Answer:
42/ 5 pi J
Explanation:
The word done by friction is given by
[tex]F=\mu Nx[/tex]where μ = coefficent of frction, N = weight of the box, and x = distance traveled.
Now in our case
μ = 0.13
N = 35.0 N
x = 0.40 * 2 pi (since it is the circumeference of the circle
Therefore,
[tex]F=(0.13)(35.0)(0.40\cdot2\pi)[/tex]which simplifies to give
[tex]F=\frac{42}{5}\pi[/tex]which is our answer!
Calculate the speed of sound in air for the following temperatures:A. 0°C B. 25°CC. 30°C D. -15°C
The speed of sound in air at a given temperature is given by:
[tex]v=331\sqrt[]{1+\frac{T}{273}}[/tex]Then we need to plug the value of the temperature to determine each case.
A.
[tex]v=331\sqrt[]{1+\frac{0}{273}}=331[/tex]B.
[tex]v=331\sqrt[]{1+\frac{25}{273}}=345.82[/tex]C.
[tex]v=331\sqrt[]{1+\frac{30}{273}}=348.71[/tex]D.
[tex]v=331\sqrt[]{1+\frac{(-15)}{273}}=321.78[/tex]Therefore the speeds for the given temperatures are:
A. 331 m/s
B. 345.82 m/s
C. 348.71 m/s
D. 321.78 m/s
The image depicts two bar magnets immersed inof that contains small needle-In particles ofIron.What is the BEST Interpretation of the patterncreated by the iron particles?A The pattern of iron particles is created by thenorth pole of one magnet facing the south poleof the other magnetB. The pattern of kroparticles is created by thesouth pols of one magnet facing the south poleof the other magnet because south poles attracteach otherC. The pattern of iron particles is created by therepulsion of like magnetic poles, we cannotdetermine if the poles are north or southpolarizationD. The pattern of fron particles is created by thenorth pole of one magnet facing the north poleof the other magnet because north poles attracteach other.
option A is correct
The pattern of iron particles is created by the
north pole of one magnet facing the south pole
of the other magnet
as the lines of force emerges from north and ends at south
"Surface tension" is a key phrase in the articleHow does the author refine the meaning of this phrase over the course of the article?(A) by providing real world examplos, citing specific statistics, and quoting expert opinions(8)by defining the torm, providing a list of materials, and contrasting two surface tension experiments(C) by defining the term, providing real world examples, and guiding the reader through ways to investigatesurface tension(D) by explaining the history of the term, defining the key components of an experiment, and thendescribing a surface tension experiment
(C) by defining the term, providing real world examples, and guiding the reader through ways to investigate surface tension
The parallel circuit at the right depicts two resistors connected to a voltage source. The voltage source (ΔVtot) is a 12-V source and the resistor values are 7.4 Ω (R1) and 3.9 Ω (R2).a. Determine the equivalent resistance of the circuit. b. Determine the current in each branch resistor:Current in R1 = Current in R2 = c. Determine the total current in the circuit.
Given:
Two resistors are connected in parallel.
The resistance of resistor 1 is
[tex]R1=\text{ 7.4}\Omega[/tex]The resistance of resistor 2 is
[tex]R2\text{ = 3.9 }\Omega[/tex]The value of the voltage source is
[tex]\Delta V_{tot}=\text{ 12 V}[/tex]Required:
(a)The equivalent resistance of the circuit.
(b) The current in each branch of the resistor.
(c) The total current in the circuit.
Explanation:
(a) In a parallel circuit, the equivalent resistance can be calculated as
[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R1}+\frac{1}{R2} \\ =\frac{1}{7.4}+\frac{1}{3.9} \\ R_{eq}=\text{ 2.55 }\Omega \end{gathered}[/tex](b) In a parallel circuit, the voltage across each resistor is the same.
The current through the resistor R1 can be calculated according to Ohm's law
[tex]\begin{gathered} I_{R1}=\frac{V}{R1} \\ =\frac{12}{7.4} \\ =1.62\text{ A} \end{gathered}[/tex]The current through the resistor R2 can be calculated as
[tex]\begin{gathered} I_{R2}=\frac{V}{R2} \\ =\frac{12}{3.9} \\ =3.08\text{ A} \end{gathered}[/tex](c) The total current in the circuit is the sum of the current through each branch.
Thus, the total current can be calculated as
[tex]\begin{gathered} I=I_{R1}+I_{R2} \\ =\frac{12}{7.4}+\frac{12}{3.9} \\ =\text{ 4.7 A} \end{gathered}[/tex]Final Answer:
(
According to Coulomb's law, how does distance affect electric force?A.The electric force is the same no matter how far apart the objects are.B.Objects must be in contact with one another for the electric force to have an effect.C.Only sometimes can the electric force act without objects touching.D.Electric force can act at a distance but is stronger when objects are closer.
According to Coulomb's law,
[tex]Force\text{ }\propto\frac{1}{(distance)^2}[/tex]Thus, electric force can act at a distance but is stronger when objects are closer.
Hence, option D is correct.
Which compound is a molecule?Question 10 options:NaClMgBr2C6H12O6AgCl3
It's important to know the difference between a molecular compound and an ionic compound.
Ionic compounds are formed by the interaction between a metal and a non-metal, the first ones have the property of easily losing electrons, while non-metals have the property of easily gaining electrons.
In this case, C6H12O6 is a molecular compound because it's not using ionic interaction.
Therefore, the third option is correct.
Emily, with a mass of 54.1 kilograms, is riding a wagon of mass 15.0 kilograms with a velocity of +4.12 meters per second. She then jumps off the wagon and lands on the ground with a horizontal velocity of +5.59 meters per second. What is the velocity of the wagon after Emily jumps off? Answer must be in 3 significant digits.
The initial momentum of the system can be expressed as,
[tex]p_i=(m_1+m_2)u[/tex]The final momentum of the system can be expressed as,
[tex]p_f=m_1v_1+m_2v_2[/tex]According to conservation of momentum,
[tex]p_i=p_f[/tex]Plug in the known expressions,
[tex]\begin{gathered} (m_1+m_2)u=m_1v_1+m_2v_2 \\ m_2v_2=(m_1+m_2)u-m_1v_1 \\ v_2=\frac{(m_1+m_2)u-m_1v_1}{m_2} \end{gathered}[/tex]Substitute the known values,
[tex]\begin{gathered} v_2=\frac{(54.1\text{ kg+15.0 kg)(4.12 m/s)-(54.1 kg)(5.59 m/s)}}{15.0\text{ kg}} \\ =\frac{284.692\text{ kgm/s-}302.419\text{ kgm/s}}{15.0\text{ kg}} \\ =-\frac{17.727\text{ kgm/s}}{15.0\text{ kg}} \\ =-1.18\text{ m/s} \end{gathered}[/tex]Thus, the final velocity of wagon is -1.18 m/s where negative indicates that the wagon moves in the opposite direction.
Hello I could really use some help answering this question thank you!
In order to find the false statement about radiation, let's analyze each one:
A.
This statement is true, all objects emit radiation.
B.
This statement is also true.
C.
This statement is false, because radiation can transfer energy in vacuum (no matter) as well.
D.
This statement is true.
Therefore the correct answer is C.
The graph below represents the variation of the velocity versus time of a moving particle moving along the x- axis. The acceleration of this particle between 2s and 4s is equal to:
ANSWER:
2nd option: 0 m/s^2
STEP-BY-STEP EXPLANATION:
From second 2 to second 4, we can see on the graph that the velocity remains constant, therefore the acceleration is 0.
The correct answer is 2nd option 0 m/s^2
a 0.040 kg bullet was traveling at 150 m/s, inelastically collided with a 5 kg pendulum that was at rest. what is the maximum height h can the pendulum go after the collision ignoring the air resistance
We will have the following:
First, using the conservation of momentum to find the velocity of the system, that is:
[tex]\begin{gathered} (0.040kg)(150m/s)+(5kg)(0m/s)=(0.040kg+5kg)v_f \\ \\ \Rightarrow v_f=\frac{(0.040kg)(150m/s)}{(0.040kg+5kg)}\Rightarrow v_f=\frac{25}{21}m/s \end{gathered}[/tex]Now that we have the initial velocity we will use conservation of energy to determine the height of the pendulum:
[tex]\begin{gathered} \frac{1}{2}(0.040kg+5kg)(\frac{25}{21}m/s)=(0.040kg+5kg)(9.8m/s^2)h \\ \\ \Rightarrow\frac{1}{2}(\frac{25}{21}m/s)=(9.8m/s^2)h\Rightarrow h=\frac{(1/2)(25/21m/s)}{(9.8m/s^2)} \\ \\ \Rightarrow h=\frac{125}{2058}m\Rightarrow h\approx0.06m \end{gathered}[/tex]So, the pendulum mover a height of exactly 125/2058 m, that is approximately 0.06m.
hello, i need help with my zearn.. Sebastian must have at least 5 feet of rope to jump rope.Which description best represents the length of rope Sebastian needs?Any value less than or equal to 5Any value equal to 5Any value greater than 5Any value greater than or equal to 5
Sebastian must have at least 5 feet of rope to jump rope.
At least means greater or equal.
Correct option:
Any value greater than or equal to 5
Calculate the recoil velocity of a 5 kg rifle that fires a 24 g bullet that travel 1.5 m in 0.0023 s
The mass of rifle is M = 5kg . It fires the bullet of mass m =24 g = 0.024 kg
It travels a distance, d = 1.5 m and the time taken is t = 0.0023 s
The speed of the bullet will be
[tex]\begin{gathered} v=\frac{d}{t} \\ =\frac{1.5}{0.0023} \\ =652.17\text{ m/s} \end{gathered}[/tex]Let the recoil velocity be V. The initial velocity will be zero as rifle and bullet are at rest.
According to the conservation of momentum
[tex]\begin{gathered} \text{Initial momentum = final momentum} \\ (M+m)\times0=MV-mv\text{ } \end{gathered}[/tex]Here, rifle and bullet travel in opposite directions.
Substituting the values, the recoil velocity will be
[tex]\begin{gathered} V=\frac{mv}{M} \\ =\frac{0.024\times652.17}{5} \\ =3.13\text{ m/s} \end{gathered}[/tex]A ball is thrown up into the air for a total of 1.25 s before it is caught at its original position . How high did the ball go
so:
The initial speed is given by:
[tex]\begin{gathered} v=v_o+at \\ _{\text{ }}where\colon \\ v=-v_o \\ a=g=-9.8 \\ t=1.25 \\ so\colon \\ -v_o=v_o+(-9.8)\cdot1.25 \\ -2v_o=-12.25 \\ v_o=\frac{-12.25}{-2} \\ v_o=6.125 \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} y_o=0 \\ v_{}=0 \\ v_o=6.125 \\ so\colon \\ y=\frac{-v^2_o}{2g} \\ y=\frac{-6.125^2}{2(-9.8)} \\ y=1.91m \end{gathered}[/tex]Answer:
1.91m
When a -3.72*10^-4 C chargemoves from point A to point B, itselectric potential energy increasesby 0.218 J. What is the potentialdifference between A and B?Include the correct sign, + or -.(Unit = V)
We know that the electrical potential formula is
[tex]V=\frac{PE}{q}[/tex]Replacing the given information, we have
[tex]V=\frac{0.218J}{-3.72\times10^{-4}C}\approx-586.02V[/tex]Hence, the potential difference between A and B is -586.02 V.What is the length contraction of an automobile 3.133 m long when it is traveling at 51.35km/h? (Hint:for x << 1,(1-v2/c2)1/2 ~ 1 - x2/2) Compare this to the diameter of a hydrogen atom by expressing your answer in femto meters.
The length L of a moving object whose rest length is L_0 is:
[tex]L=\sqrt{1-\frac{v^2}{c^2}}L_0[/tex]The length contraction can be calculated as the difference L_0-L:
[tex]L_0-L=\left(1-\sqrt{1-\frac{v^2}{c^2}}\right)L_0[/tex]For v<, we have:
[tex]\begin{gathered} L_0-L\approx\left(1-\left[1-\frac{1}{2}\left(\frac{v}{c}\right)^2\right]\right)L_0 \\ \\ =\left(\frac{v}{c}\right)^2\frac{L_0}{2} \end{gathered}[/tex]Replace v=51.35km/h, c=300,000km/s and L_0=3.133m:
[tex]\begin{gathered} \Delta L=\left(\frac{v}{c}\right)^2\frac{L_0}{2} \\ \\ =\left(\frac{51.35\frac{km}{h}\times\frac{1h}{3600s}}{300,000\frac{km}{s}}\right)^2\frac{3.133m}{2} \\ \\ =3.54...\times10^{-15}m \\ \\ =3.54...fm \end{gathered}[/tex]The diameter of a hydrogen atom is approximately 10.6*10^-11m. Then, the length contraction of th ecar is much less than the diameter of a hydrogen atom.
Therefore, the length contraction of the automobile is approximately 3.54 femtometers.
28. A 40.0-kg pack is carried up a 2500-m-high mountain in 10.0 h. How much work is done?
ANSWER
[tex]980,000J[/tex]EXPLANATION
To find the work done in carrying the bag pack, we can apply the formula:
[tex]W=F\cdot d[/tex]where F = force ; d = distance moved
The force here refers to the weight of the bag pack, given by:
[tex]\begin{gathered} m\cdot g \\ 40\cdot9.8 \\ \Rightarrow392N \end{gathered}[/tex]where m = mass; g = acceleration due to gravity
Hence, the work done is:
[tex]\begin{gathered} W=392\cdot2500 \\ W=980,000J \end{gathered}[/tex]That is the answer.
One of the tallest radio towers is in Fargo, North Dakota. The tower is 629m tall, or about 44 percent taller than the Sears Tower in Chicago. if a bird lands on top of the tower, so that the gravitational potential energy associated with the bird is 2033 J, what is the mass?
An object moving with uniform acceleration has a velocity of 11.0 cm/s in the positive x-direction when its x-coordinate is 2.79 cm. If its x-coordinate 2.05 s later is −5.00 cm, what is its acceleration? ______ cm/s2
-14.43 cm/s^2
Explanationto solve this we need to use the formula
[tex]\begin{gathered} x=v_ot+\frac{1}{2}at^2 \\ where \\ v_o\text{ is the inital velocity} \\ t\text{ is the time} \\ x\text{ is the traveled distance} \\ a\text{ is the acceleration} \end{gathered}[/tex]Step 1
given
[tex]\begin{gathered} v_1=11\frac{cm}{s} \\ x_1=2.79\text{ cm} \\ x_2=-5\text{ cm} \\ time=2.05\text{ s} \end{gathered}[/tex]a) find the traveled distance
[tex]\begin{gathered} x=\Delta x=-5-(2.79)=-7.79 \\ the\text{ negative sign indicates the opposite side} \end{gathered}[/tex]b) replace in the formula and solve for a
[tex]\begin{gathered} x=v_{o}t+\frac{1}{2}at^{2} \\ -7.79=11\frac{cm}{s}*2.05s+\frac{1}{2}a*(2.05\text{ s\rparen}^2 \\ -7.79=22.55+2.10125a \\ subtract\text{ 22.55 in both sides} \\ -7.79-22.55=22.55+2.10125a-22.55 \\ -30.34=2.10125a \\ divide\text{ both sides by 2.10125} \\ \frac{-30.34}{2.10125}=\frac{2.101,25a}{2.10125} \\ -14.43\frac{cm}{s^2}=a \end{gathered}[/tex]so, the acceleation is
-14.43 cm/s^2
I hope this helps you
What are the types of precipitation? Select four correct answers. fog sleet dew snow wind rain hail
The four main types of precipitation are Rain, sleet drizzle, and snow.
Precipitation is any product of the condensation of atmospheric water vapours.
Thus the correct answer is Rain, sleet, drizzle(dew), and snow.
Write each of the following rates as a unit rate. 16 carrots for $4$36 for 3 hours420 miles in 6 hours
Given the rates:
• 16 carrots for $4
,• $36 for 3 hours
,• 420 miles in 6 hours
Let's write the given rates as unit rates.
To find the unit rate, divide each unit by the number on the right side,
Hence, we hae the following:
• 16 carrots for $4
[tex]\text{Unit rate = }\frac{16\text{ carrots}}{\text{\$4}}=\frac{4}{1}=4\frac{carrots}{\text{ \$}}=4\text{ carrots per dollar}[/tex]• $36 for 3 hours:
[tex]\text{Unit rate = }\frac{\text{ \$36}}{3\text{ hours}}=12\frac{\text{ \$}}{\text{hour}}=12\text{ dollars/hour}[/tex]• 420 miles in 6 hours:
[tex]\text{Unit rate=}\frac{420\text{ miles}}{6\text{ hours}}=70\frac{miles}{\text{hour}}=70\text{ miles/hour}[/tex]Therefore, the rates written as unit rates are:
• 16 carrots for $4 ==,> ,4 carrots per dollar
,• $36 for 3 hours, ,==>, ,12 dollars per hour
,• 420 miles in 6 hours, ,==> ,70 miles per hour
ANSWER:
• 4 carrots per dollar
12 dollars per hour
70 miles per hour
Compute the Larmor radius for a typical electron in the K corona
The gyro radius is the radius of a charged particle's circular motion in the presence of a uniform magnetic field.
What is Larmor radius?r=v0t0= v0tmc2e|B| r=v0t0= v0tmc2e|B| The magnetic moment of the system is produced by the rotation of the charged particles in the magnetic field. The radius of the helix is known as the Larmor radius, and it is determined by the particles' charge q, mass m, perpendicular velocity, and magnetic field B:Wavelength of those magneto sonic waves is comparable to, and shorter than, the gyro radius of energetic electrons (i.e., the ratio of electron gyro radius to perpendicular wavelength, gk>1), the finite Larmor radius effect occurs. The Dynamic Loss of the Earth's Radiation Belts in 2020.Therefore,
[tex]$_g=\frac{m v_{\perp}}{|q| B}, \omega_g=\frac{|q| B}{m}, \mathrm{~m}=9.11 \times 10^{-31} \mathrm{~kg}, \mathrm{q}=1.602 \times 10^{-19} \mathrm{C}[/tex]
To learn more about Larmor radius, refer to:
https://brainly.com/question/12946099
#SPJ1
A man pulls a sled a distance of 249m. the rope attached to the sled makes an angle of 30 degrees with the ground. The man exerts a force of 540N on the rope. How much work does the man do in pulling the sled in kilojoules?
Given data
*A man pulls a sled at a distance is s = 249 m
*Man exerted a force on the rope is F = 540 N
*The given angle is
[tex]\theta=30.0^0[/tex]The formula for the work done by the man in pulling the sled is given as
[tex]W=Fs\cos \theta[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} W=(540)(249)\cos (30.0)^0 \\ =116445.77\text{ J} \\ =116.44\text{ kJ} \end{gathered}[/tex]A train travels 110 miles in the same time that a plane covers 528 miles. If the speed of the plane is 10 miles per hour less than 5 times thespeed of the train, find both speeds.
We will have the following:
First, we will have that the expressions of distance for the train and plane are respectively:
[We remember that velocity times time equals distance]
[tex]\begin{cases}t\colon v\cdot t=110mi \\ \\ p\colon(5v-10)\cdot t=528mi\end{cases}[/tex]From this we solve for time "t" in each expression:
[tex]\begin{cases}t=\frac{110mi}{v} \\ \\ t=\frac{528mi}{5v-10}\end{cases}[/tex]Now, since the time is the same, we equal both expressions, that is:
[tex]\frac{110mi}{v}=\frac{528mi}{5v-10}[/tex]Now, we solve for "v"
[tex]\Rightarrow110(5v-10)=528(v)\Rightarrow550v-1100=528v[/tex][tex]\Rightarrow22v=1100\Rightarrow v=50[/tex]Now, we determine each vehicle's velocity:
[tex]\begin{cases}t\colon50mi/h \\ \\ p\colon240mi/h\end{cases}[/tex]So, the train is moving at 50 miles per hour and the plane is moving at 240 miles per hour.
We corroborate by determining the time it takes for each to reach the distances stated, that is:
*Train:
[tex](50mi/h)\cdot t=110mi\Rightarrow t=\frac{11}{5}h\Rightarrow t=2.2h[/tex]*Plane:
[tex](240mi/h)\cdot t=528mi\Rightarrow t=\frac{11}{5}h\Rightarrow t=2.2h[/tex]Thus proving that the velocities are the ones calculated.
A car battery with a potential difference of 11.4 volts is connected to a starter motor with a resistance of 0.242 ohms. If the startup time is 1.54 seconds, how many electrons flow out of the battery during this time? Answer must be in 3 significant digits.
Given
The potential difference is V=11.4 V
Resistance is R=0.242 ohm
Time taken, t=1.54 sec
To find
The number of electron flow.
Explanation
Let the number of electron be n
We know current is charge per unit time
Thus,
[tex]I=\frac{q}{t}=\frac{ne}{t}[/tex]By Ohm's law,
[tex]\begin{gathered} V=RI \\ \Rightarrow V=\frac{Rne}{t} \\ \Rightarrow11.4=\frac{0.242\times n\times1.6\times10^{-19}}{1.54} \\ \Rightarrow n=4.53\times10^{20} \end{gathered}[/tex]Conclusion
The number of electron flow is
[tex]4.53\times10^{20}[/tex]Two thin rods of length L are rotating with the same angular speed, W (in rad/s) about axes that pass perpendicularly through one end. Rod A is massless but has particle of mass 0.63 kg attached to its free end. Rod B has mass of 0.63 kg, which is distributed uniformly along its length. The length of each rod is 0.82 m, and the angular speed is 4.2 rad/s. Find the kinetic energies of rod A with its attached particle and of rod B.
ANSWER:
Rod A: 3.74 J
Rod B: 3.74 J
STEP-BY-STEP EXPLANATION:
Rod A:
Length of the Rod = l = 0.82 m
Mass of particle attached = m = 0.63 kg
Moment of inertia of the system about given axis:
[tex]\begin{gathered} I=m\cdot l^2 \\ \text{ replacing} \\ I=0.63\cdot(0.82)^2 \\ I=0.424\text{ kg}\cdot m^2 \end{gathered}[/tex]Angular speed of the rod = 4.2 rad/s
Kinetic energy of the rod is:
[tex]\begin{gathered} KE=\frac{1}{2}\cdot I\cdot w^2 \\ \text{ replacing:} \\ KE=\frac{1}{2}\cdot0.424\cdot4.2^2 \\ KE=3.74\text{ J} \end{gathered}[/tex]Rod B:
Length of the Rod = l = 0.82 m
Mass of particle attached = m = 0.63 kg
Moment of inertia of the system about given axis:
[tex]\begin{gathered} I=m\cdot l^2 \\ \text{ replacing} \\ I=0.63\cdot(0.82)^2 \\ I=0.424\text{ kg}\cdot m^2 \end{gathered}[/tex]Angular speed of the rod = 4.2 rad/s
Kinetic energy of the rod is:
[tex]\begin{gathered} KE=\frac{1}{2}\cdot I\cdot w^2 \\ \text{ replacing:} \\ KE=\frac{1}{2}\cdot0.424\cdot4.2^2 \\ KE=3.74\text{ J} \end{gathered}[/tex]A person slams a door. How wouldincreasing the width of the door affectits acceleration?Assume that all other variables are kept constant.A. it would increaseB. it would be unaffectedC. it would decrease
C. it would decrease
Explanation
Step 1
Diagram
Torque is a measure of the force that can cause an object to rotate about an axis. Just as force is what causes an object to accelerate in linear kinematics, torque is what causes an object to acquire angular acceleration
The angular acceleration is the time rate of change of the angular velocity and is usually designated by α and expressed in radians per second per second, it is given by the expression:
[tex]\begin{gathered} a_r=\frac{v^2}{r} \\ whrere\text{ r is the radius, } \\ \end{gathered}[/tex]so, if we increase the radius the radius , the acceleration would decrease because it is inversely proportional to the radius
[tex]\begin{gathered} r_2>r_1 \\ a_r=\frac{v^2}{r_2} \\ so \\ a_2\leq a_1 \end{gathered}[/tex]therefore, the answer is
C. it would decrease
I hope this helps you
A bobsled, starting from rest, slides down an incline that is 25.1 meters long. It takes a total of 8.68 seconds to reach the bottom. How long did it take to reach the point midway down the incline? Assume the acceleration is constant the entire way.
Since the acceleration of the of the bosled is constant the entire motion this is an uniformaly accelerated motion which means that we can use the following equations:
[tex]\begin{gathered} a=\frac{v_f-v_0}{t} \\ \Delta x=v_0t+\frac{1}{2}at^2 \\ v_f^2-v_0^2=2a\Delta x \end{gathered}[/tex]We want to know how long it takes for the bobsled to reach the midway to determine this we first need to determine the acceleration of the bobsled. We know that the incline is 25.1 m long which means that we know the change in position, we also know that it started from rest and hence the initial velocity is zero; also, we know that it takes 8.68 s to move all this way then we know the time for the entire motion. This means that we can use the second equation to determine the acceleration:
[tex]\begin{gathered} 25.1=(0)(8.68)+\frac{1}{2}a(8.68)^2 \\ a=\frac{(2)(25.1)}{(8.68)^2} \\ a=0.67 \end{gathered}[/tex]Hence, the acceleration of the bobsled is 0.67 m/s².
Now that we know the acceleration of the bobsled along the incline we can use the second equation to determine how long it takes to reach the midway which would be when it travels 12.55 m, then we have:
[tex]\begin{gathered} 12.55=(0)t+\frac{1}{2}(0.67)t^2 \\ t^2=\frac{2(12.55)}{(0.67)} \\ t=\pm\sqrt{\frac{2(12.55)}{(0.67)}} \\ t=\pm6.14 \end{gathered}[/tex]Therefore, it takes 6.14 s to reach the midway
A current of 1.2 A flows through a light bulb when it is connected across a 120 V source. What power is dissipated by the bulb?
Answer:
144 Watts
Explanation:
The power dissipated is equal to
P = IV
Where I is the current and V is the voltage. Replacing I = 1.2 A and V = 120 V, we get
P = (1.2A)(120 V)
P = 144 Watts
Therefore, the power is 144 Watts
What is the current through each resistor in the parallel portion?
Given:
Two resistors each of resistance R = 60 ohms are connected in parallel.
A resistance R' = 30 ohms is connected in series with the parallel combination.
The voltage across the circuit is V = 120 V.
To find the current through each resistor in the parallel portion.
Explanation:
The equivalent resistance of the resistors connected in parallel will be
[tex]\begin{gathered} \frac{1}{R_p}=\frac{1}{R}+\frac{1}{R} \\ =\frac{1}{60}+\frac{1}{60} \\ R_p=\text{ }\frac{60}{2} \\ =30\text{ }\Omega \end{gathered}[/tex]In a series combination, the voltage will be the sum of voltages across each resistor while the current remains the same.
In parallel combination, the voltage remains the same while the current is the sum of currents through each resistor.
Voltage in series combination will be
[tex]\begin{gathered} V=V_{Rp}+V^{\prime} \\ V=V_{30}+V_{30} \\ V_{30}=\text{ }\frac{V}{2} \\ =\frac{120}{2} \\ =60\text{ V} \end{gathered}[/tex]The voltage across each resistor in the parallel combination will be 60 ohms.
The current in the series combination will be the same.
According to the Ohm's law, the current will be
[tex]\begin{gathered} V^{\prime}=I^{\prime}R^{\prime} \\ I^{\prime}=\frac{V^{\prime}}{R^{\prime}} \\ =\frac{60}{30} \\ =2\text{ A} \end{gathered}[/tex]The current in the parallel combination will be the sum of individual current
[tex]I^{\prime}=I_1+I_2[/tex]Since the voltage across the parallel combination is the same and the resistance is also the same, so
[tex]\begin{gathered} I_1=I_2 \\ =I_p \end{gathered}[/tex]Thus, the current through each resistor in parallel combination will be
[tex]\begin{gathered} I^{\prime}=I_p+I_p \\ I_p=\frac{I^{\prime}}{2} \\ =\frac{2}{2} \\ =\text{ 1 A} \end{gathered}[/tex]Hence, the current through each resistor in the parallel portion is 1 A
2 ptsAlly, Brittany, Christine, and Danica are exercising on the track field. Their masses are55 kg, 60 kg, 50 kg, and 55 kg, respectively. Ally and Christine are walking around thetrack together at the same speed. Danica is running around the track, and Brittany hasstopped to tie her shoe. Rank the girls from least to greatest momentum.
ANSWER
[tex]\text{Brittany, Christine, Ally, Danica}[/tex]EXPLANATION
To solve this problem, we have to find the momentum of each of the girls.
Momentum can be found by applying the formula:
[tex]p=mv[/tex]where m = mass; v = velocity
Let the velocity of the person running be x.
Let the velocity of the person walking be y.
Let the velocity of the person tying her shoe be z.
The person running is moving at a much higher velocity than the person walking while the person tying her shoe is not moving at all i.e. z = 0
Therefore, we have that:
[tex]x>y>z[/tex]Let us calculate the momentum of each girl:
[tex]\begin{gathered} p_{\text{Ally}}=55\cdot y=55y\text{ kgm/s} \\ p_{\text{Brittany}}=60\cdot z=60\cdot0=0\text{ kgm/s} \\ p_{\text{Christine}}=50\cdot y=50y\text{ kgm/s} \\ p_{\text{Danica}}=55\cdot x=55x\text{ kgm/s} \end{gathered}[/tex]Therefore, knowing that x > y, we have that the ranking of the momentum from least to greatest is:
[tex]0,50y,55y,55x[/tex]Therefore, the ranking of the girls from least to greatest momentum is:
[tex]\text{Brittany, Christine, Ally, Danica}[/tex]