Answer:
a = 1.1 m/s²
Explanation:
Given:
P = 650 N
m = 75 kg
g = 9.8 m/s²
____________
a - ?
Forse:
P = m*(g - a)
650 = 75*(10 - a)
10 - a = 650 /75
Acceleration:
a = 9.8 - 650 / 75 = 9.8 - 8.7 = 1.1 m/c²
What is the density of a 45.87 g golf ball with a diameter of 4.287 cm?
We are asked to determine the density of a gulf ball given its mass and volume. To do that, we will use the formula for density:
[tex]D=\frac{m}{V}[/tex]Where:
[tex]\begin{gathered} D=\text{ density} \\ m=\text{ mass} \\ V=\text{ volume} \end{gathered}[/tex]To determine the volume we will use the fact that the gulf ball can be approximated to a sphere and the volume of a sphere is given by:
[tex]V=\frac{4}{3}\pi r^3[/tex]Where:
[tex]r=\text{ radius}[/tex]We are given the diameter. We know that the diameter is twice the radius, therefore:
[tex]r=\frac{D}{2}[/tex]Substituting the value of the diameter we get:
[tex]r=\frac{4.287\operatorname{cm}}{2}[/tex]Solving the operations:
[tex]r=2.144\operatorname{cm}[/tex]Now, we substitute the value of the radius in the formula of the volume:
[tex]V=\frac{4}{3}\pi(2.144\operatorname{cm})^3[/tex]Solving the operation we get:
[tex]V=41.282\operatorname{cm}^3[/tex]Now, we substitute the value of the volume and the mass in the formula for density:
[tex]D=\frac{45.87g}{41.282\operatorname{cm}^3}[/tex]Solving the operation:
[tex]D=1.11\frac{g}{\operatorname{cm}^3}[/tex]Therefore, the density of the ball is 1.11 g/cm^3.
How is kinetic energy and pontential energy alike ?
The kinetic energy of the particle is associated with the velocity of the particle where as the potential energy of the particle depneds upon the position ofthe particle.
Both the energies can transform into each other.
For example,
When a ball is dropped from the building's top floor, the potential energy of the ball is maximum at the top of the builiding and transform into the kinetic energy while moving in the downward direction.
The kinetic energy of the ball is maximum just before hitting the ground.
This shows the tranformation of the potential energy to the kinetic energy.
Thus, the kinetic energy and potential can transform into each other is one the similarity in behaviour.
Could someone please help me ASAP?
As shown in the picture a person is swinging a yo-yo in the circle then the direction of the velocity vector is given by the vector 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 shown in the picture a person is swinging a yo-yo in a circle, we have to find the direction of the velocity vector,
As seen in the image, a person is swinging a yo-yo in a circle, and the vector D indicates the direction of the velocity vector.
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Ellipses have only one focus. Is this true or false?
In any ellipse there are two foci. This two points are fixed and are fundamental for the construction opf the ellipse. Therefore the statement is false.
A man can crack a nut by applying a force of 100 N a lever of length 0.5 m. What should be the length of the lever if he wants to use a force of 75 N to crack the nut?
Answer:
So, to apply the force of 75N length of nut cracker should be 66.6 cm
Answer:
Explanation:
Given:
F₁ = 100 N
L₁ = 0.5 m
F₂ = 75 N
__________
L₂ - ?
According to the rule of moments:
M₁ = M₂
F₁·L₁ = F₂·L₂
The length of the lever:
L₂ = F₁·L₁ / F₂
L₂ = 100·0.5 / 75 ≈ 0.67 m
Which of the following types of radiation consists of a high energy electron?A. GammaB. BetaC. DeltaD. Alpha
Explanation:
a) Gamma rays are high-energy photons and are the most energetic part of the electromagnetic spectrum.
b) Beta particles are high-energy, high-speed electrons emitted by certain radioactive nuclei.
c) There is no radiation called delta radiation.
d) Alpha particles are particles that are composed of two protons and two neutrons.
Final answer:
Thus, the correct option is (B) Beta
a piece of steel expands 10 cm when it is heated from 20 to 40 deg C. How much would it expand if it was heated from 20 to 60 deg.C?
ANSWER
20 cm
EXPLANATION
Given:
• The change in length of a piece of steel, ΔL = 10 cm, when its temperature change is ΔT = 20°C
Find:
• The change in length of the same piece of steel, ΔL, when its temperature is changed ΔT = 40°C
The change in length of a solid material is,
[tex]\Delta L=\alpha\cdot L_o\cdot\Delta T[/tex]Where L₀ is the original length, α is the linear expansion coefficient and ΔT is the change in temperature.
In this case, we know how much is the piece of steel expanded when the temperature change is 20°C, so we can find the product αL₀,
[tex]\alpha L_o=\frac{\Delta L}{\Delta T}[/tex]Replace the known values and solve,
[tex]\alpha L_o=\frac{10\text{ }cm}{20\degree C}=0.5cm/\degree C[/tex]If the temperature change now is 40°C,
[tex]\Delta L=\alpha L_o\Delta T=0.5cm/\degree C\cdot40\degree C=20cm[/tex]Hence, when the piece of steel is heated from 20°C to 60°C it will expand 20 centimeters.
A 3 kg ball is dropped from a height of 100 m above the surface of Planet Z. If the ball reaches a velocity of 45 m/s in 7 s, what is the ball’s weight on Planet Z? What is the gravitational field strength on Planet Z?
We are given the following information
Mass of ball = 3 kg
Height = 100 m
Final velocity = 45 m/s
Time = 7 s
Recall from the equations of motion
[tex]s=u\cdot t+\frac{1}{2}\cdot g\cdot t^2[/tex]Where u is the initial velocity of the ball that is zero.
[tex]\begin{gathered} s=u\cdot t+\frac{1}{2}\cdot g\cdot t^2 \\ 100=0\cdot7+\frac{1}{2}\cdot g\cdot7^2 \\ 100=\frac{1}{2}\cdot g\cdot49 \\ g=\frac{2\cdot100}{49} \\ g=4.08\; \frac{m}{s^2} \end{gathered}[/tex]So, the gravitational acceleration of the planet Z is 4.08 m/s^2
The weight o the ball is given by
[tex]\begin{gathered} W=m\cdot g \\ W=3\cdot4.08 \\ W=12.24\; N \end{gathered}[/tex]Therefore, the weight of the ball is 12.24 N
The virtual image as seen in a plane mirror is reversed both left-to-right and top-to-bottom. Is this true or false?
ANSWER:
false
STEP-BY-STEP EXPLANATION:
The virtual images in a plane mirror have a left-right investment. But not top-to-bottom, which means that what the statement says is false.
An Olympic long jumper is capable of jumping 8.0 m. Assuming his horizontal speed is 9.1 m/s as he leaves the ground, how long is he in the air and how high does he go?
Given data:
* The initial velocity of the jumper is u = 9.1 m/s.
* The horizontal range in the given case is 8 m.
Solution:
(a). By the kinematic equation, the time taken by the jumper in terms of the initial velocity and the horizontal range is,
[tex]R=ut+\frac{1}{2}at^2[/tex]where a is the acceleration of the jumper in the horizontal direction,
As there is no force acting on the jumper in the horizontal direction, thus, the value of acceleration is zero.
Substituting the known values,
[tex]\begin{gathered} 8=9.1\times t \\ t=\frac{8}{9.1} \\ t=0.88\text{ s} \end{gathered}[/tex]Thus, the time for which the jumper remains in the air is 0.88 seconds.
(b). By the kinematics equation, the initial velocity of the jumper in the upward direction is,
[tex]v_y-u_y=gt^{\prime}_{}\ldots\ldots\ldots(1)[/tex]where u_y is the initial velocity, v_y is the final velocity of the jumper at the top of vertical displacement, g is the acceleration due to gravity, and t' is the time taken by the jumper to reach the top of vertical displacement,
The jumper will come to rest at the higher position, thus, the final velocity of the jumper at the highest position is zero.
The time taken by the jumper to reach the maximum height is equal to the time taken by the jumper to reach the ground from the maximum height.
As the total time for the complete motion of the jumper is t, thus, the time taken by the jumper to reach the maximum height from the ground is,
[tex]\begin{gathered} t^{\prime}=\frac{t}{2} \\ t^{\prime}=\frac{0.88}{2} \\ t^{\prime}=0.44\text{ s} \end{gathered}[/tex]Substituting the known values in the equation (1),
[tex]\begin{gathered} 0-u_y=-9.8\times0.44_{} \\ u_y=4.312\text{ m/s} \end{gathered}[/tex]By the kinematics equation, the maximum height reached by the jumper is,
[tex]h=u_yt^{\prime}+\frac{1}{2}gt^{\prime}^2[/tex]Substituting the known values,
[tex]\begin{gathered} h=4.312\times0.44+\frac{1}{2}\times(-9.8)\times(0.44)^2 \\ h=1.9-0.95 \\ h=0.95\text{ m} \end{gathered}[/tex]Thus, the maximum height reached by the jumper is 0.95 meters.
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 density of copper is 8.94 g/cm^3. What is the mass of a rectangular sheet of copper: 10 cm wide, 45 cm long, and 0.2 cm thick? I have no idea how to solve for mass or where to start. :(
Mass = 804.6 g
Explanation:The dimension of the rectangular copper = 10 cm wide, 45 cm long, and 0.2 cm thick
The volume of the rectangular copper = 0.1m x 0.45m x 0.002m
The volume of the rectangular copper = 0.00009 m³
The density of copper = 8.94 g/cm³ = 8.94 x 1000 kg/m³
The density of copper = 8940 kg/m³
Mass = Density x Volume
Mass = 8940 x 0.00009
Mass = 0.8046 kg
Mass = 804.6 g
Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.(a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination 14 mi away?_____ min(b) How far must the faster car travel before it has a 15-min lead on the slower car?_____ mi
We will have the following:
a) We first determine the time it takes to travel the distance to both vehicles:
*
[tex]t_1=\frac{14mi\ast1h}{55}\Rightarrow t_1=\frac{14}{55}h[/tex]*
[tex]t_2=\frac{14mi\ast1h}{60mi}\Rightarrow t_2=\frac{11}{12}h[/tex]So, we determine now the difference in time:
[tex]\frac{11}{12}h-\frac{14}{55}h=\frac{437}{660}h\approx0.66h[/tex]So, the fastest car will arrive approximately 0.66 hours sooner.
b) We determmine the distance it must travel the fastest car to have a 15 minute lead on the other one as follows:
First, we determine the time difference required:
[tex]t=\frac{15min\ast1h}{60min}\Rightarrow t=0.25h[/tex]Then, since both vehicles will move relative to each other, we will have that:
[tex]d_{c1}=(60mi/h)(0.25h)\Rightarrow d_{c1}=15mi[/tex]So, the fastest car must be 15 miles ahead of the other car in order to have a 15 minute lead with respect to the second car.
The inductor in the RLC tuning circuit of an AM radio has a value of 9 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 94 kHz
We are asked to determine the capacitance of an RLC circuit given the frequency. To do that we will use the following formula:
[tex]C=\frac{1}{4\pi^2f^2L}[/tex]Where:
[tex]\begin{gathered} C=\text{ capacitance} \\ f=\text{ frequency} \\ L=\text{ inductance} \end{gathered}[/tex]Now, We plug in the values:
[tex]C=\frac{1}{4\pi^2(94\times10^3Hz)^2(9\times10^{-3}H)}[/tex]Now, we solve the operations:
[tex]C=3.19\times10^{-10}F[/tex]The capacitance is 3.19x10^-10 farads. In Picofarads this is equivalent to:
[tex]C=0.0319pF[/tex]help, Asap!!!!!!!!!!!!!!!!!!
Answer:
0
Explanation:
all of the movement and opposite movement are the same
Use the acceleration vs time graph to answer this question. The graph shows the motion with an initial velocity of -4 m/s. Each tick mark on the x-axis represents 1 second. Calculate the velocity at t = 8 seconds.
Answer:
4 m/s
Explanation:
To find the velocity at t = 8 seconds, we will use the following equation:
[tex]v_f=v_i+at[/tex]Where vf is the final velocity, vi is the initial velocity, a is the acceleration and t is the time.
From t = 0 seconds to t = 3 seconds, we have an acceleration of 6 m/s², so we can calculate the velocity at t = 3 seconds as:
[tex]\begin{gathered} v_f=-4m/s+6m/s^2(3\text{ s)} \\ v_f=-4\text{ m/s + 18 m/s} \\ v_f=14\text{ m/s} \end{gathered}[/tex]Now, from t = 3 seconds to t = 8 seconds, the acceleration is equal to -2 m/s². So we need to use the same equation but this time, the initial velocity will be 14 m/s and the time will be 5 seconds because t = 8 s - 3 s = 5s. Then, we get:
[tex]\begin{gathered} v_f=14m/s-2m/s^2(5s) \\ v_f=14\text{ m/s - 10 m/s} \\ v_f=4\text{ m/s} \end{gathered}[/tex]Therefore, the velocity at t = 8 seconds is 4 m/s
the answer to this question
The car and the delivery truck both start from rest and accelerate at the same rate. So, the final speed of the car as compared to the truck is four times as much. So, option D is correct.
What is acceleration?Acceleration is the term used in mechanics to describe how quickly an object's velocity changes in relation to time. The magnitude of accelerations as a vector. An object will accelerate in the direction that it is being pulled in by the net force.
Acceleration is a vector quantity since it has a magnitude and a direction. Velocity is another aspect of vessel quantities. The change in the vector of velocity over a time interval divided by the time interval is the definition of acceleration.
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What is the normal force of a 0.0037 kg tennis ball rolling at a constant speed of 3 m/s
across a desk?
The normal force of the rolling ball is 0.037 N
Mass of tennis ball= 0.0037 kg
Constant speed= 3m/s
we need to apply the concept of laws of motion
Since the ball is rolling at a constant speed, it is an example of uniform motion.
So,
Normal force=weight of the body
= 0.0037 x 10 ( acceleration of gravity= 10 m/s²)
Normal force= 0.037 N
Therefore the normal force on the ball is 0.037 N
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A car is being tested for safety by colliding it with a brick wall.The 1300 kg car is initially driving towards the wall at a speedof 15 m/s, and after colliding with the wall the car movesaway from the wall at 2 m/s. If the car is in contact with thewall for 0.5 s, calculate the average force exerted on the carby the wall.
Answer:
44200 N
Explanation:
To calculate the average force exerted on the car, we will use the following equation
[tex]\begin{gathered} Ft=\Delta p \\ Ft=m(v_f-v_i) \end{gathered}[/tex]Where F is the average force, t is the time, m is the mass, vf is the final velocity and vi is the initial velocity of the car.
Replacing t = 0.5s, m = 1300 kg, vf = -2 m/s, and vi = 15 m/s and solving for F, we get
[tex]\begin{gathered} F(0.5s)=(1300\text{ kg\rparen\lparen-2 m/s - 15 m/s\rparen} \\ F(0.5s)=(1300\text{ kg\rparen\lparen-17 m/s\rparen} \\ F(0.5s)=-22100\text{ kg m/s} \\ F=\frac{-22100\text{ kg m/s}}{0.5\text{ s}} \\ F=-44200\text{ N} \end{gathered}[/tex]Therefore, the average force exerted on the car by the wall was 44200 N
A particular cookie provides 54.0 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 103-kilogram weight 2.45-decimeters above the ground with an energy efficiency of 25%. How many repetitions of this exercise can the athlete do with the energy supplied from one of these cookies?
A maximum of about 229 repetitions of something like the exercise can be performed by that of the athlete utilizing the energy provided by each of the biscuits.
The proportion of input to produced energy can be used to define energy consumption.
A cookie, therefore, therefore has 54.0 kcal of calories. The 54.0 kcal throughout this croissant is used as power input by the athlete.
Efficiency = output energy / input energy
It can be written as:
Output energy = efficiency × input energy
Puting the values of efficiency and input energy.
Output energy = 0.25 × 54 kcal = 13.5 kcal.
The weightlifting exercise can be done n times for the output energy. This outgoing energy comes from mgh in the shape of potential energy. So,
Energy per repetition = [tex]mgh[/tex]
Put the values of m, g and h in above equation.
Energy per repetition = 103 kg × 9.8 m/ × (2.45 × 0.1m) = 247.303 J
Energy per repetition = 0.059 kcal.
So,
amount of repetitions = sum of output energy / energy per repetition
amount of repetitions = 13.5 kcal / 0.059 kcal = 229 repetitions.
Therefore, amount of repetition can be be 229.
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Students conducted an experiment in which a ball was thrown into the air and then caught at the same height from which it was released. They determined that the ball had 100 J of kinetic energy when it was released. They calculated the ball's energy just before it was caught and determined that it's kinetic energy was less than 100 J.Which of the following statements MOST LIKELY accounts for the difference in the ball's initial and final kinetic energy?
A ball was thrown into the air and then caught at the same height from which it was released.
The initial kinetic energy of the ball was 100 J and the ball's kinetic energy just before it was caught was less than 100 J.
Before they were decommissioned, the NASA space shuttles required two solid rocket boosters (SRBs) to launch the shuttle from Earth’s surface. Both SRBs produced 1.7 x10^7 N at liftoff. The combined mass of a shuttle and rocket boosters was about 1.5 x 10^ 6 kga) Calculate the net acceleration of a space shuttle and rockets at the time of liftoff. (b) Calculate the speed of the shuttle and rockets after 10.0 s.
Given:
The force on the booster is
[tex]F=\text{ 1.7}\times10^7\text{ N}[/tex]The mass is
[tex]m=\text{ 1.5}\times10^6\text{ kg}[/tex]Required:
(a) The net acceleration
(b) Speed of the shuttle and rockets after time t = 10 s
Explanation:
(a) The net acceleration can be calculated as
[tex]\begin{gathered} a=\frac{F}{m} \\ =\frac{1.7\times10^7}{1.5\times10^6} \\ =11.33\text{ m/s}^2 \end{gathered}[/tex](b)
The initial speed of the rocket and shuttle will be zero.
The speed of the rocket and shuttle after time t = 10 s will be
[tex]\begin{gathered} v=0\text{ m/s + 11.33}\times10 \\ =\text{ 113.3 m/s} \end{gathered}[/tex]Final Answer:
(a) The net acceleration is 11.33 m/s^2.
(b) The speed of the rocket and shuttle is 113.3 m/s
What is the car's velocity between 11h and 15h?
Equation:
df-di/ t f-ti
The velocity of the car between time 11 hr and 15 hr which is shown in the graph would be 4 m/s².
The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.
We are given that,
Displacement of the car = Δx = (20km) - (4km) = 16 km
Time interval of the car = Δt = (15h)- (11h) = 4hours
v = dx/dt
dx = v dt
∫dx = ∫v dt
Δx = v Δt
v = Δx/Δt
Therefore, for get the value of velocity between the given time interval , putting the value in in above equation,
v = 16km/4hours
v = 4 km/hours
Thus, The velocity of the car between time 11 hr and 15 hr will be given as 4km/hours.
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When a 6.0-F capacitor is connected to a generator whose rms output is 25 V, the current in the circuit is observed to be 0.40 A. What is the frequency of the source?
The frequency of the source is 1.66 Hz.
What is the impedance?Let us recall that the impedance of the circuit is the opposition that is offered to the flow of current by a circuit component that is not a resistor. Now let us find the impedance.
I = V/Z
I = current
V = voltage
Z = impedance
Z = V/I
Z = 25/0.4
Z = 62.5 ohm
Z^2 = R^2 + Xc^2
Z^2 = Xc^2
Xc= Z
Xc = 2πfC
f = frequency
C = capacitance
f= Xc/2πC
f = 62.5/2 * 3.142 * 6
f = 1.66 Hz
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What is the average velocity of a car that travels 48 km north in 2.0 h?
The average velocity of the car would be 24 kilometers per hour in the north direction if the car travels 48 kilometers in the north for 2 hours.
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, we have to find the average velocity of the car if the car travels 48 kilometers in the north for 2 hours.
The average velocity of the car = 48 kilometers / 2 hours
= 24 kilometers per hour
Thus, the average velocity of the car would be 24 kilometers per hour
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QUESTION 22As the rocket moves from position "b" to posisition "c", its speed is:constant.O continuously increasing.O continuously decreasing.increasing for a while and constant thereafter.constant for a while and decreasing thereafter.
The rocket movement due to the direction between a and b, and turning on the engine, movement is described by E image
Is continuously increasing, the force is constant and the acceleration too, which means that the speed continues increasing
A girl walks 600 m north and then 800 m east. What is the displacement from her starting point?
Displacement from her starting point is 529m
What is Displacement?
"Displacement" describes a change in an object's location. It is a vector quantity since it has a magnitude and a direction. It looks like an arrow that leads from the beginning to the end.
Given : A girl walks 600m north
She then walks 800m east
To Find : Displacement from her starting point
Solution: North
West East
South
North=600m
East =800m
We use Pythagoras Theorem
Displacement to be covered = [tex]\sqrt{800^{2} }[/tex]- [tex]\sqrt{600^{2} }[/tex]
529m
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The potential energy of a system can be changed by varying the position of objects in the system. At which point do the coaster cars have the most potential energy?
To find:
At which point the potential energy of the coaster car is the highest?
Explanation:
The potential energy of an object is the energy possessed by the object due to the position of the object. The gravitational potential energy of the object is directly proportional to the height of the object.
Thus an object will have the highest potential energy when it is at the highest point.
Final answer:
Thus the coaster cars will have the highest potential energy when it is at the highest point on the roller coaster.
Therefore, the coaster cars will have the most potential energy at point A.
The cable is drawn into the motor with an acceleration of 3 m/s2 .
Determine the time needed for the load at B to attain a speed of 10 m/s, starting from rest.
The time needed for the load at B to attain a speed of 10 m/s, starting from rest is 3.33 s
How to determine the timeAcceleraion is defined according to the following formula:
a = (v – u) / t
a is the acceleration v is the final velocity u is the initial velocity t is the timeWith the above formula, we can determine the time. Details below.
The following data were obtained from the question
Acceleration (a) = 3 m/s² Initial velocity (u) = 0 m/sFinal velocity (v) = 10 m/sTime (t) =?a = (v – u) / t
3 = (10 – 0) / t
3 = 10 / t
Cross multiply
3 × t = 10
Divide both sides by 3
t = 10 / 3
t = 3.33 s
Thus, we can say that the time required to attain the velocity of 10 m/s is 3.33 s
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Missing part:
See attached photo
The displacement (in meters) of a particle moving in a straight line is given by S= t^2 - 7t + 17 ii)iii)iv)
Given displacement of a particle moving in a straight line,
[tex]S=t^2-7t+17[/tex](i) Calculate the average velocity in the interval [3,4]
[tex]\begin{gathered} \text{Average velocity = }\frac{S(4)-S(3)}{4-3} \\ \text{Average velocity = }\frac{\lbrack(4)^2-7\times4+17\rbrack-\lbrack(3)^2-7\times3+17\rbrack}{4-3} \\ \text{Average velocity =}\frac{5-5}{1} \\ \text{Average velocity = 0 m/s} \end{gathered}[/tex](iii) Calculate the average velocity in the time interval [4,5].
[tex]\begin{gathered} \text{Average velocity = }\frac{S(5)-S(4)}{5-4} \\ \text{Average velocity = }\frac{\lbrack(5)^2-7\times5+17\rbrack-\lbrack(4)^2-7\times4+17\rbrack}{5-4} \\ \text{Average velocity = }\frac{7-5}{1} \\ \text{Average velocity = }2 \end{gathered}[/tex](iv) Calculate the average velocity in the time interval [4,4.5].
[tex]\begin{gathered} \text{Average velocity = }\frac{S(4.5)-S(4)}{4.5-4} \\ \text{Average velocity = }\frac{\lbrack(4.5)^2-7\times4.5+17\rbrack-\lbrack(4)^2-7\times4+17\rbrack}{4.5-4} \\ \text{Average velocity = }\frac{5.75-5}{0.5} \\ \text{Average velocity = }1.5 \end{gathered}[/tex]