Give some example acostic using "ELEMENTS"

A high school physics student is sitting in a seat read-

Nataly_w [17]

The equilibrium condition allows finding the result for the force that the chair exerts on the student is:

The reaction force that the chair exerts on the student's support is equal to the student's weight.

Newton's second law gives the relationship between force, mass and acceleration of bodies, in the special case that the acceleration is is zero equilibrium condition.

∑ F = 0

Where F is the external force.

The free body diagram is a diagram of the forces on bodies without the details of the shape of the body, in the attached we can see a diagram of the forces.

Let's analyze the force on the chair.

$N_{chair} - W_{chair} - W_{student} = 0 \\ \\N_{chair} = W_{chair} + W_{student}$

Let's analyze the forces on the student.

$N_{student} - W_{student} = 0 \\N_{student} = W _{student}$

In conclusion using the equilibrium condition we can find the result for the force that the chair exerts on the student is:

The reaction force that the chair exerts on the student's support is equal to the student's weight.

Learn more here: brainly.com/question/18117041

7 0

3 years ago

What is the magnitude of fs on an object lying on a flat surface without moving, on

Degger [83]

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

magnitude of force on the object, F = 10 N

angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

$\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N$

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

7 0

3 years ago

What is half-life? a. Half-life is a term that describes the time it takes for half of a particle to disintegrate c. Half-life i

Alisiya [41]

Well, half-life is the radioactivity of an identified isotope that decreases by half of the actual value.
So, your answer would be C.

4 0

4 years ago

What is the nuclear binding energy in joules for an atom of helium–4? Assume the following: Mass defect = 5.0531 × 10-29 kilogra

Romashka [77]

The equation given, E = mc², can be directly used to determine the unknown amount of nuclear binding energy. Substitute the values given to the equation,
E = (5.0531 x 10^-29 kg) x (3 x 10^8 m/s)² = 4.548 x 10^-12 J
Thus, the nuclear binding energy is equal to 4.548 x 10^-12 J.

5 0

4 years ago

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A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi

KIM [24]

The stone reaches the wall at a height of 1.62 m.

The stone lands at a point 24.5 m from the point of projection.

The stone is projected horizontally with a velocity u at a height h from the ground. The wall is located at a distance x from the point of projection. The stone takes a time t to reach the wall and in the same time the stone falls a vertical distance y.

The horizontal distance x is traveled with a constant velocity u.

$x=ut$

Calculate the time taken t.

$t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s$

The stone's initial vertical velocity is zero. It falls through a distance y in the time t under the action of acceleration due to gravity g.

$y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m$

The height h₁ of the stone above the ground when it reaches the wall is given by,

$h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m$

When the stone reaches the wall, its height from the ground is 1.62m.

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance R from the point of projection. If the time taken by the stone to reach the ground is t₁, then,

$h=\frac{1}{2} gt_1^2$

Calculate the time taken by the stone to reach the ground.

$t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s$

The horizontal distance traveled by the stone is given by,

$R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m$

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

3 0

3 years ago