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CaHeK987 [17]
4 years ago
11

For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this

is simple harmonic motion (in the vertical direction) with amplitude 5 cm and frequency 2 Hz. There is a box on the floor with mass m = 1 kg.
(a) Assuming the box remains in contact with the floor throughout, find the maximum and minimum values of the normal force exerted on it by the floor over an oscillation cycle.
(b) How large would the amplitude of the oscillations have to become for the box to lose contact with the floor, assuming the frequency remains constant? (Hint: what is the value of the normal force at the moment the box loses contact with the floor?)
Physics
1 answer:
ikadub [295]4 years ago
7 0

Answer:

no idea

Explanation:

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A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

F_g = F_c

\frac{GmM}{r^2} = \frac{mv^2}{r}

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0
3 years ago
Which graph shows an object that is dropped?
Olegator [25]

Answer:

I'm pretty sure it's the third one where velocity goes from positive to negative

Explanation:

the positive velocity is before the object hits the ground and the negative is after

8 0
3 years ago
Give examples of the one element from the alkaline earth metal group and one from the noble gases group. Include each element at
zhuklara [117]
Alkali metals: left column of your periodic table (not hydrogen, but anything below it). They have one valence electron, which they are happy to share in a reaction.

Halogens: second column from the right of your periodic table. They are one electron short of a full shell, so they are reactive in the opposite way that alkalis are--they want electrons.

Atomic number (number of protons) is the big number on the periodic table square. Hydrogen's is 1.

Atomic mass is a little number down below. For example, Hydrogen's is 1.008.

Neutrons are a tricky subject, because different isotopes of the same element can have different numbers of neutrons. You can't generally get this from the atomic mass, because the atomic mass is a weighted average of naturally occurring isotopes. Hydrogen can have 0,1, or 2 neutrons. To answer this, you'd have to choose a particular isotope from the table of isotopes (a completely different chart from the periodic table) which has a certain number of neutrons: n = weight - Z.

Valence electrons are the electrons in the outermost shell. (The column of the table).

<span> Number of principal shells is the row of the periodic table. </span>
8 0
3 years ago
A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875
mario62 [17]

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

5 0
3 years ago
Read 2 more answers
an object of mass 8 kg is whriled round in a vertical circle of radius 2m with a constant speed of 6m/s .Then the maximum and mi
algol13

Answer:

Maximum Tension=224N

Minimum tension= 64N

Explanation:

Given

mass =8 kg

constant speed = 6m/s .

g=10m/s^2

Maximum Tension= [(mv^2/ r) + (mg)]

Minimum tension= [(mv^2/ r) - (mg)]

Then substitute the values,

Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N

Minimum tension= [8 × 6^2)/2 -(8×9.8)]

=64N

Hence, Minimum tension and maximum Tension are =64N and 2224N respectively

5 0
3 years ago
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