Answer:
192
Explanation:
It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is x(t)=Ccos(ωt)+Ssin(ωt), where C, S, and ω are constants.
A) Using the general equation for x(t) given in the problem introduction, express the initial position of the block xinit in terms of C, S, and ω (Greek letter omega).
b) Find the value of S using the given condition that the initial velocity of the block is zero: v(0)=0.
c)What is the equation x(t) for the block? Express your answer in terms of t, ω, and xinit.
d)Find the equation for the block's position xnew(t) in the new coordinate system.
Express your answer in terms of L, xinit, ω (Greek letter omega), and t.
The characteristics of the expression of the simple harmonic motion allows to find the results for the expression of the mass- block system are:
A) The constant Ces: C = xinit
B) The ocsntna S is: S = 0
C) The equation of the system is: x = xinit cos wt
D) If the reference system is at some extreme, the equation is:
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
The simple harmonic movement is an oscillatory movement where the restoring force is proportional to the displacement, the general equation that describes this movement is indicated.
x = C cos wt + S sin wt
Where x is the displacement C and S are constants. W the angular velocity and t the time.
A) The initial position of the body occurs when the time is zero, t = 0
We substitute.
x = C cos 0 + S sin 0
[tex]x_{init}[/tex] = C
B) The velocity of the particle is defined.
[tex]v= \frac{dx}{dt} \\ v= C w \ sin \ wt - Sw \ cos \ wt[/tex]
The initial velocity occurred for time zero t = 0
v = - S w
It indicates that the initial velocity is zero, since the angular velocity must be different from zero, it implies that the constant is valid.
S = 0
C) The equation for the block remains.
x (t) = [tex]x_{init} \ cos \ wt[/tex]
D) In some cases it is measured with respect to another reference system, the most common are:
For maximum compression it is the zero of the system. The maximum extension is the zero of the system.
In these cases, the change that must be made is
x = [tex]L - x_{min}[/tex] t
we substitute
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
L = [tex]x_{init}[/tex] (1 + cos wt)
In conclusion, using the characteristics of the expression of the simple harmonic motion we can find the results for the expression of the mass- block system are:
A) The constant Ces: C = xinit
B) The ocsntna S is: S = 0
C) The equation of the system is: x = xinit cos wt
D) If the reference system is at some extreme, the equation is:
[tex]L - x_{init} = x_{init} \ cos \ wt[/tex]
Learn more about simple harmonic motion here: brainly.com/question/17315536
A rock is thrown off a cliff with a speed of 5 m/s downward. How far will it fall after 7 seconds have elapsed?
Free-fall Acceleration is -10 m/s^2
I also need the Formula
Answer:
Explanation:
s = s₀ + v₀t + ½at²
if the throw point is origin and UP the positive direction and ignoring air resistance.
s = 0 + (-5)(7) + ½(-10)(7²)
s = 0 - 35 - 245
s = - 280 m
For the ballistic missile aimed to achieve the maximum range of 9500 km, what is the maximum altitude reached in the trajectory
Explanation:
The range R of a projectile is given the equation
[tex]R = \dfrac{v_0^2}{g}\sin{2\theta}[/tex]
The maximum range is achieved when [tex]\theta = 45°[/tex] so our equation reduces to
[tex]R_{max} = \dfrac{v_0^2}{g}[/tex]
We can solve for the initial velocity [tex]v_0[/tex] as follows:
[tex]v_0^2 = gR_{max} \Rightarrow v_0 = \sqrt{gR_{max}}[/tex]
or
[tex]v_0 = \sqrt{(9.8\:\text{m/s}^2)(9.5×10^6\:\text{m})}[/tex]
[tex]\:\:\:\:\:\:\:=9.6×10^3\:\text{m/s}[/tex]
To find the maximum altitude H reached by the missile, we can use the equation
[tex]v_y^2 = v_{0y}^2 - 2gy = (v_0\sin{45°})^2 - 2gy[/tex]
At its maximum height H, [tex]v_y = 0[/tex] so we can write
[tex]0 = (v_0\sin{45°})^2 - 2gH[/tex]
or
[tex]H = \dfrac{(v_0\sin{45°})^2}{2g}[/tex]
[tex]\:\:\:\:\:\:= \dfrac{[(9.6×10^3\:\text{m/s})\sin{45°}]^2}{2(9.8\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:\:\:= 2.4×10^6\:\text{m}[/tex]
two billiard balls moving along the same line hit each other head-on. each has a mass of 0.220 kg; one has an initial velocity of 1.84 m/s, the other an initial velocity of 0.530 m/s. if the collision is elastic, what are their final velocities? ignore friction.
Hi there!
Since the collision is elastic, we must also satisfy the following condition:
Ei = Ef, or:
KEi = KEf
Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.
mv1 + mv2 = mvf1 + mvf2
0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)
0.2882/0.220 = vf1 + vf2
1.31 = vf1 + vf2
Now, we can express this as a conservation of energy:
1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²
Plug in values and simplify:
0.403315 = 1/2m(vf1² + vf2²)
Simplify further:
3.6665 = vf1² + vf2²
Use the equation derived from momentum above and solve for one variable:
vf2 = 1.31 - vf1
Plug in this expression for vf2:
3.6665 = vf1² + (1.31 - vf1)²
Expand:
3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²
Simplify:
1.9504 = -2.62vf1 + 2vf1²
Solve for vf1 using a graphing calculator:
vf1 = -0.53 m/s or 1.84 m/s; we must figure out which one is correct.
Since v1 is heading to the right initially with a velocity of 1.84 m/s, we know that the ball's velocity could not have stayed the same in both magnitude and direction, so the final velocity must be -0.53 m/s.
Now, we can solve for the velocity of the other ball (initial of 0.53 m/s):
vf2 = 1.31 - (-0.53) = 1.84 m/s.
Now, you could have also made the connection that when two balls of the SAME MASS experience an ELASTIC collision, the velocities are simply "exchanged" from one to another. I just used this more "extensive" method to prove this.
A flywheel having constant angular acceleration requires 4.70 s to rotate through 164 rad . Its angular velocity at the end of this time is 101 rad/s . Find the angular velocity at the beginning of the 4.70 s interval. Find the angular acceleration of the flywheel.
Answer:
A) -31.2 rad/s
B) 28.1 rad/s^2
Explanation:
a stone weight 490 N on a planet whose acceleration due to gravity is half that of the Earth. Calculate (1) the mass of a stone
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
Acceleration due to gravity on that planet is half than that of om earth ~
that is ~
[tex] \dfrac{9.8}{2} \: \: m/ {s}^{2} [/tex][tex]4.9 \: \: m/s {}^{2} [/tex]And it's weight is equal to ~
[tex]490 \: \: N[/tex]now, As we know ~
[tex]f = m \times g[/tex]here ,
f = Weight = 490 Newtonsm = mass of stone = ?g = Acceleration due to gravity on that planet = 4.9 m/s²let find the mass of stone ~
[tex]490 = m \times 4.9[/tex][tex]m = \dfrac{490}{4.9} [/tex][tex]m = 100 \: \: kg[/tex]Mass of the stone is 100 kg ~
I hope it helped
The control center of human body is * 1/1 heart
brain
liver
We have ________ sense organs * 1/1
6
4
5
Which energy we get by burning of fuels * 1/1
Solar
Heat
Light
Sources of energy that can be used again and again will never run out * 1/1 Non renewable energy Renewable energy Running water What is called structure that build on rivers to get electricity * 1/1 Dam Check dam Tunnel Which nutrients protect us from diseases * 1/1 Carbohydrate Vitamin and Mineral Protein When is the best time of the day to water your lawn * 1/1 Early morning and late evening Afternoon Night Which of these ways to wash the car saves the most water? * 1/1 Wash it in the driveway with the garden hose Drive it into the lake Take it through a car wash that recycles water Why do we keep food and vegetables in refrigerator? * 1/1
A 2 kg ball is rolling down a hill at a constant speed of 4 m/s. How much kinetic energy does the ball have?
Objects 1 and 2 attract each other with a gravitational force of 179 units. If the distance separating objects 1 and 2 is changed to four times the original value (i.e., quadrupled), then the new gravitational force will be ______ units.
Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant throughout the universe.
D is the distance between the 2 objects.
the distance is now quadrupled.
Fgravitynew = G*(mass1*mass2)/(4D)² =
= G*(mass1*mass2)/(16D²) =
= (G*(mass1*mass2)/D²) / 16 = Fgravity/16
the new gravitational force will be 179/16 = 11.1875 units
What happens when the object is placed at F? Explain
your answer.
Answer:
Sample Response: No image will be formed because the rays will not converge to or diverge from a common point.
Explanation:
PLEASE HELP!
A 9kg particle is initially at rest at x=0. It is subject to a single force Fx (N) which varies with x (m) as shown in the
diagram
F
2
1
0
ББ by
x
- 1
1
-2
The kinetic energy of the particle when it is at x = 3 m is:
Hi there!
With a Force/Displacement curve, we must take the integral (area underneath the curve) to calculate the work done.
We know that:
W = ΔKE
Calculate the work by finding the area underneath the force curve from
x = 0 to 3 m:
We can use a trapezoid:
A = 1/2(3 + 2)(3) = 7.5 J
This is the amount of work done, and since the object starts from rest:
7.5J = KEf - KEi (0 J)
7.5J = KEf = 7.5 J
The center of mass of a 1600 kg car is midway between the wheels and 0.7 m above the ground. The wheels are 2.6 m apart. (a) What is the minimum acceleration A of the car so that the front wheels just begin to lift off the ground
Answer:
Explanation:
I guess we are ASSUMING that this is a rear wheel drive car as a front wheel drive car will never get the front wheel normal force to zero
If we consider it as a statics problem and choose our moment center carefully...say 0.7 m above the rear wheel to ground contact point.
Call the traction force at the rear wheels F
The normal force on the front wheels will be zero, so no moment generated by the front wheels.
Summing moments about our chosen point to zero
1600(9.8)[2.6 / 2] - F[0.7] = 0
F = 291,200
this force will create an acceleration of
a = F/m
a = 291200/1600
a = 182 m/s²
which is about 18.6 times gravity acceleration
Pendulum makes 12 complete swings in 8 seconds, what are its frequency and period on earth
Hi there!
We can begin by finding the period of the pendulum.
[tex]T = \text{ # of complete swings / seconds} = 12 / 8 = \boxed{\text{1.5 sec}}[/tex]
The frequency is simply the reciprocal of the period, so:
[tex]f = \frac{1}{T} = \frac{1}{1.5} = \frac{2}{3}Hz \text{ or } \boxed{0.67 Hz}[/tex]
A car with an initial position of 10.0 m
and an initial velocity of 16.0 m/s accelerates at an average rate of 0.50 m/s2 for 4.0 s. What is the car’s position after 4.0 s?
Answer:
78
Explanation:
x=xi+vi(t)+1/2a(t)^2
x=10+16(4)+1/2(0.50)(4)^2
x=74+4
x=78 m
A clothes dryer in a home draws a current of 10 amps when connected on a special 220-volts household circuit.what is the resistance of the dryer?
Answer:
22Ω
Explanation:
if V ⇒ voltage
I ⇒ current
R ⇒ resistance
V = IR
220 = 10 x R
220 / 10 = R
22 = R
what does stimlus mean
Answer:
a thing or event that evokes a specific functional reaction in an organ or tissue.
or
a thing that rouses activity or energy in someone or something; a spur or incentive.
All circuits include
a battery, wires, and a switch.
an energy source, a resistor, and a battery.
a battery, a light bulb, and a switch.
an energy source, a load, and wires.
Answer:
a battery, wires, and a switch.
Explanation:
All circuits include?
If the penny is thrown horizontally at 25 m/s from the 170 meter building, how long will it take for the penny to hit the ground?
9514 1404 393
Answer:
about 5.89 seconds
Explanation:
The penny will hit the ground at the same time it would if it were simply dropped. The equation for the vertical motion is ...
h(t) = -4.9t^2 +170 . . . . . where 170 is the initial height in meters
h(t) = 0 when ...
4.9t^2 = 170
t = √(170/4.9) ≈ 5.89
The penny will hit the ground in about 5.89 seconds.
Why do you suppose Km values are so frequently standardized and published, drawing attention to the value of Vmax/2, rather than Vmax itself
Km values are standardized because half the Vmax (Vmax/2) is more informative than Vmax. This value (Km) can be used to calculate the affinity of the enzyme by a given substrate.
The Km (Michaelis constant) of the enzyme refers to the value in which the concentration of substrate is equal to half of its maximum velocity (Vmax/2).
This value (Km) is inversely proportional to the affinity of an enzyme by a given substrate.
An enzyme showing a high Km also exhibits a low affinity for its specific substrate, and thereby this enzyme requires a high concentration of the substrate to reach its maximum velocity (Vmax).
In consequence, the Km value is a more informative value than the maximum velocity (Vmax), which only indicates the concentration of an enzyme catalyzing a reaction under ideal conditions.
Learn more in:
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An object with an initial velocity of 10 m/s accelerates at a rate of 3.5 m/s2 for 8 seconds. How far will it have traveled during that time?
Free-fall Acceleration is -10 m/s^2
I also need the formula
Answer:
Explanation:
s = s₀ + v₀t + ½at²
assuming that the acceleration is in the direction of initial velocity.
(it would not have to be so)
s = 0 + 10(8) + ½(3.5)(8²)
s = 192 m
a man can jump 9meteres on the moon.how high can he jump on the earth.
Answer:
You can jump 1.5 feet on Earth.
Explanation:
Because the moons gravity is weaker that earth so it would be easier to jump further on the moon.
Objects 1 and 2 attract each other with a gravitational force of 12 units. If the mass of Object 2 is tripled, then the new gravitational force will be _____ units.
Explanation:
Fgravity = G*(mass1*mass2)/D².
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between the objects.
so, if you triple one of the masses, what does that do to our equation ?
Fgravitynew = G*(3*mass1*mass2)/D²
due to the commutative property of multiplication
Fgravitynew = 3* G*(mass1*mass2)/D² = 3* Fgravity
so, the right answer is 3×12 = 36 units.
An object is dropped from a vertical height of 1.89 m above the balcony level. What is the object’s speed when it is 2.20 m below the balcony level if 10.0% energy is lost due to the air resistance? Does it matter when to apply 10% loss before V calculations or after? [8.49m/s] [yes it does, 0.9Energy result in √0.9Velocity]
a.
The object's speed at 2.20 m below balcony level is 8.74 m/s
Let the balcony level be 0 m and the height above the balcony level be positive and height below the balcony level negative.
Using the principle of conservation of energy, the total energy at a vertical height of 1.89 m above the balcony level equals the total mechanical energy when the object is 2.20 m below the balcony level and
So, E = E'
U + K + f = U' + K' + f'
where U = initial potential energy at 1.89 m = mgh, K = initial kinetic energy at 1.89 m = 0 J(since it is released from rest), f = energy loss at 1.89 m = 0 J, U' = final potential energy at 2.20 m below balcony level = mgh', K = final kinetic energy at 2.20 m = 1/2mv², f' = energy loss at 1.89 m = 10%U = 0.10mgh(since 10% of the initial energy is lost).
So,
U + K + f = U' + K' + f'
mgh + 0 + 0 = mgh' + 1/2mv² + 0.10mgh
mgh = mgh' + 1/2mv² + 0.10mgh
Dividing through by m, we have
gh = gh' + 1/2v² + 0.10gh
So, gh - 0.10gh = gh' + 1/2v²
0.90gh = gh' + 1/2v²
1/2v² = 0.90gh - gh'
1/2v² = g(0.90h - h')
v² = 2g(0.90h - h')
Taking square-root of both sides, we have
v = √[2g(0.90h - h')]
where v = velocity of object at 2.20 m below balcony level, h = height above the balcony level = 1.89 m, h' = height below the balcony level = -2.20 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the values of the variables into the equation, we have
v = √[2g(0.90h - h')]
v = √[2 × 9.8 m/s²{0.90 × 1.89 m - (-2.20 m)}]
v = √[2 × 9.8 m/s²(1.701 m + 2.20 m)]
v = √[2 × 9.8 m/s²(3.901 m)]
v = √[76.4596 m²/s²]
v = 8.74 m/s
So, the object's speed at 2.20 m below balcony level is 8.74 m/s
b.
Yes it does matter when we apply 10% loss before V calculations
We need to apply the 10 % loss before V calculations because this would give us a proper value for V since the energy is lost before V is obtained.
So, yes it does matter when we apply 10% loss before V calculations
Learn more about speed of a falling object here:
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Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.150 rev/srev/s . As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight.
What is the radius of the circle in which Jack travels? Treat him as a point mass.
Answer:
Explanation:
At the top of the arc, 3/4 of the acceleration of gravity is use to supply the necessary centripetal acceleration.
0.75g = ω²R
R = 0.75g/ω²
R = 0.75(9.81) / (0.15 rev/s)(2π rad/rev)²
R = 8.283006...
R = 8.28 m
this is electricity in physics please help
Explanation:
a. (i) When the variable resistor is set at zero, the only resistance in the circuit is due to the lamp. So the current flowing through the circuit is
[tex]I = \dfrac{V}{R} = \dfrac{220\:\text{V}}{440\:Ω} = 0.5\:\text{A}[/tex]
(ii) The power output P of the lamp is given by
[tex]P = I^2R = (0.5\:\text{A})^2(440\:Ω) = 110\:\text{W}[/tex]
b. (i) The variable resistor is in a series connection to the lamp so when its value is set to its maximum value of 660 Ω, the total resistance of the circuit is simply the sum of the two resistances:
[tex]R_T = R_{vr} + R_L = 660\:Ω + 440\:Ω = 1100\:Ω[/tex]
Therefore, the current through the circuit is
[tex]I = \dfrac{V}{R_T} = \dfrac{220\:\text{V}}{1100\:Ω} = 0.20\:\text{A}[/tex]
(ii) Using the result in Part (ii), we can solve for the potential difference across the lamp as follows:
[tex]V_L = IR_L = (0.20\:\text{A})(440\:Ω) = 88\:\text{V}[/tex]
(iii) The power output of the lamp is
[tex]P = I^2R_L = (0.20\:\text{A})^2(440\:Ω) = 17.6\:\text{W}[/tex]
(iv) The rate at which electrical energy is supplied, i.e., the power output of the circuit is equal to the square of the current multiplied by the total resistance of the circuit:
[tex]P = I^2R_T = (0.20\:\text{A})^2(1100\:Ω) = 44\:\text{W}[/tex]
A 1.1 kg ball drops vertically onto a floor, hitting with a speed of 23 m/s. It rebounds with an initial speed of 5.0 m/s. (a) What impulse acts on the ball during the contact
Hi there!
We know that:
I = Δp = m(vf - vi)
Plug in the given values. Remember to take into account direction ⇒ let the rebound velocity be positive and initial be negative.
I = 1.1(5 - (-23)) = 30.8 Ns
plz answer the question.
Answer:
A. Determine the equation of constraints
Just before it strikes the ground, what is the watermelon's kinetic energy?
Answer:
Answer: At its lowest point, the kinetic energy of a watermelon just as it touches the ground is zero if it does not touch anything on its way down.
Explanation:
This is because that upon having been dropped from a height, an object no longer has any kinetic energy at all. Kinetic energy transforms to gravitational potential energy during the fall and there's nothing left over for kinetic once you stop accelerating anymore. Fortunately, things don't stay still until they land very often! For example, if a person catches the fruit with his hands after some air resistance slows him down - making him more similar in speed to the lag of the trajectory - then he'll be able to share some of his saved up gravitational potential with that watermelon and do some
The car on this ramp starts from rest. When released, it
accelerates at a constant rate. It has an initial position of 12 cm
from the top of the ramp, and has an average velocity of 1.20 m/s
for a total of 1.80 s. Which is the correct final position of the car?
Answer:
Explanation:
s 0.12 + 1.20(1.80) = 2.28 m from the top.
A boat is using echo-sounding equipment to measure the depth of the water underneath it, as illustrated in the first diagram.
The equipment in the boat sends a short pulse of sound downwards and detects the echo after a time interval of 0.80s. i Describe how an echo is caused. ii The speed of sound in water is 1500 m/s. Calculate the distance travelled (in metres) by the sound in 0.80 s.
Answer:
Explanation:
Echo is caused by sound energy reflecting off of "hard" surfaces. It could be as simple as a change in density of the material the sound is traveling through.
In 0.8 s, the sound has traveled 0.8(1500) = 1200 m.
That means the object that reflected the sound is 600 m below the boat. The sound took 0.4 s to reach the object and another 0.4 s to return the echo.