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3 years ago

Define , gravitational acceleration

1 answer:

3 0

The simplest answer would be "acceleration due to gravity."

The exact value of this acceleration changes depending on which planet your on (for example).

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A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

If you poke a hole into a plastic bottle, and stick a straw in it, and then let the straw connect to a glass cup, and then blow

den301095 [7]

It was decrease because the water js going from one place to another

6 0

3 years ago

As ice melts, more sunlight is absorbed. As more sunlight is absorbed, the atmosphere heats up. As the atmosphere heats up, more

Ne4ueva [31]

Answer: Conduction

Explanation: As ice melts,more sunlight is absorbed,as more sunlight is absorbed,the atmosphere heats up,as the atmosphere heats up,more ice melts. This whole process is known as CONDUCTION.

Conduction occurs when there is different temperature between two materials, when there exist a transfer energy(i.e transfer from higher-energy atoms in the hot region to the lower-energy atoms in the cold region). When the atmosphere is heated up due to sunlight,there is a rise in temperature which makes the ice to melt due to transfer of warmth it absorbed from sunlight.

3 0

3 years ago

What is the force of a 1500 kg car accelerating at 45 m/s2 ?

tiny-mole [99]

Answer:

67500

Explanation:

F=ma

F = 1500 × 45

F = 67500

3 0

2 years ago

A basketball of mass 0.23kg is thrown horizontally against a rigid vertical wall with a velocity of 20m/s. It rebounds with a ve

Anna11 [10]

Answer:

$8.1\:\mathrm{Ns}$

Explanation:

The impulse-momentum theorem gives the impulse on an object to be equal to the change in momentum of that object. Since mass is maintained, the change in momentum of the basketball is:

$\Delta p = m\Delta v$ , where $m$ is the mass of the basketball and $\Delta v$ is the change in velocity.

Since the basketball is changing direction, its total change in velocity is:

$\Delta v = 20-(-15)=35\:\mathrm{m/s}$ .

Therefore, the basketball's change in momentum is:

$\Delta p = m\Delta v = 0.23\cdot 35= 8.05=8.1\:\mathrm{kg\cdot m/s}$ .

Thus, the impulse on the basketball is $\fbox{$8.1\:\mathrm{Ns}$}$ (two significant figures).

7 0

2 years ago