Answers
We are given the following vector forces:
[tex]\begin{gathered} F_1=(-14,4) \\ F_2=(-6,1) \end{gathered}[/tex]These vector forces are expressed as:
[tex]F=(x,y)[/tex]Where the first component "x" is the horizontal component, and "y" is the vertical component. Since we are asked about the net force in the vertical component we need to add the components "y" of both vectors, we get:
[tex]\Sigma F_y=4+1=5[/tex]Therefore, the net force in the vertical direction is 5.
Related Questions
Except for the nodes on a standing wave, what is the frequency f of the points executing simple harmonic motion?
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Take into account that in a standing wave, the frequency f of the points executing simple harmonic motion, is simply a multiple of the fundamental harmonic fo, that is:
f = n·fo
where n is an integer and fo is the first harmonic or fundamental.
fo is given by the length L of a string, in the following way:
fo = v/λ = v/(L/2) = 2v/L
becasue in the fundamental harmonic, the length of th string coincides with one hal of the wavelength of the wave.
As shown in Fig. B9, a block of mass m2=2.1 kg sits on an inclined plane with angle 35 degrees .The coefficients of static and kinetic friction between the block and the plane are unknown.The block is attached to a rope, which is hung over a frictionless and massless pulley and attached to a hanging mass m1.(a) The block my is on the verge of sliding, when the hanging mass is increased to m1=1.6kg.01. find the coefficient of static friction.As shown in Fig. Bob, an identical block (with mass m2) is placed on top of the block on the inclined plane. The hanging mass m1 remains unchanged. The two blocks on the inclined plane are sliding down the incline with the same acceleration a=0.31 m/s^2Find02.the coefficient of kinetic friction.03.what is the minimum value of the coefficient of static friction between the two identical blocks (m2) so that they move together?
Answers
Question 1.
The free body diagrams of the situation are shown below:
The forces applied on m1 are in only the y direction then we only have one equation of motion to get it we use Newton's Second law. Then we have:
[tex]T-W_1=m_1a[/tex]The forces applied on m2 are in the x and y direction, this means that we have two equations of motion. We don't expect this block to move on the y-direction (which we defined); this means that the acceleration in this direction has to be zero. Furthermore, since the blocks m1 and m2 are attach thorugh the rope their acceleration has to be the same; this means that the acceleration of block m2 in the x direction has to be a. Applying this and Newton's second law we have for block 2:
[tex]\begin{gathered} (W_2)_x+f_f-T=m_2a \\ N-(W_2)_y=0 \end{gathered}[/tex]where (W2)x and (W2)y denote the component of the weight in the x and y direction, respectively. Now, since block m2 is on the verge of sliding this means that the system is in equilibrium, that is, the acceleration of the system is equal to zero. Then we have that the equations above take the form:
[tex]\begin{gathered} T-W_1=0 \\ (W_2)_x+f_f-T=0 \\ N-(W_2)_y=0 \end{gathered}[/tex]To determine the coefficient of static friction we firs need to determine the force of fricction. To do this we first solve the first equation for T:
[tex]T=W_1[/tex]Now we plug this value in the second equation and solve for the force of friction:
[tex]\begin{gathered} (W_2)_x+f_f-W_1=0 \\ f_f=W_1-(W_2)_x \end{gathered}[/tex]Now we need to remember that the force of friction can be obtain by:
[tex]f_f=\mu_{}N[/tex]where mu is the coefficient of friction (this rule applies for static and kinetic friction). Then we have:
[tex]\begin{gathered} \mu N=W_1-(W_2)_x_{} \\ \mu=\frac{W_1-(W_2)_x}{N} \end{gathered}[/tex]To find the normal force we use the third equation of motion:
[tex]N=(W_2)_y[/tex]Hence we have:
[tex]\mu=\frac{W_1-(W_2)_x}{(W_2)_y}[/tex]Now we need to remember that the components of the weight on an inclined plane are given by:
[tex]\begin{gathered} W_x=W\sin \theta \\ W_y=W\cos \theta \end{gathered}[/tex]where theta is the angle of the plane.
Plugging this on the expression for the coefficient of friction:
[tex]\mu=\frac{W_1-W_2\sin \theta}{W_2\cos \theta}[/tex]Finally we plug the values given to find the coefficient:
[tex]\begin{gathered} \mu=\frac{(1.6)(9.8)-(2.1)(9.8)\sin 35}{(2.1)(9.8)\cos 35} \\ \mu=0.23 \end{gathered}[/tex]Therefore the coefficient of static friction is 0.23
Question 2
For this question we are going to look at the two block in the inclined plane as one block with mass of 4.2 kg and we are going to name the two blocks as m2.
The free body diagram in this case is:
Applying Newton's second law in each mass lead to the system of equations:
[tex]\begin{gathered} T-W_1=m_1a \\ (W_2)_x-f_f-T=m_2a \\ N-(W_2)_y=0 \end{gathered}[/tex]From the first equation of motion we have:
[tex]T=W_1+m_1a[/tex]Plugging this in the second equation:
[tex](W_2)_x-f_f-W_1-m_1a=m_2a[/tex]but we know that:
[tex]f_f=\mu_{}N[/tex]and:
[tex]N=(W_2)_y[/tex]Then we have:
[tex](W_2)_x-\mu(W_2)_y-W_1-m_1a=m_2a[/tex]Solving for the coefficient and using the expression for the wight on an inclined plane we have:
[tex]\begin{gathered} (W_2)_x-\mu(W_2)_y-W_1-m_1a=m_2a \\ \mu=\frac{W_2\sin \theta-W_1-m_1a-m_2a}{W_2\cos \theta} \end{gathered}[/tex]Plugging the values given we have:
[tex]\begin{gathered} \mu=\frac{(4.2)(9.8)\sin 35-(1.6)(9.8)-(1.6)(0.31)-(4.2)(0.31)}{(4.2)(9.8)\cos 35} \\ \mu=0.18 \end{gathered}[/tex]Therefore the coefficient of kinetic friction is 0.18.
Question 3.
To find the coefficient of static friction between the two blocks on the inclined plane we need that the upper block to be stationary with respect to the lower block, that means that the friction has to be equal to the weight in the x-direction, that is:
[tex]\begin{gathered} f_f=W_2\sin 35 \\ \mu W_2\cos 35=W_2\sin 35 \\ \mu=\frac{\sin 35}{\cos 35} \\ \mu=0.7 \end{gathered}[/tex]Therefore the coefficient between the blocks has to be at least 0.7
A force F1 of magnitude 4.80 units acts on an object at the origin in a direction = 27.0° above the positive x-axis. (See the figure below.) A second force F2 of magnitude 5.00 units acts on the object in the direction of the positive y-axis. Find graphically the magnitude and direction of the resultant force F1 + F2.magnitude unitsdirection ° counterclockwise from the +x-axis
Answers
We will determine the magnitude of the final component as follows:
[tex]\begin{gathered} m=\sqrt{(5cos(27))^2+(5sin(27)+4.8)^2}\Rightarrow m=8.356527029... \\ \\ \Rightarrow m\approx8.36 \end{gathered}[/tex]So, the magnitude is approximately 8.36.
And the direction will be approximately 57.8° counter-clockwise.
This can be seeing as follows:
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.
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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
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
Answers
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
Answers
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.
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
Answers
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
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.
Answers
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.
Answers
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.
A helicopter flies at a constant altitude towing an airborne 65 kg crate as shown in the diagram. The helicopter and the crate only move in the horizontal direction and have an acceleration of 3.0 m/s21) find the vertical component of the tension in the cable (in Newtons). Ignore the effects of air resistance.2) find the magnitude of the tension in the cable (in Newtons). Ignore the effects of air resistance.3) find the angle(with respect to the horizontal) of the tension in the cable (in degrees). Ignore the effects of air resistance.
Answers
Given data:
* The weight of the crate is 65 kg.
* The acceleration of crate and helicopter is,
[tex]a=3ms^{-2}[/tex]Solution:
(1). The forces acting on the crate is represented as,
The y-component of tension is balancing the weight of the crate.
Thus, the y-component (vertical) of tension is,
[tex]\begin{gathered} T_y=W \\ T_y=mg \end{gathered}[/tex]Where m is the mass of crate and g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} T_y=65\times9.8 \\ T_y=637\text{ N} \end{gathered}[/tex]Thus, the vertical component of the tesnion is 637 N.
(2). The x-component of tension in the cable is,
[tex]\begin{gathered} T_x=ma \\ T_x=65\times3 \\ T_x=195\text{ N} \end{gathered}[/tex]Thus, the tension in the cable is,
[tex]\begin{gathered} T=\sqrt[]{T^2_x+T^2_y} \\ T=\sqrt[]{195^2^{}+637^2} \\ T=666.2\text{ N} \end{gathered}[/tex]Hence, the tesnion in the cable is 666.2 N.
(3). The angle of tension with the horizontal is,
[tex]\begin{gathered} \tan (\theta)=\frac{T_y}{T_x} \\ \tan (\theta)=\frac{637}{195} \\ \tan (\theta)=3.3 \\ \theta=73.14^{\circ} \end{gathered}[/tex]Thus, the angle made by the tension with the horizontal is 73.14 degree.
Derive the equation S=ut+½at²
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Average velocity = (initial velocity + final velocity )/ 2
AVG v = (u+v)/2
Also,
Distance = avg v x time
s= ([u+v]/2) x t
s= distance travelled by a body in time t
u= initial velocity
a= acceleration
From the equation of motion:
v= u +at
Replacing
s= ( [u+u+at]/2)xt
s= [( 2u+ 2at ( x t ]/2
s= (2ut+at^2)/2
s= ut+1/2at^2
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.
Answers
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.
The standard exam page is 8.50 inches by 11.0 inches. Its area in cm 2 is?
Answers
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
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.
Answers
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.
If a sloth is traveling at 0.067 mps or 0.15 mph how long does it take the sloth to travel an 11.5 meter tree
Answers
Answer:
Explanation:
Given:
V = 0.067 m/s
D = 11.5 m
___________
t - ?
t = D / V
t = 11.5 / 0.067 ≈ 172 s or t ≈ 2.87 min or t ≈ 2 min 52 s
Please do number three for me step-by-step and explain the east and west thank you
Answers
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.
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?
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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
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
Answers
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
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
Answers
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.
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
Answers
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 graduated cylinder measures ________, usually in units of ___________.
Answers
Answer:
A graduated cylinder measures in units of volume, usually in units of milliliters.
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?
Answers
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
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
Answers
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.Select the correct answerA car moves with an average speed of 45 miles/hour. How long does the car take to travel 90 miles?O A2 minutesВ.2 hoursOC.45 minutesΟ Ο ΟD45 hoursE90 minutesResetNext
Answers
Given data:
The average speed of the car is
[tex]S=45\text{ miles/hour}[/tex]The distance traveled by the car is
[tex]D=90\text{ miles}[/tex]The average speed can be expressed as,
[tex]\begin{gathered} S=\frac{D}{T} \\ T=\frac{D}{S} \end{gathered}[/tex]Here, T is the time taken by the car.
Substituting the values in the above equation, we get:
[tex]\begin{gathered} T=\frac{90\text{ miles}}{45\text{ miles/hour}} \\ =2\text{ hours} \end{gathered}[/tex]Thus, the time taken by the car is 2 hours, and option (B) is correct.
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 :
Answers
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.
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?
Answers
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
What is the current produced by a potential difference of 240 Volts through a resistance of .28 ohms
Answers
Given:
The potential difference is V = 240 Volts
The resistance is R = 0.28 Ohms
To find the current produced.
Explanation:
The current produced can be calculated by the formula
[tex]I=\text{ }\frac{V}{R}[/tex]On substituting the values, the current produced will be
[tex]\begin{gathered} I=\frac{240}{0.28} \\ =\text{ 857.14 A} \end{gathered}[/tex]Thus, the current produced is 857.14 A
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
Answers
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
A spring block system undergoes a simple harmonic oscillation with an amplitude A = This oscillations in one minute. The maximum acceleration of this oscillator is:
Answers
The maximum acceleration of the oscillator is 1.48 m/s²
Simple harmonic motion is a particular kind of periodic motion in which the restoring force on the moving item is inversely proportional to the size of the displacement and acts in the direction of the object's equilibrium position.
A periodic variable's amplitude is a gauge of its change over a single period.
The amplitude of a simple harmonic motion, A = 15 cm = 0.15 m
The frequency, f = 30 oscillations per minute = 30/60 = 1/2
The maximum acceleration of the oscillator is given as:
a = A × ω²
Now the formula for angular frequency is:
ω = 2πf
Therefore,
a = A × ( 2πf )²
Substituting the values in the equation,
a = 0.15 × ( 2 × π × 1/2 )²
a = 0.15 × π²
a = 1.48 m/s²
Learn more about acceleration here:
brainly.com/question/460763
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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
Answers
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