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3 years ago

Why do moving back and forth, and moving in a circle look the same on a graph of position versus time?

Physics

1 answer:

Tomtit [17]3 years ago

3 0

Answer:

Because in both kind of motions, the position is periodically repeated. Both are simple harmonic motions.

Explanation:

Moving back and forth means that the moving object goes through the same position every time a certain period of time elapses.

The same is true for the circular motion, every time the object completes a turn, the object is in the same position.

They are examples of simple harmonic motions. The time it takes the object to repeat a complete cycle is called period.

Therefore, in a grahp of position versus time, the curve that represents the position repeats itself every so often equal to one period, both for moving back and forth and for moving in a circle.

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The mass of a hypothetical planet is 1/100 that of Earth and its radius is 1/4 that of Earth. If a person weighs 600 N on Earth,

omeli [17]

To solve this problem we will apply the Newtonian concept of gravitational acceleration produced by a planet. This relationship is given by:

$g = \frac{GM}{r^2}$

Where,

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius

The values given are based on the constants of the earth, so they can be expressed as

$M_p = \frac{1}{100} M_e$

$r_p = \frac{1}{4} r_e$

The relationship of gravity would then be given:

$g_e = \frac{GM_e}{r_e^2}$

The relationship with the new planet, from the gravity of the earth would be given

$g_p = \frac{GM_p}{r_p^2}$

$g_p = \frac{G(1/100)M_e}{(1/4 r_e)^2}$

$g_p = \frac{GM_e 16}{100 r_e^2}$

$g_p = 0.16 \frac{GM_e}{r_e^2}$

$g_p = 0.16g_e$

The relationship with the weight of the earth would be given as:

$W_e = m*g_e = 600N$

$W_p = m*g_p = m(0.16g_p)$

$W_p = (m*g_p)(0.16)$

$W_p = 600*0.16$

$W_p = 96N$

Therefore the weigh on this planet would be 96N

3 0

3 years ago

A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w

steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as

$mgL(1 - \cos{31°}) = \frac{1}{2}mv^2$

Note that the mass gets cancelled out and then we solve for the velocity v as

$v = \sqrt{2gL(1 - \cos{31°})}$

$\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}$

$\:\:\:\:\:= 1.3\:\text{m/s}$

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3 years ago

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Cell are the ------- living part of an organism

lana66690 [7]

Answer:

3 smallest

Explanation:

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3 years ago

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A jet airplane is in level flight. The mass of the airplane is m=8950 kg. The airplane travels at a constant speed around a circ

Mice21 [21]

Answer:

The net force is 91780.8 N.

Explanation:

mass, m = 8950 kg

Radius, R = 9.33 miles = 15015.2 m

Time, T = 0.123 h = 442.8 s

There are two forces acting on the plane.

Horizontal force is the centripetal force and the vertical force is the weight.

$Fx =m R w^2\\\\Fx = m R \frac{4\pi^2}{T^2}\\\\Fx = 8950\times 15015.2\times \frac{4\times 3.14\times 3.14}{442.8\times 442.8}\\\\Fx = 27030.8 N \\\\Fy = m g \\\\ Fy = 8950\times 9.8 \\\\Fy = 87710 N$