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OleMash [197]
3 years ago
11

A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto

ps bouncing and reaches its new equilibrium position (x=0), the student notices that the spring has stretched 0.82 m. The student's friend pulls the platform down 0.32 m further and then releases it at t=0. What is the amplitude of the motion of the student on the platform?

Physics
1 answer:
Ilya [14]3 years ago
6 0
Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer: 
0.32 m (single amplitude), or
0.64 m (double amplitude)

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Question 2 of 10
makvit [3.9K]

Answer:

D.Entropy tends to increase.

Explanation:

The second of thermodynamics states that the state of entropy of the entire universe,as an isolate system, will always increase over time

7 0
2 years ago
An woman whose weight is 804 N stands on a long horizontal plank of wood 1.55 m from one end. The plank is uniform and is suppor
meriva

Answer:

Explanation:

There will be reaction force by  each vertical post on horizontal plank . Let it be R₁ and R₂ . R₁ is reaction force by the post nearer to woman

Taking torque of all forces about the end far away from the woman

Torque by reaction force = R₁ x 5.5

= 5.5 R₁ upwards

Torque by weight of woman in opposite direction , downwards

= - 804 x ( 5.5 - 1.55 )

= - 3175.8

Torque by weight of the plank in opposite direction , downwards .

= - 27 x 5.5 / 2

= - 74.25

Torque by R₂ will be zero as it passes through the point about which torque is being taken .

Total torque

= 5.5 R₁ - - 3175.8 - - 74.25  = 0 ( For equilibrium )

5.5 R₁ = 3250

R₁ = 590.9 N .

6 0
3 years ago
A hockey puck with a mass of 0.159 kg slides over the ice. The puck initially slides with a speed of 4.75 m/s, but it comes to a
Anton [14]

Answer:

The energy dissipated as the puck slides over the rough patch is 1.355 J

Explanation:

Given;

mass of the hockey puck, m = 0.159 kg

initial speed of the puck, u = 4.75 m/s

final speed of the puck, v = 2.35 m/s

The energy dissipated as the puck slides over the rough patch is given by;

ΔE = ¹/₂m(v² - u²)

ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)

ΔE = -1.355 J

the lost energy is 1.355 J

Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J

5 0
3 years ago
Calculate the moment of inertia for each scenario: (a) An 80.0 kg skater is approximated as a cylinder with a 0.140 m radius. (b
Zina [86]

Answer:

a) the moment of inertia is 0.784 Kg*m²

b) the moment of inertia is with arms extended is 1.187 Kg*m²

c) the angular velocity in scenario (b) is 4.45 rad/s

Explanation:

The moment of inertia is calculated as

I = ∫ r² dm

since

I = Ix + Iy

and since the cylinder rotates around the y-axis then Iy=0 and

I = Ix = ∫ x² dm

if we assume the cylinder has constant density then

m = ρ * V = ρ * π R²*L = ρ * π x²*L

therefore

dm = 2ρπL* x dx

and

I = ∫ x² dm = ∫ x² 2ρπL* x dx = 2ρπL∫ x³ dx = 2ρπL (R⁴/4 - 0⁴/4) = ρπL R⁴ /2 =  mR² /2

therefore

I skater = mR² /2 = 80 Kg * (0.140m)²/2 = 0.784 Kg*m²

b) since the arms can be seen as a thin rod

m = ρ * V = ρ * π R²*L = ρ * π R²*x

dm =ρ * π R² dx

I1 = ∫ x² dm = ∫ x² * ρ * π R² dx = ρ * π R²*∫ x² dx = ρ * π R²* ((L/2)³/3 - (-L/2)³/3)

= ρ * π R²*2*L³/24 = mL²/12

therefore

I skater 2 = I1 + I skater =  mL²/12 + mR² /2= 8 Kg* (0.85m)²/12 +(80-8) Kg * (0.140m)²/2 = 1.187 Kg*m²

c)  from angular momentum conservation

I s2 * ω s2 = I s1 * ω s1

thus

ω s2 = (I s1 / I s2 )* ω s1 /= (0.784 Kg*m²/1.187 Kg*m²) * 6.75 rad/s = 4.45 rad/s

4 0
3 years ago
A negative charge of -0.550 m exerts an upward <a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="5868766e
almond37 [142]

Answer:

a. +10.9μC

b. 0.600N and downward

Explanation:

To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as

F=(kq₁q₂)/r²

since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as

q₂=Fr²/kq₁

q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)

q₂=0.054/4950

q₂=1.09*10⁻⁵c

q₂=10.9μC.

Hence the second charge is +10.9μC

b. From the rule of charges which state that like charges repel and unlike charges attract, we can conclude that the two above charges will attract since they are unlike charges. Hence the direction of the force will be downward into the second charge and the magnitude of the force will remain the same as 0.600N

8 0
3 years ago
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