scalar equivalent of that velocity (only focusing on magnitude).

The explorer aims to swim across a river to reach his campsite, as depicted in the image below. The river's velocity is vr and the explorer's swimming speed in still water is ve. If he were to swim straight towards the campsite, he would end up downstream due to the river's current. Therefore, he must swim at an angle that allows him to overcome the current while still moving towards his goal. This angle relative to the shore is what we need to determine. The explorer's speed can be broken down into its horizontal (vx) and vertical (vy) components. In order to counteract the river's flow:

$v_{ey}=v_r$

We can calculate the vertical component of the explorer's swimming speed as

$v_{ey}=|v_e|cos\alpha$

Thus

$v_r=|v_e|cos\alpha$

Finding the value of $\alpha$

$\displaystyle cos\alpha=\frac{v_r}{|v_e|}$

$\displaystyle cos\alpha=\frac{0.665}{0.759}=0.876$

Then the angle is given by

$\alpha=28.82^o$

The component of the explorer's velocity that goes horizontally is

$v_{ex}=0.759sin28.82^o$

$v_{ex}=0.366\ m/s$

This represents the actual velocity directed towards the campsite

Considering that

$\displaystyle v=\frac{x}{t}$

To find t

$\displaystyle t=\frac{x}{v}$

Calculating the duration for the explorer to cross the river

$\displaystyle t=\frac{29.3}{0.366}$

$t=80\ sec$

As this time is under the hypothermia threshold (300 seconds), the conclusion is

Approved. Although he will feel cold, he should manage to cross successfully.

3 0

3 months ago

A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s

ValentinkaMS [3465]

This question is incomplete. The query is regarding a 3.00 cm diameter coin rolling up a 30.0° incline. With an initial angular speed of 60.0 rad/s, it rolls without slipping. Given that the moment of inertia of the coin is (1/2) MR², the distance the coin travels up the incline is calculated as 0.124 m.

6 0

2 months ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

kicyunya [3294]

Answer:

$tan \theta = \mu_s$

Explanation:

Aby obiekt był w spoczynku na nachyleniu, wynikowa siła działająca na niego musi wynosić zero. Równanie sił działających w kierunku równoległym do nachylenia jest następujące:

$mg sin \theta - \mu_s R =0$ (1)

gdzie

$mg sin \theta$ to składowa ciężaru równoległa do nachylenia, przy czym m oznacza masę obiektu, g oznacza przyspieszenie grawitacyjne, a $\theta$ to kąt nachylenia

$\mu_s R$ to siła tarcia, z $\mu_s$ jako współczynnikiem tarcia oraz R jako reakcją normalną nachylenia

Równanie sił w kierunku prostopadłym do nachylenia to

$R-mg cos \theta = 0$

gdzie

R to reakcja normalna

$mg cos \theta$ to składowa ciężaru prostopadła do nachylenia

Obliczając R,

$R=mg cos \theta$

I podstawiając do (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Rearanżując równanie,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

To jest warunek, przy którym równowaga jest zachowana: gdy tangens kąta staje się większy niż wartość $\mu_s$ , siła tarcia nie jest w stanie zrównoważyć składowej ciężaru równoległej do nachylenia, dlatego obiekt zaczyna zsuwać się w dół.

4 0

3 months ago