If we were able to get more work out of a machine than we put in, energy would be created in the process. According to the Law of Conservation of Energy, this is not possible.
Therefore, the answer is:
[tex]\begin{gathered} \text{ No, it is not possible to get more work} \\ \text{ out of a machine than we put in.} \end{gathered}[/tex]You need to construct a 400 pF capacitor for a science project. You plan to cut two L x L metal squares and place spacers between them. The thinnest spacers you have are 0.20 mm thick. What is the proper value of L?Express your answer in centimeters.
Consider that the formula for the capacitance of a square parallel plate capacitor is:
[tex]C=\epsilon_o\frac{A}{d}=\epsilon_o\frac{L^2}{d}[/tex]where A=L^2 is the area of each plate, d is the separation between plates and
ε0 is the dielectric permitivity of vacuum ans is equal to 8.82*10^-12 F/m.
If you solve the previous expression for L and replace the given values for d and C, you obtain:
[tex]\begin{gathered} L=\sqrt[]{\frac{dC}{\epsilon_o}} \\ d=0.20mm=0.20\cdot10^{-3}m=2.0\cdot10^{-4}m \\ C=400pF=400\cdot10^{-12}F=4.00\cdot10^{-10}F \\ L=\sqrt[]{\frac{(2.0\cdot10^{-4}m)(4.00\cdot10^{-10}F)}{8.85\cdot10^{-10}\frac{F}{m}}} \\ L\approx0.0095m=0.95cm \end{gathered}[/tex]Hence, the proper value of L to construct the required capacitor is approximately 0.95cm
An organ pipe that is closed at one end has a length of 3 m. What is the second longest harmonic wavelength for sound waves in this pipe
Given
Lenght 3m
Procedure
The relationships between the standing wave pattern for a given harmonic and the length-wavelength relationships for closed-end air columns are summarized below.
Harmonic # 3
[tex]\begin{gathered} \lambda=\frac{4}{3}\cdot L \\ \lambda=\frac{4}{3}\cdot3 \\ \lambda=4\text{ m} \end{gathered}[/tex]The second-longest wavelength would be 4m
How much heat is needed to bring 25.5 g of water from 29.3 °C to 43.87 °C.Q =m=ΔT= C=solution =
We are given the following information
Mass of water = 25.5 g
Initial temperature of water = 29.3 °C
Final temperature of water = 43.87 °C
The specific heat capacity of water is 4.186 J/g.°C
The amount of heat required is given by
[tex]\begin{gathered} Q=m\cdot c\cdot\Delta T \\ Q=m\cdot c\cdot(T_f-T_i) \end{gathered}[/tex]Let us substitute the given values into the above formula
[tex]\begin{gathered} Q=25.5\times4.186\cdot(43.87-29.3) \\ Q=1,555.25\; J \end{gathered}[/tex]Therefore, we need 1,555.25 Joules of heat to bring 25.5 g of water from 29.3 °C to 43.87 °C.
Given a material of specific heat c in Cal/gramC^o and mass 6 grams. If the material is heated by absorbing 7 calories of heat then which of these expressions yields the change in temperature of the material in Celsius degrees? A)7 divided by (c - 6) B)7 times (c - 6) C)7 divided by (6 times c)
Given:
The mass is m = 6 grams.
The heat absorbed is Q = 7 calories.
The unit of specific heat, c of the material is cal/gram degree Celsius
To find the change in temperature of the material in degrees Celsius.
Explanation:
The formula to calculate the temperature change is
[tex]\begin{gathered} Q\text{ = mc}\Delta T \\ \Delta T=\frac{Q}{mc} \end{gathered}[/tex]Substituting the values, the change in temperature will be
[tex]\Delta T\text{ = }\frac{7}{6\times c}[/tex]Final Answer: The change in temperature will be 7 divided by (6 times c).
A periodic wave has a frequency of 3.2 Hz. What is the wave period?answer in:____ s
Given:
Frequency of wave, f = 3.2 Hz.
Let's find the period of the wave.
To find the wave period, apply the formula:
[tex]T=\frac{1}{f}[/tex]Where:
T is the period in seconds
f is the frequency hertz.
Thus, we have:
[tex]\begin{gathered} T=\frac{1}{3.2} \\ \\ T=0.3125\text{ s} \end{gathered}[/tex]Therefore, the wave period is 0.3125 seconds.
ANSWER:
0.3125 s
A plant that is 4.1 cm tall is 10.3 cm from a converging lens. You observe that the image of this plant is upright and 6.2 cm tall. What is the focal length of the lens?
We know that the magnification is given by:
[tex]M=\frac{h^{\prime}}{h}[/tex]where h is the height of the object and h' is the height of the image, then in this case we have:
[tex]M=\frac{6.2}{4.1}[/tex]On the other hand we also know that the magnification is given by:
[tex]M=-\frac{i}{o}[/tex]where i is the distance of the image to lens and o is the distance of the object to the lens. From this we have:
[tex]\begin{gathered} \frac{6.2}{4.1}=-\frac{i}{10.3} \\ i=-\frac{10.3\cdot6.2}{4.1} \end{gathered}[/tex]Once we have the distance of the image we can use the lens equation to find the focal point:
[tex]\begin{gathered} \frac{1}{10.3}+\frac{1}{-\frac{10.3\cdot6.2}{4.1}}=\frac{1}{f} \\ \frac{1}{10.3}-\frac{4.1}{10.3\cdot6.2}=\frac{1}{f} \\ f=(\frac{1}{10.3}-\frac{4.1}{10.3\cdot6.2})^{-1} \\ f=30.41 \end{gathered}[/tex]Therefore the focal distance of the lens is 30.41 cm and this can be round to 30 cm, hence the answer is D
what is the wavelength of 2.6 million Hz ultrasound as it travels through human tissue ?
The wavelength of ultrasound waves can be given as,
[tex]\lambda=\frac{v}{f}[/tex]The speed of waves in human tissue is 1540 m/s.
Plug in the known values,
[tex]\begin{gathered} \lambda=\frac{1540\text{ m/s}}{(2.6\text{ million Hz)(}\frac{10^6}{1\text{ million}})}(\frac{10^3\text{ mm}}{1\text{ m}}) \\ =0.592\text{ mm} \end{gathered}[/tex]Thus, the wavelength of waves in human tissue is 0.592 m.
Question 3 of 5Molly wants to measure how hot water is when it begins to boil. What toolshould she use?O A. A graduated cylinderO B. A thermometerO C. CalipersD. A test tube
To find
Molly wants to measure how hot water is when it begins to boil. What tool should she use?
Explanation
A graduated cylinder is used to measure the volume of liquid.
Thermometer is a device which is used to measure temperature.
Calipers are used to measure dimensions of objects
A test tube is used to hold liquid during laboratory experiments
Conclusion
To measure the hotness of water she needs
B. A thermometer
Jenna manipulated the equation 4x+7=10 by adding -7 to both sides. Which of the following properties justifies this manipulation?1. The associative property of addition.2. The addition property of equality.3. The commutative property of addition.4. The multiplication property of equality.
2. The addition property of equality.
Explanation
Step 1
[tex]\begin{gathered} 4x+7=10 \\ \text{Jenna added -7 ( or subtract 7) to both sides, so} \\ 4x+7-7=10-7 \\ 4x=3 \end{gathered}[/tex]she applied the addition property for inequalities. it states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.
so, the answer is
2. The addition property of equality.
Compare the power of one motorcylce that travels twice as fast as a second indetical motorcycle.A.Same amount of powerB.Twice the powerC.4 times the powerD.Half the power
Given:
The power of one motorcycle travels twice as fast as a second motorcycle.
To compare the power of motorcycles.
Explanation:
Let the speed of one motorcycle be v.
The speed of the second motorcycle will be 2v.
Power is calculated by the formula
[tex]P=\text{ Force}\times velocity[/tex]The power of the first motorcycle will be
[tex]P_1=Fv_1[/tex]The power of the second motorcycle will be
[tex]\begin{gathered} P_2=Fv_2 \\ =F\times2v_1 \\ =2Fv_1 \\ =2P_1 \end{gathered}[/tex]Thus, the power of the second motorcycle is twice the power of the first motorcycle.
Please help me create a table and graph for the experiment below.Mass of bob: 25g, Time noted: 0.54 s, 0.54 s, 0.54 sMass of bob: 50g, Time noted: 0.54 s, 0.56 s, 0.49 sMass of bob: 100g: Time noted: 0.53 s, 0.53 s, 0.46 s
Given:
Mass of bob: 25g, Time noted: 0.54 s, 0.54 s, 0.54 s
Mass of bob: 50g, Time noted: 0.54 s, 0.56 s, 0.49 s
Mass of bob: 100g: Time noted: 0.53 s, 0.53 s, 0.46 s
To find:
The average time period for each mass of the pendulum bob.
Explanation:
As the sensor records the time it takes for the pendulum to complete one-half of the period, the time period of the pendulum can be calculated by multiplying the recorded value by 2.
For the bob of mass 25 g,
Period (s) Trial 1: 0.54 s × 2 = 1.08 s
Period (s) Trial 2: 0.54 s × 2 = 1.08 s
Period (s) Trial 3: 0.54 s × 2 = 1.08 s
Average Period (s): T = (1.08 s + 1.08 s + 1.08 s)/3 = 1.08 s
For the bob of mass 50 g,
Period (s) Trial 1: 0.54 s × 2 = 1.08 s
Period (s) Trial 2: 0.56 s × 2 = 1.12 s
Period (s) Trial 3: 0.49 s × 2 = 0.98 s
Average Period (s): T = (1.08 s + 1.12 s + 0.98 s)/3 = 1.06 s
For the bob of mass 100 g,
Period (s) Trial 1: 0.53 s × 2 = 1.06 s
Period (s) Trial 2: 0.53 s × 2 = 1.06 s
Period (s) Trial 3: 0.46 s × 2 = 0.92 s
Average Period (s): T = (1.06 s + 1.06 s + 0.92 s)/3 = 1.013 s
Final answer:
The average value of the period is
For bob of mass 25 g, Average Period = 1.08 s
For bob of mass 50 g, Average Period = 1.06 s
For bob of mass 100 g, Average Period = 1.013 s
Why doesn't an object falling from an airplane continue to accelerate? (1 point)
O Gravity's force diminishes as the object nears the ground.
O Air resistance on the object will eventually equal the force of force of gravity.
O The object's weight varies as it nears the ground.
O Hitting the ground stops the object's acceleration.
A falling object accelerates as it descends. The quantity of air resistance rises in proportion to the speed. The pull of gravity eventually is balanced by the force of air resistance as it grows. The item will cease accelerating since there is no net force at this point in time (0 Newton).
Since the upward force of air resistance eventually equals the downward force of gravity, a falling item cannot continue to accelerate indefinitely before reaching its terminal velocity.In contrast to air resistance, which operates in the opposite direction and slows acceleration, gravity causes objects to accelerate downhill. Greater surface area falling objects encounter more air resistance. In the absence of air, or in a vacuum, all objects fall with the same precise rate of acceleration.To know more about gravity
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Answer: (B). Air resistance on the object will eventually equal the force of force of gravity.
Explanation:
Question: imagine the decay of a very unusual particle that has a charge of Q = +24e. If it decays into 12 particles, 3 of which have electrical charges of +2e each, what is the collective charge of the other 9 particles?
According to the law of charge conservation,
The net charge of the system before and after the decay should remain constant.
As the charge before the decay is +24e.
Thus, the net charge after the decay should be +24e.
Given that, the charge on 3 particles out of 12 is +2e each.
Thus, collective charge of the other 9 particles is,
[tex]\begin{gathered} q=Q-3(+2e) \\ q=24e-6e \\ q=+18e \end{gathered}[/tex]Thus, the collective charge on the other 9 particles is +18e.
Jontell finds a giant spring that has a 350N/m spring constant. Jontell figures out how to compress is 1.5m. If Jontell and his spring propelled cart's mass is 85kg, how fast is he going after the push fromthe spring? PEspring1/2*k*x?^2
Given data
*The given mass of the propelled cart's is m = 85 kg
*The given spring constant is k = 350 N/m
*The spring compresses at a distance is x = 1.5 m
The formula for the speed is given by the conservation of energy as
[tex]\begin{gathered} U_{p.e}=U_k \\ \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v=\sqrt[]{\frac{kx^2}{m}} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\sqrt[]{\frac{350\times(1.5)^2}{85}} \\ =3.04\text{ m/s} \end{gathered}[/tex]Prove that the area of the parallelogram is equal to | A × B |
The area of a paralellogram with base a and height h is given by:
[tex]A=h\cdot a[/tex]If two adjacent sides of a parallelogram have lengths a and b and are separated by an angle φ, then the base of the parallelogram is a and the height is given by b*sin(φ). Then, the area of the parallelogram is given by:
[tex]A=a\cdot b\cdot\sin (\phi)[/tex]On the other hand, the cross product of two vectors is defined as:
[tex]\vec{a}\times\vec{b}=a\cdot b\cdot\sin (\phi)\hat{n}[/tex]Where the unitary vector is directed toward the direction perpendicular to a and b according to the right hand rule.
The modulus of the cross product of a and b is:
[tex]|\vec{a}\times\vec{b}|=a\cdot b\cdot\sin (\phi)[/tex]We can see that both the area of the parallelogram and the modulus of the cross product have the same expressions. Therefore:
[tex]A=|\vec{a}\times\vec{b}|[/tex]A truck covers 40.0 m in 9.45 s while uniformly slowing down to a final velocity of 2.10 m/s.(a) Find the truck's original speed._____ m/s(b) Find its acceleration._____ m/s2
(a)
In order to find the original speed, let's use the formula below to find an expression for the acceleration:
[tex]\begin{gathered} V=V_0+a\cdot t\\ \\ 2.1=V_0+a\cdot9.45\\ \\ a=\frac{2.1-V_0}{9.45} \end{gathered}[/tex]Now, we can use the following formula to find the initial speed:
[tex]\begin{gathered} \Delta S=V_0t+\frac{at^2}{2}\\ \\ 40=V_0\cdot9.45+\frac{\frac{(2.1-V_0)}{9.45}\cdot9.45^2}{2}\\ \\ 40=9.45V_0+4.725(2.1-V_0)\\ \\ 40=9.45V_0+9.9225-4.725V_0\\ \\ 4.725V_0=40-9.9225\\ \\ V_0=\frac{30.0775}{4.725}\\ \\ V_0=6.3656\text{ m/s} \end{gathered}[/tex](b)
Now, calculating the acceleration, we have:
[tex]\begin{gathered} a=\frac{2.1-6.3656}{9.45}\\ \\ a=-0.4514\text{ m/s^^b2} \end{gathered}[/tex]A girl throws a ball vertically downward at 10m/s from the roof of a building 20m high. Whatwill its speed be when it strikes the ground?
Answer:
22.18 m/s
Explanation:
We will use the following equation:
[tex]v^2_f_{}=v^2_i+2ay[/tex]Where vf is the final velocity
vi is the initial velocity, so it is -10 m/s
a is gravity, so it is -9.8 m/s²
y is the change in the height so it is -20 m
Therefore, replacing the values, we get:
[tex]\begin{gathered} v^2_f=(-10)^2+2(-9.8)(-20) \\ v^2_f=100+392 \\ v^2_f=492 \\ v_f=\sqrt[]{492}=22.18\text{ m/s} \end{gathered}[/tex]So, the ball strikes the ground at 22.18 m/s
A block has a velocity of 6 m/s to the East and 360 J of kinetic energy. The block is pushed West with a 30 N external force, while the block moves 3 m East. How much work is done by the force?
Work done will be 270 J
What is work energy theorem?
The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
according to work force theorem
Work done = Force x direction = FD Cosθ
Even if the force is applied to the opposite direction, the box will move in the direction of East. Firstly it was already 6m towards East and after applying force, the box moves further 3m towards same direction i.e East.
= 30 N x (9) cos 0⁰
= cos 0⁰ is 1
= 270 J
270 J work is done by the force.
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Joanne drives her car with a mass of 1000 kg at a speed of 16m/s. what is the
the magnitude of road friction force needed to bring her car to a halt in 14s .
The frictional force will be the product of mass and the deceleration of the car which is 1142.9 N
What is Friction ?Friction is a force that opposes motion. It depends on the surface in contact and independent on the area of the surface.
Given that Joanne drives her car with a mass of 1000 kg at a speed of 16m/s. Before we calculate the the magnitude of road friction force needed to bring her car to a halt in 14s, let us first list out all the necessary parameters
Mass m = 1000 kgInitial velocity u = 16 m/sFinal velocity v = 0 m/sTime t = 14 sAcceleration a = ?Frictional Force F = ?The Frictional Force F = ma
From first equation of motion, v = u - at
Substitute the necessary parameters into the equation
0 = 16 - (a × 14)
14a = 16
a = 16/14
a = 1.143 m/s²
Then Frictional Force F = 1000 × 1.143
F = 1142.9 N
Therefore, the magnitude of road frictional force needed to bring her car to a halt in 14s is 1142.9 N
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6. A van of mass 1200 kg was moving at a velocity of 8 m/s when it was involved in a head-on collisionwith a lorry moving in the opposite direction. Assuming that the van came to a stop after the collision...(a) calculate the momentum of the van before the collision;(b) calculate the momentum of the van after the collision(c) find the change in momentum of the van (d) if the van took .30 s to stop, calculate the force that acted on each driver
Given data:
* The mass of the van is 1200 kg.
* The velocity of the van before the collision is 8 m/s.
* The velocity of the van after the collision is 0 m/s.
Solution:
(a). The momentum of the van before the collision is,
[tex]p_i=mv_i[/tex]where m is the mass of van, p_i is the momentum of van before the collision, and v_i is the velocity of van before the collision,
[tex]\begin{gathered} p_i=1200\times8 \\ p_i=9600kgms^{-1^{}} \end{gathered}[/tex]Thus, the momentum of the van before the collision is 9600 kgm/s.
(b). The momentum of the van after the collision is,
[tex]p_f=mv_f[/tex]weere p_f is the final momentum, and v_f is the final velocity of the van,
Substituting the known values,
[tex]\begin{gathered} p_f=1200\times0 \\ p_f=0^{} \end{gathered}[/tex]Thus, the momentum of the van after the collision is 0 kgm/s.
(c). The change in the momentum of the van is,
[tex]\begin{gathered} dp=p_f-p_i \\ dp=0-9600 \\ dp=-9600kgms^{-1} \end{gathered}[/tex]Here, the negative sign indicates that the momentum of van is decreasing with time.
Thus, the change in the momentum of the van is -9600 kgm/s.
(d). According to the Newton's second law, the force acting on the van in terms of the change in momentum is,
[tex]F=\frac{dp}{dt}[/tex]where dt is the time interval in which the momentum of the van changes,
Substituting the known values,
[tex]\begin{gathered} F=-\frac{9600}{0.30} \\ F=-32000\text{ N} \\ F=-32\times10^3\text{ N} \\ F=-32\text{ kN} \end{gathered}[/tex]Here, the negative sign is indicating the direction of force acting on the van is opposite to the direction of motion of van before the collision.
Thus, the force acting on the van is -32 kN.
Please help me solve From a previous question, the index of refraction of the liquid is 1.37
ANSWER
EXPLANATION
From the previous part, we have that the index of refraction of the liquid is 1.37, so we have to replace this in the equation and solve,
[tex]\sin\theta_c=\frac{n_{air}}{n_{liquid}}=\frac{1.00}{1.37}\approx0.73[/tex]And then, take the inverse of the sine to find the critical angle,
[tex]\theta_c=\sin^{-1}0.73\approx46.9\degree[/tex]Hence, the critical angle is 46.9°, rounded to the nearest tenth.
The press box at a basketball park is 38.0ft above the ground. A reporter in the press box looks at an angle of 15 degrees below the horizontal to see second base. What is the horizontal distance from the press box to second base?
The press box at a basketball park is 38.0 ft above the ground.
A reporter in the press box looks at an angle of 15 degrees below the horizontal to see the second base.
Let us draw the diagram to better understand the problem.
Here x is the horizontal distance from the press box to the second base.
With respect to angle 15°, the opposite side is 38 ft and the adjacent side is x.
Recall from the trigonometric ratios,
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \tan 15\degree=\frac{38}{x} \\ x=\frac{38}{\tan 15\degree} \\ x=141.8\; ft \end{gathered}[/tex]Therefore, the horizontal distance from the press box to the second base is 141.8 ft.
the heat in Yravels at the of lines. This is also radiation known as and/or write one phrase below each of these terms e heat conduction
Using the concept of Heat radiation, we have described conduction, convection, and radiation,
Heat can travel from one place to another in several ways. The different modes of the heat transfer include: Conduction, Convection, Radiation
Meanwhile, if the temperature difference exists between the two systems, heat will find a way to transfer from higher to the lower system.
Conduction is defined as:
The process of transmission of energy from one particle of the medium to another with the particles being in direct contact with a each other.
Following are the examples of the conduction:
Ironing of clothes is a example of conduction where the heat is conducted from the iron to the clothes.
Heat is transferred from hands to ice cube resulting in a melting of an ice cube when held in hands.
Convection is defined as the:
The movement of fluid molecules from higher temperature regions to lower temperature regions.
Examples of the convection include:
Boiling of water, that is molecules that are denser move at bottom while the molecules which are less dense move upwards resulting in a circular motion of the molecules so that water gets heated.
Warm water around a equator moves towards the poles while cooler water at the poles moves towards the equator.
Radiation is defined as:
Radiant heat is present in some or other form in our daily lives. Thermal radiations are referred to as the radiant heat. Thermal radiation is generated by the emission of the electromagnetic waves.
Following are the examples of the radiation:
Microwave radiation emitted in the oven is an example of radiation.
UV rays coming from the sun is a example of radiation.
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An astronaut, of mass 95.0 kg, sits on a newton scale as the shuttle takes off. The newton scale reads 5500 N. What is the acceleration of the shuttle?
Given data:
* The mass of the astronaut is m = 95 kg.
* The force read by the Newton scale when the shuttle takes off is F = 5500 N.
Solution:
According to Newton's second law, the force on the astronaut in terms of the acceleration and mass is,
[tex]\begin{gathered} F=\text{ma} \\ a=\frac{F}{m} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} a=\frac{5500}{95} \\ a=57.9ms^{-2} \\ a\approx58ms^{-2} \end{gathered}[/tex]The shuttle and the astronaut are moving with the same acceleration.
Thus, the acceleration of the shuttle is 58 meters per second squared.
A train is traveling at 100 mil/hr and travels for 10 hrs . How far did it travel ?
Givens.
• The speed is 100 mi/hr.
,• The time elapsed is 10 hr.
To find the distance traveled, use the constant motion formula.
[tex]d=vt[/tex]Where, v = 100 mi/hr and t = 10 hr.
[tex]\begin{gathered} d=100\cdot\frac{mi}{hr}\cdot10hr \\ d=1000mi \end{gathered}[/tex]Therefore, the train traveled 1000 miles.
QUESTION 26If the woman in the previous question doubles the constant horizontal force that she exerts on the box to push it on the same horizontal floor,O with a constant speed that is double the speed "vo" in the previous question.with a constant speed that is greater than the speed "vo" in the previous question, but not necessarily twice as great.for a while with a speed that is constant and greater than the speed "vo" in the previous question, then with a speed that increases thereafO for a while with an increasing speed, then with a constant speed thereafter.
When the woman exerts a
Which of the following descriptions best describes all of the factors that need to be considered when
determining an object's terminal velocity? (1 point)
O air density and the object's drag coefficient
O the object's weight and area it presents
O the object's weight and area it presents, as well as air density
O the object's weight, length, and width
All of the factors that need to be considered when determining an object's terminal velocity are the object's weight and area it presents, as well as air density.
option C is the correct answer
What is terminal velocity?Terminal velocity is obtained when the speed of a moving object is no longer increasing or decreasing. That is the object's acceleration (or deceleration) is zero.
Mathematically, the formula for terminal velocity is given as;
V = √(2mg)/(ρAC)
where;
m is the mass of the falling objectg is the acceleration due to gravityρ is the density of the fluid through which the object is fallingA is the projected area of the objectC is the drag coefficientThus, the variables to consider in determining terminal velocity of an object incudes the area, density of air, mass, etc.
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The wavelength of a light wave as it passes through a material is 657.61 nm with a speed of 232,536,355.7 m/s. What is the frequency of this wave?
Given:
The wavelength of light is
[tex]\begin{gathered} \lambda\text{ = 657.61 nm} \\ =657.61\times10^{-9}\text{ m} \end{gathered}[/tex]The speed is v = 232536355.7 m/s
Required: The frequency of the wave.
Explanation:
The frequency can be calculated by the formula
[tex]f=\frac{v}{\lambda}[/tex]On substituting the values, the frequency will be
[tex]\begin{gathered} f=\text{ }\frac{232536355.7}{657.61\times10^{-9}} \\ =3.536\times10^{14}\text{ s} \end{gathered}[/tex]Final Answer: The frequency of the wave is 3.536e14 s
An ox exerts a forwards force of 7100 N. If the ox has a weight of 8000 N, what is the minimum coefficient of static friction? (HINT: if there was no static friction the ox would slip and not move forward, what friction is required to allow the ox to move without slipping)
Given data
*An ox exerts a forwards force is F = 7100 N
*An ox has weight is N = 8000 N
The formula for the minimum coefficient of static friction is given as
[tex]\begin{gathered} F=\mu_sN \\ \mu_s=\frac{F}{N} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \mu_s=\frac{7100}{8000} \\ =0.887 \end{gathered}[/tex]Hence, the minimum coefficient of static friction is 0.887
What 2 factors affect the Kinetic Energy of an object? Which of the 2 factors has more of an influence? Explain. What are the 2 units?
Kinetic energy is a form of energy that an object or a particle has by reason of its motion.
[tex]K=\frac{1}{2}mv^2[/tex]m: mass
v: velocity
velocity has more influence
m: kg
v: m/s