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m_a_m_a [10]
3 years ago
11

Increasing the mass attached to a spring will increase its vibrational period.

Physics
1 answer:
iren2701 [21]3 years ago
8 0
Increasing the mass attached to a spring will increase it's vibrational period - this is true. Imagine placing an elephant at the end of a bungee cord vs an apple, the apple will recoil faster than the elephant.
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Positive and negative charger objects affect which type of field?
Ilya [14]

Answer:-

<h2>Electric force</h2>

The interaction between two like-charged objects is repulsive. ... Positively charged objects and neutral objects attract each other; and negatively charged objects and neutral objects attract each other.

Explanation:

Hope it is helpful...

6 0
3 years ago
4.- Una vagoneta de 1000 kg de peso parte del reposo en el punto 1 y desciende, sin rozamiento, por la vía indicada en la figura
Akimi4 [234]

Answer:

A) 49,050 N

B) 16 m

Explanation:

Question:

El dibujo de la pregunta se obtiene de un documento titulado "TRABAJO DIVERSO Y ENERGÍA" que se encuentra en línea y se presenta aquí.

La masa dada del vagón, m = 1,000 kg

La altura del punto en el que descansa el vagón, punto 1, h₁ = 12 m

A) El radio en el punto 2, el punto más bajo, R = 6 m

La fuerza, 'N', que la vía ejerce sobre el vagón en el punto 1 viene dada por la siguiente relación;

N = El peso del vagón + La fuerza de movimiento del vagón

∴ N = m × g + m × a

Dónde;

g = La aceleración debida a la gravedad ≈ 9,81 m / s²

a = La aceleración del vagón

Observamos que para el movimiento circular, la fuerza de movimiento del vagón, m × a = La fuerza centrípeta que actúa sobre el vagón = m × v² / R

∴ m × a = m × v² / R

Dónde;

v² = La velocidad del vagón en el punto 2 = 2 · g · h₁

Por lo tanto;

N = m × g + m × a = m × g + m × v² / R = m × g + m × 2 · g · h₁ / R

∴ N = 1000 × 9,81 + 1000 × 2 × 9,81 × 12/6 = 49,050

La fuerza que ejerce el vagón en el punto 2, N = 49,050 N

B) En el punto 3, tenemos;

N = m · g - m · a₃

La fuerza centrípeta en el punto 3, m · a₃ = m · v₃² / R₃

∴ La altura en el punto 3, h₃ = 4 m

El cuadrado de la velocidad en el punto 3, v₃² = 2 · g · (h₁ - h₃)

Para que el vagón esté seguro en el punto 3, la fuerza de la vía sobre el vagón, N = 0 para que el vagón permanezca en la vía actuando

Por lo tanto;

N = m · g - m · a₃ = 0

m · g = m · a₃ = m · v₃² / R₃ = m · (2 ​​· g · (h₁ - h₃)) / R₃

∴ R₃ = (2 · g · (h₁ - h₃)) / g = (2 · (h₁ - h₃)) = 2 × (12 - 4) = 16

El radio de curvatura en el punto 3 para que el punto sea seguro es R₃ = 16 m.

5 0
3 years ago
A frog has a genetic mutation in its skin cells that causes part of its skin to turn orange. The frog will not pass this genetic
yaroslaw [1]
Because it is not dominant? that is what im guessing
8 0
3 years ago
Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.
finlep [7]

Answer:

\tau=(-32k)\ N-m

\theta=116.55^{\circ}

Explanation:

Given that.

Force acting on the particle, F=(-6i + 2.0j)\ N

Position of the particle, r=(4i+4j)\ m

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

\tau=F\times r

\tau=(-6i + 2.0j)\times (4i+4j)

\tau=\begin{pmatrix}0&0&-32\end{pmatrix}

\tau=(-32k)\ N-m

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r, |r|=\sqrt{4^2+4^2}=5.65\ m

Magnitude of F, |F|=\sqrt{(-6)^2+2^2}=6.324\ m

Using dot product formula,

F{\circ}\ r=|F|.|r|\ cos\theta

cos\theta=\dfrac{F{\circ} r}{|F|.|r|}

cos\theta=\dfrac{-24+8}{6.324\times 5.65}

\theta=116.55^{\circ}

Therefore, this is the required solution.

6 0
3 years ago
Read 2 more answers
in a Mercury thermometer the level of Mercury Rises when its bulb comes in contact with a hot object what is the reason for this
olasank [31]

Answer:

because thermometric liquid readily expands on heating or contracts on cooling even for a small difference in the temperature of the body.

3 0
3 years ago
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