Three masses are connected via light strings and an ideal pulley, as shown below. Mass m2 is twice as massive as m1; m3 is three times as massive as m1. Everything is initially at rest, but then m3 is released, and it falls 1.30m to the floor. If the speed of m3 just before it hits the floor is 3 m/s, what is the coefficient of kinetic friction between the table and the blocks? (Hint: Use an energy analysis)
Answers
The coefficient of the kinetic friction is 0.29
The coefficient of kinetic friction refers to the ratio of the frictional force restricting and opposing the motion of two contact surfaces to the normal force pushing the two surfaces around each other.
From the conservation of energy theorem;
The work done by friction = K.E + P.E
\(\mathbf{-\mu_k ( m_1 + m_2) g h = \dfrac{1}{2} (m_1 + m_2 + m_3)v^2 - m_3gh}\)
where;
m₂ = 2m₁m₃ = 3m₁∴
\(\mathbf{-\mu_k ( m_1 + 2m_1)\times g \times 1.3 = \dfrac{1}{2} (m_1 + 2m_1+ 3m_1)3^2 - 3m_1\times 9.8 \times 1.3}\)
\(\mathbf{-\mu_k ( 3m_1 )\times g \times 1.3 = \dfrac{1}{2} \times 6 m_1 \times 9 - 3m_1\times 9.8 \times 1.3}\)
\(\mathbf{-\mu_k ( 3m_1 )\times g \times 1.3 =27 m_1 -38.22 m_1}\)
\(\mathbf{-\mu_k38.22m_1 =27 m_1 -38.22 m_1}\)
\(\mathbf{-\mu_k38.22m_1 =-11.22m_1}\)
\(\mathbf{\mu_k =\dfrac{-11.22m_1}{38.22m_1}}\)
\(\mathbf{\mu_k = 0.29 }\)
Therefore, the coefficient of the kinetic friction is 0.29
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Related Questions
A watermelon seed has the following coordinates: x = -7.1 m, y = 5.6 m, and z = 0 m. Find its position vector as (a) a magnitude and (b)
an angle relative to the positive direction of the x axis. If the seed is moved to the xyz coordinates (4.7 m, Om, O m), what is its
displacement as (c) a magnitude and (d) an angle relative to the positive direction of the x axis? Put the angles in the range (-180°, 180°]
Answers
Who has so much time to measure the fμcking seeds?
PLEASE HELP FOR PHYSICS!
All objects exert a gravitational force on all other objects. This force is given by, F = GMm r2 , where the value of G = 6.673 × 10–11 N–m2/kg2 , M is the mass of the heavier object, m is the mass of the lighter object, and r is the distance between the two objects.
What is the force of gravity between two balls of mass 50 kg each if the distance between them is 25 m. Assume that there is no interference from any other gravitational field.
Answers
Hi there!
Recall Newton's Law of Universal Gravitation:
\(\large\boxed{F_g = G\frac{m_1m_2}{r^2}}\)
Where:
Fg = Force of gravity (N)
G = Gravitational Constant
m1, m2 = masses of objects (kg)
r = distance between objects (m)
Plug in the given values stated in the problem:
\(F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}\)
5 postulates of neil bohr atomic model
Answers
Answer:
What is Bohr’s Model of an Atom?
The Bohr model of the atom was proposed by Neil Bohr in 1915. It came into existence with the modification of Rutherford’s model of an atom. Rutherford’s model introduced the nuclear model of an atom, in which he explained that a nucleus (positively charged) is surrounded by negatively charged electrons.
7. What do the following scientists study:
a. Geologists -
b. Zoologists -
c. Anthropologists -
d. Botanists -
e. Chemists -
f. Physicists -
on line papper
Answers
These scientists are known to study different aspects of science. The study they engage are;
What do scientists studya. Geologists - study the physical structure and composition of the Earth, including rocks, minerals, and natural resources.
b. Zoologists - study animals, their behavior, physiology, evolution, and distribution.
c. Anthropologists - study human societies, cultures, and behavior across time and space.
d. Botanists - study plants, their physiology, structure, ecology, and evolution.
e. Chemists - study the composition, properties, and behavior of matter and the chemical changes it undergoes.
f. Physicists - study the fundamental principles that govern the behavior of matter and energy in the universe, including mechanics, thermodynamics, electromagnetism, quantum mechanics, and relativity.
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State the difference between positive and negative zero error of a vernier calliper
Answers
Answer:
Positive zero error: When the jaws of the caliper are closed, the zero mark on the vernier scale is to the right of the zero mark on the main scale. This means that the caliper reads a value greater than the actual value, leading to a positive zero error.
Negative zero error: When the jaws of the caliper are closed, the zero mark on the vernier scale is to the left of the zero mark on the main scale. This means that the caliper reads a value less than the actual value, leading to a negative zero error.
A marble statute is exposed to the weather, within a few years
Answers
Answer:
it might be damaged
Explanation:
depending on how harsh the weather is it might be more or less damaged
If the instantaneous voltage at a given moment in the circuit RL is V=VmaxSIN(150), then the instantaneous current at the...... same instant I=Imaxsin
Answers
The instantaneous current at the same moment in the RL circuit can be expressed as I = Imaxsin(150), where Imax represents the maximum current.
1. Given that the instantaneous voltage at a specific moment in the RL circuit is V = Vmaxsin(150).
2. We can express the current at the same moment using Ohm's Law, which states that V = IR, where V is voltage, I is current, and R is resistance.
3. In an RL circuit, the resistance is represented by the symbol R, and it is typically associated with the resistance of the wire or any resistors in the circuit.
4. However, the given equation does not explicitly mention resistance.
5. Since we are considering an RL circuit, it suggests the presence of inductance (L) along with resistance (R).
6. In an RL circuit, the voltage across the inductor (VL) can be expressed as VL = L(di/dt), where L is the inductance and di/dt represents the rate of change of current.
7. At any given instant, the total voltage across the circuit (V) can be expressed as the sum of the voltage across the resistor (VR) and the voltage across the inductor (VL).
8. Therefore, V = VR + VL.
9. Since the given equation represents the instantaneous voltage (V), we can deduce that V = VR.
10. By comparing V = VR with Ohm's Law (V = IR), we can conclude that I = Imaxsin(150), where Imax represents the maximum current.
The specific values of Vmax, Imax, and the phase angle have not been provided in the question, so we are working with the general expression.
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a plane wall is 250mm thick and it's wall area is 4.5m^2. if the thermal conductivity is 9.35w/m°c and surface temperature are steady at 150°c and 40°c. calculate the heat flow across the plane wall and the temperature gradient in the flow direction
Answers
The temperature gradient in the flow of direction is 294525 W.
What is Temperature gradient?A temperature gradient is the gradual variance in temperature with distance. The slope of the gradient is consistent within a material. A gradient is established anytime two materials at different temperatures are in physical contact with each other.
Q= T/( L/ KA)
Q= ( 1500 − 450) / 0.15 / 9.35v * 4.35)
= 294525 W
Units of measure of temperature gradients are degrees per unit distance, such as °F per inch or °C per meter.
Many temperature gradients exist naturally, while others are created. The largest temperature gradient on Earth is the Earth itself. Q= T/Ka.
Therefore, The temperature gradient in the flow of direction is 294525 W.
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c = speed of light = 3.00 × 108 m/s
A gamma ray has a very high frequency of about 1019 s−1. What is the wavelength of the gamma ray?
A.
3.00 × 10−11 m
B.
3.00 × 1027 m
C.
3.33 × 1010 m
D.
3.33 × 10−12 m
URGENT!!!!!!!!!!
Answers
Answer:
The correct answer is option A: 3.00 × 10^(-11) m.
Explanation:
To find the wavelength of a gamma ray with a frequency of about 10^19 s^(-1), we can use the equation:
wavelength = speed of light / frequency
Given:
Speed of light (c) = 3.00 × 10^8 m/s
Frequency (f) = 10^19 s^(-1)
Substituting the values into the equation:
wavelength = (3.00 × 10^8 m/s) / (10^19 s^(-1))
To simplify the expression, we can rewrite the denominator as (1 / 10^(-19)) s:
wavelength = (3.00 × 10^8 m/s) / (1 / 10^(-19)) s
To divide by a fraction, we multiply by its reciprocal:
wavelength = (3.00 × 10^8 m/s) × (10^(-19) s)
Applying the properties of exponents, we can add the exponents when multiplying with the same base:
wavelength = 3.00 × 10^(-11) m
Therefore, the wavelength of the gamma ray is approximately 3.00 × 10^(-11) m.
an immersion heater rated 2.0A,240V is used to boil water from temperature 52°c to hundred°c. if the mass of the water is 2.5kg, determine the time taken to boil water
Answers
Answer:
1.05 x 103s
Explanation:
The answer to the question is 1.05 x 103s
Calculate how much work the force of gravity does on the sphere from B to C .
Answers
(a) The work done by the force of gravity from A to B is 4.41 Joules.
(b) The work done by the force of gravity from B to C is zero.
(c) The work done by the force of gravity from A to C is 4.41 Joules.
a) To calculate the work done by the force of gravity from A to B, we need to consider the change in potential energy. The potential energy at point A is maximum due to the maximum angle of 35.0∘ to the left of vertical, while at point B, the string is vertical, and the potential energy is zero.
The change in potential energy (ΔPE) is given by:
ΔPE = m * g * h
where m is the mass of the sphere (0.500 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the change in height.
Since the potential energy at point A is maximum, the change in height is equal to the length of the string (0.900 m).
ΔPE = 0.500 kg * 9.8 m/s^2 * 0.900 m = 4.41 J
Therefore, the work done by the force of gravity from A to B is 4.41 Joules.
b) From B to C, the change in height is zero since the string is already vertical. Hence, the work done by the force of gravity from B to C is zero.
c) The total work done by the force of gravity from A to C is the sum of the work done from A to B and from B to C.
Total work = Work from A to B + Work from B to C = 4.41 J + 0 J = 4.41 J
Therefore, the work done by the force of gravity from A to C is 4.41 Joules.
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I think it is the question:
A Pendulum Is Made Up Of A Small Sphere Of Mass 0.500 Kg Attached To A String Of Length 0.900 M. The Sphere Is Swinging Back And Forth Between Point A, Where The String Is At The Maximum Angle Of 35.0∘ To The Left Of Vertical, And Point C, Where The String Is At The Maximum Angle Of 35.0∘ To The Right Of Vertical. The String Is Vertical When The Sphere Is At
A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.900 m. The sphere is swinging back and forth between point A, where the string is at the maximum angle of 35.0∘ to the left of vertical, and point C, where the string is at the maximum angle of 35.0∘ to the right of vertical. The string is vertical when the sphere is at point B.
a) Calculate how much work the force of gravity does on the sphere from A to B.
b) Calculate how much work the force of gravity does on the sphere from B to C.
c) Calculate how much work the force of gravity does on the sphere from A to C.
magnetism is an example of:
A. contact force
B. non-contact force
C.friction
Answers
Answer:
The answer is Non contract force
A 12 cm diameter piston-cylinder device contains air at a pressure of 100 kPa at 24oC. The piston is initially 20 cm from the base of the cylinder. The gas is now compressed and 0.1 kJ of boundary work is added to the gas. The temperature of the gas remains constant during this process.
a. How much heat was transferred to/from the gas?
b. What is the final volume and pressure in the cylinder?
c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?
Answers
Answer:
ΔQ = 0.1 kJ
\(\mathbf{v_f = 1.445*10^{-3} m^3}\)
\(\mathbf{P_f = 156.5 \ kPa}\)
ΔS = -0.337 J/K
The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.
Explanation:
GIven that:
Diameter of the piston-cylinder = 12 cm
Pressure of the piston-cylinder = 100 kPa
Temperature =24 °C
Length of the piston = 20 cm
Boundary work ΔW = 0.1 kJ
The gas is compressed and The temperature of the gas remains constant during this process.
We are to find ;
a. How much heat was transferred to/from the gas?
According to the first law of thermodynamics ;
ΔQ = ΔU + ΔW
Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.
Now
ΔQ = ΔU + ΔW
ΔQ = 0 + 0.1 kJ
ΔQ = 0.1 kJ
Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ
b. What is the final volume and pressure in the cylinder?
In an isothermal process;
Workdone W = \(\int dW\)
\(W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}\)
Since the gas is compressed ; then \(v_f< v_i\)
However;
\(W =- nRT \ In \dfrac{V_f}{V_i}\)
\(W =- P_1V_1 \ In \dfrac{V_f}{V_i}\)
The initial volume for the cylinder is calculated as ;
\(v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3\)
Replacing over values into the above equation; we have :
\(100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}\)
The final pressure can be calculated by using :
\(P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f\)
\(P_f =\dfrac{P_iV_i}{V_f}\)
\(P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}\)
\(P_f = 1.565*10^2 \ kPa\)
\(\mathbf{P_f = 156.5 \ kPa}\)
c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?
The change in entropy of the gas is given by the formula:
\(\Delta S=\dfrac{\Delta Q}{T}\)
where
T = 24 °C = (24+273)K
T = 297 K
\(\Delta S=\dfrac{-100 \ J}{297 \ K}\)
ΔS = -0.337 J/K
The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.
HELP right awayyy !!!
Answers
Answer:
Question 4 is actually Acceleration
If Scobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long would Scoobie take to drive around Earth's equator in days
Answers
Scobie will take 10 days to drive around Earth's equator.
To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:
\( d = 2\pi r \)
Where:
r: is the Earth's radius = 6371 km
Then, the distance is:
\( d = 2\pi r = 2\pi*6371 km = 40030.2 km \)
Now, if we divide the above distance by the speed of the car we can find the time:
\( t = \frac{d}{v} = \frac{40030.2 km}{160.0 km/h} = 250.2 h*\frac{1 d}{24 h} = 10 d \)
Therefore, Scobie will take 10 days to drive around Earth's equator.
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What happen when the rest of the container is filled with water?
Answers
Answer:
the water will float over mercury and the steel ball will be in it's respected place
Three capacitors C1 = 10.4 µF, C2 = 20.0 µF, and C3 = 29.5 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum potential difference across the series combination.
Answers
Answer:
66.75 V
Explanation:
A capacitor is a device that stores electrical energy in an electric field. It is made up of two conductor plates separated by a dielectric. It stores electrical charges which then produces a potential difference between the plates.
The equivalent capacitance for capacitors in series is given as:
\(\frac{1}{C_{eq}}=\frac{1}{C_1} +\frac{1}{C_2}+\frac{1}{C_3} \\ \\\frac{1}{C_{eq}}=\frac{1}{10.4} +\frac{1}{20}+\frac{1}{29.5}\\ \\C_{eq}=5.55\ \mu F\)
\(The\ total\ charge\ stored(Q)\ is:\\\\Q=C_{eq}V\\\\Q=5.55*10^{-6}F*125\ V=694 \mu C\)
The potential difference on each capacitor is:
\(V_1=\frac{Q}{C_1}=\frac{694\mu C}{10.4\mu F} =66.75\ V\\\\V_2=\frac{Q}{C_2}=\frac{694\mu C}{20\mu F} =34.71\ V\\\\V_3=\frac{Q}{C_3}=\frac{694\mu C}{29.5\mu F} =23.53\ V\)
The maximum potential difference is 66.75 V across the small capacitor
Use the image below to answer the following question (ruler not to scale).
If magnet B is moved closer to magnet A, how will magnetic potential energy be affected?
O. It will become zero.
O. It willincrease.
O. It will decrease.
O. It will stay the same.
Please help
Answers
Answer:
it depends on wether the + and - are facing eachother
or away from eachother
Explanation:
Answer:
Yes, they are correct, you need to include all parts of the question, don't just copy and paste the words. But in the diagram that was supposed to be there, the magnets' north poles are facing each other. Therefore they would push away from each other if they were to get any closer. So, the magnetic potential energy would decrease due to their movement away from each other. So your answer is (C. It will decrease.
Explanation:
An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of the transverse and longitudinal waves in rock are 8.8 km/s and 5.9 km/s respectively. A seismograph records the arrival of the transverse waves 69 s before that of the longitudinal waves. Assuming the waves travel in straight lines, how far away is the center of the earthquake
Answers
Answer:
1239.216 km
Explanation:
The speed of the transverse = 8.8km/s
The speed of the longitudinal = 5.9km/s
distance = speed x time,
8.8km/s x trans_time = 5.9km/s x long_time
8.8 / 5.9 = long_time / trans_time
1.49 = long_time / trans_time
long_time = 1.49 trans_time
the transverse wave was 69s faster than longitudinal,
trans_time - long_time = 69s
trans_time - 1.49trans_time = 69s
0.49 trans_time = 69
trans_time = 69 / 0.49 = 140.82s
long_time = 140.82 - 69 = 71.82s
the distance of the earthquake;
distance = 8.8 x 140.82 = 1239.216 km
To convert a Galvanometer of internal resistance RG = 200 2 into Ammeter a small resistance connected in parallel with the Galvanometer using a wire of resistance per length equal (2000/km). If the maximum current can be measured by the constructed Ammeter is 10A, assuming that the maximum current sustained by the Galvanometer is equal 0.8 mA. What is the length of the wire used to build up this ammeter?
Answers
To convert a galvanometer into an ammeter, a parallel resistance is added. With the given values, the resistance of the wire used to build the ammeter is 8 Ω/km, resulting in a wire length of 4 meters.
A galvanometer is an instrument that is used to detect and measure small amounts of electric current. It has a high resistance, which means that it only allows a small amount of current to flow through it. To convert a galvanometer into an ammeter, a small resistance is connected in parallel with the galvanometer. This reduces the overall resistance of the circuit, allowing more current to flow through it. The internal resistance of the galvanometer is given as RG = 200 Ω. Let the resistance of the wire used to build the ammeter be R. The maximum current that can be measured by the constructed ammeter is 10 A, and the maximum current sustained by the galvanometer is 0.8 mA.Using Kirchhoff's laws, we can find the total resistance of the circuit. Let this be RT: RT = RG + R, where RG is the internal resistance of the galvanometer and R is the resistance of the wire used to build the ammeter. Substituting the given values: RT = 200 Ω + R. The current through the ammeter is given by: I = V/RT, where V is the voltage across the circuit. Since the voltage is constant, we can write I = k/RT, where k is a constant.The maximum current that can be measured by the ammeter is 10 A. Substituting this value into the equation above 10 = k/(200 + R). The maximum current sustained by the galvanometer is 0.8 mA. This means that the total current through the circuit is 0.8 mA + I, where I is the current through the ammeter. Substituting the values into the equation above: 0.8 mA + k/RT = 10 mA, simplifying, we get k/RT = 9.2 mA.Substituting this value into the equation we derived earlier: 10 = (9.2 mA)/(200 Ω + R). Solving for R, we get R = 8 Ω/km. The resistance per unit length of the wire is given as 2000 Ω/km. Therefore, the length of the wire used to build the ammeter is given by:L = R/2000 = 8/2000 km = 4 m.For more such questions on ammeter:
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Can power be negative in physics? I am doing a lab in which I run down the stairs as quickly as possible. They are 5.4m tall and my work is 5321.7 J. After calculating work, I got -5321.7 J. I am supposed to calculate my power used over 3 trials. The time for trial 1 was 4.69 and the seconds for trials 2 and 3 are 4.25. After calculating my power I got -1134.69 W and -1252.16 W. I am not sure if I got my power right. Is my power really negative or is my power positive?
Answers
Answer:
Yes.
Explanation:
A negative power would just represent a loss of power. So in your case it lost -1252.16 W
explain why sound wave travel faster in liquid than gas
Answers
Answer:
Because gas contains free molecules but not liquid.
Please mark as brainliast
Our galaxy, the Milky Way, contains approximately 4.0 x 1011 stars with anaverage mass of 2.0 X 1030 kg each. How far away is the Milky Way from ournearest neighbor, the Andromeda Galaxy, if Andromeda contains roughly thesame number of stars and attracts the Milky Way with a gravitational force of2.4 x 1030 N?
Answers
We will have the following:
First, we remember:
\(F=\frac{Gm_1m_2}{r^2}\)Then, from the problem we will have that:
\(2.4\ast10^{30}N=\frac{G(2.0\ast10^{30}kg)(2.0\ast10^{30}kg)}{r^2}\)\(\Rightarrow r^2=\frac{G(2.0\ast10^{30}kg)^2}{2.4\ast10^{30}N}\Rightarrow r\approx1.1116666667\ast10^{20}m\)So, the Andromeda galaxy is approximately 1.1*10^20 meters from the milky way.
The acceleration of the Andromeda galaxy towards the milky way is:
\(2.4\ast10^{30}N=(2.0\ast10^{30}kg)a\Rightarrow a=1.2m/s^2\)So, the acceleration towards the milky ways is 1.2m/s^2.
A child throws a ball up with an initial velocity of 4
m/s. If the starting height of the ball was 2.5 meters,
what will be the height of the ball after 0.5 seconds?
Use a = -9.8 m/s/s.
Answers
The height of the ball after 0.5 seconds is 3.275m
What happens at the Maximum Height ?At maximum height reached by an object, the final velocity of that object is zero.
Given that a child throws a ball up with an initial velocity of 4m/s. If the starting height of the ball was 2.5 meters, what will be the height of the ball after 0.5 seconds?
Use a = -9.8 m/s/s.
Let us use the formula h = ut + 1/2at²
h = 4 × 0.5 - 1/2 × 9.8 × 0.5²
h = 2 + 4.9 × 0.25
h = 2 - 1.225
h = 0.775 m
The height of the ball after 0.5 seconds will be 2.5 + 0.775 = 3.275m
Therefore, the height of the ball after 0.5 seconds is 3.275m
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The following table lists the speed of sound in various materials. Use this table to answer the question.
Substance Speed (m/s)
Glass 5,200
Aluminum 5,100
Iron 4,500
Copper 3,500
Salt water 1,530
Fresh water 1,500
Mercury 1,400
Hydrogen at 0°C 1,284
Ethyl Alcohol 1,125
Helium at 0°C 965
Air at 100°C 387
Air at 0°C 331
Oxygen at 0°C 316
Sound will travel fastest in air at _____.
-5°C
0°C
10°C
15°C
Answers
Sound will travel fastest in air at 15°C.
Speed of sound in air
The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;
Air at 100°C 387 m/s
Air at 0°C 331 m/s
From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.
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why does the acceleration due to gravity have to be measured in an airless tube?
Answers
It is measured in the airless tube because there is no air to offer resistance to the fall of the object.
What is the acceleration due to gravity?The acceleration due to gravity refers to the acceleration that is imparted to a body because of its movement in the earth's gravitational field. Irrespective of mass, all objects move with the same acceleration in the earth gravitational field.
If air is present, air resistance makes it appear as if different masses are accelerated to different amounts. Thus, the acceleration due to gravity is measured in an airless tube the where there is no air to offer resistance to the fall of the object.
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Air at 273K and 1.01x10³Nm2 pressure contains 2.70x1025 molecules per cubic meter. How many molecules per cubic meter will there be at a place where the temperature is 223K and pressure is 1.33x10 Nm-2
Answers
The molecules of O2 that are present in 3.90 L flask at a temperature of 273 K and a pressure of 1.00 atm is 1.047 x 10^23 molecules of O2
Step 1: used the ideal gas equation to calculate the moles of O2
that is Pv=n RT where;
P(pressure)= 1.00 atm
V(volume) =3.90 L
n(number of moles)=?
R(gas constant) = 0.0821 L.atm/mol.K
T(temperature) = 273 k
by making n the subject of the formula by dividing both side by RT
n= Pv/RT
n=[( 1.00 atm x 3.90 L) /(0.0821 L.atm/mol.k x273)]=0.174 moles
Step 2: use the Avogadro's law constant to calculate the number of molecules
that is according to Avogadro's law
1 mole = 6.02 x10^23 molecules
0.174 moles=? molecules
by cross multiplication
the number of molecules
= (0.174 moles x 6.02 x10^23 molecules)/ 1 mole =1.047 x 10^23 molecules of O2
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A 250 kg cart is traveling at 8 m/s when it strikes a 100 kg cart at rest. After the elastic collision, the 250 kg cart continues to travel forward but at a lower velocity of 3 m/s. Determine the velocity of the 100 kg cart after the elastic collision.
Answers
Answer: In this scenario, we have two carts colliding with each other. One cart weighs 250 kg and is moving at a speed of 8 m/s, while the other cart weighs 100 kg and is initially at rest.
After the collision, the 250 kg cart continues moving forward, but at a slower speed of 3 m/s. We want to find out the speed at which the 100 kg cart moves after the collision.
To solve this, we use the principle that the total "push" or momentum before the collision should be the same as the total momentum after the collision.
Since the 100 kg cart is initially at rest, its momentum is zero. The momentum of the 250 kg cart before the collision is 250 kg * 8 m/s = 2000 kg·m/s.
After the collision, the momentum of the 250 kg cart becomes 250 kg * 3 m/s = 750 kg·m/s.
To find the momentum of the 100 kg cart after the collision, we subtract the momentum of the 250 kg cart after the collision from the total momentum before the collision: 2000 kg·m/s - 750 kg·m/s = 1250 kg·m/s.
Now, we divide this momentum by the mass of the 100 kg cart to find its velocity: 1250 kg·m/s / 100 kg = 12.5 m/s.
Therefore, the 100 kg cart moves at a velocity of 12.5 m/s after the collision, in the opposite direction of the 250 kg cart's motion.
Consider the circuit in Figure 5 with e(t) = 12sin(120pit) V. When S1 and S2 are
open, i leads e by 30°. When SI is closed and S2 is open, i lags e by 30°. When S1 and S2 are closed, i has an amplitude 0.5A. What are R, L, and C?
Answers
Based on the information, it should be noted that the resistance R is 0.5 Ω.
How to calculate the resistanceWhen S1 and S2 are open, i leads e by 30°. In this case, the circuit consists of only the inductor (L) and the capacitor (C) in series. Therefore, the impedance of the circuit can be written as:
Z = jωL - 1/(jωC)
Since i leads e by 30°, we can express the phasor relationship as:
I = k * e^(j(ωt + θ))
Z = jωL - 1/(jωC) = j(120π)L - 1/(j(120π)C)
Re(Z) = 0
By equating the real parts, we get:
0 = 0 - 1/(120πC)
Let's assume that there is a resistance (R) in series with the inductor and capacitor. The impedance equation becomes:
Z = R + jωL - 1/(jωC)
Z = R + jωL
Im(Z) = ωL > 0
Substituting the angular frequency and rearranging the inequality, we have:
120πL > 0
L > 0
This condition implies that the inductance L must be greater than zero.
When S1 and S2 are closed, i has an amplitude of 0.5 A. In this case, the impedance is:
Z = R + jωL - 1/(jωC)
Since the amplitude of i is given as 0.5 A, we can express the phasor relationship as:
I = 0.5 * e^(j(ωt + θ))
By substituting this phasor relationship into the impedance equation, we can determine the value of R. The real part of the impedance must be equal to R:
Re(Z) = R
Since the amplitude of i is 0.5 A, the real part of the impedance must be equal to 0.5 A: 0.5 = R
Therefore, the resistance R is 0.5 Ω.
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HELP! A student collected data about the distance a car travels over time. Which type of graph should he use to show the data?
A. Circle Graph
B. Bar Graph
C. Scatter plot
D. Histogram