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2 years ago

6

An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with

the surface. The astronaut falls back down to the surface after a short time interval. Which of the following forces CANNOT be neglected when analyzing the motion of the astronaut?

1 answer:

Anna71 [15]2 years ago

6 0

(D) The gravitational force between the astronaut and the asteroid.

Reason :

All the other forces given in the options, except (D), doesn't account for the motion of the astronaut. They are the forces that act between nucleons or atoms and neither of them accounts for an objects motion.

You might be interested in

How do you know the speed of an electromagnetic wave in a vacuum?

ycow [4]

Electromagnetic waves need no matter to travel - they can travel through empty space (a vacuum). In a vacuum, all electromagnetic waves travel at approximately 3 x 108 m/s - which is the fastest speed possible. ...
Light traveling value through an optical Fibre is, 2 x 108 m/s. Hope that helps.

5 0

2 years ago

An object must move in the (same or different) direction as the force is being applied.

Marta_Voda [28]

Answer:

only reason an object will move in a different direction to the net force on it is because of its prior momentum and it will always accelerate in the direction of the force if thats what u mean.. lol

Explanation:

7 0

1 year ago

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an

Nat2105 [25]

Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

8 0

2 years ago

The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 440g . The vibra

ANEK [815]

Answer:

Tension, T = 2038.09 N

Explanation:

Given that,

Frequency of the lowest note on a grand piano, f = 27.5 Hz

Length of the string, l = 2 m

Mass of the string, m = 440 g = 0.44 kg

Length of the vibrating section of the string is, L = 1.75 m

The frequency of the vibrating string in terms of tension is given by :

$f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$

$\mu=\dfrac{m}{l}$

$\mu=\dfrac{0.44}{2}=0.22\ kg/m$

$T=4L^2f\mu$

$T=4\times (1.75)^2\times (27.5)^2 \times 0.22$

T = 2038.09 N

So, the tension in the string is 2038.09 N. Hence, this is the required solution.

6 0

2 years ago

By how much will the length of a chicago concrete walkway that is 18 m long contract when the equipment drops from 24 degrees ce

xenn [34]

Answer:

Contraction of Chicago concrete walkway = 8.64 x 10⁻³ m

Explanation:

Thermal expansion or compression is given by ΔL = LαΔT

Here Length of Chicago concrete walkway, L = 18 m

Change in temperature, ΔT = (-16 - 24) = -40 °C

Coefficient of thermal expansion for concrete, α = 12 x 10⁻⁶ °C⁻¹

Substituting

ΔL = LαΔT = 18 x 12 x 10⁻⁶ x (-40) = -8.64 x 10⁻³ m

Contraction of Chicago concrete walkway = 8.64 x 10⁻³ m

4 0

2 years ago