Answer:
14.2 m/s
Explanation:
When the source and receiver are getting closer, we can use the following equation:
[tex]f_o=(\frac{v+v_o}{v-v_s})f_s[/tex]Where fo is the observed frequency, fs is the emitted frequency, vo is the speed of the observed, vs is the speed of the source, and v is the speed of the sound. Solving for vo, we get:
[tex]\begin{gathered} \frac{f_o}{f_s}=\frac{v+v_o}{v-v_s} \\ \\ \frac{f_o}{f_s}(v-v_s)=v+v_o \\ \\ \frac{f_o}{f_s}(v-v_s)-v=v_o \\ \\ v_o=\frac{f_o}{f_s}(v-v_s)-v \end{gathered}[/tex]Then, replacing v = 340 m/s, vs = 0 m/s, fs = 960 Hz, and fo = 1000 Hz, we get:
[tex]\begin{gathered} v_r=\frac{1000}{960}(340-0)-340 \\ \\ v_r=14.2\text{ m/s} \end{gathered}[/tex]Therefore, the speed of the patrol car is 14.2 m/s
What common unit is used to measure sound intensity? Answer here
Decibel (dB)
Explanation:Sound intensity is the power carried by sound waves per unit area in a perpendicular direction to that area.
The unit of sound intensity is the decibel
Kinematics question. Please note I’ve never done kinematics before so lots and lots of explanation very necessary.
Given figure is Time -velocity graph that indicate acceleration .
Now, according to problem
1) A graph line is parallel to time axis i.e. velocity is not changing with passes of time , so acceleration (a) = 0.
Here,
acceleration (a)= 0
time (t)= 20 sec
starting speed (u)= 40 m /s
final speed (v)= 40m /s
distance covered( s)= ?
Now using formula for linear motion ,we get
[tex]\begin{gathered} s=ut+\text{ }\frac{1}{2}at^2; \\ s=\text{ 40}\times20+\text{ 0}\begin{cases}a={0} \\ t={20}\end{cases} \\ s=800m; \end{gathered}[/tex]Answer is 800m
2) For graph line B---
Starting speed(u)= 25 m/s ( when time =0)
final speed (v) = 50 m/s (when time =20 s)
time (t)= 20 sec
acceleration (a)= ?
distance travelled (s)= ?
Now acceleration is given by
[tex]\begin{gathered} a=\frac{v-u}{t}; \\ a=\frac{50-25}{20}=\text{ }\frac{25}{20}=\text{ 1.25ms}^{-2} \end{gathered}[/tex]Again distance travelled in 20 s is given by
[tex]\begin{gathered} s=ut\text{ +}\frac{1}{2}at^2; \\ s=\text{ 25}\times20+\frac{1}{2}\times1.25\times20^2; \\ s=500+250=750\text{ m} \end{gathered}[/tex]Answer is a= 1.25m/s² and s= 750m
3) when t= 20 sec then distance travelled by A=800m and distance travelled by B= 750 m . Therefore A is ahead of B
4) distance travelled by car A in 40 sec is given by
[tex]\begin{gathered} s=ut\text{ +}\frac{1}{2}at^2 \\ s=\text{ 40}\times40+0\begin{cases}a={0} \\ u={40}\end{cases} \\ s=1600m \end{gathered}[/tex]Now distance travelled by B is given by
[tex]undefined[/tex]Calculate the answer to the correctnumber of significant digits.1.899+ 0.58
ANSWER
2.48
EXPLANATION
We have to add like we normally do with whole numbers and put the decimal point in the same place it is in each summand:
The significant digits start with the first digit different from zero, reading from left to right:
This number has 4 significant digits in total: 2, 4, 7 and 9
Note that the numbers added don't have the same number of decimal places. The first one has three decimal places while the second one has two decimal places. This means that we have to round the answer to the second decimal place - which is hte 3rd significant digit. SInce the 4th significant digit is more than 5, we have to add one to the third significant digit:
[tex]2.479\Rightarrow2.48[/tex]Solve using a system of linear equationsA motorboat takes 3 hours to travel 108 miles going upstream. The return trip takes 2 hours going downstream. What is the rate of the boat in still water and what is the rate of the current? Rate of the boat in still water=( )mi/h Rate of the current=( )mi/h
Let's use the constant motion equation: d = v*t, to express the equation for each case. But first, let's find the speed.
[tex]v=\frac{d}{t}=\frac{108mi}{3hr}=36\cdot\frac{mi}{hr}[/tex]The speed going upstream is 36 mi/hr.
[tex]v=\frac{108mi}{2hr}=54\cdot\frac{mi}{hr}[/tex]The speed going downstream is 54 mi/hr.
Observe that the distance traveled is the same.
Once we have the speed for each case, we can form the following system of equations.
[tex]\begin{gathered} x-y=36 \\ x+y=54 \end{gathered}[/tex]The first equation represents the difference between the motorboat speed in still water (x) and the rate of the current (y).
The second equation represents the addition between these two rates x and y. One equation represents the situation where the speeds are in opposite direction, and the other one represents when the
A. How many Coulombs are there in one fundamental/elementary charge?B. How many electrons are there in one coulomb of negative charge?
(A). The number of Coulombs in one fundamental charge of an electron is,
[tex]1e\text{ = 1.}6\times10^{-19}\text{ C}[/tex](B). The charge in terms of the number of electrons is,
[tex]q=ne[/tex]where q is the net charge, e is the fundamental charge and n is the number of electrons,
For the net charge of q = 1 Coulomb, the number of electrons is,
[tex]\begin{gathered} 1\text{ C=n}\times1.6\times10^{-19}\text{ C} \\ n=\frac{1}{1.6\times10^{-19}} \\ n=0.625\times10^{19} \\ n=6.25\times10^{18}\text{ } \end{gathered}[/tex]Thus, the number of electrons in the 1 Coulomb of charge is,
[tex]6.25\times10^{18}[/tex]
One coulomb (1 C) is equal to approximately 6.24 x 10 18 elementary charges. Thus, an elementary charge is approximately 1.60 x 10 -19 C.
The unit of electrical charge quantity in the International System of Units is the coulomb. Coulomb is the SI unit of electric charge which is equal to the amount of charge transported by a current of one ampere in one second. It can be also, a property of a matter due to which electrical and magnetic effects are produced. It is denoted by C. Mathematically, 1 Coulomb = 1 Ampere × 1 second.
To illustrate the magnitude of 1 Coulomb, an object would need an excess of 6.25 x 1018 electrons to have a total charge of -1 C. And of course, an object with a shortage of 6.25 x 1018 electrons would have a total charge of +1 C. The charge on a single electron is -1.6 x 10 -19 Coulomb.
Charge on an electron(e)= 1.6∗10^−19 C approximately.
By quantization property of charge(Smallest unit is e),
Charge (Q) = 1C where n-number of electrons.
Electron (e) = 1.6× 10^-19
A number of electrons (n) =?
Q= ne
1 C = n × 1.6×10^-19 C
n = 1/(1.6×10^-19)
Which gives, n= 6.25*10^18 electrons
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I have finished most of this question just need help on making sure it’s correct and the last 2
Given
m: mass
m = 50 kg
KE: kinetic energ
PEg: gravitational potential energy
v: speed
Procedure
Point A
KE = 0
PEg = mgh
PEg = 50 kg * 10 m/s^2 * 100 m
PEg = 50000 J
v = 0 m/s
Point B
h = 60 m
PEg = mgh
PEg = 50 kg * 10 m/s^2 * 60 m
PEg = 30000 J
KE = 20000 J
[tex]\begin{gathered} v=\sqrt[]{\frac{2\cdot KE}{m}} \\ v=\sqrt[]{\frac{2\cdot20000\text{ J}}{50\text{ kg}}} \\ v=\sqrt[]{800\text{ }} \\ v=28.28\text{ m/s} \end{gathered}[/tex]Point C
h = 30 m
PEg = mgh
PEg = 50 kg * 10 m/s^2 * 30 m
PEg = 15000 J
KE = 35000 J
[tex]\begin{gathered} v=\sqrt[]{\frac{2\cdot KE}{m}} \\ v=\sqrt[]{\frac{2\cdot35000\text{ J}}{50\operatorname{kg}}} \\ v=37.41\text{ m/s} \end{gathered}[/tex]Point D
h = 60 m
PEg = mgh
PEg = 50 kg * 10 m/s^2 * 60 m
PEg = 30000 J
KE = 20000 J
[tex]\begin{gathered} v=\sqrt[]{\frac{2\cdot KE}{m}} \\ v=\sqrt[]{\frac{2\cdot20000\text{ J}}{50\text{ kg}}} \\ v=\sqrt[]{800\text{ }} \\ v=28.28\text{ m/s} \end{gathered}[/tex]
Point E
h = 0
PEg = 50 kg * 10 m/s^2 * 0
PEg = 0 J
KE = 50000 J
[tex]\begin{gathered} v=\sqrt[]{\frac{2\cdot KE}{m}} \\ v=\sqrt[]{\frac{2\cdot50000\text{ J}}{50\text{ kg}}} \\ v=44.72\text{ m/s} \end{gathered}[/tex]If your favorite radio station is on the FM Dial at100 MHz (Mega Hertz) Calculate the wave length for this Radio station. Hint use the equation c = Lf, Where L is the wavelength and c is given at the top of the exam.
In order to calculate the wavelength, let's use the given formula, using c = 3 * 10^8 m/s (speed of light in vacuum) and f = 100*10^6 Hz:
[tex]\begin{gathered} 3\cdot10^8=L\cdot100\cdot10^6\\ \\ L=\frac{3\cdot10^8}{100\cdot10^6}=\frac{3\cdot10^8}{1\cdot10^8}=3\text{ m} \end{gathered}[/tex]Therefore the wavelength is 3 meters.
Which law of thermodynamics states that absolute zero cannot be reached? Also a hot cup of tea absorbs thermal energy from a cold room? True false
The third law of thermodynamics states that the absolute zero-temperature 0 K cannot be reached by a finite number of steps.
Thus, the answer is the third law of thermodynamics.
Also, a hot cup of tea absorbs thermal energy from a cold room. The given statement is false.
Thus, the given statement is False.
A ball of mass 1.86 kilograms is attached to a cord 1.29 meters long and swung in a vertical circle at a constant speed of 5.27 meters per second. What is the centripetal force acting on the ball? Include units in your answer. What is the tension in the cord when the ball is at the bottom of its path? Include units in your answer. What is the tension in the cord when the ball is at the top of its path? Include units in your answer. All answers must be in 3 significant digits.
Explanation
Step 1
Draw
so
a)centripetal force:
the centripetal force is given by.
[tex]\begin{gathered} F=ma \\ F=m\frac{v^2}{r} \\ \text{where } \\ F_{C\text{ }}\text{ is the centripetal force} \\ m\text{ is the mass } \\ v\text{ is the velocty } \\ r\text{ is the radius} \end{gathered}[/tex]now, replace
[tex]\begin{gathered} F=m\frac{v^2}{r} \\ F=1.86\text{ kg }\frac{(\text{ 5.27 }\frac{m}{s})^2}{1.29\text{ m}} \\ F=40.044\text{ Newtons} \end{gathered}[/tex]so, the centripetal force is 40.0446 Newtons
b) What is the tension in the cord when the ball is at the bottom of its path?
to find the tension in bottom, we need to add the weigth of the ball,so
[tex]\begin{gathered} \text{weigth}=\text{ mass}\cdot accelofgravity \\ w=mg \end{gathered}[/tex]hence, the tension would be
[tex]\begin{gathered} T_{bottom}=m\frac{v^2}{r}+mg \\ \end{gathered}[/tex]replace
[tex]\begin{gathered} T_{bottom}=m\frac{v^2}{r}+mg \\ T_{bottom}=40.044\text{ N+(1.86 kg}\cdot9.81\text{ }\frac{\text{m}}{s^2}) \\ T_{bottom}=40.044\text{ N+18.2466 N} \\ T_{bottom}=58.291\text{ N} \end{gathered}[/tex]c)What is the tension in the cord when the ball is at the top of its path?
to find the tension in the top we need to subtract the weigth, so
[tex]\begin{gathered} T_{\text{top}}=m\frac{v^2}{r}-mg \\ replace \\ T_{\text{top}}=40.044\text{ N-18.2466 N} \\ T_{\text{top}}=21.79\text{ Newtons} \end{gathered}[/tex]I hope this helps you
a block sliding on the ground where uk= 0.193 experiences a 14.7N friction force. what is the mass of the block in kg
Answer:
F = ma
14.7N = m(0.193)
m = 76.3 kg
Two positive charges of 10 C are separated by 12 meters. What is the force between the charges?
In order to calculate the force between the charges, we can use the formula below:
[tex]F=K\cdot\frac{q_1\cdot q_2}{d^2}[/tex]Where F is the force (in Newtons), K is the Coulomb constant (K = 9 * 10^9), q1 and q2 are the charges (in Coulombs) and d is the distance between them (in meters).
So, for q1 = q2 = 10 C and d = 12 m, we have:
[tex]\begin{gathered} F=9\cdot10^9\cdot\frac{10\cdot10}{12^2} \\ F=\frac{900\cdot10^9}{144} \\ F=6.25\cdot10^9\text{ N} \end{gathered}[/tex]Therefore the force between the charges is 6.25 * 10^9 N.
Question 1: Assume that the pendulum of a grandfather clock acts as one of Planck'sresonators. If it carries away an energy of 8.1 x 10-15 eV in a one-quantumchange, what is the frequency of the pendulum? (Note that an energy this smallwould not be measurable. For this reason, we do not notice quantum effects in thelarge-scale world.)
Given:
Energy = 8.1 x 10⁻¹⁵ eV.
Let's find the frequency of the pendulum.
To find the frequency, apply the formula for the energy of a light quantum:
[tex]E=hf[/tex]Where:
E is the energy
h is Planck's constant = 6.63 x 10⁻³⁴ m² kg/s
f is the frequency.
Where:
1 eV = 1.6 x 10⁻¹⁹ J.
Rewrite the formula for f and solve:
[tex]f=\frac{E}{h}[/tex]Thus, we have:
[tex]f=\frac{8.1\times10^{-15}*(1.6\times10^{-19})}{6.63\times10^{-34}}[/tex]Solving further:
[tex]\begin{gathered} f=\frac{8.1\times10^{-15}*(1.6\times10^{-19})}{6.63\times10^{^{-34}}} \\ \\ \\ f=1.95\text{ Hz.} \end{gathered}[/tex]Therefore, the frequency of the pendulum is 1.95 Hz.
ANSWER:
1.95 Hz
If frictional forces and air resistance were acting upon the falling ball in #1 would the kinetic energy of theball just prior to striking the ground be more, less, or equal to the value predicted in #1?
Given that the ball is falling in two different situations.
Kinetic energy is the energy that an object possesses due to its motion and it is proportional to the square of the velocity of the object.
The frictional force and the air resistance are the forces that will be opposing the motion of the ball. That is these forces act in the opposite direction of the motion of the ball thus reducing the net acceleration acting on the ball.
As the acceleration decreases the final velocity of the ball decreases.
Thus the kinetic energy of the ball just prior to striking the ground will be less than before.
A crane used 136.65 watts of power to raise a 16.22 N object in 4.97 seconds. Through what vertical distance was the object displaced?
Given,
Power, P = 136.65 watts
Force, F = 16.22 N
Time, t = 4.97 seconds
The work done is calculated by the given formula,
[tex]W=F\times d[/tex]Now, the formula of power is given by
[tex]\begin{gathered} \text{Power = }\frac{W\text{ork done}}{\text{Time}} \\ \text{P = }\frac{F\times d}{T} \\ d=\frac{P\times t}{F} \\ d=\frac{136.65\text{ W}\times4.97\text{ s}}{16.22\text{ N}} \end{gathered}[/tex]Thus, the vertical distance is
[tex]d=41.87\text{ m}[/tex]When an arrow is released from its bow, its energy is transformed from potential energy to kinetic energy. Determine if the statement is correct.
Let's determine if the statement ''When an arrow is released from its bow, its energy is transformed from potential energy to kinetic energy'' is correct.
Let's first define kinetic and potential energies.
Potential energy is the energy stored in an object due to its position which depends on the relative position of various parts of the system.
Kinetic energy is the energy possesed by an object in motion.
Therefore, before the arrow is released from the bow, it is at rest and the energy is stored.
Hence, it posssess potential energy.
When the arrow is now released from the bow, it is now in motion. Since it is now in motion, the potential energy is now transformed to kinetic energy.
Therefore, the statement ''When an arrow is released from its bow, its energy is transformed from potential energy to kinetic energy'' is correct.
Calculate the mass of an object with a weight of 467 N.
We will have that the mass of the object is:
[tex]\begin{gathered} m=\frac{467N}{9.8m/s^2}\Rightarrow m=\frac{2335}{49}kg \\ \\ \Rightarrow m\approx47.65kg \end{gathered}[/tex]So, the mass of the object is 2335/49 kg, that is approximately 47.65 kg.
A train accelerates uniformly from rest to a velocity of 40 m s^-1, in 5 seconds.a)Find the acceleration of the train during the 5 seconds.b) Find the distance travelled by the train during the 5 seconds.
Answer:
a) 8 m/s²
b) 100 m
Explanation:
The acceleration of the train can be calculated as
[tex]a=\frac{v_f-v_i}{t}[/tex]Where vf is the final velocity, vi is the initial velocity, and t is the time. Replacing vf = 10 m/s, vi = 0 m/s, and t = 5 s, we get:
[tex]a=\frac{40\text{ m/s - 0 m/s}}{5\text{ s}}=\frac{40\text{ m/s}}{5\text{ s}}=8\text{ m/s}^2[/tex]Then, the distance traveled can be calculated using the following equation
[tex]x=v_it+\frac{1}{2}at^2[/tex]Replacing vi = 0 m/s, t = 5 s, and a = 8 m/s², we get
[tex]\begin{gathered} x=(0\text{ m/s\rparen\lparen5 s\rparen + }\frac{1}{2}(\text{ 8 m/s}^2)(5\text{ s\rparen}^2 \\ x=0\text{ m + }\frac{1}{2}(8\text{ m/s}^2)(25\text{ s\rparen} \\ x=0\text{ m+100 m} \\ x=100\text{ m} \end{gathered}[/tex]Therefore, the distance traveled by train was 100 m.
A train car with a mass of 5 kg and speed of 5 m/s is traveling to the right. Another train car with a mass of 2 kg is standing still. After the collision, the 5 kg train car is stuck to the 2 kg train car, what is their combined Final Velocity?
From the Law of Conservation of Linear Momentum, we have:
[tex]m_1v_1+m_2v_2=m_1v_1^{\prime}+m_2v_2^{\prime}[/tex]If the two particles have the same velocity after the collision, then v₁'=v₂'.
Let v be equal to the final velocity of the particles. Then:
[tex]\begin{gathered} m_1v_1+m_2v_2=m_1v+m_2v \\ \\ \Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v \end{gathered}[/tex]Since v is unknown, isolate it from the equation:
[tex]v=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]Replace the data to find the value of v:
[tex]\begin{gathered} m_1=5kg \\ v_1=5\frac{m}{s} \\ \\ m_2=2kg \\ v_2=0 \\ \\ \Rightarrow v=\frac{(5kg)(5\frac{m}{s})+(2kg)(0)}{5kg+7kg}=\frac{25kg\frac{m}{s}}{12kg}=2.08333...\frac{m}{s} \end{gathered}[/tex]Therefore, the combined final velocity of both train cars is approximately 2.1m/s.
Which statement is not true of the reactions on the Sun?A.They are fusion reactions.B.They occur between helium atoms.C.Fast-moving nuclei collide.D.Hydrogen atoms are reacted.
Answer and explanation:
A correct option is an option (B). Option b is not true of the reaction on the sun because the reactions do not happen in between helium atoms.
All other options (A), (C), and (D) are incorrect because, in sun, the nuclear fusion reaction takes place between hydrogen atoms. Due to the sun's gravitational force and high temperature, the nuclei of helium move vary fast.
Conclusion:
The correct option is option (B).
Hello, is it possible to help me solve this question? -A 475-gram ball is traveling horizontally at 12.0 m/s to the left when it is suddenly struck horizontally by a bat, causing it to reverse direction and initially travel at 8.50 m/s to the right. If the bat produced an average force of 1275 N on the ball, for how long (in milliseconds) was it in contact with the ball
Change in momentum = ΔP
ΔP = F t
Where:
F= force
t = time
Pf - Pi = Ft
M vf - m vi = F t
m= mass = 475 g = 0.475 kg
t = M vf - M vi / F
t = M (vf-vi) / f
Replacing:
t = 0.475 ( 8.50 - (-12 ) )/ 1275 = 0.0076 s = 7.64 ms
In a smartphone, how many megabytes (MB) are there in thirty-one gigabytes (GB) of memory?
We know that for each Gb there are 1000 Mb, thus in 31 Gb there will be 31000 Mb.
A tractor travels at a constant speed of 6m/s. Find the power supplied by the engine if it can supply a maximum force of 467 kN.
speed = v = 6m/s
Force = 467 kn = 467000N
Power = Work/ time
Work = Force x distance
Speed = distance / time
Isolating Distance:
Distance = Speed * time
P =[ Force * (speed * time ) ] / time
P = Force * speed
Replacing:
P = 467000 N * 6 m/s = 2,802,000 W
A 1 kg mass has a kinetic energy of 1 Joule when its speed isA. 0.45 m/sB. 1.4 m/sC. 1 m/sD. 4.4 m/s
Kinetic energy = 1/2 x mass x velocity^2
Where:
Kinetic energy = 1 J
mass = 1 kg
Isolate v (speed)
KE = 1/2 m v^2
√(2KE/m) = v
v = √(2KE/m)
Replacing with the values given:
v = √(2*1/1)
v= √2 = 1.4 m/s
Answer: 1.4 m/s (B)
lc. A student makes a claim that states the first hill of a roller coaster is always the tallest(Point A) provided there is no mechanical assistance at any point after Point A on the track.Do you accept or reject her claim? Explain your answer. (2 points)
We accept the claim. This comes from the fact that if there is no mechanical assistance after point A (that is the tallest) then we make sure that the potential energy converts in kinetic energy and that way we make sure that the rollercoaester reaches all the other hills. If the tallest point is not the first one and there's no mechanical assitances then the rollercoester won't reach the hills that are taller than the first one (the kinetic energy won't be enough to reach it).
Find the equivalent resistance of thiscircuit.R₁www400 ΩIR₂600 ΩReq = [?] 2R3www500 Ω
We are given the following information.
Resistor: R₁ = 400 Ω
Resistor: R₂ = 600 Ω
Resistor: R₃ = 500 Ω
We are asked to find the equivalent resistance of the given circuit.
Notice that the resistors R₁ and R₂ are in parallel and this parallel combination is in series with resistor R₃.
First, let us find the parallel resistance of R₁ and R₂.
[tex]R_p=\frac{R_1\times R_2}{R_1+R_2}=\frac{400\times600}{400+600}=\frac{240000}{1000}=240\;\Omega[/tex]Finally, let us find the series resistance of Rp and R₃.
[tex]\begin{gathered} R_{eq}=R_p+R_3 \\ R_{eq}=240+500 \\ R_{eq}=740\;\Omega \end{gathered}[/tex]Therefore, the equivalent resistance of the given circuit is 740 Ω.
If a microwave is determined to use 840 W of power on a 120 V circuit, how many amperes does it need to run?
ANSWER:
7 amperes
STEP-BY-STEP EXPLANATION:
Given:
Power (P) = 840 W
Voltage (V) = 120 V
We can calculate the current value as follows:
[tex]\begin{gathered} I=\frac{P}{E} \\ \\ \text{ We replacing:} \\ \\ I=\frac{840}{120} \\ \\ I=7\text{ A} \end{gathered}[/tex]Therefore, the number of amperes is 7
Solve the inequality for x. Show each step of the solution.10x>9(2x-8)-20
Given,
[tex]10x>9(2x-8)-20[/tex]On simplifying the above equation,
[tex]\begin{gathered} 10x>18x-72-20 \\ \Rightarrow10x>18x-92 \\ \Rightarrow10x-18x>-92 \\ \Rightarrow-8x>-92 \end{gathered}[/tex]On further simplifying,
[tex]\begin{gathered} 8x>92 \\ \Rightarrow x>\frac{92}{8} \\ \Rightarrow x>11.5 \end{gathered}[/tex]Thus on solving the given inequality, we get, x>11.5
A 87 kg person sits at rest at the top of a 51m high amusement park ride. How much energy does the person have at this point
We will calculate potential energy, in this case, using the next formula
[tex]PE=\text{mgh}[/tex]where m is the mass, g is the gravity and h is the height
in our case
m=87 kg
g=9.8 m/s^2
h=51m
we substitute
[tex]PE=87(9.8)(51)=43482.6J[/tex]ANSWER
The energy is 43482.6J
A jet plane has a maximum deceleration rate of -7.30 m/s/s. It touches down on a runway with a speed of 122 m/s. From the instant it touches the runway, what is the minimum time needed before it comes to rest??
The maximum time needed before the jet plane that has a maximum deceleration of -7.30 m/s² comes to rest is 16.71 seconds.
What is time?Time can be defined as an ongoing and continuous sequence of events that occur in succession, from past through the present, and to the future.
To calculate the maximum time needed for the jet to come to rest, we use the formula below.
Formula:
t = (v-u)/a........... Equation 1Where:
t = Maximum time needed for the jet to come to restv = Final speedu = Initial speeda = Maximum deceleration of the jetFrom the question,
Given:
v = 0 m/s (to rest)u = 122 m/sa = - 7.30 m/s²Substitute these values into equation 1
t = (0-122)/(-7.3)t = -122/7.3t = 16.71 secondsHence, the maximum time needed is 16.71 seconds.
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How high should you be if you want to drop a tennis ball and have it hit the ground with a velocity of -25.0 mi/hr?
Data: Final velocity = 11.176m/s (or 25 mi/hr)
Firstly, we need to replace our values on the third given equation. It is convenient to use all the values on their SI units.
[tex]11.176^2=0^2-2*9.81*\Delta x\Rightarrow\Delta x=\frac{11.176^2}{-2*9.81}=-6.366m[/tex]Thus, the ball would need to be dropped from a height of 6.366m, or 20.89ft