7.41 m/s^2
ExplanationWhen an object is traveling in a straight line with an increase or decrease in velocity at equal intervals of time, then the object is said to be in uniform acceleration
to find the acceleration we can use the formulas
[tex]\begin{gathered} x=v_it+\frac{1}{2}at^2 \\ and \\ v_f=v_1+at \end{gathered}[/tex]so
Step 1
a)
let
[tex]\begin{gathered} initial\text{ velocity}=v_i=65\frac{mi}{h} \\ final\text{ velocity= 0 \lparen complete stop-rest\rparen} \\ distance=57\text{ m=0.0354182 miles} \end{gathered}[/tex]b) replace in equation (2)
[tex]\begin{gathered} v_{f}=v_{1}+at \\ 0=65+at \\ -65=at \\ a=-\frac{65}{t} \end{gathered}[/tex]replace in equation (1)
[tex]\begin{gathered} x=v_{i}t+\frac{1}{2}at^{2} \\ 0.0354182=65t+\frac{1}{2}(-\frac{65}{t})t^2 \\ 0.0354182=65t-\frac{1}{2}(\frac{65}{})t^ \\ 0.0354182=65t-\frac{65}{2}t \\ 0.0354182=t(65-\frac{65}{2}) \\ 0.0354182=t(32.5) \\ t=\frac{0.0354182}{32.5}=0.001089\text{ hours} \end{gathered}[/tex]now, replace in the previous equation to find a
[tex]\begin{gathered} a=-\frac{65}{t} \\ a=-\frac{65\frac{mi}{h}}{0.001089\text{ hours }} \\ a=-59687\text{ }\frac{miles}{hour^2} \end{gathered}[/tex]so, the answer is
[tex]a=-59687.78\frac{m\imaginaryI les}{hour^{2}}[/tex]finally, let's convert the acceleration from miles per squared hour into meters per square second
remember that:
[tex]\begin{gathered} 1\text{ hour = 3600 seconds } \\ 1\text{ miles=1609.34 meters} \end{gathered}[/tex]so
[tex]\begin{gathered} a=-59,687\frac{m\imaginaryI les}{hour^{2}}(\frac{1\text{ hour}}{3600\text{ s}})^2(\frac{1609.34\text{ m}}{1\text{ mile}}) \\ a=-59,687\frac{\text{m}\mathrm{i}\text{les}}{\text{hour\textasciicircum{\text{2}}}}*\frac{1\text{ hour}^2}{12960000\text{s}^2}*(\frac{1609.34\text{ m}}{1\text{ mile}}) \\ a=-7.41\text{ }\frac{m}{s^2} \end{gathered}[/tex]hence
the acceleration is -7.41 m/s^2
I hope this helps you
The energy released when 0.375 kg of uranium are converted into energyis equal toa. 2.35 x 1014 Jb. 3.38 x 1016 JC. 4.53 x 1016 Jd. 7.69 x 1016 j
Given that the mass of uranium is m = 0.375 kg = 375 g.
We have to find the energy.
First, we need to find the number of moles.
The number of moles can be calculated as
[tex]\begin{gathered} \text{Number of moles =}\frac{Given\text{ mass}}{atomic\text{ mass}} \\ =\frac{375}{235} \\ =1.59\text{ } \end{gathered}[/tex]Next, we have to convert the number of moles into the number of atoms.
The number of atoms will be
[tex]\begin{gathered} \text{Number of atoms=1.59}\times\text{6.022}\times10^{23} \\ =\text{ 9.57}\times10^{23} \end{gathered}[/tex]One atomic mass unit releases 931.5 MeV energy.
The energy can be calculated as
[tex]\begin{gathered} E=\text{ number of atoms}\times931.5MeV\times atomic\text{ mass of uranium} \\ =9.57\times10^{23}\times(931.5\times10^6eV)\times235 \\ =\text{ }2.095\text{ }\times10^{35}eV\text{ } \\ =\text{ 2.095}\times10^{35}\times1.6\times10^{-19}\text{ J} \\ =\text{ 3.35 }\times10^{16\text{ }}J \end{gathered}[/tex]
A car will skid to halt at a uniform rate of -9.4 m/s^2. If you measure skid marks that are 34 m long, with what speed was the car going just before the driver slammed the brakes?
The speed with which the car going just before the driver slammed the brakes is 25.28 m / s, if the car skids to halt at a uniform rate of -9.4 m/s^2.
v² = u² + 2 a s
v = Final velocity
u = Initial velocity
a = Acceleration
s = Displacement
v = 0
a = - 9.4 m / s²
s = 34 m
0 = u² + ( 2 * - 9.4 * 34 )
u² = 639.2
u = 25.28 m / s
Deceleration is the slowing down of an object. It can be negative acceleration or positive acceleration. Deceleration is negative if the object moves in positive direction and it is positive if the object moves in negative direction.
Therefore, the speed with which the car going just before the driver slammed the brakes is 25.28 m / s
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When a sound wave bends around a barrier, so that you can still hear thesound even though you cannot see the source, the sound wave hasundergone:O A. diffraction.O B. compression.C. refraction.D. reflectionSUBMIT
Diffraction
Explnation":The bending of waves around obstacle in its path is called diffraction.
When a sound wave bends around a barrier whose dimensinn is lesser than or equal to its wavelength, this phenomenon is referred to as diffraction.
The sound can be heard even though the source of the sound cannot be seen. e
What is the volume of a cube whose side is 10 cm?1) 10–3 m32) 10 m33) 102 m34) 103 m3
Given data:
* The side of the cube is 10 cm.
Solution:
The volume of the cube is,
[tex]V=a^3[/tex]where a is the side of cube,
Substituting the known values
[tex]\begin{gathered} V=(10\times10^{-2})^3 \\ V=(10^{-1})^3 \\ V=10^{-3}m^3 \end{gathered}[/tex]Thus, first option is the correct answer.
A wave period 10ms travels through a medium. The graph shows the variation of particle displacement with distance for the wave.
What is the average speed of a particle in the
medium during one cycle?
A wave period 10ms travels through a medium. The graph shows the variation of particle displacement with distance for the wave. The average speed of a particle in the medium during one cycle is (v)= 20 m/s.
What is velocity?The velocity is a physics term that means a matter that covers a distance in a particular time. its a vector quantity. It can be measured in m/s or cm/s.
How can we calculate the velocity of the wave?To calculate the average velocity we are using the formula is,
v=d/t
Here we are given,
According to the picture the wave start from +4.0 displacement and goes decreasing at -4.0 and then comes back at +4.0 and completes the full cycle and complete the amount of distance is (d)= 20cm = 0.2m
The period of the wave is (t)= 10ms = 10*10⁻³ s
We have to calculate the average speed of a particle in the medium during one cycle= v m/s.
Now we put the values in the above equation we get,
v=d/t
Or, v= 0.2/ 10*10⁻³
Or, v= 20 m/s
So from the calculation we can say that, The average speed of a particle in the medium during one cycle is (v)=20 m/s.
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Johnny and Francis are participating in the Roller Skating dance championship. While 73.0-kg Johnny roller skates backwards at 3.0 m/s, 57.0-kg Francis jumps into his arms with a velocity of 5.0 m/s in the same direction. How fast do they roll backwards together?
Given,
The mass of Johnny, m₁=73.0 kg
The mass of Francis, m₂=57.0 kg
The initial velocity of Johnny, u₁=3.0 m/s
The initial velocity of Francis, u₂=5.0 m/s
From the law of conservation of momentum, the total momentum of a system will always remain the same.
Therefore,
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]Where v is the velocity with which they roll backward together.
On substituting the known values in the above equation,
[tex]\begin{gathered} 73.0\times3.0+57.0+5.0=(73.0+57.0)v \\ v=\frac{219+285}{130} \\ =3.88\text{ m/s} \end{gathered}[/tex]Thus the velocity with which they roll backward together is 3.88 m/s
The force diagram shown here describes the forces that external objects (the surface andEarth) exert on a woman (in this scenario, the force diagram does not change with time). Describe three different types of motion of the woman that are consistent with the force diagram.
The three different motions are;
The upward motion of the woman is constantThe downward motion of the woman is also constantThe horizontal motion of the woman is zero.What is force diagram?
Force diagram is a pictorial or graphical illustration of different forces acting on object.
In this given question, there two forces acting on the woman as depicted in the force diagram.
The first force is surface force (Fs)The second force is force of Earth (FE)In the given force diagram, the woman is in equilibrium, this implies that the surface force and the Earth force are equal.
The three different types of motion of the woman that are consistent with the force diagram include the following;
The upward motion of the woman is constantThe downward motion of the woman is also constantThe horizontal motion of the woman is zero since there is no horizontal force on the woman.Learn more about force diagram here: https://brainly.com/question/3624253
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When conditions are difficult bacteria make new bacteria this way?A.Mitosis B. Binary fission C. Lytic cycleD. Cell cycle E. Conjunction
The situations are difficult for bacteria to make new micro organism this manner is Binary fission. alternative B.
Resistant infections are hard and from time to time not possible to deal with. Antimicrobial resistance is a clearly occurring process. however, the boom in antimicrobial resistance is resulting from a aggregate of antibiotic and antifungal publicity to micro organism and their spread and resistance mechanisms.
when resistant lines multiply they produce genetically equal copies. All of those copies are proof against antibiotics. There are two major methods that bacterial cells acquire antibiotic resistance. One is because of New bacteria mutations that arise inside the mobile's DNA during replication. every other manner bacteria accumulate resistance is through horizontal gene switch.
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An 825-kg cannon rests on a frictionless surface and fires a 1.12-kg projectile at 124 meters per second to the right. What is the velocity of the cannon immediately after the projectile is fired?(a) 0.124 m/s to the left. (b) 0.152 m/s to the left.(c) 0.168 m/s to the left.(d) 0.174 m/s to the left.
Given,
Mass of the cannon, m₁=825 kg
Mass of the projectile, m₂=1.12 kg
The velocity of the projectile, v₂=124 m/s
According to the law of conservation of momentum, the momentum of the projectile fired should be equal to the momentum of the cannon.
i.e.,
[tex]m_1v_1=m_2v_2[/tex]Therefore the velocity of the cannon is given by,
[tex]v_1=\frac{m_2v_2}{m_1}[/tex]On substituting the known values in the above equation,
[tex]v_1=\frac{1.12\times124}{825}=0.168\text{ m/s}[/tex]Therefore the velocity of the cannon immediately after the projectile is shot is 0.168 m/s
An object travels east from the starting position of (0, 0). A position versus time graph is shown below.
A Position versus Time graph is shown with y-axis labeled position in centimeters up to 5.0 and x-axis labeled time in seconds up to 8. A line connects points 0, 0 and 4, 4.0. A horizontal line connects points 4, 4.0 and 8, 4.0, through the point 6, 4.0.
This object was moving at a velocity of 1.0 m/s east at the end of 4.0 seconds. Determine the average and instantaneous velocities in m/s at 8.0 seconds.
Average = 1.0 m/s east; instantaneous = 4.0 m/s east
Average = 4.0 m/s east; instantaneous = 8.0 m/s east
Average = 1.0 m/s east; instantaneous = 0 m/s
Average = 0.50 m/s east; instantaneous = 0 m/s
The average and instantaneous velocities in m/s at 8.0 seconds would be 0.5 m/s and 0 m/s respectively, therefore the correct answer is option D.
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
As given in the problem, this object was moving at a velocity of 1.0 m/s east at the end of 4.0 seconds.
The average velocity of the object = ( 4 - 0 ) / (8 -0)
The instantaneous velocity of the object = 0 m/s
Thus, the average and instantaneous velocities in m/s at 8.0 seconds would be 0.5 m/s and 0 m/s respectively, therefore the correct answer is option D.
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The star Wolf-359 is located 7.85 light-years away from Earth.
Which statement is therefore true about Wolf-359?
Responses
Wolf-359 produced its maximum amount of light 7.85 years ago.
It would take 7.85 years to travel from the earth to Wolf-359.
It takes light from Wolf-359 7.85 years to reach the earth.
The star Wolf-359 was born 7.85 years ago.
The statement that is therefore true about the star Wolf-359 is that it takes light from Wolf-359 7.85 years to reach the earth.
A light year is a measure of distance. It can simply be defined as the distance covered by a light wave in one Earth year. Light travels at a speed of about 300000 km / s.
Since the star Wolf-359 is located 7.85 light-years away from Earth, any light wave emitted by the star Wolf-359 would take 7.85 years to reach Earth and we could see the light wave only after 7.85 years from when the light was emitted.
Therefore, the statement that is therefore true about Wolf-359 is that it takes light from Wolf-359 7.85 years to reach the earth.
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You are asked to find the volume of a cube that is 6.50 m long and 8.45m
wide and 7.00 m in height. Calculate the volume. How many significant
figures should you show in your answer?
Answer:
384.475 m³
6 significant figures
Explanation:
Volume of a cube is l x w x h where l- length, w = width, h = height
Volume of given cube = 6.5 x 8.45 x 7 = 384.475 m³
6 significant figures
A Petrol flows at 4m/a through a hose of 10mm diameter. What nozzle diameter will dispense it at the rate of 16m/s?
1cm of the nozzle Diameter will dispense it at the rate of 16m/s.
The continuity equation is what.
A continuity equation, sometimes referred to as a transport equation, is an equation that describes the movement of a certain quantity.
Although it can be applied generally to any significant quantity, it is extremely simple and useful when used with preserved quantities. Continuity equations can be used to explain a variety of physical processes since, under the right conditions, mass, energy, momentum, electric charge, and other natural quantities all remain constant.
From the continuity formula: A 1 V 1 = A 2 V 2.
π(1) 2 ×1=4×πr^ 2 ⇒ 41 =r 2 ⇒r= 21 cm
Thus, the nozzle's diameter is 21 2 cm, or 1 cm.
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28 cubic inches of a gas with an absolute pressure of 35 psi is compressed to a volume of 5 cu.in, what is the new pressure of the gas. (Assume that there's no change in temperature.
First, we will write the formula we will be using: the Ideal gas law
[tex]PV=nRT[/tex]Now, let's see what is given to us
Volume: 28 Cubic Inches
Pressure: 35 PSI
New Volume: 5 cubic inches
New Pressure?
No change in temperature
Since we know that only the volume is changing, we can simplify the ideal gas law into Boyle's law
[tex]P_1V_1=P_2V_2[/tex]So now, we can subsititute in the values we got.
(28)(35) = 5(P2)
P2 = 196 PSI
The construction of ship is based on ----------.A. Archimedes' principleB. Newton's lawC. Joule's law
A. Archimedes principle.
Archimedes' principle states that a body that is completely or partially submerged in a fluid is acted upon by an upward force, called buoyant force, the magnitude of which is equal to the weight of the fluid displaced.
From the principle, we can say that, if the weight of water displaced when the sip enters the water is at least equal to the weight of the ship, then the ship will float.
Thus the correct answer is option A.
Given a circuit consisting of a DC battery connected to a single resistor of resistance 16.0 ohms. If the electric current through the resistor is 5.00 Amperes, then which of these is the voltage of the battery in SI units?A)0.00625B)80C)0.00417D)240E)0.00313
Given:
The resistance in the circuit is R = 16 ohm
The current in the circuit is I = 5 A.
To find the voltage in the circuit.
Explanation:
The voltage in the circuit can be calculated using Ohm's law,
[tex]\begin{gathered} V=5\times16 \\ =80\text{ V} \end{gathered}[/tex]Final Answer: The voltage in the circuit is 80 V.
1. You are viewing a bug under a magnifying glass. If the focal length of the glass is 5cm and thebug is placed 2cm from the glass, how many times bigger does the bug appears to be?
Given:
The focal length of the magnifying glass is f = 5 cm
The object distance is u = -2 cm
Required:
The magnification of the bug.
Explanation:
First, we need to find the image distance using the lens formula
The image distance will be
[tex]\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\ =\frac{1}{5}+\frac{1}{-2} \\ v=-\frac{10}{3}\text{ cm} \\ =-3.33\text{ cm} \end{gathered}[/tex]The magnification can be calculated as
[tex]\begin{gathered} m\text{ = }\frac{v}{u} \\ =\frac{-10}{3}\times\frac{1}{-2} \\ =\text{ 1.67 } \end{gathered}[/tex]Thus, the magnification of the bug is 1.67.
Final Answer: The bug appears 1.67 times bigger.
Which path in the figure below would the ball most closely follow after it exits the channel at "y" and moves across the frictionless table top?(B)OBOf(C)*(D)(E)
Answer:
(B)
Explanation:
When an object ends a circular path, the next path is tangential to the circular path, in this case, the tangential path is (B), so the ball most closely follow path (B).
From which of the following motions is the displacement of the object different from the distance traveled?
A.) An infant..
The motions where the object's displacement is different than the distance traveled is option a. After a home run, a baseball player runs all the way around the bases.
What does displacement mean?An object's position changes if it moves in relation to a reference frame, such as when a passenger moves to the back of an airplane or a lecturer moves to the right in relation to a whiteboard. Displacement describes this shift in location.
Therefore, The term "displacement" refers to a shift in an object's position. It is a vector quantity with a magnitude and direction. The symbol for it is an arrow pointing from the initial location to the ending place. For instance, if an object shifts from location A to position B, its position changes.
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See full question below
For which one of the following motions is the object's displacement different than the distance traveled?
a. After a home run, a baseball player runs all the way around the bases.
b. A car accelerates from rest to a high speed.
c. A buggy car moves across the floor at a constant speed.
Is nuclear energy apart of physics ?
Yes, the nuclear energy is a part of physics
Nuclear energy:
It basically originates from the splitting of uranium atoms. This is basically done by the process of fission. It is the energy that is released in significant amounts in processes that affect atomic nuclei, the dense cores of atoms
Can you give me the answers with being very organized Ans just explain goood
Explanation
the equation is base ond the Pythagorean theorem,it states that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)
[tex]a^2+b^2=c^2[/tex]so,
Step 1
if
[tex]undefined[/tex]. dv/dt = a and dx/dt = v(t) where a is constant.Find v(t) in terms of v(0), and a.a.b. Find x(t) in terms of x(0), v(0) and a.
Given:
[tex]\begin{gathered} \frac{dv}{dt}=a \\ \frac{dx}{dt}=v \end{gathered}[/tex]a is a constant
To find:
(a) v(t) in terms of v(0), and a
(b) x(t) in terms of x(0), v(0) and a
Explanation:
(a) We can write,
[tex]dv=adt[/tex]Integrating both sides we can write,
[tex]\begin{gathered} \int_{v(0)}^{v(t)}dv=\int_0^tadt \\ v(t)-v(0)=at \\ v(t)=v(0)+at \end{gathered}[/tex]Hence, the velocity is,
[tex]v(t)=v(0)+at[/tex](b)
We can also write,
[tex]\begin{gathered} \frac{dx}{dt}=v(t) \\ dx=v(t)dt \end{gathered}[/tex]Integrating both sides we get,
[tex]\begin{gathered} \int_{x(0)}^{x(t)}dx=\int_0^tv(t)dt \\ \int_{x(0)}^{x(t)}dx=\int_0^t[v(0)+at]dt \\ x(t)-x(0)=v(0)t+\frac{1}{2}at^2 \\ x(t)=x(0)+v(0)t+\frac{1}{2}at^2 \end{gathered}[/tex]Hence,
[tex]x(t)=x(0)+v(0)t+\frac{1}{2}at^2[/tex]A person bikes 15 km south then 3km east what the displacement
Given data
*A person bikes towards the south is u = 15 km
*Also it covers the distance towards the east is v = 3 km
The formula for the displacement is given as
[tex]s=\sqrt[]{u^2+v^2}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} s=\sqrt[]{(15)^2+(3)^2} \\ =\sqrt[]{234} \\ =15.29\text{ Km} \end{gathered}[/tex]Hence, the displacement is s = 15.29 km
Which picture correctly shows the path of the reflected light ray given an object inside the focal point?
The reflective surface of a concave mirror is curved inward and away from the light source. Concave mirrors focus light inward to a single focal point.
picture B correctly shows the path of the reflected light ray given an object inside the focal point.
What is concave mirror?The reflective surface of a concave mirror is curved inward and away from the light source. Concave mirrors focus light inward to a single focal point. In contrast to convex mirrors, the image formed by a concave mirror changes depending on the distance between the object and the mirror.Concave mirrors are widely used as reflectors in automobile and motor vehicle headlights, torchlights, railway engines, and so on. The light source is placed at the focal point of the mirror, so that after reflection, the light rays travel over a large distance as parallel high-intensity light beams.To learn more about : Concave mirrors
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Ernest Rutherford's experiment of aiming a beam of alpha particles at a sheet of gold foil led to the discovery of the? A. protonB. quarkC. nucleusD. electron
To find
Ernest Rutherford's experiment of aiming a beam of alpha particles at a sheet of gold foil led to the discovery of the?
Explanation
It discovered the nucleus
Conclusion
The answer is
C.nucleus
A vector A has components Ax = −5.50 m and Ay = 7.50 m. Find the magnitude (in m) and the direction (in degrees counterclockwise from the +x-axis) of the vector.magnitude mdirection ° counterclockwise from the +x-axis
The x-component of vector A is Ax = -5.50 m
The y-component of vector A is Ay = 7.50 m
We are asked to find the magnitude and the direction of the vector (counterclockwise from the +x-axis)
The magnitude of vector A is given by
[tex]\begin{gathered} |A|=\sqrt[]{A_x+A_y} \\ |A|=\sqrt[]{(-5.50)^2+(7.50)^2_{}} \\ |A|=\sqrt[]{30.25+56.25} \\ |A|=\sqrt[]{86.50} \\ |A|=9.3\; m \end{gathered}[/tex]So, the magnitude of vector A is 9.3 m
First, let us draw a diagram to better understand the problem.
The direction of vector A is given by
[tex]\begin{gathered} \tan \theta=\frac{A_y}{A_x} \\ \tan \theta=\frac{7.50}{-5.50} \\ \theta=\tan ^{-1}(\frac{7.50}{-5.50}) \\ \theta=-53.75\degree \end{gathered}[/tex]From the diagram, notice that θ lies in the second quadrant.
So, the counterclockwise direction from the positive x-axis is given by
[tex]180\degree-53.75\degree=126.25\degree[/tex]So, the direction is 126.25° counterclockwise from the +x-axis.
Bone has a Young's modulus of about1.8 x 10^10 Pa. Under compression, it canwithstand a stress of about 1.69 x 10^8 Pa be-fore breakingAssume that a femur (thigh bone) is 0.49 mlong, and calculate the amount of compressionthis bone can withstand before breaking.Answer in units of mm.
Answer:
4.6 mm
Explanation:
The Young's Modulus is defined as
[tex]Y=P\frac{L_0}{\Delta L}[/tex]Where
Y = Young's Modulus
P = pressure (stress) applied
L0 = inital length
ΔL = change in length
Now in our case, we know that the femur is 0.49 m long (L0 = 0.49) and it can withstand max stress of 1.69 x 10^8 Pa ( P = 1.69 x 10^8 Pa), and Young's Modulus for the bone is 1.8 x 10^10 Pa ( Y = 1.8 x 10^10 Pa ); therefore, the above formula gives
[tex]1.8\times10^{10}=1.69\times10^8\times\frac{0.49}{\Delta L}[/tex]We have to solve the above equation for ΔL, the amount of compression.
Multiplying both sides by ΔL gives
[tex]\Delta L\times1.8\times10^{10}=1.69\times10^8\times0.49[/tex]dividing both sides by 1.8 x 10^10 gives
[tex]\Delta L=\frac{1.69\times10^8\times0.49}{1.8\times10^{10}}[/tex]which we evaluate to get
[tex]\Delta L=\frac{1.69\times0.49}{1.8}\times\frac{10^8}{10^{10}}[/tex][tex]\begin{gathered} \Delta L=4.6\times10^{-3}m \\ \boxed{\Delta L=4.6\; mm\text{.}} \end{gathered}[/tex]Hence, the amount of compression a femur can withstand is 4.6 x 10^-3 m or 4.6 mm.
options are 1183420 J730,575 J-452,845 J1183420 Ek730575 Ek-452,845452,845 J-730,575 J-118342p J
Given:
The initial speed of car is: u = 140 km/h = 140 × 5/18 m/s= 38.889 m/s
The final speed of the car is: v = 110 km/h = 110 × 5/18 m/s = 30.5556 m/s
The mass of the car is: m = 1565 kg
To find:
A) Initial kinetic energy of the car.
B) The final kinetic energy of the car.
C) Work done during the deceleration.
Explanation:
A) The initial kinetic energy of the car can be calculated as:
[tex]KE_i=\frac{1}{2}mu^2[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} KE_i=\frac{1}{2}\times1565\text{ kg}\times(38.889\text{ m/s})^2 \\ \\ KE_i=1183417.27\text{ J} \end{gathered}[/tex]The value of initial kinetic energy is very close to the value 1183420 J given in the options. Thus, the initial kinetic energy of the car is 1183420 J.
B) The final kinetic energy of the car can be calculated as:
[tex]KE_f=\frac{1}{2}mv^2[/tex]Substituting the values in the above equations, we get:
[tex]\begin{gathered} KE_f=\frac{1}{2}\times1565\text{ kg}\times(30.5556\text{ m/s})^2 \\ \\ KE_f=730576.971\text{ J} \end{gathered}[/tex]The value of final kinetic energy is very close to the value 730575 J given in the options. Thus, the final kinetic energy of the car is 730576 J.
C) The work done in the deceleration of the car is calculated by applying the work energy theorem. Thus, we get:
[tex]\begin{gathered} W=KE_f-KE_i \\ \\ W=730575\text{ J}-1183420\text{ J} \\ \\ W=-452845\text{ J} \end{gathered}[/tex]The work done in the deceleration of the car is -452845 J. The work done is negative because the car is decelerating.
Final answer:
A) The initial kinetic energy of the car is 1183420 J.
B) The final kinetic energy of the car is 730575 J.
C) The work done in decelerating the car is -452845 J.
What is the extension force (in lbf) of a 5 inch diameter cylinder with a pressure of 200psi?
The area of the cylinder can be given as,
[tex]A=\pi(\frac{d}{2})^2[/tex]Substitute the known value,
[tex]\begin{gathered} A=(3.14)(\frac{5\text{ in}}{2})^2 \\ =19.625in^2 \end{gathered}[/tex]The extension force acting on the cylinder is,
[tex]F=PA[/tex]Substitute the known values,
[tex]\begin{gathered} F=(200psi)(19.625in^2)(\frac{1\text{ lbf}}{1psiin^2})_{} \\ =3925\text{ lbf} \end{gathered}[/tex]Thus, the force exerted on the cylinder is 3925 lbf.
You are interested in retrieving a souvenir from a shipwreck located 115 feet below the water. The path of your dive is modeled by the equation y = 0.05x2 - 4x - 38. Assuming you can hold your breath for the duration of the dive, will you be able to retrieve your souvenir? why?
In order to find out if we will be able to retrieve the souvenir, let's use y = -115 (negative because it's below the water) in the equation and calculate the value of x.
If there is at least one real solution, we will be able to retrieve the souvenir.
To solve this quadratic equation, let's use the quadratic formula:
[tex]\begin{gathered} y=0.05x^2-4x-38 \\ -115=0.05x^2-4x-38 \\ 0.05x^2-4x-38+115=0 \\ 0.05x^2-4x+77=0 \\ a=0.05,b=-4,c=77 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_1=\frac{-(-4)+\sqrt[]{(-4)^2-4\cdot0.05\cdot77}_{}}{2\cdot0.05} \\ x_1=\frac{4+\sqrt[]{16-15.4}_{}}{0.1} \\ x_1=\frac{4+0.77}{0.1}=47.7 \\ x_2=\frac{4-0.77}{0.1}=32.3 \end{gathered}[/tex]Since there are real solutions for x, the answer is YES.
