Answer:
#_pile = 12 celdas
Explanation:
Lead acid sulfur batteries generate each cell a potential of 2 volts. By colonato to reach the voltage of 24 volts
#_pile = 24/2
#_pile = 12 cledas
serially connected
A 2.5 kg rock is released from rest at the surface of a pond 1.8 m deep. As the rock falls, a constant upward force of 5.0 N is exerted on it by water resistance. Let y=0 be at the bottom of the pond.
Required:
a. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0 m.
b. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 0.50 m.
c. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water's surface is 1.0 m.
Answer:
a) Nonconservative Work
[tex]W_{disp} = 9\,J[/tex]
Final Gravitational Potential Energy
[tex]U_{f} = 0\,J[/tex]
Final Translational Energy
[tex]K_{f} = 35.131\,J[/tex]
b) Nonconservative Work
[tex]W_{disp} = 6.5\,J[/tex]
Final Gravitational Potential Energy
[tex]U_{f} = 12.259\,J[/tex]
Final Translational Energy
[tex]K_{f} = 25.373\,J[/tex]
c) Nonconservative Work
[tex]W_{disp} = 4\,J[/tex]
Final Gravitational Potential Energy
[tex]U_{f} = 24.518\,J[/tex]
Final Translational Energy
[tex]K_{f} = 15.614\,J[/tex]
Explanation:
The nonconservative work due to water resistance is defined by definition of work:
[tex]W_{disp} = F\cdot (y_{o}-y_{f})[/tex] (1)
Where:
[tex]W_{disp}[/tex] - Dissipate work, in joules.
[tex]F[/tex] - Resistance force, in newtons.
[tex]y_{o}[/tex] - Initial height, in meters.
[tex]y_{f}[/tex] - Final height, in meters.
The final gravitational potential energy ([tex]U_{f}[/tex]), in joules, is calculated by means of the definition of gravitational potential energy:
[tex]U_{f} = m\cdot g\cdot y_{f}[/tex] (2)
Where:
[tex]m[/tex] - Mass of the rock, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
The final translational kinetic energy ([tex]K_{f}[/tex]), in joules, is obtained by means of the Principle of Energy Conservation, Work-Energy Theorem and definitions of gravitational potential energy and translational kinetic energy:
[tex]m\cdot g\cdot y_{o} = U_{f} + K_{f} + W_{disp}[/tex] (3)
[tex]K_{f} = m\cdot g\cdot y_{o} - U_{f} - W_{disp}[/tex]
Lastly, the mechanical energy of the system ([tex]E[/tex]), in joules, is the sum of final gravitational potential energy, translational kinetic energy and dissipated work due to water resistance:
[tex]E = U_{f} + K_{f} + W_{disp}[/tex] (4)
Now we proceed to solve the exercise in each case:
a) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 0\,m[/tex])
[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 0\,m)[/tex]
[tex]W_{disp} = 9\,J[/tex]
Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 0\,m[/tex])
[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0\,m)[/tex]
[tex]U_{f} = 0\,J[/tex]
Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 0\,J[/tex], [tex]W_{disp} = 9\,J[/tex])
[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -0\,J-9\,J[/tex]
[tex]K_{f} = 35.131\,J[/tex]
b) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 0.50\,m[/tex])
[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 0.5\,m)[/tex]
[tex]W_{disp} = 6.5\,J[/tex]
Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 0.5\,m[/tex])
[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (0.5\,m)[/tex]
[tex]U_{f} = 12.259\,J[/tex]
Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 12.259\,J[/tex], [tex]W_{disp} = 6.5\,J[/tex])
[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -12.259\,J-6.5\,J[/tex]
[tex]K_{f} = 25.373\,J[/tex]
c) Nonconservative Work ([tex]F = 5\,N[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]y_{f} = 1\,m[/tex])
[tex]W_{disp} = (5\,N)\cdot (1.8\,m - 1\,m)[/tex]
[tex]W_{disp} = 4\,J[/tex]
Final Gravitational Potential Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{f} = 1\,m[/tex])
[tex]U_{f} = (2.5\,kg) \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1\,m)[/tex]
[tex]U_{f} = 24.518\,J[/tex]
Final Translational Energy ([tex]m = 2.5\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 1.8\,m[/tex], [tex]U_{f} = 24.518\,J[/tex], [tex]W_{disp} = 4\,J[/tex])
[tex]K_{f} = (2.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.8\,m) -24.518\,J-4\,J[/tex]
[tex]K_{f} = 15.614\,J[/tex]
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The ball moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point Group of answer choices the work done on it by gravity is -118 J and the work done on it by the tension in the string is zero. the work done on it by gravity is -118 J and the work done on it by the tension in the string is 118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J. the work done on it by gravity is 118 J and the work done on it by the tension in the string is -118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to zero.
Answer:
The ball moves from lowest to highest point:
W = M g h = 3 * 9.8 * 4 = 118 J
This is work done "against" gravity so work done by gravity is -118 J
The tension of the string does no work because the tension does not
move thru any distance W = T * x = 0 because the length of the string is fixed.
friction between two flat surfaces can be divided into two categories. what are the two most common kinds of friction?
Answer:
kinetic and static
Explanation:
hope it helps! ^w^
the current through a wire is measured as the potential difference is varied what is the wire resistance
Answer:
Resistance, R = 0.02 Ohms
Explanation:
Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.
Mathematically, Ohm's law is given by the formula;
V = IR
Where;
V is the voltage or potential difference.
I is the current.
R is the resistance.
From the attachment, we would pick the following values on the graph of current against voltage;
Voltage, V = 0.5 V
Current = 25 A
To find resistance;
R = V/I
R = 0.5/2.5
Resistance, R = 0.02 Ohms
Note:
Resistance (R) is the inverse of slope i.e change in current with respect to change in voltage.
A current of 1 mA flows through a copper wire. How many electrons will pass a point in each second?
Answer:
A current of 1ma flows through a copper wire, how many electron will pass a given point in one second? 1 Coulomb = 6.24 x 10^18 electrons (or protons)/1Sec which is also equal to 1 Amp/1 Sec. 1mA is 1/1000th of 1A so only 1/1000th of 6.24 x 10^18 electrons will pass a given point in 1 Sec.
A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m. With what speed was it thrown, assuming negligible air resistance
The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.
What is the projectile motion?An item or particle that is propelled in a gravitational influence, such as from the crust of the Ground, and moves along a curved route while solely being affected by gravity is said to be in projectile motion.
A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m.
The initial velocity is zero. Then the time taken to reach the ground is given as,
h = ut + 1/2at²
- 24 = 0×t + 1/2 (-9.8)t²
24 = 4.9t²
t² = 4.8979
t = 2.213 seconds
Then the horizontal speed is given as,
v = 18 / 2.213
v = 8.133 meters per second
The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.
Learn more about projectile motion:
https://brainly.com/question/29761109
#SPJ6
A Man has 5o kg mass man in the earth and find his weight
Answer:
49 N
Explanation:
Given,
Mass ( m ) = 50 kg
To find : Weight ( W ) = ?
Take the value of acceleration due to gravity as 9.8 m/s^2
Formula : -
W = mg
W = 50 x 9.8
W = 49 N
Given the triangle shown below, what is the cosine of the angle 0 ?
triangle trig image 1
O Vx/V
O Vx/Vy
O Vy/Vx
O Vy/
Answer:
?
Explanation:
i am sorry but there is no triangle
Explanation: edmentum sample answer
You are cooking breakfast for yourself and a friend using a 1,140-W waffle iron and a 510-W coffeepot. Usually, you operate these appliances from a 110-V outlet for 0.500 h each day. (a) At 12 cents per kWh, how much do you spend to cook breakfast during a 30.0 day period
Answer:
The cost is 297 cents.
Explanation:
Power of iron, P = 1140 W
Power of coffee pot, P' = 510 W
Voltage, V = 110 V
Time, t = 0.5 h each day
Cost = 12 cents per kWh
(a) Total energy
E = P x t + P' x t
E = 1140 x 0.5 x 60 x 60 + 510 x 0.5 x 60 x 60
E = 2052000 + 918000 = 2970000 J
1 kWh = 3.6 x 10^6 J
E = 0.825 kWh
For 30 days
E' = 0.825 x 30 = 24.75 kWh
So, the cost is
= 12 x 24.75 = 297 cents
A free undamped spring/mass system oscillates with a period of 4 seconds. When 10 pounds are removed from the spring, the system then has a period of 2 seconds. What was the weight of the original mass on the spring? (Round your answer to one decimal place.)
Answer:
13.3 pounds.
Explanation:
For a spring of constant K, with an attached object of mass M, the period can be written as:
T = 2*π*√(M/K)
Where π = 3.14
First, we know that the period is 4 seconds, then we have:
4s = (2*π)*√(M/K)
We know that if the mass is reduced by 10lb, the period becomes 2s.
Then the new mass of the object will be: (M - 10lb)
Then the period equation becomes:
2s = (2*π)*√((M-10lb)/K)
So we have two equations:
4s = (2*π)*√(M/K)
2s = (2*π)*√((M-10lb)/K)
We want to solve this for M.
First, we need to isolate K in one of the equations.
Let's isolate K in the first one:
4s = (2*π)*√(M/K)
(4s/2*π) = √(M/K)
(2s/π)^2 = M/K
K = M/(2s/π)^2 = M*(π/2s)^2
Now we can replace it in the other equation.
2s = (2*π)*√((M-10lb)/K)
First, let's simplify the equation:
2s/(2*π) = √((M-10lb)/K)
1s/π = √((M-10lb)/K)
(1s/π)^2 = ((M-10lb)/K
K*(1s/π)^2 = M - 10lb
Now we can use the equation: K = M*(π/2s)^2
then we get:
K*(1s/π)^2 = M - 10lb
(M*(π/2s)^2)*(1s/π)^2 = M - 10lb
M/4 = M - 10lb
10lb = M - M/4
10lb = (3/4)*M
10lb*(4/3) = M
13.3 lb = M
A boy of mass 50 kg on a motor bike is moveny coith 20m/see what is hio k.E
Uranus and Neptune may have a compressed liquid water ocean beneath their atmospheres. What three pieces of evidence support this conclusion?
Answer:
Distance from sun, orbit and rotation. Presence of interior oceans, and elements that forms compressed water.
Explanation:
Both the planets are Jovian planets and have layers formed by ice such as that of Uranus does not have any surface. Such as the planet is only rotating fluids. While 80% of the mass of Neptune is made up of fluids or icy water and also consists of ammonia and methane.what does it mean to have an acceleration of 8m/s^2
A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant angular acceleration, a of 6 rad/s If the initial angular velocity is 1.5 rad/s, determine The maximum angular velocity and linear velocity of the wheel after 1 complete revolution.
Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:
[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]
where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,
[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]
ωf = 8.8 rad/s
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
v = 2.2 m/s
B.F.Skinner emphesized the importance of-----?
Answer:
BFSkinner enfatizó la importancia de creía en la importancia de desarrollar la psicología experimental y dejar atrás el psicoanálisis y las teorías acerca de la mente basadas en el simple sentido común.
Explanation:
An object moves in a direction parallel to its length with a velocity that approaches the velocity of light. The length of this object, as measured by a stationary observer:________
a. approaches infinity.
b. approaches zero.
c. increases slightly.
d. does not change.
Answer:
b. approaches zero.
Explanation:
The phenomenon is known as length contraction.
Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.
let the length of the train = L
Let the length observed when the train is in motion = L₀
Apply Einstein's special theory of relativity;
[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]
from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)
As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L). (L₀ < L)
Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero. (L₀ = 0)
Thus, the length of this object, as measured by a stationary observer approaches zero
What is the approximate radius of an equipotential spherical surface of 30 V about a point charge of +15 nC if the potential at an infinite distance from the surface is zero?
Answer:
V = k Q / R potential at distance R for a charge Q
R = k Q / V
R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m
Note: Our equation says that if R if infinite then V must be zero.
True or False: The forces applied by our muscles on our bones are usually several times larger than the forces we exert on the outside world with our limbs.
Answer:
True
Explanation:
This is because of the point where the forces are applied by our muscles and
the angle they have about the bones. Take for example the diagram I uploaded.
If we do a free body diagram and a sum of torques, we would get that:
[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]
In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:
[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]
If we solve the equation for the force of the muscle we would get that:
[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]
Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.
Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:
[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]
which yields:
[tex]F_{muscle}=4.04 F_{hand}[/tex]
so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.
HELP NEEDED FAST (last cram sessions before finals)
BRAINLIEST!
Three resistors are connected in series across a 75-V potential difference. R, is 170 and R2 is 190. The potential difference across R3 is 21 V. Find the current in the circuit.
Explanation:
The sum of the voltages of the components connected in a series circuit is equal to the voltage across the battery.
[tex]V_T = V_1 + V_2 +V_3[/tex]
From Ohm's law ([tex]V=IR[/tex]) and in a series circuit, the amount of current flowing through the components is the same for all. So we can write [tex]V_T[/tex] as
[tex]V_T= 75\:\text{V} = I(170)+I(190) + 21\:\text{V}[/tex]
[tex]I(170+190)=54\:\text{V}[/tex]
[tex]I= \dfrac{54\:\text{V}}{360\:\text{ohms}}=0.15\:\text{A}[/tex]
A supertrain with a proper length of 100 m travels at a speed of 0.950c as it passes through a tunnel having a proper length of 50.0 m. As seen by a trackside observer, is the train ever completely within the tunnel? If so, by how much do the train’s ends clear the ends of the tunnel?
Answer:
19m
Explanation:
we have proper length L = 100m
the speed of the train v = 0.95
the speed of light is given as = 3x10⁸
length of the tunnel is given as = 50 meters
we can solve for the lenght contraction as
LX√1-v²/c²
= 100 * √1-(0.95*3x10⁸)²/(3x10⁸ )
= 31.22 metres
the train would be well seen at
50 - 31.22
= 18.78
= this is approximately 19 metres
we conclude tht the trains ends clears the ends of the tunnel by 19 meters.
thank you!
A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency is 890 Hz, the wavelength is .10m, and the amplitude is 6.5 mm. The tension in the line, in SI units, is closest to
Answer:
T = 712.9 N
Explanation:
First, we will find the speed of the wave:
v = fλ
where,
v = speed of the wave = ?
f = frequency = 890 Hz
λ = wavelength = 0.1 m
Therefore,
v = (890 Hz)(0.1 m)
v = 89 m/s
Now, we will find the linear mass density of the wire:
[tex]\mu = \frac{m}{L}[/tex]
where,
μ = linear mass density of wie = ?
m = mass of wire = 90 g = 0.09 kg
L = length of wire = 1 m
Therefore,
[tex]\mu = \frac{0.09\ kg}{1\ m}[/tex]
μ = 0.09 kg/m
Now, the tension in wire (T) will be:
T = μv² = (0.09 kg/m)(89 m/s)²
T = 712.9 N
A 47-kg box is being pushed a distance of 8.0 m across the floor by a force whose magnitude is 188 N. The force is parallel to the displacement of the box. The coefficient of kinetic friction is 0.24. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.
Answer:
(a) 1504 J
(b) - 884.35 J
(c) 0 J
(d) 0 J
Explanation:
Mass, m = 47 kg
displacement, s = 8 m
Force, F = 188 N
Coefficient of friction = 0.24
(a) Work done by applied force
W = F s = 188 x 8 = 1504 J
(b) Work done by the friction force
W' = - 0.24 x 47 x 9.8 x 8 = - 884.35 J
(c) Work done by the gravitational force
W''= m g s cos 90 = 0 J
(d) Work done by the normal force
W''' = m g scos 90 = 0 J
In the graph below, why does the graph stop increasing after 30 seconds?
A. The hydrogen gas is absorbing heat to undergo a phase change.
B. A catalyst needs to be added to increase the amount of hydrogen produced.
C. No more hydrogen can be produced because all of the reactants have become products at this point.
D. It has reached the maximum amount of product it can make at this temperature. The temperature would need to increase to produce more.
Answer:
The answer is "Option C".
Explanation:
It's evident from the figure below that after thirty minutes, not no more hydrogen can be created because all of the reactants have converted into products.
hydrogen gas created in cm cubes per period x = 20 seconds, y = 45 centimeters squared, and so on.
A reaction's terminus (the graph's flat line) indicates that no further products are being created during the reaction.
A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular acceleration(assuming it was constant)
Answer:
α = 1930.2 rad/s²
Explanation:
The angular acceleration can be found by using the third equation of motion:
[tex]2\alpha \theta=\omega_f^2-\omega_i^2[/tex]
where,
α = angular acceleration = ?
θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s
Therefore,
[tex]2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}[/tex]
α = - 1930.2 rad/s²
negative sign shows deceleration
What bet force is required to stop a 2250 kg car if the decelerates at a rate of -4.3 m/s^2 please answer fast
Answer:
Force = Mass × Acceleration
[tex]{ \tt{force = 2250 \times 4.3}} \\ = { \tt{9675 \: newtons}}[/tex]
Give an example of how you could make a measurement that is, at the same time, very precise and very inaccurate
Answer:
Accuracy refers to how close a measurement is to the true or accepted value. ... Precision is independent of accuracy. That means it is possible to be very precise but not very accurate, and it is also possible to be accurate without being precise. The best quality scientific observations are both accurate and precise.
plz help me with hw A bus of mass 1000 kg moving with a speed of 90km/hr stops after 6 sec by applying brakes then calculate the distance travelled and amount of force applied.
Answer:
Mass, M = 1000 kg
Speed, v = 90 km/h = 25 m/s
time, t = 6 sec.
Distance:
[tex]{ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}[/tex]
Force:
[tex]{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}[/tex]
what is the final velocity if you have an initial velocity of 5 m/s with an acceleration of 3 m/s^2 over a 4 second interval
Answer:
initial velocity (u)=5m/s
final velocity (v)=?
acceleration (a)=3m/s^2
time (t)=4s
now,
acceleration (a)=v-u/t
3=v-5/4
3×4=v-5
12=v-5
12+5=v
17=v
v=17
The displacement x of a particle varies with time t as x = 4t 2 -15t + 25. Find the position,
velocity and acceleration of the particle at t = 0. When will the velocity of the particle becomes
zero? Can we call the motion of the particle as one with uniform acceleration?
Answer:
x = 4 t^2 - 15 t + 25 displacement of particle
dx / dt = 8 t - 15 velocity of particle
d^2x / dt^2 = 8 acceleration of particle
If 8 t -15 = o then t = 8 / 15
Since acceleration is a constant 8 then motion has uniform acceleratkon
A plane flying horizontally at an altitude of 1 mi and a speed of 480 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.
Answer:
First remember that the distance between two points (a, b) and (c, d) is given by the equation:
[tex]d = \sqrt{(a - c)^2 + (b - d)^2}[/tex]
Now let's define the position of the radar as:
(0mi, 0mi)
Then we can write the position of the plane as:
(480mi/h*t, 1mi)
where t is time in hours.
Then we can write the distance equation as:
[tex]d(t) = \sqrt{(480\frac{mi}{h}*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }[/tex]
Now we want to get:
the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.
So first we want to find the value of t such that:
d(3) = 3mi
We will look at the positive value of t, because at this point the plane is increasing its distance to the station.
[tex]3mi = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480\frac{mi}{h}*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480\frac{mi}{h}*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{\frac{8mi^2}{230,400 mi^2/h^2} } = t = 0.0059 h[/tex]
The rate of change when the plane is 3 mi away from the station is:
d'(0.0059h)
remember that:
d'(t) = dd(t)/dt
We can write:
d(t) = h( g(t) )
such that:
h(x) = √x
g(t) = (480mi/h*t)^2 + (1mi)^2
then:
d'(t) = h'(g(t))*g'(t)
This is:
[tex]d'(t) = \frac{dd(t)}{dt} = \frac{1}{2}*\frac{2*t*480mi/h}{\sqrt{(480mi/h*t)^2 + (1mi)^2} }[/tex]
The rate of change at t = 0.0059h is then:
[tex]d'(0.0059h) = \frac{1}{2}*\frac{2*0.0059h*(480mi/h)^2}{\sqrt{(480mi/h*0.0059h)^2 + (1mi)^2} } =452.6 mi/h^2[/tex]
