Explanation:
B. Choose will (‘ll) or (be) going to, whichever is correct or more likely, and one of these verbs (7 marks- half each).
Get out of the building! It sounds like the generator isgoing toexplode.
Roman _____________ early before she reaches 65. She mentioned it at the meeting recently.
A: I think _____________ home across the park.
B: That’s a good idea.
Next year, no doubt, more people _____________ the competition as the prize money increases.
A: Can we meet at 10.00 outside the ILS building?
B: Okay. I_____________ you there.
Don’t sit on that bench, I_____________.
I’m not feeling well. In fact, I think I_____________.
‘Closed over the new year period. This office _____________ on 2nd January’ (a sign on an office window).
I’m sure you _____________ a good time staying with Merga.
We _____________ with Kedija tonight. She’s asked us to be there at 7:00.
‘The 2.35 to Arba Minch_____________ from gate 5’ (announcement at the airport).
I wouldn’t walk across the old bridge if I were you. It looks like it ____________.
I read in the paper that they _____________the price of the gas again.
Do you like my new solar watch? Here, I_____________you how it works.
A: AtoMessay isn’t in his office at the moment.
B: In that case, I_____________him at home.
Suppose a magnetic field exists with the north pole at the top of the computer monitor and the south pole at the bottom of the monitor screen. If a positively charged particle entered the field moving from your face to the other side of the monitor screen, which way would the path of the particle bend? Select one:a.leftb.rightc.upd.downe.none of the above
Given:
The direction of magnetic field is from the top to bottom.
The direction of current is into the screen.
To find the direction of particle.
Explanation:
According to Fleming's left hand rule,
First finger indicates the direction of magnetic field,
Central finger indicates the direction of current,
Thumb indicates the direction of motion.
On applying Fleming left hand rule, the direction of particle bend will be towards left.
Thus, the correct option is a
How do i solve this problem? Hint: The following are the three relevant equations:a = G M / r2a = v2/rT = 2πr/vEliminating a and v and solving for r gives:r = 3√(GMT2/(4π2))Be sure to plug in the period in seconds. This will provide the orbital radius from the center of the earth. Using this value and the radius of the earth, we can determine the required height above the earth's surface. Finally, we can use v = 2πr/T to find the orbital speed.
An engineer wants a satellite to orbit the Earth with a period of 48 hours.
The acceleration due to gravity at the surface of the Earth is given by
[tex]a=\frac{GM}{r^2^{}}[/tex]Where G is the gravitational constant (G = 6.67430×10⁻¹¹ Nm²/kg²), M is the mass of the Earth (m = 5.97x10²⁴ kg)
The acceleration of the satellite in a circular motion (centripetal acceleration) is given by
[tex]a=\frac{v^2}{r}[/tex]Where v is the speed of the satellite.
Equating both equations, we get
[tex]\begin{gathered} \frac{GM}{r^2}=\frac{v^2}{r} \\ \frac{GM}{r}=v^2 \end{gathered}[/tex]Substitute v = 2πr/T into the above equation
[tex]\begin{gathered} \frac{GM}{r}=(\frac{2\pi r}{T})^2 \\ \frac{GM}{r}=\frac{4\pi^2r^2}{T^2} \\ GM=\frac{4\pi^2r^2\cdot r}{T^2} \\ GM=\frac{4\pi^2r^3}{T^2} \\ r^3=\frac{GMT^2}{4\pi^2} \\ r=\sqrt[3]{\frac{GMT^2}{4\pi^2}} \end{gathered}[/tex]So, we have got the equation for radius r.
Let us first convert the period from hours to seconds
[tex]T=48\times60\times60=172800\; s[/tex]Substitute it into the above equation and find the radius (r).
[tex]\begin{gathered} r=\sqrt[3]{\frac{6.67430\times10^{-11}\cdot5.97\times10^{24}\cdot(172800)^2}{4\pi^2}} \\ r=67045443\; m \\ r=6.7045443\times10^7\; m \end{gathered}[/tex]Therefore, the satellite must orbit at an orbital radius of 67,045,443 m
(b) How far is this above the Earth's surface?
The required height above the earth's surface can be found by subtracting the radius of the earth from the orbital radius from the center of the earth that we calculated in the previous part.
The radius of Earth is 6.37×10⁶ m
[tex]\begin{gathered} h=6.7045443\times10^7-6.37\times10^6 \\ h=60675443\; m \\ h=6.0675443\times10^7\; m \end{gathered}[/tex]Therefore, the required height above the earth's surface is 60,675,443 m
(c) Speed of the satellite
The speed of the satellite is given by
[tex]\begin{gathered} v=\frac{2\pi r}{T} \\ v=\frac{2\cdot\pi\cdot67045443}{172800} \\ v=2,437.84\; \; \frac{m}{s} \end{gathered}[/tex]Therefore, the orbital speed of the satellite is 2,437.84 m/s
A rock tied to the end of string swings at a constant angular rate. If you are told thatthe string can support a total of 210 N of force before breaking, what is the maximumangular velocity the rock can rotate if the rock has a mass of 0.23 kg, and the length ofthe string is 0.35 m? Give your answer in units of radian per second.
Given data
The magnitude of the total force is F = 210 N
The mass of the rock is m = 0.23 kg
The length of the string is L = 0.35 m
The expression for the maximum angualr velocity is given as:
[tex]\begin{gathered} F=m\omega^2L \\ \omega=\sqrt{\frac{F}{mL}} \end{gathered}[/tex]Substitute the value in the above equation.
[tex]\begin{gathered} \omega=\sqrt[]{\frac{210\text{ N}}{0.23\text{ kg }\times\text{0.35 m}}} \\ \omega=51.07\text{ rad/s} \end{gathered}[/tex]Thus, the maximum angular velocity of the rock is 51.07 rad/s.
A car moves with a constant speed of 220 km/h. Express this speed in m/s and calculate the distance covered in 20 seconds
Hello..!
We first apply the data to the problem.
Data:
V = 220km/hT = 20sD = ?Now, we convert the "km/h" to "m/s".
Conversion:
220km/h • (1000m/1km) • (1h/3600s)V = 61.1m/sThen, we apply the formula that is.
Formula:
D = V • TFinally we develop the problem.
Developing:
D = (61.1m/s) • (20s)D = 1222mThe distance covered is 1222 meters.
Answer:
Formula:
D = v • tResolver:D = (61.1m/s) • (20s)D = 1222mHope this helped
A 8,707 newton car is initiallyat rest. How much force (inNewton) is required to movethe car by 16.73 meters, with afinal velocity of 5.84 m/s?
Given the weight of the car, W = 8707 N, and the car moves a distance, s= 16.73 m, and final velocity, v = 5.84 m/s
Let the mass of the car be m and acceleration due to gravity g = 9.8 m/s^2
Also, weight is given by the formula,
[tex]W=mg\text{ }[/tex]Then, the mass of the car will be
[tex]\begin{gathered} m=\frac{W}{g} \\ =\frac{8707}{9.8} \\ =888.46\text{ kg} \end{gathered}[/tex]The acceleration, a can be calculated by the formula
[tex]\begin{gathered} v^2-u^2=2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}[/tex]Here, u is the initial velocity, u=0.
[tex]\begin{gathered} a=\frac{(5.84)^2}{2\times16.73} \\ =\frac{34.10}{33.46} \\ =1.019m/s^2 \end{gathered}[/tex]The force will be
[tex]\begin{gathered} F=\text{ ma} \\ =888.46\times1.019 \\ =\text{ 905.34 N} \end{gathered}[/tex]Thus, the force is 905.34 N
On July 19, 1969, the lunar orbit of Apollo 11 was adjusted to an average height of 122 kilometers above the Moon's surface. The radius of the Moon is 1840 kilometers, and the mass of the Moon is 7.3 x 1022 kilograms. How long did it take to orbit once? Include units in your answer. Answer must be in 3 significant digits.
Given data:
* The mass of the Moon is,
[tex]m=7.3\times10^{22}\text{ kg}[/tex]* The radius of the Moon is,
[tex]\begin{gathered} R=1840\text{ km} \\ R=1840\times10^3\text{ m} \end{gathered}[/tex]* The height of the Apollo 11 is,
[tex]\begin{gathered} h=122\operatorname{km} \\ h=122\times10^3\text{ m} \end{gathered}[/tex]Solution:
The period of revolution of the Apollo 11 around the Moon is,
[tex]T=2\pi\sqrt[]{\frac{(R+h)^3}{Gm}}[/tex]where G is the gravitational constant,
Substituting the known values,
[tex]\begin{gathered} T=2\pi\sqrt[]{\frac{(1840\times10^3+122\times10^3)^3}{6.67\times10^{-11}\times7.3\times10^{22}}} \\ T=2\pi\times\sqrt[]{\frac{(1962\times10^3)^3}{48.69\times10^{11}}} \\ T=2\pi\times1245.46 \end{gathered}[/tex]Thus, the value of time period is,
[tex]\begin{gathered} T=7825.46\text{ s} \\ T=\frac{7825.46}{60\times60}\text{ hr} \\ T=2.17\text{ hr} \end{gathered}[/tex]Thus, Apollo 11 takes 2.17 hours to complete orbit once around the Moon.
Two pieces of window glass are separated by a distance, d. If a beam of light of wavelength l=469 nm passes through the first piece of glass. What is the minimum distance, in nm, such that the light intensity transmitted through the right is a maximum?
The distance needs to be optimized to cause constructive interference, which can be seen on the following drawing:
So the distance needs to be exactly of a single wavelength.
Then, our final answer is d=469nm
An astronaut floating at rest in space has run out of fuel in her jetpack. Sherealizes that throwing tools from her toolkit in the opposite direction will helppropel her back toward the space station. If the astronaut has a mass of 91kg and she throws a hammer of mass 4 kg at a speed of 3.5 m/s, what will bethe approximate resultant velocity that carries her back to the space station?Astronautmass01 kg-X-Hammermass4 kgoOA. 0.10 m/sB. 0.15 m/sO C. 0.34 m/sOD. 0.05 m/s
ANSWER
B. 0.15 m/s
EXPLANATION
If we consider the system astronaut-hammer as an islotated system and we apply the law of conservation of momentum we have:
[tex]p=0=m_hv_h+m_Av_A[/tex]Where mh is the mass of the hammer, mA is the mass of the astronaut, vh is the hammer's final velocity and vA is the astronaut's final velocity.
In this system the velocities will be constant, because there are no other forces acting and the initial velocity for both objects is zero (because they are at rest).
Let's solve the equation above for vA:
[tex]v_A=\frac{-m_hv_h}{m_A}[/tex]The astronaut's velocity is:
[tex]v_A=\frac{-4kg\cdot3.5m/s}{91\operatorname{kg}}\approx-0.15m/s[/tex]The minus sign indicates that the astronaut is moving in the opposite direction of the hammer's motion. Therefore the correct answer is option B. 0.15m/s
In the image below, a worker is pushing a crate with a mass of 20 kg up aramp at a constant rate. Ignoring friction, how much force must the workerapply so that the crate continues to move at the same speed? (Recall that g =9.8 m/s2WeightO A. 37.3 NOB. 62.5 NO C. 50.7NO D. 48.6N
Free diagram of the crate:
The free body equation for the crate is given as,
[tex]F=mg\sin \theta[/tex]Here, F is the force applied, m is the mass of the crate, g is the acceleration due to gravity and θ is the angle of inclination.
Substituting all known values,
[tex]\begin{gathered} F=20\text{ kg}\times9.8\text{ m/s}^2\times\sin (15^{\circ}) \\ =50.7\text{ N} \end{gathered}[/tex]T
Draw the following components of a circuitCellBatteryLight Bulb Fuse
ANSWER and EXPLANATION
We want to draw the circuit diagram of the given components.
The fuse will be placed in between the battery (and cell) and the load
To do this, use the appropriate symbols as shown below:
That is the answer.
A water molecule consists of an oxygen atom with two hydrogen atoms bound to it. The angle between the two bonds is 106. If the bonds are 0.10nm long, where is the center of mass of the molecule?
The diagram representing this scenario is shown below
From the information given,
Number of molecules in hydrogen, NH = 1
Number of molecules in oxygen, NO = 15.99
y = 0.1cos53
y = 0.06 nm
x = 0.1Sin53
x = 0.08 nm
r1 = xi + yj = 0.08i + 0.06j
r2 = = 0.08i - 0.06j
r3 = 0
rcm = 2miri/2mi = (m1r1 + m2r2 + m3r3)/(m1 + m2 + m3)
r = 1(0.06i + 0.08j + 0.06i - 0.08j)/2(1 + 1 + 15.99)
r = (6.7 x 10^-3i) nm
the center of mass of the molecule = (6.7 x 10^-3i) nm
What is force acting on an object having a mass of 560 g on Earth?
Given,
[tex]{ \blue{ \tt{m = 560g}}}[/tex]
[tex]{ \blue{ \tt{g = 9.8 m/s²}}}[/tex]
[tex]{ \blue{ \tt{W = ?}}}[/tex]
[tex]{ \purple{ \tt{W = mg}}}[/tex]
[tex]{ \purple{ \sf{W= 560 × 9.8}}}[/tex]
[tex]{ \boxed{ \red{ \sf{W = 5488N}}}}[/tex]
Rashad and Carlos measure the length, width and height of a textbook at 6 cm, 4 cm, and 3respectively. What is the volume of the textbook?
Answer:
72 cm³
Explanation:
The volume of a textbook can be calculated as
Volume = Length x Width x Height
Replacing the length by 6 cm, the width by 4 cm, and the height by 3 cm, we get:
Volume = 6 cm x 4 cm x 3 cm
Volume = 72 cm³
Therefore, the volume of the textbook is 72 cm³
Jason is pulling a box across the room. He is pulling with a force of 16 newtons and his arm is making a 71 angle with the horizontal, if the box weighs 15 newtons what is the netforce on the box in the vertical direction? Treat up as the positive direction, and down as the negative direction
We know that
• The force is 16 Newtons.
,• The angle of application is 71 degrees.
,• The box weighs 15 Newtons.
First, let's make a free-body diagram.
As you can observe, the vertical vectors are the weight of the box and the vertical component of the applied force.
[tex]\Sigma F_y=F_y-W[/tex]Where each force is defined as follows
[tex]\begin{gathered} F_y=F\cdot\sin \theta \\ W=mg \end{gathered}[/tex]Let's use the expressions above to find the net vertical force.
[tex]\begin{gathered} \Sigma F_y=F\cdot\sin \theta-mg=16N\cdot\sin 71-15 \\ \Sigma F_y\approx0.13N \end{gathered}[/tex]Therefore, the net vertical force is 0.13 Newtons.B) what is the frequency of the sound wave;i.e, the tuning fork? C) the water continues to leak out the bottom of the tube When the tube next resonates with tuning fork, what is length of the air column?
Answer:
60 cm
Explanation:
The wavelength of the sound wave in the air column closed at one end is given by
[tex]\lambda=\frac{4}{n}L[/tex]where n = 1, 3, 5, 7,9, 11, 13, 15 are all odd numbers.
Now in our case, we know that the length of the pipe is 225 or 2.25 m. The 8th resonance corresponds to n = 15; therefore, the wavelength of the sound wave is
[tex]\lambda=\frac{4}{15}\times225[/tex][tex]\boxed{\lambda=60\; cm\text{.}}[/tex]Hence, the wavelength of the sound wave is 60 cm.
2. A student claims that the normal force acting on the cart is equal in magnitude to the weight of the cart.Is the student correct?A. Yes, the student is correct because the normal force is directly proportional to the mass of the cart,as described in Newton's Second Law of Motion.B. No, the student is incorrect because the normal force is not the equal and opposite reaction to thecart's weight being applied on a surface, as described in Newton's Third Law of Motion.C. Yes, the student is correct because the normal force is the equal and opposite reaction to thecart's weight being applied on a surface, as described in Newton's Third Law of Motion.D. No, the student is incorrect because the normal force is inversely proportional to the mass of the cart,as described in Newton's Second Law of Motion.
Explanation:
The first law of Newton's says that the velocity of an object will remain unless there is a force acting on the object, the second law of newton says that the force is proportional to the mass and the acceleration, and the third law of newton says that there is always an equal an opposite forces acting on two objects that interact.
So, the normal force can be described as a result of the third law of Newton. It means that it is equal in magnitude to the weight of the cart.
Therefore, the answer is:
C. Yes, the student is correct because the normal force is the equal and opposite reaction to the cart's weight being applied on a surface, as described in Newton's Third Law of Motion.
A small moon is in a circular orbit around an exoplanet. It orbits at an altitude of 700 km. This exoplanet is the same size as Earth, but is only 1/2 as dense.Does this exoplanet exert a torque on its moon? Why or why not?No, because the force exerted by the planet on the meteoroid has a negligible magnitude.Yes, because the meteoroid's direction of motion is constantly changing.Yes, because the planet exerts a centripetal force on the meteoroid.No, because the planet exerts a force on the meteoroid parallel to its position vector relative to the center of mass of the planet.
According to kepler's 2nd law a planet sweeps equal area in equal time around a planet.
here the angular momentum of planet is constant
if angular momentum is constant then according to conservation of angular momentum the net torque on the moon due to the exoplanet will be equal to 0.
So here the net torque = 0.
That's why last option is correct.
Suppose a force F, = (1/3)x acts on anobject. Assume at x = 0 the force pushesthe object and it starts moving to the right.What happens to the magnitude of the workdone as the object moves to the right?A. More work is done B. Less work is doneC. A constant amount of work is doneD. It is impossible to predict
We are given that a variable force acts on an object given by the function:
[tex]F_x=\frac{1}{3}x[/tex]This force will increase as the object moves since the value of "x" will increase. The work done is defined as:
[tex]W=Fd[/tex]Where "F" is the force and "d" is the total distance. This magnitude will also increase therefore, more work is done.
Given that the charge on an electron is -1.6 x 10^-19 Coulombs, what is the magnitude of the electric force on an electron in Newtons when it is placed in an electric field of strength 12,000 Volts/m? A)1.6 x 10^-19 B)3.2 x 10^-19 C)1.92 x 10^-15 D)6.4 x 10^-19 E)4.5 x 10^22
Given,
The charge on an electron is
[tex]q=-1.6\times10^{-19}\text{ C}[/tex]The electric field strength is
[tex]E\text{ = 12,000 Volts/m}[/tex]Solution:
The electric force is calculated by the formula,
[tex]F=qE[/tex]Put the given values in the formula.
[tex]\begin{gathered} F=(1.6\times10^{-19}\text{ C\rparen}\times(12,000\text{ Volts/m\rparen} \\ F=19.2\times10^{-16}\text{ N} \\ F=1.92\times10^{-15}\text{ N} \end{gathered}[/tex]Thus, the given options c is correct.
Please do number three for me step-by-step and explain the east and west thank you
the Given that the distance from mall to home is 1000 m.
The distance from home to the library is 1200 m.
So the total displacement is 1000 +1200 = 2200 m.
The displacement is from the library to the mall, so the displacement is towards the west.
So the displacement is 2200 m towards west.
The standard exam page is 8.50 inches by 11.0 inches. Its area in cm 2 is?
We know that one inch is equal to 2.54 cm, this means that:
[tex]\begin{gathered} 8.5\text{ in=}21.59\text{ cm} \\ \text{and} \\ 11\text{ in=27.94 cm} \end{gathered}[/tex]To obtain the area of the page we multiply the width by the length then:
[tex](21.59)(27.94)=603.22[/tex]Therefore the area of the page is 603.22 squared cm
A graduated cylinder measures ________, usually in units of ___________.
Answer:
A graduated cylinder measures in units of volume, usually in units of milliliters.
A student is sitting in a chair. The student's mass is 55 kg. Find the normal force exerted on the student by thechair, in N.Round your answer to one decimal place
We have the next diagram
Where N is the normal and W is the weight
Therefore
N=W
The formula to calculate the W is
[tex]W=mg[/tex]where m is the mass and g is the gravity
In our case,
m=55
g=9.8 m/s^2
[tex]W=55(9.8)=539N[/tex]Therefore the normal force
N=539N
ANSWER
the normal force is 539N
Bob drops a rock from the roof of a building (5m tall) onto Joe's head (1.5m tall). The mass of the rock is 0.5kg.a) What is the speed of the rock as it his Joe's head?b) If the rock is only travelling 8.0 m/s, how much work was done by air resistance?
Given:
The initial height of the rock was,
[tex]h_i=5\text{ m}[/tex]The final height of the rock is,
[tex]h_f=1.5\text{ m}[/tex]The mass of the rock is,
[tex]m=0.5\text{ kg}[/tex]The travelling speed of the rock is,
[tex]v^{\prime}=8.0\text{ m/s}[/tex]To find:
a) The speed of the rock at Joe's head
b) how much work was done by air resistance
Explanation:
The displacement of the rock is,
[tex]\begin{gathered} h=h_i-h_f \\ =5-1.5 \\ =3.5\text{ m} \end{gathered}[/tex]The final speed at Joe's head is,
[tex]\begin{gathered} v=\sqrt{u^2+2gh} \\ =\sqrt{0^2+2\times9.8\times3.5} \\ =\sqrt{68.6} \\ =8.28\text{ m/s} \end{gathered}[/tex]Hence, the speed of the rock at Joe's head is 8.28 m/s.
b)
The speed at Joe's head was 8.0 m/s, and the loss of kinetic energy is,
[tex]\begin{gathered} \frac{1}{2}\times m[(8.23)^2-(8.0)^2] \\ =\frac{1}{2}\times0.5\times[3.73] \\ =0.93\text{ J} \end{gathered}[/tex]This loss of energy is the work done by the air resistance.
Hence, the work done by the air resistance is 0.93 J.
If Adam rides his bicycle in a straight line for 21 min with an average velocity of 9.98 km/h south how far has he ridden
Given data:
* The time taken by the Adam is,
[tex]\begin{gathered} t=21\text{ min} \\ t=1260\text{ s} \end{gathered}[/tex]* The average velocity of the Adam is,
[tex]\begin{gathered} v=9.98kmh^{-1} \\ v=9.98\times\frac{10^3}{60\times60}ms^{-1} \\ v=2.77ms^{-1} \end{gathered}[/tex]Solution:
The distance traveled by the Adam is,
[tex]\begin{gathered} d=vt \\ d=2.77\times1260 \\ d=3490.2\text{ m} \\ d=3.49\text{ km} \end{gathered}[/tex]Thus, the distance traveled by the Adam on the bicycle is 3.49 km.
A box of mass m = 2 kg is kicked on a rough horizontal plane with an initial velocity v_0 . If the net work done on the crate during its entire motion , until it comes to rest , is -36 J , then the initial velocity v_0 of the box is equal to :
Given:
The mass of the box is m = 2 kg
The work done is W = -36 J
The final velocity of the object is
[tex]v_f=\text{ 0 m/s}[/tex]To find the initial velocity of the box.
Explanation:
The initial velocity of the box can be calculated as
[tex]\begin{gathered} Work\text{ done = change in kinetic energy} \\ W=\frac{1}{2}m(v_f^2-v_o^2) \\ -36=\frac{1}{2}\times2(0^2-v_o^2) \\ -v_{_0}^2=-36 \\ v_o=6\text{ m/s} \end{gathered}[/tex]Hence, the initial velocity of the box is 6 m/s.
A coyote was holding a large bird cage a few meters above the ground. He lowers it at a slow constant speed onto the X he has drawn on the ground. Which statement best describes the change in the total mechanical energy of the Earth-bird cage system?Question 5 options:The total mechanical energy decreases, because the coyote does negative work on the bird cage by exerting a force in the direction opposite to its displacement.The total mechanical energy is unchanged, because there is no change in the bird cage’s kinetic energy as it is lowered to the table.The total mechanical energy is unchanged, because no work is done on the Earth-bird cage system while the book is lowered.The total mechanical energy decreases, because the coyote does positive work on the bird cage by exerting a force that opposes the gravitational force.
Total mechanical energy is conserved in a system.
Thus, the total mechanical energy is unchanged, because no work is done on the Earth-bird cage system while the book is lowered.
A blue car weighing 1,302 kg is accelerating forward at a rate of 4 m/s². What is the forward force of the car?
Answer:
5208N
Explanation:
force=mass×acceleration
mass(kg)=1,302
acceleration(m/s²)=4
f= m × a
f= 1302 × 4
f=5208N
You are wanting to walk on an icy road without falling over. You have a ruler, some cotton wool and some salt and vinegar crisps. Devise a plan to increase the friction so that you can walk safely.
Using the concept of friction, we got salt > vinegar crisps >cotton wool > ruler is the order of friction which they provide.
We know very well that the frictional force is directly proportional to the surface of the area and the surface of the icy road is very smooth. So the contact area of man's foot with the surface of the icy road is very small and the fraction between man's foot and the surface of the icy road is also small so this is the reason why man can not walk on the icy road properly and safely.
So in order to walk safely on icy roads we have to increase friction so in order to increase friction we have to first use salt then vinegar crisps then cotton wool and then a ruler.
Hence the order of friction which they provide is salt > vinegar crisps >cotton wool > ruler
To Know more about friction, visit here:
https://brainly.com/question/13000653
#SPJ9
Two skydivers jump from an airplane at an altitude of 5000m. Suppose one is 57 kg and the other is 68 kg. Using the data given in the example in class to find the amount of time each takes to get to the ground.Area = 0.18 m² air density = 1.21 kg/m³ drag coefficient C = .070
Given:
Two skydivers jump from an airplane at an altitude of: d = 5000 m
The mass of the first skydiver is: m1 = 57 kg
The mass of the second skydiver is: m2 = 68 kg
Area = 0.18 m²
Air density = 1.21 kg/m³
Drag Coefficient: C = 0.070
To find:
The amount of time each skydiver takes to get to the ground.
Explanation:
The magnitude of drag force which acts opposite in the opposite direction is equal to the weight of the skydiver. Thus, the magnitude of the drag force can be calculated as:
[tex]\begin{gathered} F_1=m_1g=57\text{ kg}\times9.8\text{ m/s}^2=558.6\text{ kg.m/s}^2 \\ \\ F_2=m_2g=68\text{ kg}\times9.8\text{ m/s}^2=666.4\text{ kg.m/s}^2 \end{gathered}[/tex]Here, F1 is the magnitude of the drag force on the first skydiver and F2 is the drag force on the second skydiver.
The expression for drag force relating to the velocity is given as:
[tex]F=\frac{1}{2}C\rho Av^2[/tex]For the first skydiver the drag force is given as:
[tex]\begin{gathered} F_1=\frac{1}{2}C\rho Av_1^2 \\ \\ \text{ Substituting the values in the above equation, we get:} \\ \\ 558.6\text{ kg.m/s}^2=\frac{1}{2}\times0.070\times1.21\text{ kg/m}^3\times0.18\text{ m}^2\times v_1^2 \\ \\ 558.6\text{ kg.m/s}^2=7.623\times10^{-3}\text{ kg/m}\times v_1^2 \\ \\ v_1^2=\frac{558.6\text{ kg.m/s}^2}{7.623\times10^{-3}\text{ kg/m}} \\ \\ v_1^2=73278.24\text{ m}^2\text{/s}^2 \\ \\ v_1=\sqrt{73278.24\text{ m}^2\text{/s}^2} \\ \\ v_1=270.70\text{ m/s} \end{gathered}[/tex]The velocity v1 of the first skydiver is 270.70 m/s.
The time t1 taken by the first skydiver to get to the ground can be calculated as:
[tex]\begin{gathered} v_1=\frac{d}{t_1} \\ \\ t_1=\frac{d}{v_1} \\ \\ \text{ Substituting the values in the above equation, we get:} \\ \\ t_1=\frac{5000\text{ m}}{270.70\text{ m/s}} \\ \\ t_1=18.47\text{ s} \end{gathered}[/tex]The first skydiver takes 18.47 seconds to get to the ground.
For the second skydiver, the drag force is given as:
[tex]\begin{gathered} F_2=\frac{1}{2}C\rho Av_2^2 \\ \\ \text{ Substituting the values in the above equation, we get:} \\ \\ 666.4\text{ kg.m/s}^2=\frac{1}{2}\times0.070\times1.21\text{ kg/m}^3\times0.18\text{ m}^2\times v_2^2 \\ \\ 666.4\text{ kg.m/s}^2=7.623\times10^{-3}\text{ kg/m}\times v_2^2 \\ \\ v_2^2_{{}}=\frac{666.4\text{ kg.m/s}^2}{7.623\times10^{-3}\text{ kg/m}} \\ \\ v_2^2=87419.65\text{ m}^2\text{/s}^2 \\ \\ v_2=\sqrt{87419.65\text{ m}^2\text{/s}^2} \\ \\ v_2=295.67\text{ m/s} \end{gathered}[/tex]The velocity v2 of the second skydiver is 259.67 m/s.
The time t2 taken by the second skydiver to get to the ground can be calculated as:
[tex]\begin{gathered} v_2=\frac{d}{t_2} \\ \\ t_2=\frac{d}{v_2} \\ \\ \text{ Substituting the values in the above equation, we get:} \\ \\ t_2=\frac{5000\text{ m}}{295.67\text{ m/s}} \\ \\ t_2=16.91\text{ s} \end{gathered}[/tex]The second skydiver takes 16.91 seconds to get to the ground.
Final answer:
The first skydiver takes 18.47 seconds to get to the ground.
The second skydiver takes 16.91 seconds to get to the ground.