The perfectly straight lines of reflection demonstrated in the nature is due to particle nature of the light. Therefore, Newton described that the light consists of 'particles' which means second option is correct.
What is the image distance if a 5.00 cm tall object is placed 2.33 cm from a converging lens with a focal length of 5.75 cm?0.603cm1.66cm-0.255cm-3.92cm
We will have the following:
First, we will recall that:
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]That is:
[tex]\begin{gathered} \frac{1}{5.75}=\frac{1}{2.33}+\frac{1}{u}\Rightarrow\frac{1}{u}=-\frac{1368}{5359} \\ \\ \Rightarrow u=-\frac{5359}{1368}\Rightarrow u\approx-3.92 \end{gathered}[/tex]So, the image distance is approximately -3.92 cm.
a 2403 kg racecar has a total momentum of 9.912*10^4kgm/s at one point in the race. calculate the speed of the racecar at that point
In order to calculate the speed, we can use the formula for the momentum:
[tex]p=m\cdot v[/tex]Where p is the momentum (in kg m/s), m is the mass (in kg) and v is the speed (in m/s).
So, using p = 99120 kg m/s and m = 2403 kg, we have:
[tex]\begin{gathered} 99120=2403\cdot v\\ \\ v=\frac{99120}{2403}\\ \\ v=41.25\text{ m/s} \end{gathered}[/tex]an ice has a volume of 8975 ft^3. what is the mass in kilograms of the iceberg? the density of ice 0.917 g/cm^3
The density is given by:
[tex]\rho=\frac{m}{V}[/tex]where V is the volume and m is the mass.
To determine the mass we have to solve the equation for m:
[tex]m=\rho V[/tex]Now, before we can calculate the mass we have to convert the volume given to cubic meter, this comes from the fact that the density is given in g/cm^3 units. We have to remember that a ft is equal to 30.48 cm, then we have:
[tex]8975ft^3(\frac{30.48\text{ cm}}{1\text{ ft}})(\frac{30.48\text{ cm}}{1\text{ ft}})(\frac{30.48\text{ cm}}{1\text{ ft}})=2.54\times10^8[/tex]Hence the volume of the iceberg is:
[tex]2.54\times10^8cm^3[/tex]Now that we have the volume in the correct units we plug its value and the density in the equation for the mass above:
[tex]\begin{gathered} m=2.54\times10^8(0.917) \\ m=2.32\times10^8 \end{gathered}[/tex]Hence the mass of the iceber is 2.32x10^8 g.
Therefore the mass of the iceberg in kilograms is:
[tex]2.32\times10^5\text{ kg}[/tex]PLEASE HELP
A planet's distance from _____ and its _____ both determine its overall gravity.
A) the sun; mass
B) the Kuiper belt; diameter
C) Mars; temperature
D) the Milky Way; perimeter
How to do Projectile Motion?
A thrown ball undergoes projectile motion so throwing a ball in the air is an example of projectile motion.
What is Projectile Motion?Projectile motion is the motion of an object pitched (projected) into the air. After the starting force that launches the object, the only occurrence of the force of gravity in the object is called a projectile motion, and its path is called its trajectory. Projectile motion is a form of motion in which an object or particle ( called a projectile, is thrown near the earth's surface and moves along a curved path under the action of gravity only. Throwing a ball or a cannonball. The motion of a billiard ball on the billiard table. t. The motion of the earth around the un-projectile motion is a special case of two-dimensional motion. A particle in motion at a vertical level with an initial velocity and experiencing a free-fall (downward) acceleration, displays projectile motion.
So we can conclude that Projectile motion is applicable in both throwing and hitting. A thrown ball undergoes projectile motion.
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An object is projected or flung into the air, and only gravity's acceleration affects the object's velocity. A projectile is what it is, and its trajectory is what it took to get there.
What is Projectile motion?An item or particle that is projected toward the surface of the Earth and moves along a curved path only under the influence of gravity is said to be experiencing projectile motion. Galileo demonstrated that this curving path was a parabola, however it can also be a straight line in the unique situation where it is hurled straight up.
Ballistics is the study of such motions, and this trajectory is a ballistic trajectory. Gravity, which works downward and gives the item a downward acceleration toward the Earth's center of mass, is the sole force of mathematical significance that is actively acting on it.
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Question 24 of 25What disadvantage of analog signals is overcome by sending digital signals?A. The waves used to transmit analog signals carry more energy.B. The waves used to transmit analog signals are more dangerous.dC. Noise decreases the quality of analog signals.O0D. Noise decreases the loudness of analog signals.SUBMIT
The correct answer is option C, "Noise decreases the quality of the analog signals."
The anlog signals q
Please help me, it's asking me here how does static energy work?
Static energy is the energy due to motionless state of a particle. If a particle is at rest it possess static energy also known as potential energy. When some force is applied to the particle then the static energy gets converted into kinetic energy of the particle.
A car initially at rest travels with a uniform acceleration of 8 m / s^2 . Calculate the distance covered by the car in 3 s .
We know that
• The initial velocity is zero because it starts from rest.
,• The acceleration is 8 m/s^2.
,• The time elapsed is 3 seconds.
Use a formula that relates initial velocity, acceleration, time, and distance.
[tex]d=v_0t+\frac{1}{2}at^2[/tex]Use the given magnitudes to find d.
[tex]\begin{gathered} d=0\cdot3\sec +\frac{1}{2}\cdot(8\cdot\frac{m}{s^2})(3\sec )^2 \\ d=4\cdot\frac{m}{s^2}\cdot9s^2 \\ d=36m \end{gathered}[/tex]Therefore, the distance covered is 36 meters.A 244 kg motorcycle is travelling with aspeed of 14.7 m-s-1A) Calculate the kinetic energy (in J) of themotorcycle.B) If the speed of the motorcycle is increasedby a factor of 1.6, by what factor does itskinetic energy change?C) Calculate the speed (in m-s-1) of themotorcycle if its kinetic energy is 1/3 of thevaluefound in (a).
Given data:
* The mass of the motorcycle is m = 244 kg.
* The speed of the motorcycle is u = 14.7 m/s.
Solution:
(A). The kinetic energy of the motorcycle is,
[tex]K_1=\frac{1}{2}mu^2[/tex]Substituting the known values,
[tex]\begin{gathered} K_1=\frac{1}{2}\times244\times(14.7)^2_{} \\ K_1=26362.98\text{ J} \end{gathered}[/tex]Thus, the value of kinetic energy is 26362.98 J.
(B). If the speed of the motorcycle is increased by a factor of 1.6,
[tex]\begin{gathered} v=14.7\times1.6 \\ v=23.52\text{ m/s} \end{gathered}[/tex]Thus, the kinetic energy of the motorcycle becomes,
[tex]\begin{gathered} K_2=\frac{1}{2}mv^2 \\ K_2=\frac{1}{2}\times244\times(23.52)^2 \\ K_2=67489.23\text{ m/s} \end{gathered}[/tex]Dividing K_2 by K_1,
[tex]\begin{gathered} \frac{K_2}{K_1}=\frac{67489.23}{26362.98} \\ \frac{K_2}{K_1}=2.56 \end{gathered}[/tex]Thus, the kinetic energy is increased by the factor of 2.56.
(C). The 1/3 of the kinetic energy in the first part is,
[tex]\begin{gathered} K=\frac{1}{3}\times K_1 \\ K=\frac{1}{3}\times26362.98 \\ K=8787.66\text{ J} \end{gathered}[/tex]Thus, the speed of the motorcycle with the kinetic energy K is,
[tex]\begin{gathered} K=\frac{1}{2}mv^2_{}_{} \\ 8787.66=\frac{1}{2}\times244\times v^2 \\ 8787.66=122\times v^2 \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} v^2=\frac{8787.66}{122} \\ v^2=72.03 \\ v\approx8.5\text{ m/s} \end{gathered}[/tex]Thus, the speed of the motorcycle is 8.5 m/s.
7. What is the velocity of a 850kg car after starting at rest when 13,000J of work is done to it.
Answer:
5.53 m/s
Explanation:
The work is equal to the change in the kinetic energy, so
[tex]\begin{gathered} W=\Delta KE \\ W=\frac{1}{2}m(v^2_f-v^2_i)^{}^{} \end{gathered}[/tex]Since the car starts at rest, the initial velocity vi = 0 m/s, so we can solve for the final velocity vf as follows
[tex]\begin{gathered} W=\frac{1}{2}mv^2_f \\ 2W=mv^2_f \\ \frac{2W}{m}=v^2_f \\ v_f=\sqrt[]{\frac{2W}{m}} \end{gathered}[/tex]So, replacing the work W = 13,000J and the mass m = 850kg, we get:
[tex]\begin{gathered} v_f=\sqrt[]{\frac{2(13,000J)}{850\operatorname{kg}}} \\ v_f=5.53\text{ m/s} \end{gathered}[/tex]Therefore, the velocity is 5.53 m/s
Find the magnitude of the sumof these two vectors:B63.5 m101 m57.0°
Vector diagram:
The resultant vector is given as,
[tex]R=\sqrt[]{A^2+B^2+2AB\cos \theta}[/tex]Here, θ is the angle between vector A and B.
Substituting all known values,
[tex]\begin{gathered} R=\sqrt[]{(63.5)^2+(101)^2+2\times101\times63.5\times\cos (33^{\circ})} \\ =158.08\text{ m} \end{gathered}[/tex]Therefore, the resultant magnitue of the sum of these two vectors are 158.08 m.
The x-component of the magnitude is given as,
[tex]\begin{gathered} R_x=101\cos (57^{\circ})+63.5\cos (90^{\circ}) \\ =55.0\text{ m} \end{gathered}[/tex]The y- component of the magnitude is given as,
[tex]\begin{gathered} R_y=63.5\sin (90^{\circ})+101\sin (57^{\circ}) \\ =148.2\text{ m} \end{gathered}[/tex]Therefore, the direction is given as,
[tex]\begin{gathered} \phi=\tan ^{-1}(\frac{R_y}{R_x}) \\ =\tan ^{-1}(\frac{148.2\text{ m}}{55.0\text{ m}}) \\ =69.63^{\circ} \end{gathered}[/tex]Therefore, the direction of the resultant vector is 69.63°.
A raindrop has a mass of 7.7 × 10-7 kg and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop.(a)Fraindrop= _________________ units ________(b)Fearth= _________________ units_____________
ANSWER:
a) Fraindrop
[tex]F=7.546\cdot10^{-6}N[/tex](b) Fearth
[tex]F=-7.546\cdot10^{-6}N[/tex]STEP-BY-STEP EXPLANATION:
(a)
We calculate the force, multiplying the value of the mass by gravity, just like this:
[tex]\begin{gathered} F=m\cdot a \\ F=7.7\cdot10^{-7}\cdot9.8 \\ F=7.546\cdot10^{-6}N \end{gathered}[/tex](b)
by newton's 3rd law they are are equal and opposite so:
[tex]F=-7.546\cdot10^{-6}N[/tex]Which of the following scientists discovered that atoms contain electrons?a. Daltonb. RutherfordC. Thomsond. Bohr
Until 1897, atom was thought as the fundemental particle. But in 1897 J.J Thompson discovered that the atoms contains electrons. He discovered during his experiment with cathod ray tube.
Thus the correct answer is option C.
At what point, if any, during a dive is a skydiver experiencing complete freefal? Explain. (1 point)•skydiver will experience complete freefall the moment right before they jump out of the plane because they are free to start fallingat any moment.•A skydiver will experience complete freefall when they first jump out of the plane because they only experience air resistance oncethey deploy their parachute.•A skydiver will never experience complete freefall until after they have deployed their parachute because they are now falling at asafe speed for their landingA skydiver will never experience complete freefall because as soon as they start their dive, they will experience air resistance.
To find:
Which of the given statements is true.
Explanation:
The free fall is defined as the motion of an object under the influence of only gravitational force. In free fall, there will be no forces, except gravitational force acting on the object.
When the skydiver jumps, gravity will pull the diver downwards. And the air resistance due to the air in the atmosphere will oppose this motion of the diver. Thus there will be two forces acting on the diver, gravity and the air resistance. Thus the skydiver will never experience free fall.
Final answer:
Thus the correct answer is option D.
A spring with spring constant 40 N/m is compressed .1m past it natural length. A mass of .5kg is attached to the spring. A. What is the elastic potential energy stored in the spring?B. The spring is released. What is the speed of the masses as it reaches the natural length of the spring?
Given data
*The given spring constant is k = 40 N/m
*The given compressed length is x = 0.1 m
*The given mass is m = 0.5 kg
(a)
The formula for the elastic potential energy stored in the spring is given as
[tex]U_p=\frac{1}{2}kx^2[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} U_p=\frac{1}{2}(40)(0.1)^2 \\ =0.2\text{ J} \end{gathered}[/tex]Hence, the elastic potential energy stored in the spring is 0.2 J
(b)
The formula for the speed of the masses is given by the conservation of energy as
[tex]\begin{gathered} U_p=U_k \\ \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v=x\sqrt[]{\frac{k}{m}} \end{gathered}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} v=(0.1)\sqrt[]{\frac{40}{0.5}} \\ =0.89\text{ m/s} \end{gathered}[/tex]Hence, the speed of the masses as it reaches the length of the spring is v = 0.89 m/s
When the buoyant force on an object is equal to or greater than its weight, the object __
When the buoyant force on an object is equal to or greater than its weight, the object accelerates upwards and floats.
What is buoyant force?
Buoyant force is the upward force exerted on an object that is fully or partly immersed in a fluid.
This upward force is also called Upthrust.
According to Archimedes' principle which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object.
An object will accelerate if its upthrust is greater than its weight, but will reach an upward terminal velocity when upthrust is equal to weight plus drag force.
Thus, when the buoyant force on an object is equal to or greater than its weight, the object accelerates upwards and floats.
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Problem Try to answer the following questions:(a) What is the maximum height above ground reached by the ball?(b) What are the magnitude and the direction of the velocity of the ball just before it hits the ground? Show Your Problem Solving Steps: Show these below:1) Draw a Sketch2) Choose origin, coordinate direction3) Inventory List – What is known?4) Write the kinematics equation(s) and solution of Part (a):5) Write the kinematics equation(s) and solution of Part (b):Problem 3 A small ball is launched at an angle of 30.0 degrees above the horizontal. It reaches a maximum height of 2.5 m with respect to the launch position. Find (a) the initial velocity of the ball when it’s launched and (b) its range, defined as the horizontal distance traveled until it returns to his original height. As always you can ignore air resistance.(a) Initial velocity [Hints: How is v0 related to vx0 and vy0. How can you use the information given to calculate either or both of the components of the initial velocity?](b) Range [Hints: This problem is very similar to today’s Lab Challenge except that for the challenge the ball will land at a different height.]
3.
[tex]\begin{gathered} \theta=30^{\circ} \\ y_{\max }=2.5m \end{gathered}[/tex]a)
[tex]\begin{gathered} y_{\max }=\frac{v^2\sin ^2(\theta)}{g} \\ \end{gathered}[/tex]Solve for v:
[tex]\begin{gathered} v=\sqrt[]{\frac{y_{\max }\cdot g}{\sin ^2(\theta)}} \\ v=98\cdot\frac{m}{s} \end{gathered}[/tex]b)
[tex]\begin{gathered} r=\frac{v^2}{9}\sin (2\theta) \\ r=\frac{98^2}{9.8}\cdot\sin (2\cdot30) \\ r=\frac{98^2}{9.8}\sin (60) \\ r=848.7m \end{gathered}[/tex]a car goes from 32 m/s to a complete stop in 4.8 seconds. calculate the average stopping force of the car if has a mass of 2500 kg
The average stopping force is 16,500 N
Initial velocity of car (v₁)= 32m/s
Final velocity (v₂) = 0m/s
Time to stop= 4.8 seconds
Mass of car= 2500 kg
we need to apply the concept of laws of motion
Acceleration of car (a)= Change in velocity/time
a= v₂-v₁/t
a= 0-32/4.8
a= -6.6 m/s² ( deceleration)
Force= mass x acceleration
Force= 2500x 6.6
Force= 16500 N
Therefore the average stopping force is 16,500 N.
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I’m confused about which electromagnetic waves have the lowest frequency
The eletromagnetic wave that has the highest frequency is the gamma rays. It also has the highest energy and shortest wavelengths.
On the other hand, the type of eletromagnetic wave that has the lowest frequency, lowest energy and longest wavelength is radio waves.
PLEASE HELPPPPPPPPPPPPPPPPPPPPPPP
Answer:
right answer is death valley
Explanation:
because it is close to surface gravitational field
mexico
cus its the farthest
You turn a corner and are driving up asteep hill. Suddenly, a small boy runs out on the street chasing a ball. You slam on the brakes and skid to astop leaving a 50-foot-long skid mark on the street. The boy calmly walks away but a policemen watching fromthe sidewalk walks over and gives you a speeding ticket. He points out that the speed limit on this street is 25mph. After you recover your wits, you begin to examine the situation. You determine that the street makes anangle of 25◦with the horizontal and that the coefficient of static friction between your tires and the street is0.80. You also find that the coefficient of kinetic friction between your tires and the street is 0.60. Your car’sinformation book tells you that the mass of your car is 1600 kg. You weigh 140 lbs.
Given data:
Total displacement of the car;
[tex]s=50\text{ ft}[/tex]Speed limit;
[tex]v_m=25\text{ mph}[/tex]The angle of street from horizontal;
[tex]\theta=25\degree[/tex]Coefficient of static friction;
[tex]\mu_s=0.80[/tex]Coefficient of kinetic friction;
[tex]\mu_k=0.60[/tex]Mass of the car;
[tex]M=1600\text{ kg}[/tex]Weight of the man;
[tex]W=140\text{ lbs}[/tex]The kinetic friction force is given as,
[tex]F_k=\mu_k(M+m)g\cos \theta[/tex]Here, m is the mass of the man and g is the acceleration due to gravity.
The acceleration of the car driving up a steep hill is given as,
[tex]\begin{gathered} (M+m)g\sin \theta+F_k=(M+m)a \\ (M+m)g\sin \theta+\mu_k(M+m)g\cos \theta=(M+m)a \\ g\sin \theta+\mu_kg\cos \theta=a \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} (32\text{ ft/s}^2)\times\sin (25\degree)+0.6\times(32\text{ ft/s}^2)\times\cos (25\degree)=a \\ \approx30.92\text{ ft/s}^2 \end{gathered}[/tex]The velocity of the car is given as,
[tex]v^2=u^2-2as[/tex]Here, v is the final velocity (v=0, as the car stops), and u is the initial velocity.
The initial velocity of the car is given as,
[tex]u=\sqrt[]{v^2+2as}[/tex]Substituting all known values,
[tex]\begin{gathered} u=\sqrt[]{0^2+2\times(30.92\text{ ft/s}^2)\times(50\text{ ft})} \\ \approx55.61\text{ ft/s} \\ \approx37.91\text{ mph} \end{gathered}[/tex]Therefore, your speed is greater than the speed limit. Thus, you can not fight the ticket in the court.
What is the energy of a proton accelerated through a potential difference of 500,000 V?
ANSWER
[tex]8.01\cdot10^{-14}J[/tex]EXPLANATION
We want to find the energy of the proton accelerated through the given potential.
To do this, apply the relationship between energy and potential:
[tex]V=\frac{E}{q}[/tex]where q = charge
V = potential
The charge of a proton is:
[tex]1.602\cdot10^{-19}C[/tex]Therefore, we have that the energy of the proton is:
[tex]\begin{gathered} E=V\cdot q \\ E=500000\cdot1.602\cdot10^{-19} \\ E=8.01\cdot10^{-14}J \end{gathered}[/tex]That is the answer.
Two 4.587 cm by 4.587 cm plates that form a parallel-plate capacitor are charged to +/- 0.671 nC. What is the electric field strength inside the capacitor if the spacing between the plates is 1.257 mm?
ANSWER:
3.6 x 10^6 N/C
STEP-BY-STEP EXPLANATION:
Given:
Charge (q) = 0.671 nC = 0.671 x 10^-9 C
Side (s) = 4.587 cm = 4.587 x 10^-3 m
Vacuum permittivity (ε0) = 8.85 x 10^-12 F/m
We can calculate the electric field using the following formula:
[tex]\begin{gathered} E=\frac{q}{ε_0\cdot A} \\ \\ \text{ We replacing:} \\ \\ E=\frac{0.671\cdot10^{-9}}{(8.85\cdot10^{-12})(4.587\cdot10^{-3})(4.587\cdot10^{-3})} \\ \\ E=\:3603477.12=3.6\cdot10^6\text{ N/C} \end{gathered}[/tex]The electric field is equal to 3.6 x 10^6 N/C
A helicopter takes off and travels forward at an angle of 59.4 above horizontal. After following this path for 294 meters, the pilot changes the angle of flight to 10.5 degrees above horizontal and follows this path for 849 meters. After these two legs, what is the helicopter’s horizontal distance from the point of take off?
985 m
447 m
408 m
964 m
After the two legs, the helicopter’s horizontal distance from the point of take off is 979 m
For the first leg,
d = 294 m
θ = 59.4°
[tex]d_{x}[/tex] = d cos θ
[tex]d_{x}[/tex] = 294 * cos 59.4°
[tex]d_{x}[/tex] = 147 m
For the second leg,
d = 849 m
θ = 10.5°
[tex]d_{x}[/tex] = d cos θ
[tex]d_{x}[/tex] = 849 * cos 10.5°
[tex]d_{x}[/tex] = 832 m
Total horizontal distance = [tex]d_{x}[/tex] ( 1st leg ) + [tex]d_{x}[/tex] ( 2nd leg )
Total horizontal distance = 147 + 832
Total horizontal distance = 979 m
Therefore, after the two legs, the helicopter’s horizontal distance from the point of take off is 979 m
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The wiring in a house must be thick enough so it does not become so hot as to start a fire.part aWhat diameter must a copper wire be if it is to carry a maximum current of 34 A and produce no more than 1.6 W of heat per meter of length?
Given:
The maximum current in the circuit is,
[tex]i=34\text{ A}[/tex]The power per length is,
[tex]\frac{P}{l}=1.6\text{ W/m}[/tex]To find:
The diameter of the copper wire
Explanation:
The power (P) produced by current i, through a copper wire of resistance R and length l is given by,
[tex]\begin{gathered} Pl=i^2R \\ \frac{R}{l}=\frac{P}{i^2} \\ \frac{R}{l}=\frac{1.6}{34\times34} \end{gathered}[/tex]Now,
[tex]\begin{gathered} R=\frac{\rho l}{A} \\ R=\frac{\rho l}{\pi r^2} \end{gathered}[/tex]The resistivity of copper is,
[tex]\rho=1.72\times10^{-8}\text{ ohm.m}[/tex]So, we can write,
[tex]\begin{gathered} \frac{R}{l}=\frac{\rho}{\pi r^2} \\ \frac{1.6}{34\times34}=\frac{1.72\times10^{-8}}{\pi r^2} \\ r^2=\frac{1.72\times10^{-8}\times34\times34}{1.6} \\ r=3.5\times10^{-3}\text{ m} \\ diamer\text{ is,} \\ 2r=7.0\times10^{-3}\text{ m} \end{gathered}[/tex]Hence, the diameter is,
[tex]7.0\times10^{-3}\text{ m}[/tex]A sea wave propagates with a speed of 12 cm/s and a length of 0.4 meters,find its period.A:0,1 minB:10sC:1sD:3,33s
Given:
The speed of the wave is v = 12 cm/s = 0.12 m/s
The wavelength of the wave is
[tex]\lambda\text{ = 0.4 m}[/tex]To find the period.
Explanation:
The time period can be calculated by the formula
[tex]T=\frac{\lambda}{v}[/tex]On substituting the values, the time period will be
[tex]\begin{gathered} T=\frac{0.4}{0.12} \\ =3.33\text{ s} \end{gathered}[/tex]Thus, the time period of the sea wave is 3.33 s
A cat chases a mouse across a 0.66 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor taylor (jdt3899) – Homework 3, 2d motion 22-23 – tejeda – (LermaHPHY1 1) 3 2.4 m from the edge of the table. The acceleration of gravity is 9.81 m/s 2 . What was the cat’s speed when it slid off the table?
The cat’s speed when it slid off the table will be 6.552 m/s
The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Kinematics equations are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.
a = -g = 9.8 m[tex]/s^{2}[/tex]
using equation of motion
x = u(horizontal )*t + 1/2 * a (horizontal) * [tex]t^{2}[/tex]
since , a (horizontal) = 0
x = u(horizontal )*t
u = x / t equation 1
similarly
y = u(vertical)*t + 1/2 * a (vertical) * [tex]t^{2}[/tex]
u(vertical) = 0
t = [tex]\sqrt{2y / a}[/tex] equation 2
substituting the value of equation 2 in equation 1
u = x / [tex]\sqrt{2y / a}[/tex]
= [tex]\sqrt{\frac{-9.81}{2*-0.66} } * 2.4[/tex]
= 6.552 m/s
The cat’s speed when it slid off the table will be 6.552 m/s
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3. A rescuer jumped from an airship in the ocean 1.20 x 102 m above the water's surface. Whatwas her kinetic energy at the moment she was 30.0 m from the water's surface? What was herspeed at that moment assuming her mass is 60.0 kg?
Given data,
The initial velocity of the body is zero.
The distance travelled by the rescuer upto the height of 30 m from the water surface is,
[tex]\begin{gathered} S=102-30 \\ S=72\text{ m} \end{gathered}[/tex]The final velocity of the rescuer at the height 30 m is,
[tex]v^2-u^2=2gS[/tex]where g is the acceleration due to gravity.
Substituting the known values,
[tex]\begin{gathered} v^2=2\times9.8\times72 \\ v^2=1411.2 \\ v=37.6ms^2 \end{gathered}[/tex]Thus, the kinetic energy of the rescuer is,
[tex]K=\frac{1}{2}mv^2[/tex]Substituting the known values,
[tex]\begin{gathered} K=\frac{1}{2}\times60\times1411.2 \\ K=42336 \\ K=42.3\text{ KJ} \end{gathered}[/tex]Thus, the kinetic energy of the rescuer is 42.3 KJ and speed of the rescuer is 37.6 meter per second square.
A baseball is rolling along a tabletop with avelocity of 3.9 m/s to the right. The tabletopis 1.1 m above the ground. The ball rolls offthe edge of the table and falls to theground.A.) What is the ball's final vertical Velocity?B.) How long does the ball take to fall?C.) how far from the table does the ball land?
To answer this question we need to notice that once the ball starts falling we have a projectile motion; which means that horizontally we have a rectilinear motion and vertically we have an uniformly accelerated motion.
Then we can use the following equations for each direction:
[tex]\begin{gathered} \text{ Horizontal motion:} \\ x=x_0+v_{0x}t \\ \text{ Vertical motion:} \\ a=\frac{v_f-v_0}{t} \\ y=y_0+v_0t+\frac{1}{2}at^2 \\ v_f^2-v_0^2=2a(y-y_0) \end{gathered}[/tex]Since the ball is moving down in the vertical direction we will think that down is the positive direction vertically.
a)
We know that the ball is rolling to the right when it rolls off the edge of the table, this means that vertically the initial velocity is zero; we also know that the ball will fall for 1.1 m and that the acceleration is the gravitational acceleration. Then we can use the third vertical motion equation to find the final velocity, plugging the values we know we have that:
[tex]\begin{gathered} v_f^2-0^2=2(9.8)(1.1) \\ v_f=\sqrt{2(9.8)(1.1)} \\ v_f=4.64 \end{gathered}[/tex]Therefore, the final vertical velocity is 4.64 m/s.
b)
To determine the time we can use the second vertical equation with the values we know:
[tex]\begin{gathered} 1.1=0+0t+\frac{1}{2}(9.8)t^2 \\ 4.9t^2=1.1 \\ t^2=\frac{1.1}{4.9} \\ t=\sqrt{\frac{1.1}{4.9}} \\ t=0.474 \end{gathered}[/tex]Therefore, it takes 0.474 s for the ball to fall.
c)
While the ball is falling it is also moving horizontally, in this direction we know the initial velocity is 3.9 m/s; using the horizontal equations we have:
[tex]\begin{gathered} x=0+(3.9)(0.474) \\ x=1.85 \end{gathered}[/tex]Therefore, the ball lads 1.85 m from the table.
A truck covers 40.0 m in 9.00 s while uniformly slowing down to a final velocity of 2.20 m/s.(a) Find the truck's original speed. m/s(b) Find its acceleration. m/s2
Given:
The distance covered by truck: d = 40.0 m
The time taken to cover the distance is: t = 9.00 s
The final velocity of the truck is: v2 = 2.20 m/s
To find:
a) the speed of the truck.
b) the acceleration
Explanation:
a)
The speed of the truck before it slows down can be calculated as:
[tex]d=\frac{1}{2}(v_2+v_1)t[/tex]Substituting the values in the above equation, we get
[tex]\begin{gathered} 40=\frac{1}{2}(2.20+v_1)\times9 \\ \\ \frac{40\times2}{9}-2.20=v_1 \\ \\ v_1=6.69\text{ m/s} \end{gathered}[/tex]b)
The truck is initially moving at a speed of 6.69 m/s. It then slows down to the final velocity of 2.20 m/s. The acceleration of the truck can be determined as:
[tex]d=v_1t+\frac{1}{2}at^2[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} 40=6.69\times9+\frac{1}{2}\times a\times9^2 \\ \\ 40=60.21+40.5a \\ \\ a=\frac{40-60.21}{40.5} \\ \\ a=-0.499 \\ \\ a\approx-0.5\text{ m/s}^2 \end{gathered}[/tex]Final answer:
a) The original speed of the truck is 6.69 m/s.
b) The acceleration of the truck is - 0.5 m/s^2.
