We will have the following:
a) We first determine the time it takes to travel the distance to both vehicles:
*
[tex]t_1=\frac{14mi\ast1h}{55}\Rightarrow t_1=\frac{14}{55}h[/tex]*
[tex]t_2=\frac{14mi\ast1h}{60mi}\Rightarrow t_2=\frac{11}{12}h[/tex]So, we determine now the difference in time:
[tex]\frac{11}{12}h-\frac{14}{55}h=\frac{437}{660}h\approx0.66h[/tex]So, the fastest car will arrive approximately 0.66 hours sooner.
b) We determmine the distance it must travel the fastest car to have a 15 minute lead on the other one as follows:
First, we determine the time difference required:
[tex]t=\frac{15min\ast1h}{60min}\Rightarrow t=0.25h[/tex]Then, since both vehicles will move relative to each other, we will have that:
[tex]d_{c1}=(60mi/h)(0.25h)\Rightarrow d_{c1}=15mi[/tex]So, the fastest car must be 15 miles ahead of the other car in order to have a 15 minute lead with respect to the second car.
A 72.5 kg student sits at a desk 1.25 m away from a 80.0 kg student. What is the magnitude of the gravitational force between the two students?
Given:
The mass of one student is,
[tex]m_1=72.5\text{ kg}[/tex]The mass of the other student is,
[tex]m_2=80.0\text{ kg}[/tex]The distance between them is,
[tex]d=1.25\text{ m}[/tex]The gravitational force between them is,
[tex]F=G\frac{m_1m_2}{d^2}[/tex]Here the gravitational constant is,
[tex]G=6.6\times10^{-11}\text{ }\frac{N.m^2}{\operatorname{kg}}[/tex]Substituting the values we get,
[tex]\begin{gathered} F=\frac{(6.6\times10^{-11})\times72.5\times80.0}{(1.25)^2} \\ =2.5\times10^{-7}\text{N} \end{gathered}[/tex]Hence the second option is correct.
A penny is dropped from a building and it takes 7.00 seconds to hit the ground.
Based on the given question, the final velocity of the penny is 68.6m/s
Calculations and ParametersBased on the definition of velocity, we can see that it has to do with the speed at which an object moves with at a particular direction.
If we observe the 1st law of motion:
v = u + gt
Where:
v = final velocity
u = initial velocity
g = acceleration due to gravity
t = time taken
We need to find the velocity as it was not given.
We already know that
u = 0 (it was at rest),
t = 7s
v = unknown
Let us put in the given values based on the equation and solve for the answer.
v = u + gt
v = 0 + ( 9.8m / s ² × 7.00s )
v = 9.8m / s ² × 7.00s
v = 68.6m/s
Therefore, the final velocity of the penny is 68.6m/s
Read more about velocity here:
https://brainly.com/question/25749514
#SPJ1
A penny is dropped from a building, and it takes 7 seconds to hit the ground, so the final velocity will be 68.6 m/s.
What is Velocity?The ratio of displacement to time is referred to as velocity of the object. It has SI unit meter per second or m/s and has a dimension formula LT⁻¹.
We already know that
Initial speed, u = 0
Time, t = 7 seconds
Use the equation of motion to find the final velocity,
v = u + gt
v = 0 + ( 9.8 m / s ² × 7.00s )
v = 9.8 m / s ² × 7.00s
v = 68.6 m/s
Hence, the penny hit the ground with a final velocity of 68.6 m/s.
To get more information about Velocity :
https://brainly.com/question/18084516
#SPJ1
On a standard day the speed of sound is 345 meters per second. A whistle whose frequency is 725 Hz is movingtoward an observer at a speed of 25.2 meters per second. What is the wavelength of the sound at the observer?(a) 0.367 m(b) 0.441 m(c) 0.511 m(d) 0.623 m
Take into account that this is a situation where the source moves presenting the Doppler effect.
In order to determine the wavelength of the sound generatd by the whistle at the observer, first calculate the frequency at the observer by using the following formula:
[tex]f=\frac{v}{v-v_s}f_s[/tex]where:
f: frequency at the observer = ?
fo: source frequency = 725 Hz
vs: source speed = 25.2 m/s
v: speed of sound = 345 m/s
replace the previous values of the parameters into the fomrula for f:
[tex]\begin{gathered} f=\frac{345m/s}{345m/s-25.2m/s}725Hz \\ f=782.1Hz \end{gathered}[/tex]Next, use the following formula to determine the wavelength of the sound at observer, by using the previous result:
[tex]\lambda=\frac{v}{f_s}=\frac{345m/s}{782.1Hz}=0.441m[/tex]Hence, the wavelength of the sound at the observer is 0.441 m
A 27.6 kg block is pulled along a rough level surface at constant velocity by the force . The figure shows the free-body diagram for the block. FN represents the normal force on the block; and f represents the force of kinetic friction. The coefficient of kinetic friction is 0.30, what is the strength of P?
The strength of P, the pulling force is 352 N.
What is friction?Friction is the force that acts to oppose the motion of an object moving over another object at their surfaces of contact.
The force of friction depends on the following:
mass/weight of the objectnature of the surface of contactConsidering the free-body diagram of the given object, the forces acting on the object are as follows:
FN = normal force acting in the opposite direction and equal to the weight of the object, mgmg = the weight of the object acting downwardsP = the pulling force in the forward directionf = frictional force acting in an opposite direction to PSolving for P:
P = f + mg
μ = f/N
f = μ * N
N = mg
f = μ * m * g
f = 0.3 * 27.6 * 9.81
f = 81.2 N
P = 81.2 N + 27.6 * 9.81
P = 352 N
Learn more about friction at: https://brainly.com/question/4618599
#SPJ1
A PVC pipe has a length of 45.132 centimeters.a. What are the frequencies of the first three harmonics when the pipe is open at both ends? Include units in your answers.b. What are the frequencies of the first three harmonics when the pipe is closed at one end and open at the other? Include units in your answers.
ANSWERS
a. f₁ = 380 Hz; f₂ = 760 Hz; f₃ = 1140 Hz
b. f₁ = 190 Hz; f₃ = 570 Hz; f₅ = 950 Hz
EXPLANATION
a. For a pipe of length L open at both ends, the frequencies of the first three harmonics are:
[tex]\begin{cases}f_1=\frac{v}{2L} \\ \\ f_2=2f_1=\frac{v}{L} \\ \\ f_3=3f_1=\frac{3v}{2L}\end{cases}[/tex]Assuming that the speed of the wave is the speed of sound: 343 m/s and knowing that the length of the pipe is L = 45.132 cm = 0.45132 m we can find the frequencies of the first three harmonics:
[tex]\begin{cases}f_1=\frac{343m/s}{2\cdot0.45132m}\approx380Hz \\ \\ f_2=2f_1=2\cdot380Hz\approx760Hz \\ \\ f_3=3f_1=3\cdot380Hz\approx1140Hz\end{cases}[/tex]b. For a pipe of length L closed at one end and open at the other, the frequencies of the first three harmonics are:
[tex]\begin{cases}f_1=\frac{v}{4L} \\ \\ f_2=DNE \\ \\ f_3=3f_1=\frac{3v}{4L}\end{cases}[/tex]In a closed pipe, there can only be odd harmonics (1, 3, 5...). Therefore, the second harmonic does not exist and the "third harmonic" would be the 5th,
[tex]\begin{cases}f_1=\frac{v}{4L} \\ \\ f_3=3f_1=\frac{3v}{4L} \\ \\ f_5=5f_1=\frac{5v}{4L}\end{cases}[/tex]Again, the length of the pipe is 45.132 cm = 0.45132 m, so the first three harmonics are:
[tex]\begin{cases}f_1=\frac{343m/s}{4\cdot0.45132m}\approx190Hz \\ \\ f_3=3f_1=3\cdot190Hz=570Hz \\ \\ f_5=5f_1=5\cdot190Hz=950Hz\end{cases}[/tex]Two parallel wires carry currents of 20 A and 10 A, respectively, in the opposite direction and have a length of 5 m each. Determine the force between the wires if the distance between the wires is 1.0 cm.0.010 N, attraction0.010 N, repulsion0.020 N, attraction0.020 N, repulsion
Given:
The currents, I₁=20 A
I₂=10 A
The length of the wires, L=5 m
The distance between two wires, d=1.0 cm=0.01 m
To find:
The magnetic force between the wires.
Explanation:
As the wires are conducting the current in opposite directions, the nature of the force between the wires is repulsive.
The magnitude of the force between two wires is given by,
[tex]F=\frac{\mu_0I_1I_2L}{2\pi d}[/tex]Where μ₀=4π×10⁻⁷ H/m is the permeability of free space.
On substituting the known values,
[tex]\begin{gathered} F=\frac{4\pi\times10^{-7}\times20\times10\times5}{2\pi\times1\times10^{-2}} \\ =0.02\text{ N} \end{gathered}[/tex]Final answer:
The magnetic force between wires is 0.020 N, repulsive.
Austin does his Power lifting every morning to stay in shape. He lifts a 90 kg barbell, 2.3 m above the ground.a) How much energy does it have when it was on the ground? Jb)How much energy does it have after being lifted 2.3 m? Jc) What kind of energy does it have after being lifted? d) How much work did Austin do to lift the barbell? Je) If he lifted it in 1.9s, what was his power? W
ANSWER
[tex]\begin{gathered} (a)0J \\ (b)2030J \\ (c)\text{ Potential energy} \\ (d)2030J \\ (e)1070W \end{gathered}[/tex]EXPLANATION
Parameters given:
Mass of barbell, m = 90 kg
Height above ground, h = 2.3 m
(a) We want to find the energy the barbell has on the ground. #
When it is on the ground, the barbell is stationary, which means its velocity is 0 m/s, hence, its kinetic energy is also 0 J, since kinetic energy is given as:
[tex]\begin{gathered} KE=\frac{1}{2}mv^2 \\ KE=\frac{1}{2}\cdot m\cdot0=0J \end{gathered}[/tex]Also, on the ground, it is at a height of 0 m, hence, its potential energy is 0 J:
[tex]\begin{gathered} PE=mgh \\ PE=m\cdot g\cdot0=0J \end{gathered}[/tex]where g = acceleration due to gravity
Therefore, on the ground, the energy the barbell had was 0 J.
(b) After it had been lifted 2.3 m, its height above the ground became 2.3 m.
Now, we can find the potential energy possessed by the barbell:
[tex]\begin{gathered} PE=90\cdot9.8\cdot2.3 \\ PE=2028.6J\approx2030J \end{gathered}[/tex]After it is lifted, it is once again stationary, hence, it has no kinetic energy.
Therefore, the energy the barbell has after it has been lifted 2.3 m is 2070J.
(c) As stated in (b) above, after being lifted, the barbell only possesses potential energy since it is at a height above the ground and it is not moving.
(d) The work done in lifting the barbell is equal to the force applied multiplied by the height moved by the barbell.
That is:
[tex]W=F\cdot d[/tex]The force applied is equal to the weight of the barbell:
[tex]\begin{gathered} F=W=mg \\ F=90\cdot9.8 \\ F=882N \end{gathered}[/tex]Therefore, the work done is:
[tex]\begin{gathered} W=882\cdot2.3 \\ W=2028.6J\approx2030J \end{gathered}[/tex](e) He lifted the barbell in 1.9 seconds. To find his power, we have to divide the work done by the time taken to do the work.
That is:
[tex]\begin{gathered} P=\frac{W}{t} \\ P=\frac{2030}{1.9} \\ P=1068.4W\approx1070W \end{gathered}[/tex]That was his power.
The train above is traveling at a constant velocity because the forces acting on it are in equilibrium. Therefore, the missing force must have a magnitude (blank) of newtons to the (blank).
The missing force has a magnitude of 800 N to the right
Explanation:The forces acting on the train are in equilibrium.
This means that the sum of all the forces acting in the right direction equals the sum of all the forces acting in the left direction
Let the missing force be represented by F
1700 = 900 + F
F = 1700 - 900
F = 800 N
Therefore, the missing force has a magnitude of 800 N to the right
What is the displacement of the particle in the time interval 7 seconds to 8 seconds?OA. O metersОВ.1.5 metersOC. 3 metersOD. 7 meters
Given,
A velocity-time graph.
The area under the curve of a velocity-time graph of an object gives us the displacement of the object.
From the graph, we can see that the area under the curve from 7 seconds to 8 seconds is a triangle.
The height of the triangle is h=6 m/s.
And the base of the triangle is b=1 s.
The area of a triangle is given by,
[tex]A=\frac{1}{2}bh[/tex]On substituting the known values,
[tex]\begin{gathered} A=0.5\times6\times1 \\ =3\text{ m} \end{gathered}[/tex]Therefore the displacement of the particle in the time interval 7 s to 8 s is 3 meters.
Thus, the correct answer is option C.
You decide to roll a 0.10 kg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelength. If you roll it at 1.4x1^-3 m/s, what is its wavelength? Express your answer to two significant figures and include the appropriate units. How will the answer from above compare with the de Broglie wavelength of the high-speed electron that strikes the back face of one of the early models of a TV screen at 1/10 the speed of light (2x10^-11 m) ?
ANSWER:
(a) 4.73*10^-30 m
(b) 2.37*10^-19 times smaller
STEP-BY-STEP EXPLANATION:
Given:
Mass of ball = m = 0.10 kg
Speed of ball = v = 1.4x10^-3 m/s
(a)
Since, de Broglie wavelength is given by:
[tex]\lambda=\frac{h}{mv}[/tex]Where, h is the Plank's Constant ( h = 6.626x10^-34 kg m^2/s). Therefore, de Broglie wavelength of the ball will be:
[tex]\begin{gathered} \lambda=\frac{6.626\cdot10^{-34}}{0.10\cdot1.4\cdot10^{-3}} \\ \lambda_{\text{ball}}=4.73\cdot10^{-30} \end{gathered}[/tex](b)
[tex]\begin{gathered} \lambda_{electron}=2\cdot10^{-11} \\ \frac{\lambda_{\text{ball}}}{\lambda_{electron}}=\frac{4.73\cdot10^{-30}}{2\cdot10^{-11}} \\ \frac{\lambda_{\text{ball}}}{\lambda_{electron}}=2.37\cdot10^{-19} \end{gathered}[/tex]It means that the wavelength of the ball is 2.37*10^-19 times smaller
In order to hear a sound, even though there is an obstacle between you and the source, the sound wave must:A.diffract.B.refract.C.shorten.D.reflect.
We will have the following:
The sound must diffract. [Option A]
Convert each quantity to the indicated units.
a. 3.01 g to cg
b. 6200 m to km
c. 0.13 cal/g to kcal/g
Answer:
Explanation:
a) 301 cg
b) 6.2 km
c) 0.00013
A jet engine applies a force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m, the kinetic energy gained by the rocket is
We will have the following:
[tex](300N)(2m)+(-200N)(2m)=200J[/tex]Then, we from the work-kinetic force theorem we will have that the total kinetic energy gained by the rocket was 200 Joules.
An object has a position function x(t) = 5t m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.
Considering the given position function, it is found that:
a) The velocity function is: v(t) = 5 m/s.
b) The functions are graphed at the end of the answer.
Position and velocity functionThe position function in this problem, after t seconds, is defined according to the following rule:
s(t) = 5t.
The velocity function is the derivative of the position function, hence it is calculated as follows:
v(t) = s'(t) = [5t]' = 5 m/s. (position in meters, hence velocity in meters per second).
The derivative rule applied was the power rule, [5t]' = 5[t'] = 5 x 1 x t^(1 - 1) = 5t^0 = 5.
These two functions are graphed at the end of the answer, considering a domain of t ≥ 0, as time cannot assume negative values.
More can be learned about position and velocity at https://brainly.com/question/26150488
#SPJ1
Multiple Choice: A car with a mass of 825 kg moves along the roadway at aspeed of 15 m/s to the east. What impulse is required to decrease the speed of theboat to 10 m/s east?
Given:
The mass of the car is m = 825 kg
The intial speed of the car is
[tex]v_i\text{ = 15 m/s}[/tex]towards east.
The final speed of the car is
[tex]v_f\text{ = 10 m/s}[/tex]To find the impulse of the car.
Explanation:
The impulse can be calculated by the formula
[tex]Impulse\text{ = mv}_f-mv_i[/tex]On substituting the values, the impulse will be
[tex]\begin{gathered} Impulse\text{ = \lparen825}\times10\text{\rparen-\lparen825}\times15\text{\rparen} \\ =\text{ -4125 kg m/s} \end{gathered}[/tex]The impulse will be 4125 kg m/s due
The inductor in the RLC tuning circuit of an AM radio has a value of 9 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 94 kHz
We are asked to determine the capacitance of an RLC circuit given the frequency. To do that we will use the following formula:
[tex]C=\frac{1}{4\pi^2f^2L}[/tex]Where:
[tex]\begin{gathered} C=\text{ capacitance} \\ f=\text{ frequency} \\ L=\text{ inductance} \end{gathered}[/tex]Now, We plug in the values:
[tex]C=\frac{1}{4\pi^2(94\times10^3Hz)^2(9\times10^{-3}H)}[/tex]Now, we solve the operations:
[tex]C=3.19\times10^{-10}F[/tex]The capacitance is 3.19x10^-10 farads. In Picofarads this is equivalent to:
[tex]C=0.0319pF[/tex]part B:
Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to the ramp surface and directed up the ramp, moving the box up the ramp.
The magnitude of the acceleration of the box is 9.65 m/s².
What is the net force of the box?
The net force on the box is calculated as follows;
F(net) = F - Ff
where;
F is the applied forceFf is the force of frictionF(net) = F - μmgcosθ
where;
μ is the coefficient of friction given as 0.3θ is the angle of inclination of the plane = 55⁰m is the mass of the box = 15 kgF(net) = 170 - (0.3 x 15 x 9.8 x cos55)
F(net) = 144.71 N
The magnitude of the acceleration of the box is calculated as;
a = F(net) / m
a = (144.71) / (15)
a = 9.65 m/s²
Learn more about net force here: https://brainly.com/question/14361879
#SPJ1
What does a scientist mean when he or she says that an object is at rest
Answer:
I does not change position with respect to its surroundings with time
A 4.0 kW clothes dryer is connected to a 220 V circuit. How much current does the dryer use?
Given:
The power take by the dryer is
[tex]\begin{gathered} P=4\text{ kW} \\ P=4000\text{ W} \end{gathered}[/tex]
The dryer is connected to source of
[tex]V=220\text{ V}[/tex]Required: the current use by the dryer
Explanation:
we know that power taken by any appliance is given by
[tex]P=VI[/tex]here, V is the voltage of the source and I is the current drawn by the electric appliance.
Plugging all the values in the above formula, we get
[tex]\begin{gathered} 4000\text{ W}=220\text{ V}I \\ I=\frac{4000\text{ W}}{220\text{ V}} \\ I=18.18\text{ A} \end{gathered}[/tex]Thus, the current used by the dryer is
[tex]18.18\text{ A}[/tex]what is electric power
Answer:
Definition- Electric power is the rate at which electrical energy is transferred by an electric circuit.
Answer:
the rate at which electrical energy is transferred by an electric circuit.
Explanation:
Which optical instrument produces a magnified, virtual, and inverted image of small objects?1) a refracting telescope2) a single lens reflex camera3) a microscope4) a pair of binoculars
Microscope
Explanations:Microscopes are known for magnifying tiny objects
The images produced by a microscope are:
virtual (formed behind the screen)
Inverted
Magnified or enlarged
Therefore, the optical instrument which produces a magnified, virtual, and inverted image is the microscope
Example of a balanced force
1.In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is 2 Amps.2.If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of 6Ω. The current in the circuit is then 1 Amps.3.If a third identical lamp is connected in series, the total resistance is now 9Ω.4.The current through all three lamps in series is now _________ Amps. The current through each individual lamp is __________ Amps.
ANSWER
The current through all three lamps in series is now 0.67 Amps. The current through each individual lamp is 0.67 Amps.
EXPLANATION
There are three lamps connected in series, each with a resistance of 3 Ω, resulting in a total resistance of 9 Ω.
By Ohm's law, if the voltage from the battery is 6 V, then the current through all three lamps - i.e. the total current in the circuit is,
[tex]I=\frac{V}{R_{eq}}=\frac{6V}{9\Omega}=\frac{2}{3}Amps\approx0.67Amps[/tex]And, since the three lamps are connected in series - which means there are no dividing paths, the current through each individual lamp is the same as the total current of the circuit, 0.67 Amps.
Hence, the current through all three lamps and through each individual lamp is 0.67 Amps, rounded to the nearest hundredth.
A car is driving around a turn with a radius of 20.7 m. If the coefficient of friction between the tires and the road is 2.07, what is the maximum speed the car can maintain around the turn without slipping?
Answer:
A car completes a turn that has a radius of 20 meters. The coefficient of friction between the tires and road is 0.50. What maximum speed can the car safely maintain in order to complete the turn without skidding? (A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s (E) 25 m/s
Consider the graph shown. Which of the motions is consistent with the graph?a) The object has a constant velocity in the negative direction.b) The object is moving in the negative direction with a changing speed.c) The object is moving in the positive direction and slowing down.
Given:
The graph of velocity vs time of an object
To find:
Which of the motions is consistent with the graph?
Explanation:
We see here that the object's initial velocity is positive, and the final velocity is zero. So, the object is slowing down. as the initial velocity is positive, the object's direction of movement is positive.
Hence, The object is moving in a positive direction and slowing down.
Thermal equilibrium implies that:A:the state of restB:absolute zero temperatureC:the maximum temperatureD:equilibrium temperature
Explanation
Heat is the flow of energy from a high temperature to a low temperature. When these temperatures balance out, heat stops flowing
so, after some time both regions reach thermal equilibrium and no more energy is transfered.so we can conclude that
thermal equilibrium inplies that there is equilibrium temperature
D.equilibrium temperature
I hope this helps you
On which of the following does the speed of a falling object depend?a.) v ∝ mb.)v ∝ mc.)v ∝ Δh
Given:
The falling object
To find:
The dependence on the speed of the falling object
Explanation:
For an object falling freely, the total mechanical energy remains always constant. So, we can write, that the decrease in potential energy will be equal to the increase in the kinetic energy that is
[tex]\begin{gathered} \frac{1}{2}mv^2=mgh \\ v^2=2gh \end{gathered}[/tex]Hence, the speed of the falling object does not depend on the mass it depends on the height difference.
Please do this step-by-step how do you do it when it’s between
Given:
• Mass of block A = 6.0 kg
,• Mass of block B = 7.0 kg
,• Mass of block C = 13.0 kg
,• Force, F = 13.0 N
Let's find the magnitude of the tension in the rope between B and C.
Let's first find the acceleration.
We have:
[tex]13-T_B+T_B-T_A+T_A=6a+7a+13a[/tex]Thus, we have:
[tex]\begin{gathered} 13=26a \\ \\ a=\frac{13}{26} \\ \\ a=0.5\text{ m/s}^2 \end{gathered}[/tex]To find the tension between blocks B and C, we have the equation:
[tex]\begin{gathered} F-T_B=M_C*a \\ \\ T_B=F-M_c*a \end{gathered}[/tex]Where:
F = 13 N
Mc is the mass of block C = 13 kg
a is the acceleration = 0.5 m/s²
Thus, we have:
[tex]\begin{gathered} T_B=13-13*0.5 \\ \\ T_B=13-6.5 \\ \\ T_B=6.5\text{ N} \end{gathered}[/tex]Therefore, the magnitude of the tension in the rope between blocks B and C is 6.5 N
ANSWER:
6.5 N
A pottery wheel with rotational inertia 40 kgm^2 rotates at 10 rev/s. 4 kg of clay is dropped onto the wheel 1.2 m from the axis. What angular speed will the wheel have after this?1. 55 rad/s2. 8.7 rad/s3. 70 rad/s4. 0 rad/s
Given:
• Rotational inertia = 40 kg.m²
,• Initial angula speed = 10 rev/s
,• Mass, m = 4 kg
,• Diameter, d = 1.2 m
Let's find the angular speed of the wheel.
To find the angular speed, apply the formula:
[tex]L_i=(I+md^2)*w_f[/tex]Where wf is the final angular speed
I is the rotational inertia
m is the mass
d = 1.2
Li is the angular momentum.
To find the angular momentum, we have:
[tex]\begin{gathered} L_i=40*10*2\pi \\ L_i=2513.27\text{ kg.m}^2\text{ rad/s} \end{gathered}[/tex]Now, to find the final angular speed, wf, plug in values in the first equation and solve for wf:
[tex]\begin{gathered} Li=(I+md^2)w_f \\ \\ 2513.27=(40+4*1.2^2)w_f \\ \\ 2513.27=45.76w_f \\ \\ w_f=\frac{2513.27}{45.76} \\ \\ w_f=54.9\approx55\text{ rad/s} \end{gathered}[/tex]Therefore, the final angular speed is 55 rad/s.
ANSWER:
1.) 55 rad/s
Bart, mass 32.4 kilograms, and Milhouse, mass 27.6 kilograms, play on the schoolyard seesaw. If Bart and Milhouse want to sit 4.0 meters apart, how far from the center of the seesaw should Bart sit? Include units in your answer. Answer must be in 3 significant digits.
Given data:
* The mass of the Bart is m_1 = 32.4 kg.
* The mass of the Milhouse is m_2 = 27.6 kg.
* The distance between the Millhouse and Bart is d = 4 m.
Solution:
To balance the seesaw, the net moment about the center should be zero.
The diagrammatic representation of the given system is,
The distance between the Bart and Milhouse can be written as,
[tex]\begin{gathered} d=d_1+d_2 \\ 4=d_1+d_2 \\ d_2=4-d_1 \end{gathered}[/tex]where d_2 is the distance of Milhouse from the center and d_1 is the distance of Bart from the center,
Consider the moment as positive if it is in an anticlockwise direction and negative if it is a clockwise direction.
Thus, the net moment about the center is,
[tex]M=m_1d_1-m_2d_2[/tex]Substituting the known values,
[tex]\begin{gathered} 0=32.4\times d_1-27.6\times(4-d_1) \\ 0=32.4\times d_1-110.4+27.6d_1 \\ 0=60d_1-110.4 \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} 60d_1=110.4 \\ d_1=\frac{110.4}{60} \\ d_1=1.84\text{ m} \end{gathered}[/tex]Thus, the distance of the Bart from the center is 1.84 meters.
