Answers
Take into account that the impulse is given by:
[tex]I=m\Delta v=m(v-v_o)[/tex]where,
m: mas of the ball = 3.0kg
v: final speed = 3.0m/s
vo: initial speed = 9.0m/s
replace the previous values into the formula for I and simplify:
Hence, the impulse is -18 kg*m/s (it's negative due to the force is opposite to the motion of the ball)
Related Questions
Which of the following is an example of increasing energy efficiency?Question 13 options:A nuclear power plant that stores its radioactive waste more safely than an older powerplantA compact fluorescent light bulb that uses less energy than a regular incandescentlight bulbWalking to work instead of driving a carTurning off the lights when no one is in the room instead of leaving all the lights on
Answers
Energy efficiency means using less input energy and producing same output energy.
A regular incandescent light bulb uses input energy to produce the same output as the compact fluorescent light bulb.
Thus, using a compact fluorescent light bulb instead of regular incandescent bulb is an example of energy efficiency.
Radians are units for
Answers
The radian is the unit for the angular displacement.
The symbol for the radians is 'rad'. One radian is equal to the
9. Which of the following best explain what happens when an object with a negative charge comes near another object with a negative charge?A-The objects will not move.B-The objects will repel each other.C-The objects will attract each other.D-The objects will cancel each other.
Answers
In Physics, charges of same nature repel each other, and charges of different nature attract each other. In this situation, the objects will repel each other because both charges are negative.
Therefore, the answer is B.Gravity is a contact force. Is this true or false?
Answers
A contact force is a force that requires physical contact in order to interact with an object. A massive object created something called a "Gravitational field", and therefore, in order for the force of gravity to interact with an object physical contact is not required. Therefore, the statement is false.
If 100. g of gold-198 decays to 6.25g in 10.8 days, what is the half-life of gold-198?
Answers
Given data
*The amount of gold-198 is N = 100 g
*The amount of gold decays is n = 6.25 g
*The given number of days is T = 10.8 days
The expression for the radioactivity decay is given as
[tex]N=n(0.5)^{\frac{T}{t}}[/tex]*Here t is the half-life period
Substitute the values in the above expression as
[tex]\begin{gathered} 100=6.25(0.5)^{\frac{10.8}{t}} \\ t=2.7\text{ days} \end{gathered}[/tex]The pilot of a military cargo plane flying horizontally at 160m/s is 250m above a base camp.The pilot releases a hatch and a box of supplies drops to the ground. A) How long does it take the box to fall? B) how far from the base did the box fall? Hint represent the base with a dot.
Answers
Given,
The initial velocity of the box, u=160 m/s
The height at which the plane is flying, h=250 m
The angle at which the box is dropped, θ=0°
A)
The time of flight of a projectile motion when projected with zero launch angle is given by
[tex]t=\sqrt[]{\frac{2H}{g}}[/tex]Where g is the acceleration due to gravity and g=9.8 m/s²
On substituting the known values in the above equation,
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times250}{9.8}} \\ =7.14\text{ s} \end{gathered}[/tex]Therefore the box falls to the ground in 7.14 seconds.
B)
The range of a projectile is given by,
[tex]R=ut[/tex]On substituting the known values,
[tex]\begin{gathered} R=160\times7.14 \\ =1142.4\text{ m} \end{gathered}[/tex]Therefore the box falls to the ground at 1142.4 m away from the base camp.
A child and sled with a combined mass of 25.2 kg slide down a frictionless hill, starting at the top. The sled starts from rest and acquires a speed of 6 m/s by the time it reaches the bottom of the hill. What is the height (in m) of the hill?
Answers
Given there is no friction, the law of conservation of energy can be applied. This is:
[tex]\Delta E=0[/tex]For our case, the energy in the beginning will be simply potential, and by the end will be only kinetic. We can write this as:
[tex]mgh=\frac{mv^2}{2}\Rightarrow h=\frac{v^2}{2g}[/tex]For our case, this will be:
[tex]h=\frac{6^2}{2*10}=1.8m[/tex]Thus, the hill has 1.8m
The direction of the force on a positive test charge near another positive charge is away from the other charge. Is this true or false?
Answers
The given statement 'The direction of the force on a positive test charge near another positive charge is away from the other charge' is True. The magnitude of force between two charges depends upon the magnitude of charges whereas the direction depends upon the polarity of charges.
A carmaker has designed a car that can reach a maximum acceleration of 24 meters/second. The car's mass is 3,030 kilograms. Assuming the same engine is used, what should the car's mass be if the carmaker wants to reach an acceleration of 30 meters/second?
Answers
ANSWER:
2,424 kilograms
STEP-BY-STEP EXPLANATION:
The first thing we must do is calculate the ratio between the accelerations, like this:
[tex]r=\frac{30}{24}=1.25[/tex]We know that mass and acceleration are inversely proportional, that is, if one goes up, the other must go down. Since it would be the same force, we can calculate the mass of the car like this:
[tex]\frac{3030}{1.25}=2424\text{ kg}[/tex]The mass of the car is 2,424 kilograms
Abigail was scheduled to repay a loan 100 days from now with a payment of $3,650.00, but will get some extra money and be able to pay it off 40 days from now with an equivalent payment of $3,596.79. Calulate the rate of interest ?
Answers
Answer:
0.025%
Explanation:
Suppose that you apply a horizontal force of 60 N to a box and youdisplace it for a distance of 5 m. How much work did you do on the box? *
Answers
Given:
The horizontal force applied is F = 60 N
The distance displaced is d = 5 m
To find the work done.
Explanation:
The work done will be
[tex]\begin{gathered} W=F\times d \\ =60\times5 \\ =\text{ 300 J} \end{gathered}[/tex]Thus, the work done is 300 J.
Consider the graph below that represents the variation of the velocity with respect to time of an object moving along the x - axis . The distance traveled during the second part of the trip ( 10s to 18s ) is equal to :
Answers
In a velocity-time graph, the area under the curve represents the distance.
The distance traveled from 10s to 18 s is
[tex]\begin{gathered} d\text{ = }\frac{1}{2}\times10\times(18-10) \\ =\text{ 40 m} \end{gathered}[/tex]Final Answer: The distance traveled is 40 m from time 10 s to 18 s.
While moving out of her dorm room, Bridget carries a 12-kg box to her car,holding it in both arms. a) How much force must be exerted by each of herarms to support the box? b)How will this force change if Bridget holds thebox with only one arm?
Answers
weight = mass x gravity
mass = 12 kg
gravity = 9.8 m/s^2
Weight = 12 x 9.8 = 117.6 N
The force (F) needed to support the box:
F = W
FF= = 117.6 N
Divide by 2 arms:
117.6 / 2= 58.8 N
If she holds with one arm
117.6 N
a) 58.8 N
b) 117.6 N
A 2.0 µC particle with a kinetic energy of 0.10 J enters a uniform magnetic field of magnitude 0.10 T. In the field, the particle moves in a circular path of a radius 3.0 m. Find the mass of the particle.3.7 x 10-12 kg3.1 x 10-12 kg4.2 x 10-12 kg1.3 x 10-12 kg1.8 x 10-12 kg
Answers
Given:
The charge of the particle, q=2.0 μC
The kinetic energy of the particle, K=0.10 J
The magnetic field, B=0.10 T
The radius of the path of the particle, r=3.0 m
To find:
The mass of the particle.
Explanation:
The magnetic force acting on the particle is providing the particle with the necessary centripetal force.
Therefore the centripetal force acting on the particle is equal to the magnetic force acting on it.
Thus,
[tex]\frac{mv^2}{r}=\text{qvB}[/tex]Where m is the mass of the particle and v is its velocity.
On simplifying the above equation,
[tex]\begin{gathered} \frac{mv}{r}=qB \\ \Rightarrow v=\frac{qrB}{m} \end{gathered}[/tex]The kinetic energy of the particle is given by,
[tex]K=\frac{1}{2}mv^2[/tex]On substituting the value of v in the above equation,
[tex]\begin{gathered} K=\frac{1}{2}m\times(\frac{qrB}{m})^2 \\ \Rightarrow K=\frac{1}{2m}(qrB)^2 \\ \Rightarrow m=\frac{1}{2K}(qrB)^2_{} \end{gathered}[/tex]On substituting the known values in the above equation,
[tex]\begin{gathered} m=\frac{1}{2\times0.10}\times(2.0\times10^{-6}\times3.0\times0.10)^2 \\ =1.8\times10^{-12}\text{ kg} \end{gathered}[/tex]Final answer:
The mass of the particle is 1.8×10⁻¹² kg
A projectile is fired straight up with a positive velocity of 45 m/s. When the projectile returns to its starting position, the magnitude of its velocity is...?
Answers
Projectile is fired straight up with a positive velocity of 45 m/s. As when the projectile returns to its starting position, the magnitude of its velocity remains same.
Esteban hits a 0.145kg baseball traveling at him at +27.0m/s. the ball is in contact with the bat for 0.133 seconds with an average force of 51.2n. cakculate the final velocity of the baseball.m=u-u=f=v=
Answers
In order to calculate the final velocity of the ball, first let's calculate the initial momentum:
[tex]\begin{gathered} p=mv\\ \\ p=0.145\cdot27\\ \\ p=3.915\text{ kg m/s} \end{gathered}[/tex]Now, using the impulse theorem, let's calculate the change in momentum:
[tex]\begin{gathered} \Delta p=I\\ \\ \Delta p=F\cdot\Delta t\\ \\ \Delta p=51.2\cdot0.133\\ \\ \Delta p=6.8096\text{ kg m/s} \end{gathered}[/tex]Adding the change in momentum to the initial momentum, we can find the final momentum:
[tex]\begin{gathered} p_f=p+\Delta p\\ \\ p_f=3.915+6.8096\\ \\ p_f=10.7246\text{ kg m/s} \end{gathered}[/tex]Using the final momentum, let's calculate the final velocity:
[tex]\begin{gathered} p=mv\\ \\ 10.7246=0.145\operatorname{\cdot}v\\ \\ v=\frac{10.7246}{0.145}\\ \\ v=73.96\text{ m/s} \end{gathered}[/tex]The 2000 Belmont Stakes winner, Commendable, ran the horse race at anaverage speed of 15.98 m/s. If Commendable and jockey Pat Day had acombined mass of 550.0 kg, what was their KE as they crossed the finish line?
Answers
Given:
The combined mass is
[tex]m=550.0\text{ kg}[/tex]The speed is
[tex]v=15.98\text{ m/s}[/tex]Required: find the kinetic energy.
Explanation:
To find the kinetic energy we will use the formula that is given as
[tex]K.E=\frac{1}{2}mv^2[/tex]Here,
[tex]m[/tex]is the mass and
[tex]v[/tex]is the speed.
Plugging all the values in the above relation, we get
[tex]\begin{gathered} K.E=\frac{1}{2}mv^2 \\ K.E=\frac{1}{2}\times550.0\text{ kg}\times(15.98\text{ m/s})^2 \\ K.E=0.5\times550.0\text{ kg}\times255.36\text{ }(\text{m/s}^)^2 \\ K.E=70224.11\text{ J} \end{gathered}[/tex]Thus, the kinetic energy is 70224.11 J.
A 3.20-kilogram toy car is attached to a 125-centimeter string ofnegligible mass. The other end of the string is attached to a ringthat is slipped over a peg. The tensile strength of the string is144 newions. The car revolves about the peg in a circular path ona frictionless table. Find the maximum angular speed that the carcan travel in revolutions per minute without breaking the string.
Answers
Given data:
The tensile strength of the string is,
[tex]F^{\prime}=144\text{ N}[/tex]The mass of the car is,
[tex]m=3.2\text{ kg}[/tex]The length of the string is,
[tex]\begin{gathered} r=125\text{ cm} \\ r=1.25\text{ m} \end{gathered}[/tex]The force acting on the car is,
[tex]F^{\prime}=\frac{mv}{r}^2[/tex]where a is the linear acceleration and m is the mass of the car.
Substituting the known values,
[tex]\begin{gathered} 144=\frac{3.2\times v^2}{1.25} \\ v^2=56.25 \\ v=7.5ms^{-1} \end{gathered}[/tex]Thus, the angular speed of the car is,
[tex]\begin{gathered} \omega=\frac{v}{r} \\ \omega=\frac{7.5}{1.25} \\ \omega=6 \end{gathered}[/tex]Each distance of each revolution is,
[tex]\begin{gathered} d=2\pi r \\ d=2\pi\times1.25 \\ d=7.85 \end{gathered}[/tex]Thus, the value of angular speed in terms of revolution is,
[tex]\begin{gathered} \omega=\frac{6}{7.85} \\ \omega=0.76\text{ revolution per second} \\ \omega=0.76\times60\text{ revolutions per minute} \\ \omega=45.6\text{ revolutions per minute} \end{gathered}[/tex]Thus, the maximum angular speed of the toy car is 45.6 revolutions per minute.
In 1 to 2 sentences, Explain how a decrease in the resistance caused by friction change the amount of force needed to move an object?
Answers
Whenever two objects rub against each other, they cause friction. Friction works against the motion and acts in the opposite direction.
Using newton's second law we can describe a generic case of movement:
[tex]\begin{gathered} \Sigma F=ma \\ \Sigma F=F-Ff \end{gathered}[/tex]Where:
F = Amount of force need to move an object
Ff = Friction force
As we can see, if:
[tex]\begin{gathered} Ff\rightarrow0 \\ \Sigma F\rightarrow F \end{gathered}[/tex]In another words, The smaller the friction force, the smaller the opposition to motion.
A car accelerated from rest to 72km/hr in 5secs given that r=0.25m, find the number of revolution
Answers
The number of revolution = 31.83
Explanations:The car starts from rest
The initial velocity, u = 0 m/s
The final velocity, v = 72 km/hr
v = 72 x 1000/3600
v = 20 m/s
The time, t = 5 secs
The radius of the wheel of the car, r = 0.25 m
Find the acceleration using the formula
v = u + at
20 = 0 + 5a
5a = 20
a = 20/5
a = 4 m/s
Find the distance using the formula
s = ut + 0.5at²
s = 0(5) + 0.5(4)(5²)
s = 0 + 50
s = 50m
The number of revolution is calculated using the formula below:
[tex]n\text{ = }\frac{s}{2\pi r}[/tex][tex]\begin{gathered} n\text{ = }\frac{50}{2\times3.142\times0.25} \\ n\text{ = }31.83 \end{gathered}[/tex]The number of revolution = 31.83
Hallie tripped over a rock, tumbling down a small but steep hill, she was holding two cans of Sprite (which got tossed during the fall). She picked up the Sprites and continued on, later, she opened the drinks and every drop spewed out onto Hallies clothes and her floor.Explain why this happened/how
Answers
Sprite is a carbonated drink. This means that it contains C02 under pressure which gives it a refreshing taste. When the bottle of sprite is shaken, small bubbles are formed. These bubbles make it easier for the dissolved C02 to escape by joining the existing bubbles.
how much pressure is exterted by the water at the bottom of well if the depth of the wall is 6m.
Answers
Hydrostatic pressure = q g h
Density of water (q) = 1000kgm^-3 , gravity (g)= ms^2
q = 10^3 kg/m^3
1g= 10ms^-2
h= 6m
Pressure = 10^3kg/m3 x 10m/s^2 x 6
P = 60000pa
this is a practice question not a graded assingment.The friction force is 30 N
Answers
First, let's calculate the horizontal component of the applied force:
[tex]\begin{gathered} F_x=F\cdot\cos35°\\ \\ F_x=70\cdot0.819\\ \\ F_x=57.33\text{ N} \end{gathered}[/tex]Now, calculating the works, we have:
a.
To calculate this work, we use the horizontal component of the applied force:
[tex]\begin{gathered} W_=F_x\cdot d\\ \\ W=57.33\cdot10\\ \\ W=573.3\text{ J} \end{gathered}[/tex]b.
To calculate this work, we use the friction force only:
[tex]\begin{gathered} W_f=F_f\cdot d\\ \\ W_f=-30\cdot10\\ \\ W_f=-300\text{ J} \end{gathered}[/tex](We use a negative force because it is against the movement)
c.
The total work on the wagon is given by:
[tex]\begin{gathered} W_{net}=W+W_f\\ \\ W_{net}=573.3-300\\ \\ W_{net}=273.3\text{ J} \end{gathered}[/tex]Some amount of energy is transferred by heat into a system. The net work done on the system is 56 J, while the increase in its internal energy is 12 JWhat is the amount of net heat?
Answers
Given,
The net-work done on the system is
[tex]\Delta w=56\text{ J}[/tex]The increase in its internal energy is
[tex]\Delta U=12\text{ J}[/tex]By the first law of thermodynamics.
[tex]Q=\Delta U-\Delta w[/tex]Put the given values,
[tex]\begin{gathered} Q=12J-56\text{ J} \\ Q=-44\text{ J} \end{gathered}[/tex]Thus, the amount of net heat is -44 J.
Calculate final velocities assuming perfectly inelastic collisions. The bodies stick together in the final state.
Answers
In the inelastic collision, the momentum of the system is conserved before and after the collision.
The objects stick together in the final state, thus, the velocity of both the object in the final state is the same.
In the second case,
According to the law of conservation of momentum, the velocity of each object is,
[tex]\begin{gathered} M_1v_{1i}+M_2v_{2i}=(M_1+M_2)v \\ 10\times2+5\times0=(10+5)\times v \\ 20+0=15v \\ 20=15v \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} v=\frac{20}{15} \\ v=1.33\text{ m/s} \end{gathered}[/tex]Thus, the final velocity of both the objects is 1.33 m/s.
In the third case,
According to the law of conservation of momentum, the velocity of each object is,
[tex]\begin{gathered} M_1v_{1i}+M_2v_{2i}=(M_1+M_2)v \\ 50\times2+5\times0=(50+5)\times v \\ 100+0=55v \\ 100=55v \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} v=\frac{100}{55} \\ v=1.82\text{ m/s} \end{gathered}[/tex]Thus, the final velocity of both the objects is 1.82 m/s.
What is the percentage difference between the correlation and the elongation?
Answers
The percentage difference between correlation and first value of elongation can be calculated as,
[tex]\begin{gathered} P_1=\frac{0.9878-0.056}{0.9878}\times100 \\ =\frac{0.9318}{0.9878}\times100 \\ \approx94.3 \end{gathered}[/tex]Thus, the percentage difference between correlation and first value of elongation is 94.3%.
The percentage difference between correlation and second value of elongation can be calculated as,
[tex]\begin{gathered} P_2=\frac{0.9878-0.078}{0.9878}\times100 \\ =\frac{0.9098}{0.9878}\times100 \\ \approx92.1 \end{gathered}[/tex]Thus, the percentage difference between correlation and second value of elongation is 92.1%.
The percentage difference between correlation and third value of elongation can be calculated as,
[tex]\begin{gathered} P_3=\frac{0.9878-0.164}{0.9878}\times100 \\ =\frac{0.8238}{0.9878}\times100 \\ \approx83.4 \end{gathered}[/tex]Thus, the percentage difference between correlation and third value of elongation is 83.4%.
The percentage difference between correlation and fourth value of elongation can be calculated as,
[tex]\begin{gathered} P_4=\frac{0.9878-0.196}{0.9878}\times100 \\ =\frac{0.7918}{0.9878}\times100 \\ \approx80.2 \end{gathered}[/tex]Thus, the percentage difference between correlation and fourth value of elongation is 80.2%.
The percentage difference between correlation and fifth value of elongation can be calculated as,
[tex]\begin{gathered} P_5=\frac{0.9878-0.237}{0.9878}\times100 \\ =\frac{0.7508}{0.9878}\times100 \\ \approx76 \end{gathered}[/tex]Thus, the percentage difference between correlation and fifth value of elongation is 76%.
The percentage difference between correlation and sixth value of elongation can be calculated as,
[tex]\begin{gathered} P_6=\frac{0.9878-0.286}{0.9878}\times100 \\ =\frac{0.7018}{0.9878}\times100 \\ \approx71 \end{gathered}[/tex]Thus, the percentage difference between correlation and sixth value of elongation is 71%.
The percentage difference between correlation and seventh value of elongation can be calculated as,
[tex]\begin{gathered} P_7=\frac{0.9878-0.319}{0.9878}\times100 \\ =\frac{0.6688}{0.9878}\times100 \\ \approx67.7 \end{gathered}[/tex]Thus, the percentage difference between correlation and seventh value of elongation is 67.7%.
A lamp with a resistance of 60 Ohms is plugged into a 110-Volt household circuit. How much current flows through the lamp?Group of answer choices0.55 [A]1.83 [A]170 [A]6,600 [A]
Answers
Given:
The resistance of the lamp is,
[tex]R=60\text{ ohm}[/tex]The voltage in the household circuit is,
[tex]V=110\text{ V}[/tex]To find:
The current through the lamp
Explanation:
Ohm's law gives the relation between the current, voltage, and resistance of the circuit.
The current is,
[tex]i=\frac{V}{R}[/tex]So, the current is,
[tex]\begin{gathered} i=\frac{110}{60} \\ =1.83\text{ A} \end{gathered}[/tex]Hence, the current through the lamp is 1.83 A.
explain briefly the Newton's law of cooling, give 3 examples about this law and add one picture with each example
Answers
Newton's law of cooling states that the rate of cooling of an object is directly proportional to the difference in temperature between the object and its surrounding.
This can be expressed mathematically as:
[tex]\begin{gathered} Q=h_cA\triangle T \\ \\ where: \\ Q\text{ = rate of heat transfer} \\ A\text{ is the area of the surface} \\ h_c\text{ is the convection coefficient} \\ \triangle T\text{ is the change in temperature} \end{gathered}[/tex]Three examples in which Newton's law of cooling is applied:
• Melting of Ice-cream
,• Cooling of heated water
,• Forensic science
A picture to illustrate the melting of ice cream:
A picture to illustrate the cooling of heated water
If an object is 16.76 cm in front of a convex mirror that has a focal length of 67.1 cm, how far behind the mirror will the image appear to an observer ?
Answers
F = focal lenght = 67.1 cm
u= distance between mirror and object = 16.76 cm
1/f = 1/v + 1/u
1/67.1 = -1/16.76 + 1/u
1/67.1 + 1/16.76 = 1/u
u = 13.41
Electrons are pushed apart from one another because of:the force of attraction.voltage difference.the force of repulsion.magnetism.
Answers
Force of repulsion.
Explanation:Note that:
• Electrons are negatively charged
,• Like charges repel (Move away from each other)
,• Unlike charges attract
Therefore, the reason why electrons are pushed away from one another is because of the force of repulsion that exists between them. The force of repulsion exists between them because the electrons have similar charge (negative), and like charges are meant to repel.