Answer:
The mechanical advantage is 1.
Explanation:
Given;
input force, F₁ = 10 N
output force, F₀ = 10 N
mechanical advantage is called force ratio and it is given as;
M.A = output force / input force
M.A = 10 N / 10 N
M.A = 1
This is an ideal machine scenario.
Therefore, the mechanical advantage is 1.
if the mass of an object were doubled, its weight would be
Weight of a body is defined as the force with which it is attached to the earth.
mass is the total quantity of matter present.
the weight of something is given as mass * acceleration
weight of an object is directly proportional to the mass of the object. so, if mass doubles, weight will also double.
hope this helps :)
- jeron
At 30.0 m below the surface of the sea (density = 1 025 kg/m3), where the temperature is 5.00°C, a diver exhales an air bubble having a volume of 0.95 cm3. If the surface temperature of the sea is 20.0°C, what is the volume of the bubble just before it breaks the surface?
Answer:
The volume is \(V_a = 1.510 *10^{-5} m^3\)
Explanation:
From the question we are told that
The depth below the see is \(d_1 = 30.0 \ m\)
The density of the sea is \(\rho_s = 1025 \ kg /m^3\)
The temperature at this level is \(T_d = 5.00 ^oC = 278 \ K\)
The volume of the air bubble at this depth is \(V_d = 0.95 \ cm^3 = 0.95 *0^{-6}\ m\)
The temperature at the surface is \(T_a = 20^oC =293\ K\)
Generally the pressure at the given depth is mathematically evaluated as
\(P_d = P_o + \rho_s * g * d\)
Where \(P_o\) is the atmospheric pressure with a constant value
\(P_o = 1.013 *10^{5} \ Pa\)
substituting values
\(P_d = 1.013 * 10^{5} * + (1025 * 9.8 * 30 )\)
\(P_d = 4.02650 * 10^{5} \ Pa\)
According to the combined gas law
\(\frac{P_a * V_a }{T_a } = \frac{P_d * V_d }{T_d }\)
=> \(V_a = \frac{4.026650 *10^{5} * 0.95 *10^{-6} * 293 }{278 * 1.013*10^{5} }\)
=> \(V_a = 1.510 *10^{-5} m^3\)
Physics Question Work & Energy
Answer:
19,200j
Explanation:
formular W=f×d
a) W=240N×20m
= 4,800j
b) W=200N×20m
=4000j
C) Td= 4800×4000\20×50
td= 19,200,000/1000
td= 19,200j
sorry if am wrong ☹️
What are the names and number of of atoms in a molecule of nitrous oxide,N2O?
Answer:
2 nitrogen and 1 oxygen,
Explanation:
N2=2 nitrogen
O= a single element
The Greek root astro means "stars." Use the context clues to choose the correct definition of the word astronomer. According to the astronomer, the bigger the telescope, the more detail we can see in the images. O A. a person who studies astronauts O B. a person who studies the stars O C. a person who tells the future O D. a person who studies other people
Answer:
b, it's a pretty easy explanation
rocket fuel contains 50j of energy in its chemical store. the rocket has a mass of 1kg. what is the maximum speed the rocket could reach
Why do planets speed up as they get closer to the sun?:
A. air resistance
B. friction
C. gravity
D. tension
Answer:
C
Explanation:
Gravity is the main reason that make our planets to pull each other
Scientists monitor the ozone layer by taking air samples by airplane or weather balloons. What atmospheric layer do the scientists collect the ozone samples from?
Scientists typically collect ozone samples from the stratosphere, which is the atmospheric layer located between about 10 and 50 kilometers (6 to 30 miles) above the Earth's surface.
What is the stratosphere?The stratosphere is the layer of the Earth's atmosphere located above the troposphere and below the mesosphere. It extends from about 10 kilometers (6 miles) to about 50 kilometers (30 miles) above the Earth's surface.
The stratosphere is characterized by a gradual increase in temperature with altitude, due to the absorption of ultraviolet radiation by ozone in the stratosphere.
This is where most of the Earth's ozone is found and where the ozone layer is located. The ozone layer absorbs harmful ultraviolet (UV) radiation from the sun, protecting life on Earth from its harmful effects. Scientists collect air samples from this layer using airplanes or weather balloons equipped with specialized instruments.
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which is more bussin? mcdonald’s or taco bell
Answer:
Coming from my perspective sence I see the lines are super long during lunch I would have to say McDonald’s
What does the cosmological principle allow cosmologists to assume?
A. Exactly what the specific fate of the universe will be and when
B. The galaxies are organized clusters of billions of stars, gas, dust, and matter in all other forms.
C. That the small portion of the space we can see is truly representative of all the rest of the universe that we cannot see.
D. How the helium clouds impact human health and life expectancy.
Answer:
C. That the small portion of the space we can see is truly representative of all the rest of the universe that we cannot see.
Explanation:
The Cosmological Principle assumes that the small portion of the universe that we can see is representative of the entire universe, even though we can only directly observe a tiny fraction of it. It's an assumption used by Cosmologists to simplify their models of the universe.
A spider walks a distance of 0.2 m at a constant speed of 0.1 m/s. The time interval is
A.2 s
B.0.02s
C.0.5s
D.20s
Answer:
A. 2s
Explanation:
time = distance ÷ speed
T = .2 ÷.1
T= 2
The time interval will be Option B (0.02 s).
The given values in the question are:
Distance,
D = 0.2 m
Speed,
S = 0.1 m/s
As we know the formula,
→ \(Speed = \frac{Distance}{Time}\)
or,
→ \(Time = \frac{Distance}{Speed}\)
By substituting the values, we get
→ \(= \frac{0.2}{0.1}\)
→ \(= 0.02 \ s\)
Thus the above answer is right.
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. A 5cm tall object is placed perpendicular to the principal axis of a convex lens of focal
length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position
and size of the image. Also find its magnification
The nature of the image formed by the convex lens is virtual, the position of the image is 30 cm away from the lens on the same side as the object, and the size of the image is twice the size of the object. The magnification is 2, meaning the image is magnified.
Given:
Object height (h) = 5 cm
Focal length of the convex lens (f) = 10 cm
Object distance (u) = 15 cm (positive since it's on the same side as the incident light)
To determine the nature, position, and size of the image, we can use the lens formula:
1/f = 1/v - 1/u
Substituting the given values:
1/10 = 1/v - 1/15
To simplify the equation, we find the common denominator:
3v - 2v = 2v/3
Simplifying further:
v = 30 cm
The image distance (v) is 30 cm. Since the image distance is positive, the image is formed on the opposite side of the lens from the object.
To find the magnification (M), we use the formula:
M = -v/u
Substituting the values:
M = -30 / 15 = -2
The magnification is -2, indicating that the image is inverted and twice the size of the object.
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Can someone explain how to do the algebra for this question? I know everything else, I just don’t know how to rearrange the question to solve for v.
Answer:
Refer to the step-by-step Explanation.
Step-by-step Explanation:
Simplify the equation with given substitutions,
Given Equation:
\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)
Given Substitutions:
\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)
Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)
Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the left-hand side of the equation:
\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
Simplifying the third term.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)
"R²'s" cancel, we are left with:
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)
We have like terms, combine them.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)
Each term has an "m" in common, factor it out.
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)
Now we have the following equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the right-hand side of the equation:
\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)
\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)
Now we have the equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
\(\hrulefill\)
Now solving the equation for the variable "v":
\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
Dividing each side by "m," this will cancel the "m" variable on each side.
\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)
Subtract the term "gh" from either side of the equation.
\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)
Multiply each side of the equation by "10/7."
\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)
Now squaring both sides.
\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)
Thus, the simplified equation above matches the simplified equation that was given.
a hockey puck sliding at constant velocity across the ice is
A hockey puck sliding at constant velocity across the ice is neither accelerating nor decelerating.
A hockey puck sliding at constant velocity across the ice is an example of an object in motion that is experiencing constant velocity. This means that the object is moving in a straight line at a constant speed, and its velocity is not changing. In this case, the hockey puck is sliding across the ice at a constant speed, without accelerating or decelerating.
Constant velocity is a key concept in physics, as it relates to the study of motion and how objects move. It is important to understand that an object will continue to move at a constant velocity unless acted upon by an external force. In the case of the hockey puck, the external force, which is the friction between the ice and the puck, will work to slow down the puck and eventually bring it to a stop.
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describe the human condition before Science and technology was practice.
A rate law for which the units of rate are equal to the units of the rate constant is the zero order kinetics?
A. True
B. False
Answer: b
Explanation:
Do this question by electric flux
A conducting sphere of radius 10cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is \(1.5*10^{3}\) N/C and points radially inwards what is the net charge on the sphere.
Answer:
To find the net charge on the sphere using electric flux, we can use the formula:
Φ = Q/ε0
Where Φ is the electric flux, Q is the charge, and ε0 is the permittivity of free space.
Given that the electric field 20 cm from the center of the sphere is N/C and points radially inwards, we can use the formula for electric field due to a charged sphere to find the charge on the sphere:
E = kQ/r^2
Where E is the electric field, k is Coulomb's constant, Q is the charge, and r is the distance from the center of the sphere.
Substituting the given values, we get:
20 = (1/4πε0)(Q)/(0.2)^2
Solving for Q, we get:
Q = (20)(0.2)^2(4πε0)
Q = 0.64πε0 C
Now, substituting this value of Q in the formula for electric flux, we get: Φ = Q/ε0 = (0.64πε0)/(ε0) = 0.64π C
Therefore, the net charge on the sphere is 0.64π C.
The system is released from rest at θ = 0° when a constant couple moment M = 100 N.m is applied. If the mass of the cable and links AB and BC can be neglected, and each pulley can be treated as a disk having a mass of 6 kg, determine the speed of the 10-kg block at the instant link AB has rotated θ = 90°. Note that point C moves along the vertical guide. Also, the cable does not slip on the pulleys.
For a system released from rest at θ = 0° when a constant couple moment M = 100 N.m is applied, the speed of the 10-kg block is mathematically given as
V=4.33m/s
What is the speed of the 10-kg block?Generally, the equation for the workdone is mathematically given as
W=T tehta<dtheta
W=100(90*\pi/180)
W=1.5707*100
W=1.57Nm
The change in potential energy across the pulley
dP=mgh
dp=10*9.81*111.8
dp=10.97J
For the thrid position, potential energy is
dP=mg(0.3)
dP=17.658J
dP'=17.658J-13.125
dP'=-4.532J
For 2nd position dP=0
The change in Kinectic energy across the pulley\
dK.E=0.5mv^2
For 1st
dk.E=0.5m(10)^2
2nd
dK.E=0.5Iw^2
dK.E=0.5(7.5*10^-3)(v^2/0.05)^2
3rd
2nd=3rd
In conclusion,
157.07=dKE+dP.E
157.07=5v^2+ (7.5*10^-3)(v^2/0.05)^2+10.97-4.53
V=4.33m/s
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John decided to cycle to his friend's house at a speed of 5km/h and the journey took 2
hours. How far did John cycle?
Answer:100 miles
Explanation:
Answer:
10 Km
Explanation:
He is going 5km per hour and he arrived at his friend's house in 2 hours. You multiply 5 by 2 and you get 10.
Which is a characteristic of diatoms?
A)whip-like tail
B)heterotrophic
C)multicellular
D)glass-ive cell wall
Answer:
b
Explanation:
Answer:
The correct answer is D, diatoms have glass-like cell walls.
A rocket weighing 220,000 N is taking off from Earth with a total thrust of
500,000 N at an angle of 20 degrees, as shown in the image below. What is
the approximate vertical component of the net force that is moving the rocket
away from Earth?
20°
Vertical
Component
of Net Force
Thrust
500,000 N
Weight
220,000 N
The vertical component of the net force that is moving the rocket away from Earth is determined as 171,010.1 N.
Vertical component of the lift force
The vertical component of the net force that is moving the rocket away from Earth is determined as follows;
Fy = Fsinθ
where;
F is applied forceθ is the angle of inclinationFy = 500,000 x sin(20)
Fy = 171,010.1 N
Thus, the vertical component of the net force that is moving the rocket away from Earth is determined as 171,010.1 N.
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Answer:
250,000
Explanation:
Jake pulls a couch of mass 38 kg across his apartment. He pulls on the couch at an angle of 25° with a force of 300 N. The free-body diagram is shown below. What is the x component, and what is the y component?
The x component of the force would be 271.89 N and The y component of the force would be 126.78 N
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
The mathematical expression for Newton's second law is as follows
F = ma
As given in the problem Jake pulls a couch of mass 38 kg across his apartment. He pull on the couch at an angle of 25° with a force of 300 N.
The x component of the force would be
Fx = F cos 25°
= 271.89 N
The y component of the force would be
Fy = F sin25°
= 126.78 N
Thus, The x component of the force would be 271.89 N and The y component of the force would be 126.78 N
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Which example represents a class 2 lever? (1 point)
O salad tongs picking up salad
O tweezers plucking hairs
O a car door being opened and shut
a screwdriver opening a paint can
The example that represents a class 2 lever is: D. a screwdriver opening a paint can.
Which example represents a class 2 lever?The load is situated in a class 2 lever halfway between the fulcrum and the effort. The screwdriver's resting place against the paint can's rim serves as the fulcrum in this scenario.
The paint can's lid serves as the load, and the force exerted by the hand on the screwdriver's handle serves as the effort. A class 2 lever can be identified by the load being situated between the fulcrum and the effort.
Tweezers, salad tongs, and opening and closing a car door are not examples of class 2 levers. Class 1 levers include things like tweezers and salad tongs, whereas class 3 levers include things like car doors that open and close.
Therefore the correct option is D.
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which of the following can cause skin cancer and which can't?
Campfire
Flashlight
Lamp
Sun
Tanning bed
fluorescent light bulbs
A cyclist and his bicycle have a combined mass of 88 kg and a combined
weight of 862.4 N. The cyclist accelerates at 1.2 m/s2. After 2 seconds he
reaches a speed of 2.4 m/s. What is his momentum at this point?
A. 36.7 kg m/s
B. 359.3 kg:m/s
C. 105.6 kg-m/s
D. 211.2 kg:m/s
The cyclist accelerates at 1.2 m / s² after 2 seconds he reaches a speed of 2.4 m / s, then the momentum at this point would be 211.2 kg-m/s, therefore the correct answer is option D.
What is momentum?It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.
As given in the problem a cyclist and his bicycle has a combined mass of 88 kg and a combined weight of 862.4 N. The cyclist accelerates at 1.2 m/s2. After 2 seconds he reaches a speed of 2.4 m/s.
The momentum of the cyclist = 88 × 2.4
= 211.0 kgm/s
Thus, the momentum of the cyclist would be 211.0 kgm/s.
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8. An airplane is flying at 200 m/s when it touches the ground at the airport. It has a constant negative acceleration, and slows
down to a speed of 4 m/s after it traveled a length of 4,000 meters. What was the acceleration?
Answer:
Explanation:
Given
Initial velocity u = 200m/s
Final velocity = 4m/s
Distance S = 4000m
Required
Acceleration
Substitute the given parameters into the formula
v² = u²+2as
4² = 200²+2a(4000)
16 = 40000+8000a
8000a = 16-40000
8000a = -39,984
a = - 39,984/8000
a = -4.998m/s²
Hence the acceleration is -4.998m/s²
In which situation should a parent be proactive and act to assume responsibility?
Answer: Patsy is eager to learn how to bake a cake but does not know how to do it.
Explanation: i picked this and it is correct, you’re welcome:)
carmeron drives his car 15 km north. he stops for lunch and then drives 12 km south. what is the distance carmeron traveled?
Answer: 27 kilometers
Cameron drove 15 kilometers north.
Cameron also drove 12 kilometers south.
To find the total number, we must add.
15 + 12 = 27km.
Therefore, the answer is 27 km.
Can you help with these physics questions about motion? Please!!!!
Answer:
Explanation:
5-) in the 1st system the right side is heavier, the system is moving in that way with a 9.8k m^2/s
in the 2nd system right side is heavier too, the system is moving in that way with a 9.8k m^2/s
in the 3rd system both sides got the same weight so the system isn't moving in anyway. 0 m^2/s
An unhappy 0.400 kg rodent, moving on the end of a spring with force constant 3.50 N/m , is acted on by a damping force Fx=−bvx .
The equation of motion for the rodent is x(t) = -1.12cos(3.20t), and the damping force is Fd = -0.62*vx(t). The damping force will cause the amplitude of the motion to decrease over time, and the rodent will eventually come to rest at the equilibrium position.
We can use the following equations to solve this problem:
F = -kx (Hooke's Law)
F = ma (Newton's Second Law)
a = d^2x/dt^2 (Definition of Acceleration)
Fd = -bv (Definition of Damping Force)
x(t) = A*cos(ωt + φ) (Equation of Motion for Simple Harmonic Motion)
We will need to use these equations to find the displacement, velocity, and acceleration of the rodent as a function of time, and then use that information to calculate the damping force and solve for the parameters of the motion.
First, let's find the natural frequency of the system:
ω = sqrt(k/m) = sqrt(3.50 N/m / 0.400 kg) = 3.20 rad/s
Next, let's assume that the rodent starts at its maximum displacement and moves in simple harmonic motion. We can use the equation of motion for simple harmonic motion to write:
x(t) = A*cos(ωt + φ)
where A is the amplitude of the motion and φ is the phase angle.
To find A and φ, we need to use the initial conditions. We know that at t=0, the rodent is at its maximum displacement, so x(0) = A. We also know that at t=0, the velocity of the rodent is zero, so vx(0) = -Aωsin(φ) = 0. This means that either A=0 (the rodent is not moving) or sin(φ) = 0 (the rodent is moving with maximum velocity). We will assume that the latter is true, so sin(φ) = 0 and cos(φ) = 1.
Now we can write:
x(t) = A*cos(ωt)
To find A, we use the fact that the rodent has a mass of 0.400 kg and is moving on a spring with force constant 3.50 N/m. The force on the rodent is given by:
F = -kx = -3.50 N/m * A*cos(ωt)
At maximum displacement, the force is equal to the weight of the rodent:
mg = 0.400 kg * 9.81 m/s^2 = 3.92 N
So we can write:
3.92 N = -3.50 N/m * A
A = -1.12 m
Therefore, the equation of motion for the rodent is:
x(t) = -1.12cos(3.20t)
To find the velocity and acceleration of the rodent, we take the derivative of the displacement with respect to time:
vx(t) = dx/dt = 3.58sin(3.20t)
ax(t) = d^2x/dt^2 = -11.46cos(3.20t)
To find the damping force, we use the equation:
Fd = -bv = -bdx/dt = -b3.58sin(3.20t)
We don't know the value of b, so we can't solve for it directly. However, we can use the fact that the damping force is equal to the work done by the damping force over one cycle of motion. This work is equal to the energy lost by the system due to damping. Since the system is losing energy at a rate proportional to its velocity, we can write:
Energy lost per cycle = Average damping force * Distance traveled per cycle
The distance traveled per cycle is equal to 2piA = 7.04 m, since the rodent moves from its maximum displacement to its minimum displacement and back again in one cycle.
The average damping force over one cycle is equal to the time average of the damping force:
<Fd> = (1/T)∫[0,T] -bdx/dt dt
where T = 2*pi/ω is the period of the motion. Evaluating the integral gives:
<Fd> = (1/T)∫[0,T] -b(-1.12)3.20sin(3.20*t) dt
<Fd> = 3.58*b
Since the energy lost per cycle is also equal to (1/2)kA^2, we can write:
(1/2)kA^2 = <Fd>2pi*A
Solving for b, we get:
b = (kA)/(2pi)
Substituting the given values, we get:
b = (3.50 N/m * 1.12 m)/(2*pi) = 0.62 Ns/m
Therefore, the equation of motion for the rodent is:
x(t) = -1.12cos(3.20t)
vx(t) = 3.58sin(3.20t)
ax(t) = -11.46cos(3.20t)
and the damping force is given by:
Fd = -0.62*vx(t)
Note that the negative sign indicates that the damping force acts in the opposite direction to the velocity of the rodent. This means that the damping force will cause the amplitude of the motion to decrease over time, and the rodent will eventually come to rest at the equilibrium position.
Therefore,The equation of motion for the rodent is x(t) = -1.12cos(3.20t), and the damping force is Fd = -0.62*vx(t).
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