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laiz [17]
2 years ago
9

During the deceleration of an ascending elevator, the normal force on the feet of a passenger is _____ her weight. During the de

celeration of a descending elevator, the normal force on the feet of a passenger is _____ her weight.
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
7 0

Answer: Smaller than ; larger than

Explanation:

When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of

N= mg +ma, (g is gravity and a is acceleration)

here ma is negative so the N= mg-ma

Hence, it feels smaller than its original weight.

When the elevator is moving downward , then the force acting will be positive in nature

N= mg+ma,

here ma will be positive so it feels larger the original weight of passenger.

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Which of the four fundamental forces is responsible for holding together the molecules of the pizza dough as it is spinning in t
defon

Answer:

None.

Explanation:

Molecules are formed by an element's need or excess of electrons. For example, in nature oxygen generally exists as 02. Other molecules are formed via chemical reaction. The example here is the burning of gasoline. Gasoline's two main byproducts are water and carbon dioxide.

Hydrogen as an atom has one electron making it unstable. Put a second hydrogen atom next to the first and the two atoms will share electrons to fill the first energy level the atom needs to be stable.

5 0
2 years ago
Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0
3 years ago
Questions:<br>(a) Why should the amplitude of oscillation be small on a pendulum experiment​
Snowcat [4.5K]
The formula for the pendulum experiment is based on the assumption that the amplitude is small so that the angle is approximately equal to the Sine of the angle.
5 0
3 years ago
When a board with a box on it is slowly tilted to a larger and larger angle, common experience shows that the box will at some p
Savatey [412]

Answer:

c. The coefficient of kinetic friction is less than the coefficient of static friction

Explanation:

When the box finally does break loose. Then the component of the box weight which is parallel to the board weight parallel component, is equal to the \mu_gn.

w_{box}=\mu_gn

For the box to acce;erate thee must be non-zero net force acting on the box parallel to the board. Or we can say,

w_1>f_g\\w_1>\mu_gn

Therefore the force of kinetic friction must be less than the force of static friction. Thus,

\mu_g

3 0
3 years ago
Una persona cierra una puerta de 1 metro de ancho aplicando una fuerza de 40 [N], perpendicular a ella, a 90 [cm] de su eje de r
Damm [24]

Answer:

El módulo del torque aplicado es 36 Nm

Explanation:

En los movimientos rotatorios, la cantidad de fuerza aplicada no depende de la acción gravitacional sino del momento inercial, que es el equivalente angular de la inercia (masa) y representa la resistencia que un objeto ofrece al rotar alrededor de su eje. Cuando un cuerpo rígido rota alrededor de su eje debe considerarse , además de la masa, el radio de giro ya que estos dos factores determinan la resistencia del cuerpo a los cambios de movimiento rotatorio a través de un eje determinado.

De esta manera, se llama torque o momento de una fuerza a la capacidad de dicha fuerza para producir un giro o rotación alrededor de un punto.

En muchas ocasiones el punto de aplicación de la fuerza no coincide con el punto de aplicación en el cuerpo. En este caso la fuerza actúa sobre el objeto y su estructura a cierta distancia, mediante un  elemento que traslada esa acción de esta fuerza hasta el objeto. Entonces, el momento de una fuerza  es, matemáticamente,  igual al producto de la intensidad de la fuerza (módulo) por la distancia desde el punto de aplicación de la fuerza hasta el eje de giro:

M=F*d*sen θ

donde F es la fuerza en Newton (N), d la distancia en metros (m), θ el ángulo que forma la fuerza con el objeto al cual se le aplica la fuerza y M el momento, que se mide en Newton por metro (Nm).

En este caso:

  • F= 40 N
  • d= 90 cm= 0.9 m (siendo 100 cm= 1 m)
  • θ= 90° ya que la fuerza se aplica de forma perpendicular. Entonces sen θ= sen 90= 1

Reemplazando:

M=40 N*0.9 m* 1

Resolviendo:

M= 36 Nm

<u><em>El módulo del torque aplicado es 36 Nm</em></u>

3 0
3 years ago
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