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3 years ago

What is the relationship between the applied force of a hanging mass on a spring and the spring force of the spring?

Physics

1 answer:

zaharov [31]3 years ago

7 0

Answer:

elastic force and weight are related to the acceleration of the System.

Explanation:

The relationship between these two forces can be found with Newton's second law.

$F_{e}$ - W = m a

K x - m g = m a

We see that elastic force and weight are related to the acceleration of the System.

If a harmonic movement is desired, an extra force that increases the elastic force is applied, but to begin the movement this force is eliminated, in general , if the relationship between this external and elastic force is desired, the only requirement is that it be small for harmonic movement to occur

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a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleratio

Harman [31]

Answer:

The angular acceleration is $\alpha = 3.235 \ rad/s ^2$

Explanation:

From the question we are told that

The moment of inertia is $I = 0.034\ kg \cdot m^2$

The net torque is $\tau = 0.11\ N \cdot m$

Generally the net torque is mathematically represented as

$\tau = I * \alpha$

Where $\alpha$ is the angular acceleration so

$\alpha = \frac{\tau }{I}$

substituting values

$\alpha = \frac{0.1 1}{ 0.034}$

$\alpha = 3.235 \ rad/s ^2$

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3 years ago

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Answer:

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2 years ago

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Answer:

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Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This

vova2212 [387]

Answer:

$1.554\times 10^{32}\ \text{kg}$

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = $6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2$

Radius of orbit is given by

$R=\dfrac{vT}{2\pi}$

We have the relation

$\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}$

The mass of each star is $1.554\times 10^{32}\ \text{kg}$

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2 years ago