The drag force on a car traveling at 15 m/s. assume that the cars' cross-sectional area is 3.5 m^2, and that air has a density of 1.2 kg/m^3 will be 472.5 C(d) Newton
Drag force = 1/2 * density of fluid * [tex](speed)^{2}[/tex] * drag coefficient * cross sectional area
= 1/2 * 1.2 * [tex]15^{2}[/tex] * C (d) * 3.5
= 472.5 C(d) Newton
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A company is constructing a building for a recordings tudio. Which of the following would be the LEAST HELPFUL in reducing road noise in the studio?
The lest helpful solution is reduncing the number of trees on the property. Therefore, the answer is A.
This comes from the fact that all the other option will better absorb sound waves on the building itself.
The belief that for every particle there is a force particle is referred to in modern physics as which of the following?Select one:a.unification theoryb.string theoryc.supersymmetryd.quantum tunneling
We will have that the one would be supersymmetry.
Question 1: A skateboarder is traveling at 2 m/s when she reaches a downwardincline and accelerates at 3 m/s/s for 4 seconds. What speed is she traveling atthe end of those 4 seconds?O 14 m/s6 m/sI do not know how to solve this question.24 m/sO 10 m/s
Given:
The initial velocity of the skateboarder is
[tex]v_i=\text{ 2 m/s}[/tex]The acceleration is
[tex]a=\text{ 3 m/s}^2[/tex]The time taken is t = 4 s
Required: The final speed of the skateboarder.
Explanation:
The final velocity can be calculated by the formula,
[tex]v_f=v_i+at[/tex]On substituting the values, the final velocity will be
[tex]\begin{gathered} v_f=2+(3\times4) \\ =14\text{ m/s} \end{gathered}[/tex]Final Answer: The final velocity of the skateboarder after 4 s is 14 m/s.
I’m not sure how to even begin this problem. I know it’s an equilibrium problem
m= 27 kg
angle = 40°
mg = 27 x 9.8 = 264.6 N
Fty . x1 - mg (x1/2 ) = 0
Fty =
3. The mass of an electron is 9.1x10^-31 kg. The mass of a proton is 1.7x10^-27 kg.The gravitational force between them in the hydrogen atom is 1.0x10^-47 N.What is the separation distance between them?
Given:
The mass of the electron is,
[tex]m_e=9.1\times10^{-31}\text{ kg}[/tex]The mass of the proton is,
[tex]m_p=1.7\times10^{-27}\text{ kg}[/tex]The gravitational force between the electron and the proton in the hydrogen atom is,
[tex]1.0\times10^{-47}\text{ N}[/tex]To find:
the separation between them
Explanation:
The gravitational force between two masses is,
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]Here, the universal gravitational constant is,
[tex]G=6.67\times10^{-11}\text{ Nm}^2kg^{-2}[/tex]Substituting the values we get,
[tex]\begin{gathered} 1.0\times10^{-47}=\frac{6.67\times10^{-11}\times9.1\times10^{-31}\times1.7\times10^{-27}}{r^2} \\ r^2=\frac{6.67\times10^{-11}\times9.1\times10^{-31}\times1.7\times10^{-27}}{1.0\times10^{-47}} \\ r^2=1.03\times10^{-20} \\ r=\sqrt{1.01\times10^{-20}} \\ r=1.01\times10^{-5}\text{ m} \end{gathered}[/tex]A pitcher throws a 0.14-kg baseball with a speed of 28.8 m/s. The batter strikes it with an average force of 5000 N, which results in the ball traveling with an initial speed of 37.0 m/s toward the pitcher. For how long we’re the bat and ball in contact?
Given data:
* The velocity of the ball before hitting is,
[tex]u=28.8\text{ m/s}[/tex]* The velocity of the ball after hitting the bat is,
[tex]v=-37\text{ m/s}[/tex]Here, the negative sign is indicating the direction of motion of ball is in opposite direction to the direction of motion before hitting by the bat.
* The mass of the ball is,
[tex]m=0.14\text{ kg}[/tex]* The average force on the ball by the bat is,
[tex]F=5000\text{ N}[/tex]Solution:
The change in the momentum of the ball after hitting is,
[tex]\Delta p=p_2-p_1[/tex]where p2 is the momentum of the ball after hitting from the bat and p1 is the momentum of the ball before hitting from the bat,
The change in momentum in terms of mass is,
[tex]\begin{gathered} \Delta p=mv-m\text{u} \\ =m\mleft(v-u\mright) \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} \Delta p=0.14(-37-28.8) \\ \Delta p=-0.14\times65.8 \\ \Delta p=-9.212\text{ kgm/s} \end{gathered}[/tex]Thus, the time at which the bat and ball were in contact is,
[tex]\begin{gathered} F=\frac{|\Delta p|}{\Delta t} \\ \Delta t=\frac{|\Delta p|}{F} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} \Delta t=\frac{9.212}{5000} \\ \Delta t=0.0018\text{ s} \\ \Delta t=1.8\times10^{-3}\text{ s} \end{gathered}[/tex]Thus, the ball and bat were in contact for 0.0018 seconds.
A uniform disk of radius 0.473 m and unknown mass is constrained to rotate about a perpendicular axis through its center. A ring with the same mass as the disk is attached around the disk's rim. A tangential force of 0.203 N applied at the rim causes an angular acceleration of 0.125 rad/s2. Find the mass of the disk.
From the calculations, the mass of the object is 3.43 Kg.
What is the mass of the rod?We know that the term circular motion has to do with motion along a circular path. The force that acts on the body in this case is tangential to the circle. This tangential force can be obtained from;
F = mrα
F = tangential force
m = mass of the object
r = radius of the object
α = angular acceleration
When we make m the subject of the formula;
m = F/rα
m = 0.203 N/0.473 m * 0.125 rad/s2
m = 3.43 Kg
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A satellite is in orbit around Earth. How does the force exerted by Earth on the satellite compare to the force exerted by the satellite on Earth? Note: >> means much greater than; << means much less than.
The formula for the gravitational force between two bodies is given as
[tex]F=\frac{GmM}{r^2}[/tex]*Here m is the mass of the first object
*Here M is the mass of the second object
*Here r is the distance between the two objects
*Here G is the gravitational force
The force exerted by the earth on the satellite is equal to the force exerted by the satellite on the earth. Hence, the correct option is (b)
Part 1) If a force of magnitude 125 N is exerted horizontally as shown, find the force exerted by the hammer claws on the nail. (Assume that the force the hammer exerts on the nail is parallel to the nail).Answer in units of N. Part 2) Find the force exerted by the surface on the point of contact with the hammer head. Assume that the force the hammer exerts on the nail is parallel to the nail. Answer in units of N.
Part 1)
Recall, for a body to remain in rotational and translational equilibrium, the net external torque and force acting on a system must be zero. The formula for calculating torque is
Torque = Force x moment arm
Let P be the force exerted by the hammer on the nail. For rotational equilibrium, the torque about the point of contact would be equal. Thus,
For the nail,
distance = 5.86 cm = 5.86/100 = 0.0586 m
moment arm = 0.0586mSin26 = 0.0257
Torque = P x 0.0257 = 0.0257P
For the handle,
Force = 125
perpendicular distance = 28 cm = 28/100 = 0.28
Torque = 125 x 0.28 = 35
By equation both torques,
0.0257P = 35
P = 35/0.0257
P = 1361.87 N
the force exerted by the hammer claws on the nail is 1361.87 N
Part 2)
The force exerted by the surface on the point of contact is the normal force, N
It would be equal to the total downward force. Since the claw is inclined,
Normal force = Downward force x Sinθ
Downward force = P = 1361.87
θ = 26
N = 1361.87Sin26
N = 597 N
the force exerted by the surface on the point of contact with the hammer head is 597 N
A man attempts to move a truck by pushing it, but he can't move it. Describe the work done by the man.
If a man tried to move a truck by pushing it, but he is not able to move it, then the work done by the man will be equal to zero.
What is Work?In physics, the word "work" involves measurement of energy transfer that takes place when an item is moved over a range by an externally applied, at least a portion of which is applied with in the direction of the displacement. The length of the path is multiplied by the element of a force acting all along the path to calculate work if the force is constant. The work W is theoretically equivalent towards the force f times the length d, or W = fd, to portray this concept.
As per the given question,
The man tries to move the truck, but he is not able to move. It means that the total displacement is zero, which means according to the formula of work done the work is also zero.
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What are some simple machines that decrease mechanical advantage used to do?
Omultiply distance to increase speed
Odivide distance to decrease speed
O decrease work
O increase work
Some simple machines decrease mechanical advantage and are used to A) multiply distance to increase speed.
Define Mechanical Advantage.
When analyzing the forces in straightforward machines like levers and pulleys, mechanical advantage—a measure of the ratio of output force to input force in a system—is utilized. The conservation of energy still holds true and the output energy is equal to the input energy even when the forces being applied change.
The mechanical advantage is typically described in terms of perfect systems, also known as 100% efficient systems, where there are no energy losses between the input and output times.Through energy efficiency:Fi×di=Fo×do
The input energy (or work put into a machine) is on the left, and the output energy (or work that comes out of a machine, for no-loss energy) is on the right.
By extending the distance across which the effort force is applied, the equations demonstrate how a basic machine can produce the same amount of work while using less effort force.
Hence, Option A) is correct.
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Three liquids are at temperatures of 10 ◦C, 22◦C, and 29◦C, respectively. Equal masses of the first two liquids are mixed, and the equi- librium temperature is 12◦C. Equal masses of the second and third are then mixed, and theequilibrium temperature is 25.9◦C.Find the equilibrium temperature when equal masses of the first and third are mixed.Answer in units of ◦C.
Which statement given BEST describes what happens to light as it passes from air into a piece of glass?a) The speed increases, its wavelength becomes longer, and its frequency decreases.b)The speed increases, its wavelength becomes longer, and its frequency remains the same.c) The speed decreases, its wavelength becomes shorter, and its frequency remains the same.d) The speed decreases, its wavelength becomes shorter, and its frequency increases.
Given that the light passes from air to glass, that means it enters to denser medium from rarer medium.
Here, air is the rarer medium and glass is the denser medium.
Speed decreases when light enters from rarer to densr medium , while the frequency remains the same and its wavelength becomes shorter.
Thus, option C is correct.
how do i convert 5.6 kg in hg?
5.6 kg can be converted to hg using unitary method .
The unitary method is a process of finding the value of a single unit, and based on this value
here
kg = kilogram
hg = hectogram
1 kg = 10 hg
using unitary method , we can write
5.6 kg = 5.6 * 10 hg = 56 hg
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The carnival ride, the Round Up, has a radius of 1 m and rotates once each .9 s. What coefficient of static friction is required to keep the riders from slipping?
The coefficient of static friction is required to keep the riders from slipping is 0.05.
What is coefficient of static friction?The coefficient of static friction is the ratio of the maximum static friction force (F) between the surfaces in contact before movement commences to the normal (N) force.
The coefficient of static friction is required to keep the riders from slipping is calculated as follows;
static friction force = centripetal force
μmg = ma
μg = a
μg = ω²r
μ = ω²r /g
where;
ω is the angular speed of the carnival rider is the radius of the pathg is acceleration due to gravityThe given parameters;
angular speed of the carnival ride, ω = 1 rev/9s = 2π rad/9s = 0.698 rad/s
μ = (0.698² x 1) / (9.8)
μ = 0.05
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If an alarm clock is ringing what is the medium for the sound waves
Sound energy travels through air in waves. When an alamr clock rings, nearby air molecules vibrate.
Assume that the ammeter in the figure below is removed and the current that flows through the 4.0Ω path, I3, is unknown. Determine all the currents in the circuit.
The circuit is in parallel connection
Equivalent resistance = 1/Req = 1/R1 + 1/R2 + 1/R3
From the information given,
R1 = 5
R2 = 2
R3 = 4
1/Req = 1/5 + 1/2 + 1/4 = (4 + 10 + 5)/20 = 19/20
Req = 20/19 = 1.053 ohms
I = V/R
Given that V = 12,
Current flow through circuit = 12/1.053 = 11.4 A
I1 + I2 + I3 = 11.4
I1 = 12/5 = 2.4 A
I2 = 12/2 = 6 A
I3 = 12/4 = 3A
Question 5 of 10Which phrase is the best definition of matter?A. A substance that cannot be divided into smaller piecesB. The smallest piece of a chemical compound that retains theproperties of the compoundC. Something that occupies a volume of space and also has massD. A substance that can change in both volume and shape
Something that occupies a volume of space and also has mass is the best definition of matter.
everything that occupies space and has mass. Matter, in the form of atoms, which are made up of protons, neutrons, and electrons, makes up all physical objects. Greek philosophers Democritus (470–380 BC) and Leucippus are credited with originating the notion that matter was composed of constituent parts or particles (490 BC).
Atoms form the basis of matter. One proton makes up the most fundamental atom, the protium isotope of hydrogen. Atoms are composed of protons, neutrons, and electrons, however these particles are based on fermions. Despite fitting certain definitions of matter, quarks and leptons aren't commonly regarded as types of matter. It's easiest to just declare that matter is made up of atoms at the majority of levels.
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On Earth, the gravitational force of a robotic helicopter is 1.8 kilograms. What is the helicopter's gravitational
force on Mars?
On Earth, g = 9.8m/s²
On Mars, g = 3.71 m/s².
(1 point)
O 6.68 N
O 0.49 N
O 17.64 N
O2.64 N
The gravitational force of the robotic helicopter of mass 1.8 kg on Mars is 6.68 N.
What is gravitational force?Gravitational force is the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface.
To calculate the helicopter's gravitational force, we use the formula below.
Formula:
F = mg'............ Equation 1Where:
F = Gravitational force of the helicopterm = Mass of the helicopterg' = Acceleration due to gravity of MarsFrom the question,
Given:
m = 1.8 kgg' = 3.71 m/s²Substitute these values into equation 1
F = 1.8×3.71F = 6.678F ≈ 6.68 NHence, the helicopter's gravitational force on Mars is 6.68 N.
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A jet is goes from 180km/ to 139km/s in. 22 seconds what is its acceleration Vi-Vf-T-A-
Given
Initial velocity,
[tex]v_i=180\text{ km/s}[/tex]Final velocity,
[tex]v_f=139\text{ km/s}[/tex][tex]Time\text{ taken , t=22 s}[/tex]To find
The acceleration
Explanation
Acceleration is the ratio of change in velocity to that of the time taken
Thus,
[tex]\begin{gathered} a=\frac{v_f-v_i}{t} \\ \Rightarrow a=\frac{180-139}{22} \\ \Rightarrow a=1.86\text{ km/s}^2 \end{gathered}[/tex]Conclusion
The acceleration is
[tex]1.86\text{ km./s}^2[/tex]A battery does 1.922 J of work to transfer 0.089 C of charge from the negative to the positive terminal. What is the emf of this battery?
We can find the EMF as follows:
[tex]\begin{gathered} EMF=\frac{W}{q} \\ where: \\ W=1.922J \\ q=0.089C \\ so: \\ EMF=\frac{1.922}{0.089} \\ EMF=21.5955V \end{gathered}[/tex]Answer:
21.5955V
A wheel of radius 30.0 cm is rotating at a rate of 2.30 revolutions every 0.0810 s.Through what angle does the wheel rotate in 1.00 s?What is the linear speed of a point on the wheel’s rim?What is the wheel’s frequency of rotation?
Given data
*The given radius of the wheel is r = 30.0 cm = 0.30 m
*Rate of rotation is 2.30 revolutions every 0.0810 s
The angle of the wheel rotates in one second is calculated as
[tex]\begin{gathered} \theta=(2.30\times2\pi)\times\frac{1}{0.0810} \\ =178.32\text{ radian} \end{gathered}[/tex]Hence, the angle of the wheel rotating in one second is 178.32 radian
The formula for the linear speed of a point on the wheel's rim is given as
[tex]v=r\omega[/tex]Substitute the known values in the above expression as
[tex]undefined[/tex]Charge q1 = 2.00 x 10-9 C is located +0.020 m from the origin along the x-axis. Charge q2 =-3.00 x 10-9 C is located +0.040 m from the origin along the x-axis. What is the electric force exerted by these two charges on a third charge, q3 = 5.00 x 10-9 C, located at the origin?
Given,
Charge q1 = 2.00 x 10-9 C is located +0.020 m from the origin along the x-axis. Charge q2 =-3.00 x 10-9 C is located +0.040 m from the origin along the x-axis.
There is a third charge q3 = 5.00 x 10-9 C, located at the origin.
Now the force acting on the third charge due to other two charge is:
[tex]\begin{gathered} F=k\frac{2\times10^{-9}\times5\times10^{-9}}{0.02^2}+k\frac{3\times10^{-9}\times5\times10^{-9}}{0.04^2} \\ \Rightarrow F=0.0002225+0.0000834=0.000305C \end{gathered}[/tex]
Identify the Action and Reaction Forces
A pair of forces that will always move in conflict with one another is referred to as an action-reaction force.
What is newton's third law?The third law of Newton asserts because when bodies interact, they exert forces that are equal in size and directed in the reverse way. Another name for the newton's third is the law of action and reaction. This equation is crucial for understanding issues with static balance, in which all forces are all in balance, but it also holds true for bodies moving at a steady or accelerated speed.
Hammer hitting a nail : In this, the hitting by the hammer is the action and the nail is going into the surface is the reaction force.
Book placed on the table : In this book is applying gravitational force that is action force and the reaction force is applied by the table.
Hitting a tennis ball : In this, hitting the ball is the action and the ball is going in the direction of hitting is the reaction force.
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I have this homework i need help with ?Part (a)
Given that the mass of train is
[tex]m=\text{ }8.12\times10^6\text{ kg}[/tex]The deceleration of the train is
[tex]\begin{gathered} a\text{ = -104 km/h} \\ =-104\times\frac{5}{18} \\ =\text{ -28.88 m/s} \end{gathered}[/tex]The braking force can be calculated by the formula
[tex]F=ma[/tex]Substituting the values, the braking force will be
[tex]\begin{gathered} F=8.12\times10^6\times(-28.88) \\ =2.348\times10^8\text{ N} \end{gathered}[/tex]Thus, the braking force is 2.348 x 10^8 N.
A racecar travels around a circular track of radius 400 meters at constant speed. If the car's acceleration is 4 m/s2, what is its speed?
We are given that a car travels in a circular motion with an acceleration of 4m/s^2. We are asked to determine the speed of the motion. To do that we will use the following formula:
[tex]a=\frac{v^2}{r}[/tex]Where:
[tex]\begin{gathered} a=\text{ acceleration} \\ v=\text{ velocity} \\ r=\text{ radius} \end{gathered}[/tex]We will solve for the velocity first by multiplying both sides by "r":
[tex]ar=v^2[/tex]Now we take the square root to both sides:
[tex]\sqrt[]{ar}=v[/tex]Now we plug in the values:
[tex]\sqrt[]{(4\frac{m}{s^2})(400m)}=v[/tex]Now we solve the operations:
[tex]40\frac{m}{s}=v[/tex]Therefore, the velocity of the car is 40 meters per second.
b. How many square meters does an 80. ft2 rug occupy? (1 ft = 12 in.)
b.
First squared the following conversion factor: 1 ft = 12 in, as follow:
[tex]\begin{gathered} 1ft=12in \\ (1ft)^2=(12in)^2 \\ 1ft=144in^2 \end{gathered}[/tex]Then, use the previous result as a conversion factor to convert 80.0ft^2, as follow:
[tex]80.0ft^2\cdot\frac{144in^2}{1ft^2}=11520in^2[/tex]Now, consider that 1 m = 39.3701 in, then, if you squared you obtain:
[tex]\begin{gathered} 1m=39.3701in \\ (1m)^2=(39.37in)^2 \\ 1m^2=1550.004774in^2 \end{gathered}[/tex]use the previous result as a conversion factor:
[tex]11520in^2\cdot\frac{1m^2}{1550.004774in^2}\approx7.43m^2[/tex]Hence, 80.0ft^2 is approximately 7.43 m^2
A comet is known to be traveling at 236 km per second. It is observed to give light with a frequency of 641 x 1014 Hz. What frequency of light would the come be emitting if it were motionless? Express your answer to the nearest hundred trillion Hz.
ANSWER
6.4 x 10¹⁶ Hz
EXPLANATION
Given:
• The speed of the source, vs = 236 km/s
,• The speed of light, c = 300,000 km/s
,• The observed frequency of the light, fo = 641 x 10¹⁴ Hz
Find:
• The frequency of the source, fs
To solve this problem, we have to apply the Doppler Effect formula for light,
[tex]f_o=f_s\sqrt{\frac{1-v_s/c}{1+v_s/c}}[/tex]Solving for fs,
[tex]f_s=f_o\sqrt{\frac{1+v_s/c}{1-v_s/c}}[/tex]Replace the known values and solve,
[tex]f_s=641\cdot10^{14}Hz\cdot\sqrt{\frac{1+(236km/s)/(300,000km/s)}{1-(236km/s)/(300,000km/s)}}\approx6.4\cdot10^{16}Hz[/tex]Hence, the emitted frequency of the comet's light is 6.4 x 10¹⁶ Hz.
A coil of 17.771 H carries a current of 11.121 A. Compute the energy stored .
Given
The inductance is H=17.771 H
Current is , I=11.121A
To find
The energy stored
Explanation
The energy stored is given by
[tex]E=\frac{1}{2}LI^2[/tex]Putting the values,
[tex]\begin{gathered} E=\frac{1}{2}\times17.771\times(11.121)^2 \\ \Rightarrow E=1098.92J \end{gathered}[/tex]Conclusion
The energy stored is 1098.92J
Hello, i just need to check this problem, thanks in advance!
Given:
The mass of the bucket is m = 0.25 kg
The length of the string from the center is L = 0.4 m
The speed of the bucket is v = 5 m/s
To find The tension in the string at the bottom of the circle.
Explanation:
Forces acting on the bottom of the circle are tension due to the string and force due to gravity.
The tension in the string at the bottom of the circle can be calculated by the formula
[tex]T=\frac{mv^2}{L}+mg[/tex]Here, g = 9.8 m/s^2 is the acceleration due to gravity.
Substituting the values, the tension in the string at the bottom of the circle is
[tex]\begin{gathered} T\text{ = }\frac{0.25\times(5)^2}{0.4}+0.25\times9.8 \\ =18.075\text{ N} \end{gathered}[/tex]
Final Answer: Thus the tension in the string at the bottom of the circle is 18.075 N