Answer:
A) v = 31.8 m/s
B) v_f = 18.1 m/s
C) F_avg = 4130.7 N
Explanation:
A) From law of conservation of energy, we know that;
mgh = W_f + ½mv²
v is the speed at which she is going at the bottom of the slope.
Thus, making v the subject, we have;
v = √[2gh - (2W_f)/m)]
We are given;
h = 70 m
m = 58 kg
W_f = 1.04 × 10⁴ J = 10400 J
Thus;
v = √[(2 × 9.8 × 70) - (2 × 10400)/58)]
v = 31.8 m/s
B) Total force will be given by the formula;
F_t = F_k + F_r
Where;
F_k is force of kinetic friction = mg•μ_k
μ_k = 0.25
So, F_k = 58 × 9.8 × 0.25
F_k = 142.1 N
We are given force of air resistance(F_r) as 150 N
Thus;
F_t = 142.1 + 150
F_t = 292.1 N
Final velocity is gotten from the formula;
v_i² - v_f² = 2F_t•L/m
Thus;
v_f = √[v_i² - (2F_t•L/m)]
Where, v_i = 31.8 m/s, F_t = 292.1, m = 58 kg, L = 68 m
Thus;
v_f = √[31.8² - (2 × 292.1 × 68/58)]
v_f = 18.1 m/s
C) the average force exerted on her by the snowdrift as it stops her is given by the formula;
F_avg = m•v_f²/2l
F_avg = 58 × 18.1²/(2 × 2.3)
F_avg = 4130.7 N
Select the correct answer.
Proper technique can help prevent injuries.
A.
True
B.
False
Answer:
b
Explanation:
Answer:
false
Explanation:
Sara pushes her 45 kg younger sister with a force of 300 N to the right and encounters 120 N of friction. Calculate the little sister’s acceleration. Show steps please:)
A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attached to the end of the unstretched spring and then released. The fish moves downward until the spring is fully stretched, then starts to move back up as the spring begins to contract.
What is the maximum distance through which the fish falls?
Answer:
Explanation:
The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.
Change in gravity potential energy = change in spring potential energy
mgh = 1/2kh^2
Assume gravity constant g is 10m/s^2
2.6*10*h = 1/2*200*h^2
100h^2 - 26h = 0
2h(50h - 13) = 0
h = 0 or h = 13/50 = 0.65m
h = 0 is before the spring is stretched
So the maximum distance is 0.65m.
The maximum distance through which the fish falls is 0.25m.
What is law of conservation of energy?Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.
Given parameters:
Spring constant: k = 200 N/m.
Mass of the fish: m =2.6 kg.
Let the maximum distance through which the fish falls, i.e. the maximum stretched distance of the spring = x.
From law of conservation of energy:
Change in gravity potential energy = change in potential energy
mgx = 1/2kx^2
2.6×9.8×h = 1/2*200*h^2
h(100h - 2.6×9.8) = 0
h = 0 or h = 2.6×9.8/100 = 0.25m
As h = 0 is the equilibrium position; the maximum distance through which the fish falls is 0.25m.
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\A thin hoop with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 3.5 rad/s. What is its kinetic energy
The rotational or kinetic energy of the thin hoop with the given radius, mass and angular speed is 0.092J
Given the data in the question;
Mass of hoop; [tex]m = 3.0kg[/tex]Radius; [tex]r = 10cm = 0.1m[/tex]Angular speed; [tex]w = 3.5rad/s[/tex]Rotational or kinetic energy; [tex]E_{rotational} = \ ?[/tex]
Rotational energy or angular kinetic energyRotational energy or angular kinetic energy is simply kinetic energy due to the rotation of a rigid body.
It is expressed as;
[tex]E_{rotational} = \frac{1}{2}Iw^2[/tex]
Where [tex]I[/tex] is the moment of inertia around the axis of rotation and [tex]w[/tex] is the angular speed or velocity.
For the moment of inertia around the axis of rotation.
[tex]I = \frac{1}{2}mr^2[/tex]
Hence
[tex]E_{rotational} = \frac{1}{2}(\frac{1}{2}mr^2)w^2 \\\\E_{rotational} = (\frac{1}{4}mr^2)w^2[/tex]
Now, we substitute our given values into the above equation to find the rotational or kinetic energy.
[tex]E_{rotational} = (\frac{1}{4}*3.0kg * (0.1m)^2) * (3.5rad/s)^2 \\\\E_{rotational} = 0.0075kgm^2 * 12.25rad/s^2\\\\E_{rotational} = 0.092kg.m^2/s^2\\\\E_{rotational} = 0.092J[/tex]
Therefore, the rotational or kinetic energy of the thin hoop with the given radius, mass and angular speed is 0.092J
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A student placed a ladder up against a wall as shown below. The normal force applied by the wall in the ladder will be directed:
Up and to the right
To the left
Down and to the left
Straight up
The normal force applied by the wall in the ladder will be directed to the left. Hence, option (B) is correct.
What is normal force?The component of a contact force in mechanics known as the normal force is perpendicular to the surface that an item encounters.
In this context, normal refers to perpendicular in a geometric sense rather than "usual" or "expected" as it does in everyday parlance. Gravity acts on a person standing stationary on a platform; gravity would otherwise draw the person down into the Earth's core absent a countervailing force from the resistance of the platform's molecules, known as the "normal force."
As normal force acts perpendicular to the surface, the normal force applied by the wall in the ladder will be directed to the left.
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Electric field lines moves away from positive to wards negative?
The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.
Mrs. LaCross leaves school and accidentally leaves her coffee mug on the roof of her car as shown in the picture below.
She was traveling at a constant pace until a student rushes in front of her not using the sidewalks and crosswalks and she had to slam on her brakes. The mug will then move foward
Why will this happen to her coffee mug?
Answer:
When she stops at a fast pace the energy and wind will take the cup forward and it will most likeley brake
Explanation:
I'm not entirely sure this is what you were looking for but I hope this helped!
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The system is released from rest with the spring initially stretched 3 in. Calculate the velocity of the cylinder after it has dropped 0.5 in.
The velocity of the cylinder after it has dropped 0.5 in is 0.497 m/s.
Conservation of energy
The velocity of the cylinder can be determined by applying the principle of conservation of energy as shown below;
Ei = Ef
¹/₂k(xi)² + mgh = ¹/₂mv² + ¹/₂k(xt)²
where;
k is spring constant = 6 lb/in = 1050 N/mm is mass of the system = 100 lb = 45.35 kgxi is initial extension = 3 in = 0.076 mxt is the total extension = 2xi + 0.5 in = 6.5 in = 0.165 mh is height = 0.5 in = 0.0127 m¹/₂k(xi)² = -mgh + ¹/₂mv² + ¹/₂k(xt)²
¹/₂(1050)(0.076)² = -(45.35)(9.8)(0.0127) + ¹/₂(45.35)v² + ¹/₂(1050)(0.165)²
3.032 = -5.64 + 22.68v² + 14.29
3.032 = 8.65 + 22.68v²
-5.62 = 22.68v²
|v| = 0.497 m/s
Thus, the velocity of the cylinder after it has dropped 0.5 in is 0.497 m/s.
The complete question is below:
The spring has a stiffness of 6 lb/in and the mass of the load is 100 lb.
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Total internal reflection will occur when:
light goes from high to low density above the critical angle
light goes from low to high density above the critical angle
light goes from high to low density below the critical angle
light goes from low to high density below the critical angle
Answer:
Try B or C if I'm wrong sorry
Explanation:
Do you agree or disagree with Kinsey that sexuality is a fluid concept, ranging on a seven-point scale from homosexual to heterosexual? Or, do you think these are discrete constructs that have finite points where they begin and end sexuality?
Answer:
kinsey sexuailty doesn't matter because that is his or her
Explanation:
A woman 1.2 m tall lies along the axis of a space vehicle traveling at 0.87c. What is her height as measured by a stationary observer?
Answer:
L = L0 (1 - v^2/c^2) where L0 is proper length and L the measured length
L = 1.2 (1 - .87^2)^1/2 m = .59 m
5. How long does it take a car to cross a 30 m wide intersection after the light turns green after waiting at a traffic light, assuming the car accelerates at a constant 2.0 m.s2. Ignore the driver’s reaction time
Answer:
S = Vo t + 1/2 a t^2 = 1/2 a t^2 V0 = 0 for a standing start
t = (2 S / a)^1.2
t = (2 * 30 / (2.0) = (30)^1/2 = 5.5 sec
A train has a mass of 1.50 x 107 kg. If the engine can
exert a net force of 7.50 x 105 N on the train, how much
time is required for the train to reach a speed of 80.0
km/h, if the train begins from rest?
vi=0, vf=80km/hr=22m/s, m=1.5x107kg, F = 7.5x105N.
a=F/m, t=(vf-vi)/a= (vf-vi)m/F = (22m/s)(1.5x107kg)/7.5x105N = 440 kg m/sN = 440 s or 7mins
Which two elements have the same number of valence electrons?
A. C and o
B. Na and Mg
C. Cl and F
D. Ga and Ge
1) A pile driver with mass 1x10⁴ kg strikes a pile with velocity 10.0 m/s. What is the kinetic energy of the driver as it strikers the pile?
Explanation:
please mark me as brainlest
With what force will a car hit a tree if the car has a mass of 3,454 kg and it is accelerating at a rate of 5 m/s2?
Answer:
F = 17270 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by the acceleration.
ΣF = m*a
where:
F = force [N]
m = mass = 3454 [kg]
a = acceleration = 5[m/s^2]
Now replacing:
F = 3454*5
F = 17270 [N]
why can't a transformer operate on a DC signal
Answer:
Transformers do not allow DC input to flow through. This is known as DC isolation. This is because a change in current cannot be generated by DC
Explanation:
Hope this helps
Answer:
a change in current cannot be generated by DC;
If the period of a wave is 20 s, then what is its frequency?
When a moving object collides with an object that isn’t moving , what happens to the kinetic energy of each object?
Answer:
The moving object transfers kinetic energy with the object that isn't moving.
Explanation:
Kinetic energy is a form of energy that a object or particle has by the reason of motion. Potential energy is a form of energy that a object or particle has by the reason of no movement. The object that is moving knocks down or collides with an object that isn't moving, and the object that is moving transfers kinetic energy to potential energy, and the object that wasn't moving transfers potential energy to kinetic energy.
PHYSICS 1403
Lab Homework - Friction on a Ramp
A laborer wants to move crates containing bottles of olive oil from a truck to the ground by sliding them
along a ramp. The ramp is 6 m long and is at an angle of 25º. There is friction on the ramp for the first
crate. The laborer doesn't know that there is a small leak in one of the bottles. The leak leaves a layer of
oil on the ramp. The oil creates a frictionless surface for the second crate Wayne sends down the ramp. At the bottom of the ramp, the speed of the second crate (without friction) is 2.5 the speed of the first crate (with friction). Find the coefficient of kinetic friction. Hint: this is a multistep problem that is
be solved using only energy equations. Do not use kinematics or you will not receive full
credit, even if your answer is correct. Use conservation of energy and start with the frictionless case.
Hi there!
Hi there!
We can begin by simplifying the work-energy theorem for Crate 2.
Since there is no friction, there is no energy dissipated. Thus, the initial energy is equal to the final energy.
Initially, we only have gravitational potential energy (U = mgh), and when the box has fully slid down, it only has kinetic energy (KE = 1/2mv²), therefore:
[tex]E_i = E_f\\\\mgh = \frac{1}{2}mv^2[/tex]
We can cancel out the mass and solve for velocity.
[tex]gh = \frac{1}{2}v^2\\\\v^2 = 2gh \\\\v = \sqrt{2gh}[/tex]
We must use right triangle trigonometry to solve for the HEIGHT given the ramp's length (hypotenuse).
We can use sine:
[tex]sin\theta = \frac{\text{h}}{L} \\\\Lsin\theta = h = 6 * sin(25) = 2.5357 m[/tex]
Now, solve for velocity.
[tex]v = \sqrt{2(9.8)(2.5357)} = 7.05 \frac{m}{s}[/tex]
Since this is 2.5 times the speed of the first crate, we know that the final velocity of crate 1 is:
[tex]v_1 = \frac{v}{2.5} = 2.82 \frac{m}{s}[/tex]
Crate 1:
In this instance, we have friction. Recall the following.
[tex]F_f = \mu N[/tex]
On an incline, the normal force is equivalent to the cosine of the force of gravity, so:
[tex]N = mgcos\theta[/tex]
Now, create an equation for the force due to friction.
[tex]F_f = \mu mgcos\theta[/tex]
The work done by any force is:
[tex]W = F \cdot d\\\\W_f = \mu mgdcos\theta[/tex]
In this instance, d = the ramp's length, or 6 m.
Now, we can use the work-energy theorem.
Ei = Ef
However, there is energy dissipated; we can call this Wf (Work due to friction). Therefore:
Ei - Wf = Ef
Now, we can rearrange to solve for Wf:
Ei - Ef = Wf
Like above, there is initially only GPE (U = mgh) and finally only KE (K = 1/2mv²), so:
[tex]mgh - \frac{1}{2}mv^2 = \mu mgdcos\theta[/tex]
Solve for the coefficient of friction. Begin by canceling out the mass and multiplying all terms by 2:
[tex]2mgh - mv^2 = 2\mu mgdcos\theta\\\\2gh - v^2 = 2\mu gdcos\theta\\\\\mu = \frac{2gh - v^2}{2gdcos\theta}\\\\\mu = \frac{2(9.8)( 2.5357)- (2.82)^2}{2(9.8)(6)cos(25)}[/tex]
Evaluate:
[tex]\boxed{\mu = 0.39}[/tex]
The equation of state of a certain gas is p(V −b) = RT, where b is a constant. What order of magnitude do you expect b to be? Show that the internal energy of this gas is a function of temperature only
Answer:
LT 23 don't ask how I got it
what is a everyday activities examples of newtons 1 law of motion
Answer:
bouncing a baskletball
Explanation:
Answer:someone kicking a soccer ball.
Explanation:
Because the ball isn’t in motion until acted on by another object (the foot)
which vector is the sum of the vectors shown below ?
Answer:
B
Explanation:
1. While cruising along a dark sketch of highway at a speed of 25 m/s, you see that a bridge ahead has been washed out. You apply the brakes and uniformly slow down to a stop in 5.0 seconds.
The velocity values are ________. (Positive/Negative)
The acceleration value is _______. (Positive/Negative)
Therefore the object must be _________. (Speeding Up/Slowing Down)
2. A poorly tuned Yugo can accelerate from rest to a speed of 28 m/s in 20 s.
The velocity values are ________. (Positive/Negative)
The acceleration value is _______. (Positive/Negative)
Therefore the object must be _________. (Speeding Up/Slowing Down)
3. A bear is sitting at rest at t=0s. At t=5s, the bear notices honey 16 m away and takes off from rest accelerating at a rate of 2 m/s2 for 4 seconds to reach the honey.
The velocity values are ________. (Positive/Negative)
The acceleration value is _______. (Positive/Negative)
Therefore the object must be _________. (Speeding Up/Slowing Down)
4. A dog runs down his driveway with an initial velocity of -5 m/s for 8 seconds, then uniformly increases his speed to -10 m/s in 5 seconds.
The velocity values are ________. (Positive/Negative)
The acceleration value is _______. (Positive/Negative)
Therefore the object must be _________. (Speeding Up/Slowing Down)
5. You are driving on the highway at a rate of 40 m/s for 10 seconds when you notice a cop in front of you. Over the next 5 seconds you uniformly slow down to 35 m/s to avoid getting a speeding ticket.
The velocity values are ________. (Positive/Negative)
The acceleration value is _______. (Positive/Negative)
Therefore the object must be _________. (Speeding Up/Slowing Down)
6. You are traveling 20 m/s when the stoplight in front of you turns red. You step on your break to uniformly slow down to a rest in 5 seconds. You are stopped at the red light for 3 seconds when the light turns green. You speed back up to 20 m/s over the next 5 seconds.
The velocity values are ________. (Positive/Negative)
The acceleration value is _______. (Positive/Negative)
Therefore the object must be _________. (Speeding Up/Slowing Down)
The velocity values are ________. (Positive/Negative)
The acceleration value is _______. (Positive/Negative)
Therefore the object must be _________. (Speeding Up/Slowing Down)
1.
The velocity values are Negative.
The acceleration value is Negative.
Therefore the object must be Speeding Up.
2.
The velocity values are Positive.
The acceleration value is Positive.
Therefore the object must be Slowing Down.
3.
The velocity values are Positive.
The acceleration value is Positive.
Therefore the object must be Slowing Down.
4.
The velocity values are Negative.
The acceleration value is Negative.
Therefore the object must be Speeding Up.
5.
The velocity values are Positive.
The acceleration value is Positive.
Therefore the object must be Slowing Down.
6.
The velocity values are Negative.
The acceleration value is Negative.
Therefore the object must be Speeding Up.
((I tried my very best I am sorry if any answers I have given are incorrect, but I wish you good luck on your Test/Quiz or whatever you may be currently working on at the time. This took a while.))Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite A orbits at an altitude of 4500 km , and satellite B orbits at an altitude of 11100 km .
c) How much work would it require to change the orbit of satellite A to match that of satellite B?
Answer:
The required work done is [tex]6.5\times10^{9}\ J[/tex]
Explanation:
Given that,
Mass of each satellites = 940 kg
Altitude of A = 4500 km
Altitude of B = 11100 km
We need to calculate the potential energy
Using formula of potential
[tex]U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}[/tex]
Put the value into the formula
[tex]U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}[/tex]
[tex]U_{A}=-3.44\times10^{10}\ J[/tex]
We need to calculate the potential energy
Using formula of potential
[tex]U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}[/tex]
Put the value into the formula
[tex]U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}[/tex]
[tex]U_{B}=-2.14\times10^{10}\ J[/tex]
We need to calculate the value of [tex]k_{A}[/tex]
Using formula of [tex]k_{A}[/tex]
[tex]k_{A}=-\dfrac{1}{2}U_{A}[/tex]
Put the value into the formula
[tex]k_{A}=\dfrac{1}{2}\times3.44\times10^{10}[/tex]
[tex]k_{A}=1.72\times10^{10}\ J[/tex]
We need to calculate the value of [tex]k_{B}[/tex]
Using formula of [tex]k_{B}[/tex]
[tex]k_{B}=-\dfrac{1}{2}U_{B}[/tex]
Put the value into the formula
[tex]k_{B}=\dfrac{1}{2}\times2.14\times10^{10}[/tex]
[tex]k_{B}=1.07\times10^{10}\ J[/tex]
We need to calculate the work done
Using formula of work done
[tex]W=\Delta K+\Delta U[/tex]
[tex]W=(k_{B}-k_{A})+(U_{B}-U_{A})[/tex]
[tex]W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})[/tex]
[tex]W=\dfrac{1}{2}(U_{B}-U_{A})[/tex]
Put the value into the formula
[tex]W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})[/tex]
[tex]W=6.5\times10^{9}\ J[/tex]
Hence, The required work done is [tex]6.5\times10^{9}\ J[/tex]
5. A capacitor is discharging through a resistor. The initial voltage across the capacitor is 3.2 V at t = 0. The voltage across the capacitor at time t = 20 ms is 0.8 V. The time it takes for the voltage across the capacitor to drop from 0.8 V to 0.2 V is
Hi there!
Recall the equation for the voltage of a discharging capacitor:
[tex]V_C(t) = V_0e^{-\frac{t}{\tau}}[/tex]
V₀ = Initial voltage of the capacitor (V)
t = Time (s)
τ or RC = Time Constant (s)
With the given information, we can plug in the value for V₀:
[tex]V_C(t) = 3.2e^{-\frac{t}{\tau}}[/tex]
We are given that at t = 20 ms (0.02 s), the voltage of the capacitor is 0.8V. We can use this to solve for the time constant (τ).
[tex]0.8 = 3.2e^{-\frac{0.02}{\tau}}\\\\0.25 = e^{-\frac{0.02}{\tau}}[/tex]
Take the natural log of both sides and solve.
[tex]ln(0.25) = -\frac{0.02}{\tau}\\\\-1.3863 = -\frac{0.02}{\tau}\\\\\tau = \frac{0.02}{1.3863} = 0.0144 s[/tex]
Now, we can use this time constant to solve for the time taken for the voltage to drop from 0.8 V to 0.2 V. Solve for the time taken for the capacitor's voltage to drop to 0.2 V:
[tex]0.2= 3.2e^{-\frac{t}{0.0144}}\\\\0.0625 = e^{-\frac{t}{0.0144}}\\\\ln(0.0625) = -\frac{t}{0.0144}\\\\t = (-2.773)(-0.0144) = 0.04 s[/tex]
Now, subtract the times:
[tex]0.04 - 0.02 = 0.02 = \boxed{20 ms}[/tex]
g A satellite is orbiting planet Earth with a linear speed of 3,914 m/s. The orbital radius of the satellite in km is:
The orbital radius of the satellite orbiting planet Earth at the given linear speed is 26,000 km.
Orbital radius of the planetThe orbital radius of the planet is the distance of the planet from the center of the Earth. The orbital radius of the planet is calculated as follows;
[tex]v = \sqrt{\frac{GM}{r} }[/tex]
where;
M is mass of EarthG is gravitational constantr is the orbital radiusr = GM/v²
r = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴)/(3,914²)
r = 2.6 x 10⁷ m
r = 26,000 km
Thus, the orbital radius of the planet is 26,000 km.
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Please help I only have 20 min left!!!!
Power: The work performed as a function of time for a process is given by W = at3, where a = 2.4 J/s3. What is the instantaneous power output at t = 3.7 s?
Group of answer choices
99 W
139 W
208 W
69 W
The instantaneous power output at t = 3.7 s right answer is 99 W.
Power is defined by the amount of energy transferred over a period of time. Instantaneous power, on the other hand, refers to the power consumed at a point in time. Instantaneous power is an important metric in electronics. Instantaneous power is the power measured at a specific point in time.
The basic difference between average effort and instantaneous effort is that average effort is the ratio of total work time to total time. Although the instantaneous power is the limit of the average power. The statement of work does not state that the same amount of work he will complete in a second or an hour. Instantaneous power can be positive or negative. Positive instantaneous power means that energy is flowing from the source to the load, and negative instantaneous power means that energy is flowing from the load to the source.
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Please Help Me
When you apply brakes on a bicycle, the bicycle’s ________________________
energy is not destroyed; instead, the bicycle’s _____________________energy is
transformed into thermal energy. The __________________amount of energy remains the same.
Answer:
1. Acceleration begins to slow down quickly
2. Kenetic
3. Mass
Answer:
Acceleration slows down
Kinetic
Mass
Explanation:
A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionless surface below. Suppose you pull the mass so that the spring is stretched out (initial state) and then you release it, so that the mass starts moving towards the spring is unstretched position (final state). The impulse imparted on the spring-mass system by the force that the wall exerts on the spring is zero, since the wall does not move during this process.
Required:
What total percentage of the period does the mass lie in these regions?
Answer:
a) x=0 %T=0, b) x= A %T=100%, c) x=-A %T=50%
Explanation:
This is a simple harmonic movement exercise, which is explained by the expression
x = A cos (wt + Ф)
where angular velocity is related to frequency and period
w = 2π f = 2π / T
we can write the equation of the oscillation
x = A cos θ
When seeing the two equations they are equivalent, so what happens with the angle will also happen with time
We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching
The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return
a) for the unstretched spring point x = 0
In general, both distance and time are measured from this point, so the percentage of time is zero.
% T = 0
b) for x = A
let's find the angle
cos tea = x / A = 1
therefore the angles tea = 2π rad
when the movement reaches the point of 2π radians it begins to repeat so the period is complete
% T = 100%
c) the point of maximum compression x = -A
let's look for the angles
cos tea = x / A = -1
therefore the angles tea = π rad
at this point the movement is halfway so it should take half the time
% T = 50%