At the top of the ramp:
The height, h, is maximum, hence the ball has a maximum potential energy
Since the ball is not moving, the speed = 0 m/s. Hence the ball has zero kinetic energy
As the rubber ball moves down the ramp:
The ball will be in motion and have a certain amount of speed
Therefore, some of the Potential Energy is transformed to Kinetic Energy. That is, the rubber ball has both Kinetic and Potential energy
The Kinetic energy increases while the Potential energy decreases
At the bottom of the ramp:
The height = 0
The Potential energy of the rubber ball will be zero
The law of conservation of energy states that energy can neither be created nor destroyed, but can be transformed from one form to another. This is an evidence that the energy of the ball remains constant at all the points on the ramp.
[tex]\text{Total Energy = Kinetic Energy + Potential Energy}[/tex]Line D represents movement that starts out toward the south, slows down, and stops. Why is the slope of the line positive?1) The velocity is positive.2) The velocity is negative.3) The acceleration is negative.4) The acceleration is positive.
Given that line D represents the movement that start out towards the south, slow down, and stops. Let's determine why the slope of the line is positive.
The slope of a velocity-time graph is said to represent the acceleration of the moving object.
In this case, the slope starts out toward the south, slows down and stops, this means the acceleration of the object that moved is positive.
Since the acceleration is positive, the slope of the line is aslo positive.
Therefore, we can say the slope of the line is positive because the acceleration is positive.
ANSWER:
4) The acceleration is positive
A proton is accelerated by a constant electric field of the magnitude 600. N/C. Find: a) the force acting on the proton by the electric field;b) the acceleration of the particle due to the electric force.
ANSWER
[tex]\begin{gathered} (a)9.61\cdot10^{-17}N \\ (b)5.75\cdot10^{10}m\/s^2 \end{gathered}[/tex]EXPLANATION
(a) To find the force acting on the proton, we have to apply the formula representing the relationship between electric field and electric force:
[tex]F=qE[/tex]where q = charge of the proton
E = electric field
The charge of a proton is:
[tex]q=1.602\cdot10^{-19}C[/tex]Hence, the force acting on the proton by the electric field is:
[tex]\begin{gathered} F=1.602\cdot10^{-19}\cdot600 \\ F=9.61\cdot10^{-17}N \end{gathered}[/tex](b) To find the acceleration of the particle, apply the relationship between force and acceleration:
[tex]F=ma[/tex]where m = mass; a = acceleration
The mass of a proton is:
[tex]m=1.67\cdot10^{-27}\operatorname{kg}[/tex]Hence, the acceleration of the proton is:
[tex]\begin{gathered} 9.61\cdot10^{-17}=1.67\cdot10^{-27}\cdot a \\ \Rightarrow a=\frac{9.61\cdot10^{-17}}{1.67\cdot10^{-27}} \\ a=5.75\cdot10^{10}m\/s^2 \end{gathered}[/tex]QUESTION 3 Jessica and Julie are twins, with a mass of 40.0 kg each. They are both riding a merry go round, with Jessica sitting 8.0 m from the center and Julie sitting 6.0 m from the center. If Jessica is traveling at 1.3 m/s and Julie is traveling at 0.95 m/s, who is experiencing the greatest force?
Which object is the greatest inertia A. 1500kg elephant B. 5kg rock moving at 300 m/sC. 100 kg football player moving at 5 m/sD. 1000 kg car moving at 40 m/s
Answer:
A. 1500 kg elephant
Explanation:
The inertia only depends on the mass and not on the velocity, so the greater the mass, the greater the inertia. It means that the object with the greaterst inertia is:
A. 1500 kg elephant
Because it has more mass.
Hello could you please help me with this question? is parallel to the y-axis and has an x-intercept of 3
Answer:
[tex]x=3[/tex]Explanation: We need to find the equation of a line that is parallel to the y-axis and has an x-intercept of 3:
This simply implies that the slope is infinite, and so the equation must be:
[tex]x=3[/tex]The plot of the equation:
NOTE! The plot matches the description of the problem.
Find the following for path C in the figure below (a) the total distance traveledm(b) the distance from start to finishm(c) the displacement from start to finishm
(a). The total distance traveled in the path C is,
[tex]\begin{gathered} D=\text{ From 2 m to 10 m + from 10 m to 8 m + from 8 m to 10 m} \\ =\mleft(10-2\mright)+\mleft(10-8\mright)+\mleft(10-8\mright) \\ =8+2+2 \\ =12\text{ m} \end{gathered}[/tex]Thus, the total distance traveled in the path C is 12 m.
(b). The distance from the start (2 m) to the finish (10 m) is,
[tex]\begin{gathered} d=\text{from 2 m to 10 m} \\ d=10-2 \\ d=8\text{ m} \end{gathered}[/tex]Thus, the distance traveled from start to finish is 8 m.
(c). The displacement from the start (2 m) to the finish (10 m) by neglecting the return travel in the path C is,
[tex]\begin{gathered} d^{\prime}=10-2 \\ d^{\prime}=8\text{ m} \end{gathered}[/tex]Thus, the displacement for the path C is 8 m.
An iron railroad railis 750 ft long when the temperature is 34°C. What is its length (in ft) when the temperature is -18°C? (Round your answer to at least 3 decimal placeFt
Answer: Final length = 749.998 ft
Explanation:
We would assume that the coefficient of linear expansion of iron is 36 x 10^- 6/C
We would apply the formula for calculating linear thermal expansion which is expressed as
L1 = aL(θ2 - θ1)
where
L1 = change in length
L = original length
a = coefficient of linear expansion of iron
θ2 = final temperature
θ1 = initial temperature
From the information given,
L = 750
θ2 = - 18
θ1 = 34
a = 12 x 10^- 6/C
By substituting these values into the formula,
L1 = 36 x 10^- 6(- 18 - 34)
L1 = 36 x 10^- 6(- 52)
L1 = - 0.001872
Thus,
Final length - initial length = - 0.001872
Final length - 750 = - 0.001872
Final length = - 0.001872 + 750
Final length = 749.998 ft
2. Kevin is a member of the school's track and field team. His favorite event is throwing the shotput. In his last competition, Kevin exerted 165 N of force on the shot put, which accelerated28.5 m/s2. What was the mass of the shot put?
Given data
*Kevin exerted a force is F = 165 N
*The given acceleration is
[tex]a=28.5m/s^2[/tex]The formula for the mass of the shot put is given as
[tex]m=\frac{F}{a}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} m=\frac{165}{28.5} \\ =5.78\text{ kg} \end{gathered}[/tex]Thus, the mass of the shot put is m = 5.78 kg
-4, 0, -2/3, 4.11111…, 2, π, √6 'which members of this set are irrartional?
An irrational number is a real number that cannot be written as a simple fraction.
4 = 4/1 = rational
0 = 0/1 = rational
-2/3 = rational
4.11111 = irrational
2 = 2/1 = rational
π = 3.141592... = irrational
√6= irrational
Answer:
4.1111, π , √6
Standing on a bridge, you throw a stone straight upward. The stone hits a stream, 32.5 m below the point atwhich you release it, 3.10 s later. What is the speed of the stone (in m/s) just after it leaves your hand? Pleasedo not include any units in your answer below. Type in only the numerical result. If you include units, youranswer will be marked as incorrect.
Given:
Distance the stone hits the stream = 32.5m below the released point
Time = 3.10 seconds
Let's find the speed of the stone just after it leaves your hand.
To find the speed of the stone, apply the kinematic formula:
[tex]\Delta y=v_{iy}\ast t-\frac{1}{2}g\ast t^2[/tex]Since the point the stone hits the stream is below the released point is, the change in distance is:
[tex]\Delta y=0-32.5=-32.5m[/tex]Where:
a = -g = -9.8 m/s^2
t = 3.10 s
Substituet values into the formula and solve for the speed of the stone (vy).
We have:
[tex]\begin{gathered} -32.5=v_{iy}\ast3.10-\frac{1}{2}(9.8)\ast3.10^2 \\ \\ -32.5=v_{iy}\ast3.10-4.9\ast9.61 \\ \\ -32.5=v_{iy}\ast3.10-47.089 \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} v_{iy}\ast3.10=-32.5+47.089 \\ \\ v_{iy}\ast3.10=14.589 \end{gathered}[/tex]Divide both sides by 3.10:
[tex]\begin{gathered} \frac{v_{iy}\ast3.10}{3.10}=\frac{14.589}{3.10} \\ \\ v_{iy}=4.706\text{ m/s} \end{gathered}[/tex]Therefore, the speed of the stone just after it leaves your hand is 4.706 m/s
ANSWER:
4.706 m/s
How much force would someone need to move a 200kg fridge to accelerate at 5m/s^2
Given that the mass of fridge, m = 200 kg and acceleration is, a = 5 m/s^2
Force can be calculated using, Newton's law of motion,
[tex]F=ma[/tex]Substituting the values, the force will be
[tex]\begin{gathered} F=200\times5 \\ =1000\text{ N} \end{gathered}[/tex]Thus, the force needed to apply to the fridge is 1000 N.
The current through a motor is 210 A. If a battery keeps a 12.0 V potential difference across the motor, what electric energy is delivered to the motor in 10.0 s?
Given:
the current through motor is
[tex]I=210\text{ A}[/tex]The potential of the battery is
[tex]V=12\text{ V}[/tex]The time is
[tex]t=10\text{ s}[/tex]Required: the energy delivered to the motor in 10 sec is?
Explanation:
first, we calculate the power delivered to the motor.
The power is given by
[tex]P=VI[/tex]Plugging all the values in the above relation, we get:
[tex]\begin{gathered} P=12\text{ V}\times210\text{ A} \\ P=2520\text{ watt} \end{gathered}[/tex]The energy delivered to the motor is given by
[tex]E=Pt[/tex]Plugging all the values in the above relation, we get:
[tex]\begin{gathered} E=2520\text{ w}\times10\text{ s} \\ E=25200\text{ J} \\ \end{gathered}[/tex]Thus, the energy delivered to the motor is 25200 J.
5. Object A has a mass of 3 kg and is moving at 6 m/s before the collision with the 2 kg object B which is moving at -5 m/s. After the collision, object A has a velocity of 2 m/s. a. What is the total initial momentum of object A and B?b. What is the final velocity of object B after the collision?
Given,
The mass of object A is
[tex]m_1=3kg[/tex]The mass of object B is
[tex]m_2=2kg[/tex]The velocity of the object A before the collision is
[tex]u_1=6\text{ m/s}[/tex]The velocity of the object B before the collision is
[tex]u_2=-5\text{ m/s}[/tex]The velocity of object A after the collision is
[tex]v_1=2\text{ m/s}[/tex]Let the velocity of object B after the collision is v.
(a) The total initial momentum of objects A and B.
[tex]\begin{gathered} m_1u_1+m_2u_2=(3kg)\times(6\text{ m/s)+(2 }kg)\times(-5\text{ m/s)} \\ =18-10 \\ =8 \end{gathered}[/tex]The total initial momentum of objects A and B is 8.
(b) From the conservation of momentum,
[tex]\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ 8=(3kg)\times(2\text{ m/s)+(2 }kg)\times v \\ 2v=8-6 \\ v=1\text{ m/s} \end{gathered}[/tex]The final velocity of object B after the collision is 1 m/s.
Question 1001 10The device shown demonstrates which energy transformation?A Blectrical energy - electromagnetic energyB. Besteomaretic energy - Kinetic energyC. Yinetic energy-electrical energyD. Chemical energy-electromagnetic energyPlease answer this question!!
ANSWER
A. electrical energy → electromagnetic energy
EXPLANATION
The device shown is a lamp. To turn on the lamp we have to connect it to the electric power; there we have electrical energy. The lamp has a light bulb, which when it's on it emits light. Remember that light is an electromagnetic wave. Therefore the electrical energy is transformed into electromagnetic energy.
Explain in terms of IMF and KE why liquids assume the shape of any container into which they are poured, whereas solids are rigid and retain their shapes.
According to the Kinetic Molecular Theory, the kinetic energy of the particles of a solid is low enough as for the intermolecular forces to be enough to keep the atoms of a substance roughly at the same place from each other. On the other hand, the kinetic energy of the particles of a liquid is high enough as to overcome the intermolecular forces but not as high as to free the particles completely. Then, the molecules of a liquid can move away from each other, and they assume the shape of a container into which they are poured provided that there is a gravitational force pulling the particles of the liquid toward one side of the container.
To summarize:
The kinetic energy of the particles of a liquid is enough to allow them to move with respect to each other, whereas the kinetic energy of the particles of a solid is not enough to overcome the intermolecular forces, so they remain at the same place with respect to each other.
A quarterback back pedals 2.2 meters southward and then runs 6.8 meters northward. What is the magnitude and direction of the displacement?
ANSWER
[tex]4.6m\text{ Northward}[/tex]EXPLANATION
Displacement is the shortest distance between the final position and the initial position of a body.
Taking the southward direction to be the negative direction and the northward direction to be positive, to find the displacement, we have to add the distances traveled by the quarterback.
Therefore, we have:
[tex]\begin{gathered} D=-2.2+6.8 \\ D=4.6m \end{gathered}[/tex]Since the displacement is positive, its direction is Northward.
Hence, his displacement is 4.6 m Northward.
Two boats start together and race across a 42-km-wide lake and back. Boat A goes across at 42 km/h and returns at 42 km/h. Boat B goes across at 21 km/h, and its crew, realizing how far behind it is getting, returns at 63 km/h. Turnaround times are negligible, and the boat that completes the round trip first wins.(a) Which boat wins?o Boat Ao Boat Bo It's a tie(b) By how much?_____ km(c) What is the average velocity of the winning boat?_____ km/h
ANSWER
[tex]\begin{gathered} (a)\text{ Boat A} \\ \\ (b)\text{ }42\text{ km} \\ \\ (c)\text{ }0\text{ km/h} \end{gathered}[/tex]EXPLANATION
(a) To determine which boat wins, we have to find the total time the journey takes for each boat.
To do that, apply the formula for speed:
[tex]\begin{gathered} s=\frac{d}{t} \\ \\ t=\frac{d}{s} \end{gathered}[/tex]where d = distance traveled
t = time taken
For boat A, the time taken to go across and return is:
[tex]\begin{gathered} t=\frac{42}{42}+\frac{42}{42}=1+1 \\ \\ t=2\text{ hrs} \end{gathered}[/tex]For boat B, the time taken to go across and return is:
[tex]\begin{gathered} t=\frac{42}{21}+\frac{42}{63}=2+0.667 \\ \\ t=2.667\text{ hrs} \end{gathered}[/tex]We see that it takes boat B longer to complete the race. Hence, boat A will win the race.
(b) To determine how much boat A will win, we can look at the distances both boats cover. Notice that while getting across the lake, boat A travels twice as fast as boat B and travels at the same speed while returning.
This implies that in the time that boat A travels both laps of the race, boat B is just reaching the other end of the lake.
In other words, in the time that boat A travels 84 km, boat B travels 42 km.
Hence, boat A will beat boat B by a distance of:
[tex]\begin{gathered} 84-42\text{ km} \\ \\ 42\text{ km} \end{gathered}[/tex](c) In the race, the winning boat starts from one end of the lake, gets to the other end, and then, returns to the starting point.
This implies that the displacement of the boat is 0 km.
Velocity is given by dividing displacement by the time taken. Since the displacement is 0 km, the velocity of the winning boat, Boat A is 0 km/h
1) Which of the following is not true about vectors.a) Must have magnitude and direction.b) When doing math with vectors, we can usually treat the x and y components independently.c) If several vectors are added together, their order does not matter.d) The magnitude of a vector is the sum of the magnitude of its x and y components.2) Explain briefly your argument or reasoning.
a) A vector must have a magnitude and direction. A physical quantity with only the magnitude is called a scalar.
b) When doing the math we can treat the x and y components independently. We can use only x-component or only y-component for the necessary calculations. For example, the projectile motion. In projectile motion, we use the components of velocities independently.
c) We can add or subtract more than two vectors in any order. For example, If there are 3 vectors, A, B, and C then, from the associative law of vector addition,
[tex]\vec{A}+(\vec{B}+\vec{C})=(\vec{A}+\vec{B})+\vec{C}[/tex]d) The magnitude of the vectors is not the sum of their x and y components. Let vector A be represented as
[tex]\vec{A}=x\hat{i}+y\hat{j}[/tex]Where x and y are the x and y components of vector A respectively. And î and j cap are the unit vectors along the x and y direction respectively.
Then, the magnitude of vector A is given by,
[tex]A=\sqrt[]{x^2+y^2}[/tex]Thus the correct answer is option d. That is the statement in option d is not true.
If you have a mass of 65kg and the earth has a mass of 6 x 10^24. You are 6.4 x 10^6m from the center of earth. What is the Force you feel from the earth? G (the universal gravity constant) is 6.67 x 10^-11.
According to Newton's Law of Universal Gravitation, the gravitational force F between two particles with masses M and m separated by a distance r is given by:
[tex]F=G\frac{Mm}{r^2}[/tex]To find the force that you feel from the Earth, replace the value of the gravitational constant G, as well as M=6*10^24kg, m=65kg and r=6.4*10^6m:
[tex]\begin{gathered} F=(6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2})\times\frac{(6\times10^{24}\operatorname{kg})(65\operatorname{kg})}{(6.4\times10^6m)^2} \\ =635N \end{gathered}[/tex]Therefore, the magnitude of the force that you feel due to your gravitational interaction with the Earth is 635 N.
A 0.45 kg spike is hammered into a railroad tie. The initial speed of the spike is equal to1.9 m/s.If the tie and spike together absorb 22 per- cent of the spike’s initial kinetic energy as internal energy, calculate the increase in in- ternal energy of the tie and spike.Answer in units of J.
Given:
mass of the spike is
[tex]0.45\text{ kg}[/tex]initial speed od the spike is
[tex]v_=\text{ 1.9 m/s}[/tex]if the tie and spike together absorb 22 percent of the spike's initial kinetic energy.
Required:
calculate the increase in internal energy of the tie and spike.
Explanation:
here we apply conservation of energy.
[tex]\Delta U+\Delta K+\Delta P=0[/tex]total change in energy is zero.
here potential energy is not given so neglect that part. we have only
[tex]\Delta U+\Delta K=0[/tex][tex]\Delta U=-\Delta K[/tex]Here,
[tex]\Delta U=-(K_2-K_1)[/tex]K1 is initial kinetic energy and K2 is final kinetic energy that is equal to zero.
now we have
[tex]\Delta U=K_1[/tex]we know that kinetic energy is
[tex]K=\frac{1}{2}mv^2[/tex]then
[tex]\Delta U=\frac{1}{2}mv^2[/tex]plugging all the values in the above relation. we get
[tex]\begin{gathered} \Delta U=\frac{1}{2}\times0.45\text{ kg }\times(1.9\text{ m/s})^2\times0.22 \\ \Delta U=0.18\text{ J} \end{gathered}[/tex]Thus, the change in internal energy is
[tex]0.18\text{ J}[/tex]The change in angular velocity divided by time.Also equal to a/r
The angular acceleration of the body in terms of the angular velocity is,
[tex]\alpha=\frac{\text{angular velocity}}{\text{time}}[/tex]The angular velocity in terms of the linear velocity is,
[tex]\text{angular velocity=}\frac{v}{r}[/tex]where v is the linear velocity and r is the radius,
Thus, the angular acceleration becomes,
[tex]\begin{gathered} \alpha=\frac{1}{r}\times\frac{v}{t} \\ \alpha=\frac{a}{r} \end{gathered}[/tex]where a is the linear acceleration of the body,
E. O-8.237 kg m/s4. An object of mass 0.2 kg falling vertically downwardhits the ground with a speed of 13 m/s and bouncesback vertically upward with a speed 4 m/s. If the object was incontact with the ground for 0.5 seconds, calculate theaverage force exerted by the ground on the object. (1 point)A. 06.8 NB. 10.563 NC. 09.048 ND. 12.856 NE. 03.149 N5. An object was acted upon by a force of 24 for 0.8seconds. Calculate the change in its momentum. (1 point)
4)
We woud apply the formula for calculating rate of change of momentum which is expressed as
F = m(v2 - v1)/t
where
F is the force
m is the mass of the object
v2 is the final velocity
v1 is the initial velocity
t is the time
From the information given,
m = 0.2
v1 = -13
v2 = 4
t = 0.5
F = 0.2(4 - -13)/0.5
F = 6.8 N
an7. Pressure is calculated by dividing force by theover which the force is exerted.Which state of matter will hold its shape without a container?QPressure is calculated by dividing force by the “what”over which the force is exerted?
The pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed.
Then, the missing word is area.
Perform the indicated operation & simplify. Express the answer in terms of i (as a complex number).(2+4i)⋅(2+4i) =
You have the following expression;
[tex](2+4i)(2+4i)[/tex]to perform the previous operation, expand the previous factors as follow:
[tex]\begin{gathered} (2+4i)(2+4i)=(2)(2)+(2)(4i)+(4i)(2)+(4i)(4i) \\ =4+8i+8i+16i^2 \\ =4+16i+16i^2 \end{gathered}[/tex]take into account that i^2 = -1, then, you obtain:
[tex]4+16i+16(-1)=4+16i-16=16i-12[/tex]Hence, the result is 16i - 12
What does it mean that energy is quantized?
Electrons can posses only certain values of energy.
Explanation:
because of where the particle lies it. may be able to access more than that but it will be held at that position due to size. at least I think that's what this is. I took chemistry last year.
Question 1 of 10Which is an alpha particle?O A. A particle with an atomic number of 2 and an atomic mass of 4O B. A particle with an atomic number of 4 and an atomic mass of 2O C. A particle with an atomic number of 2 and an atomic mass of 3D. A particle with an atomic number of 1 and an atomic mass of 3SUBMIT
Answer:
Explanation:
An alpha particle is a helium atom. It has a mass number of 4 and an atomic number of 2. Thus, the correct option is
the earth moves at a speed of 2.95*10^4m/s and has a mass of 6.0*10^24. calculate the momentum of the earth
Given:
the speed of the earth is
[tex]v=2.95\times10^4\text{ m/s}[/tex]mass of the earth is
[tex]m=6.0\times10^{24}\text{ kg}[/tex]Required:
momentum of the earth needs to be calculated.
Explanation:
To calculate the momentum of the earth we will use momentum formula that is given as
[tex]P=mv[/tex]here P is momentum, m is the mass of the earth and v is the velocity of the earth.
plugging all the values in the above relation. we get,
[tex]\begin{gathered} P=6.0\times10^{24}\text{ kg }\times2.95\times10^4\text{ m/s} \\ P=\text{ 17.7 }\times10^{28}\text{ kg m/s} \end{gathered}[/tex]Thus, the momentum of the earth is
[tex]P=\text{17.7}\times10^{28}\text{kg\frac{m}{s}}[/tex]In an elastic collision, bumper cars 1 and 2 are moving in the same direction when bumper car 1 rear-end bumper car 2. The initial speed of car 1 is 7.73 m/s and car 2 is 4.01 m/s. The mass of car 2 is 57% greater than that of car 1.
(a) what is the final velocity in m/s of bumper car 1?
(b) what is the final velocity in m/s of bumper car 2?
a ) The final velocity after an elastic collision of bumper car 1 = 9.01 m / s
b ) The final velocity after an elastic collision of bumper car 2 = 1.73 m / s
According to law of conservation of momentum
m1 u1 + m2 u2 = m1 v1 + m2 v2
u1 = 7.73 m / s
u2 = 4.01 m / s
m2 = 0.57 m1
7.73 m1 + ( 4.01 * 0.57 * m1 ) = m1 v1 + 0.57 m1 v2
7.73 + 2.29 = v1 + 0.57 v2
v1 = 10.02 - 0.57 v2
According to law of conservation of energy
1 / 2 m1 u1² + 1 / 2 m2 u2² = 1 / 2 m1 v1² + 1 / 2 m2 v2²
7.73² m1 + ( 4.01² * 0.57 * m1 ) = m1 v1² + ( 0.57 * m1 * v2² )
59.75 + 9.17 = ( 10.02 - 0.57 v2 )² + 0.57 v2²
68.92 = 100.4 + 0.33 v2² - 11.42 v2² + 0.57v2²
10.52 v2² = 31.48
v2² = 2.99
v2 = 1.73 m / s
v1 = 10.02 - ( 0.57 * 1.72 )
v1 = 10.02 - 0.99
v1 = 9.01 m / s
Therefore,
a ) The final velocity after an elastic collision of bumper car 1 = 9.01 m / s
b ) The final velocity after an elastic collision of bumper car 2 = 1.73 m / s
To know more about law of conservation of momentum
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No tutor is helping with this difficult question. Please someone help
The labelled diagram of the situation is shown below
Recall the kinetic energy formula,
work done = 1/2mv^2
where
m is the mass
v is the velocity
Recall, work done = force x distance
Thus,
force x distance = 1/2mv^2
From the information given,
Force = 2000
distance = 0.1
m = 5
Thus,
2000 x 0.1 = 1/2 x 5 x v^2
200 = 2.5v^2
v^2 = 200/2.5 = 80
v = √80
v = 8.94 m/s
The ball will collide with the the crate with a velocity of 8.94. The velocity at which the crate will move is v2
We would apply the law of momentum.
Initial momentum before collision = final momentum after collision
m1u1 + m2u2 = m1v1 + m2v2
where
m1 and m2 are the masses of the ball and crate respectively
u1 and u2 are the initial velocities of the ball and crate respectively
v1 and v2 are the final velocities of the ball and crate respectively
From the information given,
m1 = 5
m2 = 31
u1 = 8.94
u2 = 0 (because it was stationary)
v1 = - 1.44(because it moved in the opposite direction
Thus,
5 x 8.94 + 3.1 x 0 = 5 x - 1.44 + 31 x v2
44.7 = - 7.2 + 31v2
31v2 = 44.7 + 7.2 = 51.9
v2 = 51.9/31
v2 = 1.67 m/s
This means that the velocity with which the crate will start moving is 1.67 m/s
With respect to springs, conservation of energy is expressed as
1/2kx^2 = 1/2mv^2
where
k is the spring constant
x is the compression of the spring
m is the mass of the crate
v is the velocity of the crate
From the information given,
k = 250
Thus,
1/2 x 250 x x^2 = 1/2 x 31 x 1.67^2
125x^2 = 43.22795
x^2 = 43.22795/125 = 0.3458236
x = √0.3458236
x = 0.59 m
The spring will compress by 0.59 m
Extra credit
With the presence of friction, the ball will exert lesser force on the crate and this would cause the compression of the spring to be reduced. The spring will stretch less.
If a ball is dropped from a bridge that is 36.4 meters high, how long will it take until the ball hits the ground below the bridge?
2.72 sec
Explanation
Step 1
to solve this we need to use the formula
[tex]h=v_0t+\frac{1}{2}gt^2[/tex]let
[tex]\begin{gathered} v_0=0 \\ h=36.4\text{ m} \\ g=9.8\text{ }\frac{\text{m}}{s^2} \\ t=\text{ unknown= t} \end{gathered}[/tex]replace
[tex]\begin{gathered} h=v_0t+\frac{1}{2}gt^2 \\ 36.4=0\cdot t+\frac{1}{2}(9.8\text{ }\frac{m}{s^2})t^2 \\ 36.4=+(4.9\frac{m}{s^2})t^2 \\ \text{divide both sides by 4.9} \\ \frac{36.4}{4.9}=t^2 \\ t=\sqrt[]{7.42} \\ t=2.72 \end{gathered}[/tex]therefore, the answer is
2.72 sec
I hope this helps you