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sergij07 [2.7K]

3 years ago

11

A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m

. If the spring is stretched to a length of 0.5m and released with initial velocity 0, find the position of the mass at any time t. Here damping constant is zero.

1 answer:

aleksandrvk [35]3 years ago

8 0

Answer:

Explanation:

force constant of spring k = force / extension

= 35.6 / 0.5

k = 71.2 N / m

angular frequency ω of oscillation by spring mass system

$\omega = \sqrt{\frac{k}{m} }$

where m is mass of the body attached with spring

Putting the values

$\omega = \sqrt{\frac{71.2}{5} }$

ω = 3.77 radian / s

The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2

so the equation for displacement from equilibrium position that is middle point can be given as follows

x = .5 sin ( ω t + π / 2 )

= 0.5 cos ω t

= 0.5 cos 3.77 t .

x = 0.5 cos 3.77 t .

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b) the net force on the car is zero.

Explanation:

Let's analyze each option one by one:

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a = 0

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$\sum F = ma = 0$

where $\sum F$ is the net force on the car and m is its mass. This means that the net force on the car is zero: so, the force from the engine cannot be greater than all the forces of friction, otherwise the net force cannot be zero.

b) the net force on the car is zero. --> TRUE, for what we said at point A)

c) the inertia is changing. --> FALSE. The inertia of an object just depend on the mass and the velocity of the object: as neither the mass nor the velocity are changing in this problem, then the inertia of the car is not changing.

d) the forces of friction are proportional to the acceleration of the car. --> FALSE. Generally, the force of friction acting on an object moving on a flat surface is

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where $\mu$ is the coefficient of friction, m is the mass, and g the acceleration of gravity. Therefore, the force of friction does not depend on the acceleration of the car.

e) All of the above. --> FALSE

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3 years ago

A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

kupik [55]

Answer:

The acceleration of the satellite is $0.87 m/s^{2}$

Explanation:

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ( $1.50x10^{7} m$ ) and the Earth radius ( $6.38x10^{6} m$ ) :

$r = 1.50x10^{7} m+6.38x10^{6}m$

$r = 21.38x10^{6}m$

Then, equation 2 can be used:

$T = 8.65 hrs \cdot \frac{3600 s}{1hrs}$ ⇒ $31140 s$

$v = \frac{2 \pi (21.38x10^{6}m)}{31140s}$

$v = 4313 m/s$

Finally equation 1 can be used:

$a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}$

$a = 0.87 m/s^{2}$

Hence, the acceleration of the satellite is $0.87 m/s^{2}$

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Photons are particles of electromagnetic radiation.

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kakasveta [241]

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Ede4ka [16]

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Explanation:

The equation of motion in this situation is:

$y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}$ (1)

Where:

$y=0$ is the final height of the ball

$y_{o}=h$ is the initial height of the ball

$V_{oy}=V_{o}sin(0\°)=0$ is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)

$t=0.5 s$ is the time at which the ball lands

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