a) 1 foot = 12 inches
159 feet = 159 x 12 = 1908
The distance in inches is 1908 inches
b) 1 inch = 2.54 cm
1908 inches = 1908 x 2.54 = 4846.32
The distance in centimeters is 4846.32 cm
c) 100 cm = 1 m
4846.32 cm = 4846.32/100 = 48.4632
The distance in meters is 48.4632 m
d) 1000m = 1 km
48.4632 m = 48.4632/1000 = 0.0484632
The distance in kilometers is 0.0484632 km
When an object falls through air, there is a drag force that depends on the product
of the surface area of the object, and the square of its velocity, i.e. F=CAv2
, where C
is a constant, determine the dimension of C and its SI unit
The dimensions and unit of C from the expression above is [ML^-3] and kgm^-3
What is dimension and units ?
Despite the fact that the solution to any engineering problem must incorporate units, dimensions and units are frequently misunderstood. While units are arbitrary names that are correlated to certain dimensions to make it relative, dimensions are tangible quantities that can be measured (e.g., a dimension is length, whereas a metre is a relative unit that describes length). Through a conversion factor, all units for the same dimension are related to one another.
given in question drag force depends upon the product of the cross sectional area of the object and the square of its velocity
and drag force can be given by
F = CAv^2 ----------- (1)
It is given that
Dimension of mass = [M]
Dimension of length = [L]
Dimension of time = [T]
So, by using above dimension we can write
the dimension of force,
F = [ MLT^-2]
dimension of cross-section area,
A = [L^2]
and dimension of velocity
v = [LT^-1]
now, by putting these values in equation (1), we will get
F = CAv^2
[ MLT^-2] = C[L^2][LT^-1]^2
C = [ML^-3T^0]
C = [ML^-3]
The dimensions and unit of C from the expression above is [ML^-3] and kgm^-3
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Hence, the dimension of constant C will be,
Need help with this question Short straight forward answers please :)
We will have the following:
a. The gravitational potential energy will be:
[tex]P_C=(15kg)(9.8m/s^2)(6m)\Rightarrow P_C=882J[/tex]So, the gravitational potential energy of C is 882 J.
b. The velocity of C right before it hits the ground will be:
[tex]\begin{gathered} 882J=\frac{1}{2}(15kg)v^2\Rightarrow\frac{1764J}{15kg}=v^2 \\ \\ \Rightarrow v=\frac{14\sqrt{15}}{5}m/s\Rightarrow v\approx10.84m/s \end{gathered}[/tex]So, the velocity will be approximately 10.84 m/s.
c.
1. We will have that Eg at the initial position will be: B < C
2. Vfinal upon impact with ground: B = C
3. Ek right before hitting he ground: B < C
f.
1. Eg: A > B
2. V final: A > B
3. Ek: A > B
4. V at 2 meters above the ground: A > B
5. Total energy at 2 m above the ground: A > B.
Using an electric current, you can split liquid water to form two new substances, hydrogen and oxygen gases. Is this a change in state. Explain your answer.
This process (called electrolysis) is not have a change of state. This comes from the fact that the substance (in this case water) does not keep its identity during the change, this means that we don't end up with the same substance; rather we have two new ones, hydrogen and oxygen.
Which of these is a property of an electromagnetic wave? A)magnetic and electric fields oscillate perpendicular to each other but not to the velocity of the wave B)transports energy C)has a magnetic wave but no electric wave
Electromagnetic waves :
- are transverse waves
- Can travel through a vacuum
- Tranport energy from one place to another
- can be reflected
-can be refracted
Correct options:
B)transports energy
A string with both ends fixed is vibrating in its second harmonic. The waves have a speed of 36m/s and a frequency of 60Hz. The amplitude of the standing wave at an antinode is 0.6cm. calculate the amplitude of the motion of points on the string a distance of 30cm,15cmand 7.5cm on the left hand end of the string.
The amplitude of the motion of points on the string a distance of 30cm,15cmand 7.5cm is 0.0547m, 0.027m and 0.0137m respectively.
Amplitude is the maximum range of vibration or oscillation, measured from the equilibrium position
According to the equation of the second harmonic motion
A = sin (kx)
A = Amplitude
k = [tex]\frac{2\pi f}{v}[/tex] = [tex]\frac{2*\pi * 60}{36}[/tex] = 10.467
x = distance of the point
For x = 30 cm = 0.3 m
A = sin (kx)
A = Sin (10.467 * 0.3)
A = 0.0547 m
For x = 15 cm = 0.15 m
A = sin (kx)
A = Sin (10.467 * 0.15)
A = 0.027 m
For x = 7.5 cm = 0.075 m
A = sin (kx)
A = Sin (10.467 * 0.075)
A = 0.0137 m
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8. What would the shape of the graph be if the speed of the object decreased from 50.0 km/hr at 20 s to 30 km/hr at 40 s?9. What is the acceleration in Problem 8?
We will have the following:
8. The graph will be the following:
9. The acceleration of part 8 will be determined as follows:
First, we determine the equivalency of the velocities, that is
[tex]\begin{gathered} \frac{50km}{h}\ast\frac{1000m}{1km}\ast\frac{1h}{3600s}=\frac{125}{9}m/s \\ \\ \frac{30km}{h}\ast\frac{1000m}{1km}\ast\frac{1h}{3600s}=\frac{25}{3}m/s \end{gathered}[/tex]Then, the acceleration will be:
[tex]a=\frac{(25/3\text{ }m/s)-(125/9\text{ }m/s)}{40s-20s}\Rightarrow a=-\frac{5}{18}m/s^2[/tex]So, the acceleration was of -5/18 m/s^2.
A 0.95 kg stone attached to a string is whirled in a horizontal circle of radius 38 cm as a conical pendulum. The string makes an angle of 40° with the vertical. (a) Find the speed of the stone. (b) Find the tension in the string.
The vertical component of the tension will be responsible for the weight of the stone while the horizontal component will be responsible for centripetal force.
a.) Speed v = 1.8 m/s
b.) Tension T = 12.2 N
Types of Circular MotionThere are motion of a body in a vertical circle and motion of a body in horizontal circle in which a conical pendulum is a good example.
Given that a 0.95 kg stone attached to a string is whirled in a horizontal circle of radius 38 cm as a conical pendulum. The string makes an angle of 40° with the vertical.
The parameters given are;
mass m = 0.95kgRadius r = 38 cm = 0.38mAngle Ф = 40°Speed v = ?Tension T = ?To find the speed of the stone, we will use the formula
TsinФ = mv²/r ...... (1)
Let us first find the tension by using the formula
TcosФ = mg
Tcos40 = 0.95 × 9.8
Tcos40 = 9.31
T = 9.31/cos40
T = 12.15 N
Substitute T in equation 1 to find the speed v
12.15sin40 = 0.95v²/0.38
2.97 = 0.95v²
v² = 2.97/0.95
v² = 3.125
v = √3.125
v = 1.77 m/s
Therefore, the tension in the string is 12.15 N and the speed of the stone is 1.8 m/s approximately
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The owner of a recycling company wants to reduce his electrical consumption and costs. The electromagnet used in his operation uses 12 A of current, has 7000 loops and a lifting force of 9800 N. If the lifting force needs to remain the same but the owner would like to reduce the current to only 5 A, how many loops would the electromagnet have?
Given:
Current, I = 12 A
Loops, B = 7000
Force, F = 9800 N
Let's determine the loops if the force remains the same but the current redudces to 5A.
Apply the formula:
[tex]F=\frac{I\times N}{L}[/tex]Let's solve for L.
[tex]\begin{gathered} L=\frac{I\times N}{F} \\ \\ L=\frac{12\times7000}{9800} \\ \\ L=8.57\text{ m} \end{gathered}[/tex]If the current reduces to 5 A, we have:
[tex]\begin{gathered} N=\frac{F\times L}{I} \\ \\ \text{Where I = 5 A} \\ \\ N=\frac{9800\times8.57}{5} \\ \\ N=16800\text{ } \end{gathered}[/tex]The number of loops the electromagnet would have is 16800 loops.
Shown here are astronomical objects located at different distances from earth. rank the objects based on their distances from earth, from farthest to nearest.
- star on far side of Andromeda Galaxy
- star on near side of Andromeda Galaxy
- star on far side of Milky Way Galaxy
- star near center of Milky Way Galaxy
- Orion Nebula
- Alpha Centauri
- Pluto- The Sun
The distance of astronomical objects is measure very carefully. These are having a different unit. This is the astronomical unit. The distances are very huge.
The distance between objects in space is vast and very difficult to calculate. These are learned under solar system mathematics. The values for these distances are cumbersome for astronomers and scientists to manipulate. Therefore, scientists use a unit of measurement called an astronomical unit.
Let us understand the distances first.
To know the distance of stars in Andromeda Galaxy, we should first know the distance of Andromeda Galaxy. The distance of Andromeda Galaxy from Earth is 2.5 million light years away. The astronomical unit used is the light years.
Thus, from this we can conclude that,
The star near to Andromeda Galaxy must be at a distance of 2.5 million light years away.The star far side from Andromeda Galaxy will be more than 2.5 million light years away.Now to know about the stars in Milky Way Galaxy, the distance of milky way galaxy from Earth is approximately 9 light years away.
So,
The star far from Milky Way Galaxy should be more than 9 light years away from Earth.The star near to the Milky Way Galaxy should be close to 9 light years away from Earth.Orion Nebula is 1,344 light years away.Alpha Centauri is approximately 4.3 light years away from the Earth.The Pluto is approximately 5 billion km away from the Earth.The Sun is approximately 148 million km away from the Earth.Thus, from this we can conclude that,
The farthest is the star on far side of Andromeda Galaxy, then the star on near side of Andromeda Galaxy, then comes the star on far side of Milky Way Galaxy, then the star near center of Milky Way Galaxy, then it is the Alpha Centauri, then the Orion Nebula and then it is the Pluto with the Sun being the nearest one.
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Alnico is _____.an alloy of metals with strong magnetic propertiesa brittle mixture of substances containing ferromagnetic elementsany material containing ironan element found in nature that behaves like a magnet
Alnico is an alloy made of iron combined with other metals, aluminum, nickel, and cobalt.
The alnico is a permanent magnet
Fill in the blank: 66 in. = yd
Answer: 1.833
Explanation:
Multiply the value in inches by 0.027777777777727 (the conversion factor)
So 66 inches*.027777777777727=1.833 yards
Answer:
66inches = 1.833 yards
Explanation:
Be the action of a force of 51N, a spring measures 39cm. Its length becomes 40.8 cm when subjected to another force of 61N. 1)Determine the empty length of the spring 2)Determine an elongation which will correspond to a force of 32N.3) So what is its length
Answer:
1) 29.82 cm
2) 5.76 cm
3) 35.58 cm
Explanation:
Part 1)
The force of a spring is equal to:
F = kΔx
Where k is the constant of the spring and Δx is the elongation. Δx = xf - xi, where xf is the length of the spring when the force is applied and xi is the empty length. Then
F = k(xf - xi)
Now, by the action of a force of 51N, a spring measures 39 cm, so
51 = k(39 - xi)
And by the action of a force of 61N, the spring length is 40.8 cm, so
61 = k(40.8 - xi)
To find the empty length, we need to solve the system of equations
51 = k(39 - xi)
61 = k(40.8 - xi)
First, solve the first equation for k
[tex]k=\frac{51}{39-x_i}[/tex]Then, replace this on the second equation and solve for xi
[tex]\begin{gathered} 61=k(40.8-x_i) \\ 61=\frac{51}{(39-x_i)}(40.8-x_i) \\ 61(39-x_i)=51(40.8-x_i) \\ 61(39)-61(x_i)=51(40.8)-51(x_i) \\ 2379-61x_i=2080.8-51x_i \\ 2379-2080.8=61x_i-51x_i \\ 298.2=10x_i \\ \frac{298.2}{10}=x_i \\ 29.82=x_i \end{gathered}[/tex]Therefore, the empty length of the spring is 29.82 cm
Part 2)
Now, we need to calculate the value of k, so replacing xi = 29.82, we get:
[tex]k=\frac{51}{39-29.82}=5.556[/tex]Therefore, the equation for the force is
F = 5.556Δx
Solving for Δx, we get:
Δx = F/5.556
Replacing the force by 32N, we can calculate the elongation as
Δx = 32/5.556 = 5.76 cm
Part 3)
Then, the length can be calculated by solving the following equation for xf
Δx = xf - xi
xf = Δx + xi
Replacing Δx = 5.76 cm and xi = 29.82 cm, we get:
xf = 5.76 cm + 29.82 cm
xf = 35.58 cm
So, its length is 35.58 cm
Therefore, the answers are
1) 29.82 cm
2) 5.76 cm
3) 35.58 cm
What is the electric field amplitude of an electromagnetic wave whose magnetic field amplitude is 7.9 mT ?
Given:
The amplitude of the magnetic field of the electromagnetic wave is,
[tex]B_0=7.9\text{ mT}[/tex]To find:
The amplitude of the electric field
Explanation:
Let, the amplitude of the electric field is
[tex]E_0[/tex]As we know,
[tex]\begin{gathered} \frac{E_0}{B_0}=c \\ c=3\times10^8\text{ m/s} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} \frac{E_0}{7.9\times10^{-3}}=3\times10^8 \\ E_0=3\times10^8\times7.9\times10^{-3} \\ E_0=2.37\times10^6\text{ N/C} \end{gathered}[/tex]Hence, the amplitude of the electric field is,
[tex]2.37\times10^6\text{ N/C}[/tex]Jake did an experiment in his science class. He found two of the same kind of beaker, and he filled them with different liquids. Then, he measured the volume and mass of each. The image below shows the two beakers.Based on the image, which of the following is the best conclusion? A. The liquids have the same volume and different masses. B. The liquids have different volumes and the same mass. C. The liquids have different volumes and different masses. D. The liquids have the same volume and the same mass.
Yellow liquid:
Mass = 470 g
Volume = 450ml
Pink liquid
Mass= 570 g
Volume= 450ml
Volumes are equal (450ml) and masses are different (470g and 570g)
A. The liquids have the same volume and different masses.
Can you please tell me the definition and give a example/formula of the first law of thermodynamics
We will have the following:
The first law of Thermodynamics:
Energy cannot be created or destroyed, it can only be transformed.
An example is when a car suddenly breaks in order to avoid hitting a person, here the velocity of the car, all the energy that is acumulated is transformed into thermal energy and it disipated by the friction of the breaks and the tires. The total energy of the system will remain the same, but it will change the way the energy is present. This as per our understanding of the universe so far.
One mole of an ideal gas at 1.00 atm and 0.00°C occupies 22.4 L. How many molecules of an ideal gas are in one cm^3 under these conditions?a. 28.9 b. 22 400 c. 2.69 × 1019 d. 6.02 × 1023
Given:
Atm = 1.00 atm
Temperature = 0.00°C
Amount of gas = 1 mole which occupues 22.4 L
Let;s find the number of molecules of an ideal gas are in one cm^3 under these conditions.
We have:
1 mole of ideal gas = 22.4 L
This is called the molar volume of gas.
To find the amount of molecules, apply the avogrado's constant:
Number of molecules in 1 mol = 6.023 x 10²³
Hence, for 1 cm³, we have:
[tex]\text{ No. of molecules in 1 cm}^3=\frac{6.023\times10^{23}}{22.4\times10^3}[/tex]Solving further:
[tex]\text{ No. of molecules in 1 cm}^3=\frac{6.023\times10^{23}}{22.4\times10^3}=2.69\times10^{19}mol/cm^3[/tex]Therefore, the amount of molecules of an ideal gas in one cm^3 under these conditions is:
2.69 x 10¹⁹ mol/cm^3
ANSWER:
C. 2.69 x 10¹⁹
The role or purpose of the battery in a circuit is to ____. Choose three.Group of answer choicesa) supply electric charge so that a current can existb) supply energy to the chargec) move the charge from the - to the + terminal of the batteryd) transform energy from electrical energy into light energye) establish an electric potential difference between the + and - terminalsf) replenish the charge which is lost in the light bulbg) offer resistance to the flow of charge so that the light bulb can get hot
Given:
The battery is connected to a circuit.
To find:
The purpose of the battery in the circuit
Explanation:
A battery is a source of energy in a circuit. The battery provides a push or a voltage to get the current flowing in a circuit.
Hence, the purposes are
b) supply energy to the charge
e) establish an electric potential difference between the + and - terminals
f) replenish the charge which is lost in the light bulb
What is the volume of a piece of iron ( = 7.9 g/cm3) that has a mass of 0.75 kg? (Enter your answer in cm3.) answer in: cm3
We will have the following:
[tex]\begin{gathered} \frac{7.9g}{cm^3}\ast\frac{1kg}{1000g}=\frac{0.75kg}{V}\Rightarrow0.0079kg/cm^3=\frac{0.75kg}{V} \\ \\ \Rightarrow V=\frac{0.75kg}{0.0079kg/cm^3}\Rightarrow V=\frac{7500}{79}cm^3 \\ \\ \Rightarrow V\approx94.9cm^3 \end{gathered}[/tex]So, the volume is 7500/79 cm^3, that is approximately 94.9 cm^3.
What is the force of gravity between two 50.0kg masses that are separated by 0.300m?3.71x10-8N5.59x10-7N2.78x104N1.85x10-6N
We will have the following:
[tex]\begin{gathered} F=G\frac{m_1m_2}{r^2}\Rightarrow F=\frac{(6.67\ast10^{-11}m^3\ast kg^{-1}\ast s^{-2})(50kg)(50kg)}{(0.3m)^2} \\ \\ \Rightarrow F=1.852777778...\ast10^{-6}N\Rightarrow F\approx1.85\ast10^{-6}N \end{gathered}[/tex]So, the force is approximately 1.85*10^-6 N.
A/An _____ is described as a device that is used to measure potential difference across any part of a circuit.ammeterfuseground fault interruptervoltmeter
Answer:
[tex]\text{Voltmeter}[/tex]Explanation: We need to find an instrument that measures potential difference across any part of the circuit, a potential difference is basically voltage difference across a circuit difference and the device used to measure this difference is known as:
[tex]\text{ Voltmeter}[/tex]Voltmeter used in a circuit
If a hammer is dropped from a height of 52 m and there is no air resistance, what is the acceleration the hammer experiences while it is falling towards the ground?
The hammer is fallingtowards the ground under the action of only one force which is gravity.
Thus, the acceleration of the hammer while falling towards the ground is the acceleration due to gravity.
Hence, the acceleration of the hammer in the given case is,
[tex]g=9.81ms^{-2\text{ }}[/tex]A pottery wheel with rotational inertia 40 kgm^2 rotates at 10 rev/s. 4 kg of clay is dropped onto the wheel 1.2 m from the axis. What angular speed will the wheel have after this?1. 55 rad/s2. 8.7 rad/s3. 70 rad/s4. 0 rad/s
Given:
• Rotational inertia = 40 kg.m²
,• Initial angula speed = 10 rev/s
,• Mass, m = 4 kg
,• Diameter, d = 1.2 m
Let's find the angular speed of the wheel.
To find the angular speed, apply the formula:
[tex]L_i=(I+md^2)*w_f[/tex]Where wf is the final angular speed
I is the rotational inertia
m is the mass
d = 1.2
Li is the angular momentum.
To find the angular momentum, we have:
[tex]\begin{gathered} L_i=40*10*2\pi \\ L_i=2513.27\text{ kg.m}^2\text{ rad/s} \end{gathered}[/tex]Now, to find the final angular speed, wf, plug in values in the first equation and solve for wf:
[tex]\begin{gathered} Li=(I+md^2)w_f \\ \\ 2513.27=(40+4*1.2^2)w_f \\ \\ 2513.27=45.76w_f \\ \\ w_f=\frac{2513.27}{45.76} \\ \\ w_f=54.9\approx55\text{ rad/s} \end{gathered}[/tex]Therefore, the final angular speed is 55 rad/s.
ANSWER:
1.) 55 rad/s
A parallel plate capacitor is charged by connecting it to a battery. After reaching steady state, the electric energy stored in the capacitor is UE . Select the correct claim about the amount of work needed to charge the capacitor.1) The amount of work needed to charge the capacitor is UE, because the work done is equal to the final potential energy of the system.2) No work is needed to charge the capacitor, because the charges will naturally move to their equilibrium position.3) The amount of work needed to charge the capacitor is 2UE , because it takes UE of work to remove the charges from the battery and and additional UE to place the charges on the capacitor.4) The amount of work needed to charge the capacitor is UE , because integrating the equation W=∫q dV yields the equation for the energy stored on a capacitor, Ue= 1/2qV
4. IS correct, that is the for mula used to calculate the potential energy stored
1. Is more or lees correct, but is less explained than the 4
2. Is false, is needed some energy
3. Is missunderstand, cause you are transfering energy from the battery to the capacitor, you dont have to count it twice
An unbanked asphalt highway has turns of 40m radii. How fast should the speed limit be if cars may be traveling in the rain? (Us for wet asphalt on rubber is .755)
Given data:
* The radius of the turn is r = 40 m.
* The coefficient of friction is,
[tex]\mu_s=0.755[/tex]Solution:
The centripetal force acting on the car is,
[tex]F=\frac{mv^2}{r}[/tex]where m is the mass of the car,
The frictional force acting on the car is,
[tex]F_r=\mu_smg[/tex]where g is the acceleration due to gravity,
In order to travel a car in rain, the centripetal force acting on the car must be equal to the frictional force on the same car.
Thus,
[tex]\begin{gathered} F_r=F_{} \\ \mu_smg=\frac{mv^2}{r} \\ \mu_sg=\frac{v^2}{r} \\ v^2=r\mu_sg \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} v^2=0.755\times40\times10 \\ v^2=302 \\ v=17.4\text{ m/s} \end{gathered}[/tex]Thus, the maximum speed limit of the car in rain is 17.4 m/s.
Hence, the nearest possible correct answer is option b.
Calculate the height of an image in a converging lens with a height of 6.1 cm, an image location of 6.2 cm, and a distance of 3.6 cm for the object's placement from the lens. Final answer will have 2 decimal places and might not follow sig fig rules.
the formula we will use is
1/f = 1/p + 1/q
where f is = focal length
p is the distance away (distance from the lens to the object)
and q is the image distance
It is not possible to derive an equation of motion for uniform acceleration without a time variable. Is this true or false?
The third equation of motion is given as,
[tex]v^2=u^2+2as[/tex]Here, v is the final velocity, u is the initial velocity, a is the acceleration and s is displacement.
The expression for the acceleration is given as,
[tex]a=\frac{v^2-u^2}{2s}[/tex]In the above expression, the acceleration of the body is not a function of the time variable. Therefore, it is possible to derive an equation of motion for uniform acceleration without a time variable. Hence, the given statement is false.
Four masses are arranged as shown. They are connected by rigid, massless rods of lengths 0.780 m and 0.500 m. What torque must be applied to cause an angular acceleration of 0.750 rad/s2 about the axis shown?
Given,
The length of the rods;
L=0.780 m
l=0.500 m
The angular acceleration, α=0.750 rad/s²
The masses;
m_A=4.00 kg
m_B=3.00 kg
m_C=5.00 kg
m_D=2.00 kg
The moment of inertia of the given system of masses is given by,
[tex]\begin{gathered} I=\Sigma mr^2 \\ =m_A(\frac{L}{2})^2+m_B(\frac{L}{2})^2+m_C(\frac{L}{2})^2+m_D(\frac{L}{2})^2 \\ =(\frac{L}{2})^2(m_A+m_B+m_C+m_D) \end{gathered}[/tex]Where r is the distance between each mass and the axis of rotation.
On substituting the known values,
[tex]\begin{gathered} I=(\frac{0.780}{2})^2(4.00+3.00+5.00+2.00) \\ =2.13\text{ kg}\cdot\text{m}^2 \end{gathered}[/tex]The torque required is given by,
[tex]\tau=I\alpha[/tex]On substituting the known values,
[tex]\begin{gathered} \tau=2.13\times0.750 \\ =1.6\text{ Nm} \end{gathered}[/tex]Thus the torque that must be applied to cause the required acceleration is 1.6 Nm
The inductor in the RLC tuning circuit of an AM radio has a value of 9 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 66 kHz
Given:
The inductance is,
[tex]\begin{gathered} L=9\text{ mH} \\ =9\times10^{-3}\text{ H} \end{gathered}[/tex]The radio frequency is,
[tex]\begin{gathered} f=66\text{ kHz} \\ =66\times10^3\text{ Hz} \end{gathered}[/tex]To find:
value of the variable capacitor, in picofarads
Explanation:
The frequency of the AM is,
[tex]\begin{gathered} f=\frac{1}{2\pi\sqrt{LC}} \\ \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} 66\times10^3=\frac{1}{2\pi\sqrt{9\times10^{-3}\times C}} \\ \sqrt{9\times10^{-3}\times C}=\frac{1}{2\pi\times66\times10^3} \\ \sqrt{9\times10^{-3}\times C}=2.41\times10^{-6} \\ 9\times10^{-3}\times C=5.81\times10^{-12} \\ C=6.45\times10^{-10} \\ C=645\times10^{-12}\text{ F} \\ C=645\text{ pF} \end{gathered}[/tex]Hence, the capacitance is 645 pF.
A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 4.0cm. If a timer is started when it's displacement is a maximum (hence x=4cm when t=0), what is the speed of the mass when t=3s ?
Since the frequency is 4 Hz, it completes one cycle in 1/4 seconds (its period, T, is 1/4 sec).
f = 1/T
Let's draw a graph of the displacement of the mass along time.
This pattern repeats, with each crest (high point) at every 1/4 second. This means that at t = 3, which is a multiple of 1/4, the mass is at its high point.
Notice that when the spring reaches its high or low point, it changes direction.
At this instant when it changes direction, its speed momentarily becomes 0.
That means that each multiple of 1/8 seconds, the speed is momentarily 0.
The speed of the mass is 0 at t = 3.
A 4-kg ball traveling westward at 25 m/s hits a 15-kg ball at rest. The 4-kg ball bounces east at 8.0 m/s. What is the speed and direction of the 15-kg ball? What is the impulse of the second ball?
Given:
The mass of the first ball is,
[tex]m_1=4\text{ kg}[/tex]The initial velocity of the first ball towards West is,
[tex]u_1=25\text{ m/s}[/tex]The mass of thr second ball is,
[tex]m_2=15\text{ kg}[/tex]the second object is initially at rest.
The final velocity of the first ball is,
[tex]v_1=-8.0\text{ m/s}[/tex]we are taking West as positive.
Applying momentum conservation principle we can write,
[tex]m_1u_1+m_2\times0=m_1v_1+m_2v_2[/tex]Substituting the values we get,
[tex]\begin{gathered} 4\times25+0=4\times(-8.0)+15\times v_2 \\ v_2=\frac{100+32}{15} \\ v_2=8.8\text{ m/s} \end{gathered}[/tex]THe final velocity of the second ball is towards East and the magnitude is 8.8 m/s.
The impulse of the Second ball is,
[tex]\begin{gathered} I=m_2v_2-m_2\times0 \\ =15\times8.8 \\ =132\text{ kg.m/s} \end{gathered}[/tex]