Answer:
The constant force exerted by the wall is F=11.4N
Explanation:
The problem bothers on the impulse of a force
Which is given as
Ft=mv-mu
Ft=m(v-u)
Given data
mass of ball m =2.9kg
Final speed v=8.6m/s
Initial speed u=7.3m/s
Time t= 0.33s
Substituting to find F
F*0.33=2.9(8.6-7.3)
F*0.33=2.9*1.3
F=3.77/0.33
F=11.4N
An ideal gas with k 5 1.4 is flowing through a nozzle such that the Mach number is 1.8 where the flow area is 36 cm2. Approximating the flow as isentropic, determine the flow area at the location where the Mach number is 0.9.
Answer:
The flow area at the location where the Mach number is 0.9 is 25.24 cm²
Explanation:
Here we have for isentropic flow;
[tex]\frac{A}{A^*} = \frac{1}{M}(\frac{2}{k+1} (1+\frac{k-1}{2}M^2))^{(\frac{k+1}{2(k-1)} )[/tex]
Where:
A = Area of flow = 36 cm²
M = Mach number at section of = 1.8
k = Specific heat ratio = 1.4
A* = Area at the throat
Therefore, plugging the values we get
[tex]\frac{36}{A^*} = \frac{1}{1.8}(\frac{2}{1.4+1} (1+\frac{1.4-1}{2}1.8^2))^{(\frac{1.4+1}{2(1.4-1)} ) = 1.439[/tex]
Therefore, A* = 36/1.439 = 25.01769 cm²
Where the Mach number is 0.9, we have
[tex]\frac{A}{25.02} = \frac{1}{0.9}(\frac{2}{1.4+1} (1+\frac{1.4-1}{2}0.9^2))^{(\frac{1.4+1}{2(1.4-1)} ) = 1.009[/tex]
Therefore A = 25.020× 1.009 = 25.24 cm²
The flow area at the location where the Mach number is 0.9 = 25.24 cm².
A car passes the Eiffel tower driving at a constant rate of 60 km per hour. A second car, traveling at a constant rate of 75 km per hour, passes the Eiffel tower a while later and then follows the first car. It catches the first car after traveling a distance of 75 km past the Eiffel tower. How many minutes after the first car passed the Eiffel tower did the second car pass the Eiffel tower
Answer:
300 minutes.
Explanation:
Given,
Speed of car 1 = 60 Km/h
Speed of car 2 = 75 Km/h
distance travel by car 2 to catch car 1 = 75 Km
time taken to catch the car = ?
Relative velocity between the car = 75 - 60 = 15 Km/h
[tex]time = \dfrac{distance}{speed}[/tex]
[tex]time = \dfrac{75}{15}[/tex]
t = 5 hrs
t = 5 x 60 = 300 minutes.
Two cars passing a point at different speeds can be calculated using time, distance, and speed formulas.
A car passes the Eiffel Tower driving at a constant rate of 60 km per hour. A second car, traveling at a constant rate of 75 km per hour, passes the Eiffel Tower later and catches the first car after traveling 75 km past the tower.
To calculate the time difference between the two cars passing the Eiffel Tower, we can use the formula: Time = Distance / Speed. The second car passes the tower 15 minutes after the first car.
7. What is the acceleration of the box?
a. 2.5 m/s2
b. 4 m/s2
c. 6 m/s2
d. 10 m/s2
A 3.70-mm-long, 520 kgkg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 73.0 kgkg construction worker stands at the far end of the beam.What is the magnitude of the torque about the point where the beamis bolted into place?
Answer:
the magnitude of the torque about the point where the beam is bolted into place 12.32kNm
Explanation:
N/B The units given in the question are meant to be m, kg and kg respectively
Given data
Length of beam 3.7m
Mass of beam 520kg
Mass of construction worker 73kg
Torque is expressed as the moment or turning effect brought about by a couple
T= F*d
Where d=perpendicular distance
F= force
Converting The mass of the beam and the worker to force we have
520*10= 5200N
73*10=730N
Note that the force due to the mass of the beam is acting at the center of the beam
See the free body diagram attached
Taking moments about point O
Torque =Σ Fxd
Torque =5200x1.85+7x3.7
=9620+2701
=12321 Nm
=12.32kNm
symbols for components of electrical circuit
Answer:
an electronic symbol is a pictogram used to represent various electrical and electronic devices or functions such as wires batteries resistors and transistors in a symptomatic diagram of an electrical or electronic circuit.
Explanation:
Hope this helps! please vote me Brainliest
Final answer:
Electrical circuit diagrams use specific standardized symbols to represent various components like wires, batteries, resistors, ground, and capacitors. Symbols for wire, resistor, and battery are straightforward, while capacitors have variations depending on type. ANSI and IEC provide different standard symbol guidelines.
Explanation:
Symbols for Components of Electrical Circuits
Electrical circuit diagrams use standardized symbols to represent different components. The wire is depicted as a thin black line and is assumed to be a perfect conductor in diagrams, although in reality it is not perfect but typically close enough for practical purposes. A battery or voltage source is represented by a set of parallel lines, with the longer line denoting the positive terminal and the shorter line the negative terminal. The resistor is illustrated with a zigzag symbol, which includes any resistance within the circuit connections. Ground is typically shown as a set of one to three horizontal lines, sometimes with one of the lines angled towards the bottom.
When it comes to alternating current (AC) sources, the symbol is a sine wave enclosed in a circle. This shows the oscillating nature of the AC voltage. Capacitors are represented with two parallel lines for a fixed capacitor; an electrolytic capacitor is shown with one curved line indicating the negative terminal and one straight line for the positive terminal. A variable-capacitance capacitor is represented with an arrow through the parallel lines, signifying adjustability.
Standards such as the American National Standard Institute (ANSI) and the International Electrotechnical Commission (IEC) provide guidelines for these symbols. While the ANSI standard is widely used for its recognizability, the IEC standard might be preferred for more complex circuits due to its cleaner presentation.
A car is moving with speed 20 m/s and acceleration 2 m/s2 at a given instant. Using a second-degree Taylor polynomial, estimate how far the car moves in the next second. Do you think it would be reasonable to use this polynomial to estimate the distance traveled during the next minute? Why or why not?
Answer:
[tex]T(1)=21 [/tex]
Explanation:
The equation of the position in kinematics is given:
[tex]x(t)=x_{0}+v_{0}t+0.5at^{2}[/tex]
x(0) is the initial position, in this it is 0 v(0) is the initial velocity (20 m/s)a is the acceleration (2 m/s²)So the equation will be:
[tex]x(t)=20t+0.5*2*t^{2}[/tex]
[tex]x(t)=20t+t^{2}[/tex]
Now, the Taylor polynomial equation is:
[tex]f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...[/tex]
Using our position equation we can find f'(t)=v(t) and f''(x)=a(t). In our case a=0, so let's find each derivative.
[tex]f(t)=x(t)=20t+t^{2}[/tex]
[tex]f'(t)=\frac{dx(t)}{dt}=v(t)=20+2t[/tex]
[tex]f''(t)=\frac{dv(t)}{dt}=a(t)=2[/tex]
Using the Taylor polynomial with a = 0 and take just the second order of the derivative.
[tex]f(0)+\frac{f'(0)}{1!}(x)+\frac{f''(0)}{2!}(x)^{2}[/tex]
[tex]f(0)=x(0)=0[/tex]
[tex]f'(0)=v(0)=20[/tex]
[tex]f''(0)=a(0)=2[/tex]
[tex]T(t)=f(0)+\frac{f'(0)}{1!}(t)+\frac{f''(0)}{2!}(t)^{2}[/tex]
[tex]T(t)=\frac{20}{1!}(t)+\frac{2}{2!}(t)^{2}[/tex]
[tex]T(t)=20t+t^{2}[/tex]
Let's put t=1 so find the how far the car moves in the next second:
[tex]T(1)=20*1+1^{2}[/tex]
[tex]T(1)=21 [/tex]
Therefore, the position in the next second is 21 m.
We need to know if the acceleration remains at this value to use this polynomial in the next minute, so I suggest that it would be reasonable to use this method just under this condition.
I hope it helps you!
What is the slit spacing of a diffraction necessary for a 600 nm light to have a first order principal maximum at 25.0°?
Explanation:
Given that,
Wavelength of light, [tex]\lambda=600\ nm=6\times 10^{-7}\ m[/tex]
Angle, [tex]\theta=25^{\circ}[/tex]
We need to find the slit spacing for diffraction. For a diffraction, the first order principal maximum is given by :
[tex]d\sin\theta=n\lambda[/tex]
n is 1 here
d is slit spacing
[tex]d=\dfrac{\lambda}{\sin\theta}\\\\d=\dfrac{6\times 10^{-7}}{\sin(25)}\\\\d=1.41\times 10^{-6}\ m\\\\d=1.41\ \mu m[/tex]
So, the slit spacing is [tex]1.41\ \mu m[/tex].
Why do some quasars have red shifts greater than 1?
A) They are made of tachyons and thus are able to move away from us faster than the speed of light.
B) They are made of different elements than any other objects in the universe.
C) They are moving towards us, which throws off our spectrometers.
D) Einstein proved it is impossible to have a redshift greater than 1; these are all due to gravitational lensing tricking us
Answer:
D) Einstein proved it is impossible to have a redshift greater than 1; these are all due to gravitational lensing tricking us
Explanation:
By Einstein's theory of relativity the maximum value of the red shift is 1. This correspond to a case in which the observed frequency equals the emitted frequency:
[tex]z=\frac{f_e-f_o}{f_o}[/tex]
in other cases the red shift is lower than one.
Hence, the gravitational lens change the measurements of the observed frequency and because of that we calculate a greater red shift.
Hence, the answer is:
D) Einstein proved it is impossible to have a redshift greater than 1; these are all due to gravitational lensing tricking us
hope this helps!!
What color of light has the most energy?
A. Blue
B. Red
C. Green
A Swinging Monkey A 8.72 kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume that the radial distance between the branch and the point where the monkey's mass is located to be 84.1 cm. As it swings through the lowest point on the circle, the monkey has a speed of 2.53 m/s. Calculate the magnitude of the centripetal force acting on the monkey.
Answer:
Centripetal force, F = 66.36 N
Explanation:
Given that,
Mass of the monkey, m = 8.72 kg
The radial distance between the branch and the point where the monkey's mass is located to be 84.1 cm, r = 84.1 cm
The speed of the monkey, v = 2.53 m/s
We need to find the magnitude of the centripetal force acting on the monkey. It is given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{8.72\times (2.53)^2}{84.1\times 10^{-2}}\\\\F=66.36\ N[/tex]
So, the magnitude of the centripetal force acting on the monkey is 66.36 N.
Which electromagnet is the strongest
Answer:
Picture C
Explanation:
Calculate the Kinetic Energy when the mass of the car is 1,200kg and the speed
is 2.78 m/s.
Answer:
Kinetic energy of the car = 4637.04 J
Explanation:
Mass of the car (m) =1200 kg
speed of the car (v) = 2.78 m/sec
Kinetic energy =[tex]\frac{1}{2}mv^{2}[/tex]
Putting the values of m and v,
Hence, Kinetic Energy of the car =[tex]\frac{1}{2}1200\cdot 2.78^{2}[/tex] = 4637.04 J
The kinetic energy of a 1,200kg car traveling at a speed of 2.78 m/s is calculated using the formula KE = ½ mv², yielding a kinetic energy of 4,634.88 joules.
To calculate the kinetic energy (KE) of a car with a mass of 1,200kg traveling at a speed of 2.78 m/s, we need to use the kinetic energy formula KE = ½ mv². Plugging in the given values, we have:
KE = ½ (1,200 kg) × (2.78 m/s)²
KE = ½ × 1,200 kg × 7.7284 m²/s²
KE = 4,634.88 J
Therefore, the kinetic energy of the car is calculated to be 4,634.88 joules.
The High Speed Industrial Drill With Diameter Of 98 Cm Develops 5.85hp At 1900 Rpm. What Torque And Force Is Applied To The Drill Bit? This problem has been solved! See the answer. The high speed industrial drill with diameter of 98 cm develops 5.85hp at 1900 Rpm. What torque and force is applied to the drill bit?
Answer:
The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Explanation:
Given:-
- The diameter of the drill bit, d = 98 cm
- The power at which drill works, P = 5.85 hp
- The rotational speed of drill, N = 1900 rpm
Find:-
What Torque And Force Is Applied To The Drill Bit?
Solution:-
- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).
- The relation between these quantities is given:
T = 5252*P / N
T = 5252*5.85 / 1900
T = 16.171 Nm
- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):
T = F*r
Where, r = d / 2
F = 2T / d
F = 2*16.171 / 0.98
F = 33 N
Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Answer:
1) Torque = 21.934 Nm
2) force = 44.76 N
Explanation:
Power = 5.88 hp
1 hp = 746 W
Power = 5.88 x 746 = 4364.1 W
Angular speed in Rpm = 1900 rpm
But angular speed w = (2¶N)/60 rad/s
= (2 x 3.142 x 1900)/60 = 198.968 rad/s
From,
1) Power P = T x w
Where T = torque
T = P/W = 4364.1/198.968 = 21.934 Nm
2) diameter of drill = 98 cm
Radius = 98/2 = 49 cm = 49x10^-2 m
From torque T = Force x radius
Force = Torque /radius
F = 21.934/49x10^-2 = 44.76 N
Which part of the motor is responsible for reversing the flow of current?
Answer:
Commutator.
Explanation:
It is electrical switch and consist of cylinder. Split rings are two halves of a ring that act as a Commutator. When current in the coil is Zero and is about to change the direction.
How did the lab activities help you answer the lesson question: How do environmental changes affect both living and nonliving things in a watershed? What conclusions can you make based on your observations?
Answer:
In the lab activity, I used a virtual watershed to demonstrate how pollutants enter and flow through the watershed. I constructed food chains to show how biotic factors are related. I also predicted and modeled how human activity affected the biotic factors in an ecosystem. Based on my observations, I can conclude that biotic and abiotic factors are closely linked and that human activity that changes the environment, like pollution, can affect all the other parts of an ecosystem in a negative way.
Explanation:
Answer:
In the lab activity, I used a virtual watershed to demonstrate how pollutants enter and flow through the watershed. I constructed food chains to show how biotic factors are related. I also predicted and modeled how human activity affected the biotic factors in an ecosystem. Based on my observations, I can conclude that biotic and abiotic factors are closely linked and that human activity that changes the environment, like pollution, can affect all the other parts of an ecosystem in a negative way.
A falling skydiver opens his parachute. A short time later, the weight of the skydiver-parachute system and the drag force exerted on the system are equal in magnitude. Which of the following statements predicts the motion of the skydiver at this time? a.The skydiver is at rest. b.A The skydiver is moving downward with constant speed. c.The skydiver is moving downward with an upward acceleration. d.The skydiver is moving downward with a downward acceleration.
A falling skydiver opens his parachute. A short time later, the weight of the skydiver-parachute system and the drag force exerted on the system are equal in magnitude. The following statements predicts the motion of the skydiver at this time
The skydiver is moving downward with constant speed.
Explanation:
Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.
The forces acting on a falling leaf are : gravity and air resistance.
The net force and the acceleration on the falling skydiver is upward.
An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.
As the speed decreases, the amount of air resistance also decreases until once more the skydiver reaches a terminal velocity.
A skydiver falling at a constant speed opens his parachute. When the skydiver is falling, the forces are unbalanced.
The drag force opposes the relative motion of an object in a fluid
The option that gives the statement that predicts the motion of the skydiver when the magnitude of the weight of the skydiver-parachute system and the drag force are equal is option b.
b. The skydiver is moving downward with constant speed
The reason the above option is correct is as follows:
The skydiver is falling through the air which is a fluid and is therefore subject to drag force, [tex]F_D[/tex], which is given as follows;
[tex]F_D = \dfrac{1}{2} \times \rho \times v^2 \times C_D \times A[/tex]
Where;
[tex]F_D[/tex] = The drag force
ρ = The fluid density
v = The relative object speed compared to the fluid
A = Cross sectional area
[tex]C_D[/tex] = The drag coefficient
Therefore, given that the surface area of the falling skydiver is largely increased when the parachute is opened and the drag force is a function of the square of the velocity, we have that the drag force rapidly equals the weight of the skydiver, such that the net force becomes zero, and therefore, the speed of the skydiver becomes constant
Therefore, the correct option is that the skydiver is moving downward with constant speed
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In an inelastic collision, a steel ball of mass 200 g was hit hard into a large ball of dough of mass 700 g. The velocity of the steel ball was 25 m/s. After the collision, the steel ball was stuck in the dough, and this combined system rolled for some time. What was the velocity of the system of the steel ball and dough after the collision?A) 5.6 m/sB) 7.5 m/sC) 8.8 m/sD) 11.1 m/s
Answer:
Velocity after collision will be 5.56 m/sec
So option (a) will be correct answer
Explanation:
Mass of steel ball [tex]m_1=200gram=0.2kg[/tex]
Speed of steel ball before collision [tex]v_1=25m/sec[/tex]
Mass of large ball [tex]m_2=700gram=0.7kg[/tex]
Velocity of large ball [tex]v_2=0m/sec[/tex]
According to conservation of momentum
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]0.2\times 25+0.7\times 0=(0.7+0.2)v[/tex]
[tex]v=5.56m/sec[/tex]
So velocity after collision will be 5.56 m/sec
So option (a) will be correct answer
Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence of dark energy. Rank each model from left to right based on the ratio of its actual mass density to the critical density, from smallest ratio (mass density much smaller than critical density) to largest ratio (mass density much greater than critical density).Smallest to largesta.coasting universeb.critical universec.recollapsing universe
Answer:
From smallest ratio to the largest ratio:
Coasting Universe - Critical Universe - Recollapsing Universe(From left to right)
Explanation:
The coasting universe is one that expands at a constant rate given by the Hubble constant throughout all of cosmic time. It has a ratio of actual density to critical density that is less than 1
The critical universe is one that is at balance with no expansion .I.e. the actual density and the critical density are equal, which makes the ratio of actual density to critical density to be equal to 1
Recollapsing Universe: The expansion of the universe reverses in the future and the universe eventually recollapses. The recollapsing universe has the ratio of the actual density to the critical density to be greater than 1
From smallest to largest ratio of actual mass density to critical density, the models are: coasting universe, critical universe, and recollapsing universe. In a coasting universe, expansion continues at a decreasing rate. A critical universe's expansion slows to an eventual stop. A recollapsing universe's expansion ultimately reverses, leading to contraction.
Explanation:The long-term expansion and possible contraction of the universe under different scenarios or models, in the absence of dark energy, can be ranked based on the ratio of its actual mass density to the critical density.
Continuing from the smallest mass density ratio to the largest, it would be: the coasting universe, the critical universe, and finally the recollapsing universe.
In a coasting universe, the actual density is lower than the critical density. Thus, it would continue to expand forever but at a decreasing rate. In a critical universe, the actual density equals the critical density, which means the universe's expansion will gradually slow to a stop in an infinitely far future. Lastly, in a recollapsing universe, the actual density is greater than the critical density. This leads to an eventual reversal of the expansion, causing the universe to start contracting.
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Two objects are dropped from rest from the same height. Object A falls through a distance [tex]d_A[/tex] during a time [tex]t[/tex], and object [tex]B[/tex] falls through a distance [tex]d_B[/tex] during a time [tex]2t[/tex]. If air resistance is negligible, what is the relationship between [tex]d_A[/tex] and [tex]d_B[/tex]?
Answer is [tex]d_A=\frac{1}{4} *d_B[/tex]
Please show me how I can get the answer. Thank you.
Answer:
The answer to your question is given below
Explanation:
Since both object A and B were dropped from the same height and the air resistance is negligible, both object A and B will get to the ground at the same time.
From the question, we were told that object A falls through a distance to dA at time t and object B falls through a distance of dB at time 2t.
Remember, both objects must get to the ground at the same time..!
Let the time taken for both objects to get to the ground be t.
Time A = Time B = t
But B falls through time 2t
Therefore,
Time A = Time B = 2t
Height = 1/2gt^2
For A:
Time = 2t
dA = 1/2 x g x (2t)^2
dA = 1/2g x 4t^2
For B
Time = t
dB = 1/2 x g x t^2
Equating dA and dB
dA = dB
1/2g x 4t^2 = 1/2 x g x t^2
Cancel out 1/2, g and t^2
4 = 1
4dA = dB
Divide both side by 4
dA = 1/4 dB
A textbook of mass 2.02 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.190 m , to a hanging book with mass 3.05 kg . The system is released from rest, and the books are observed to move a distance 1.19 m over a time interval of 0.800 s .
A) What is the tension in the part of the cord attached to the textbook?
B) What is the tension in the part of the cord attached to the book?
C) What is the moment of inertia of the pulley about its rotation axis?
Answer:
a. 7.38 N b. 40.87 N c. 0.113 kg-m²
Explanation:
At 20 ◦C, a rod is exactly 20.05 cm long on a steel ruler. Both the rod and the ruler are placed in an oven at 270 ◦C, where the rod now measures 20.11 cm on the same ruler. What is the coefficient of linear expansion for the material of which the rod is made? Take αsteel = 1.2 × 10−5 K−1 .
The coefficient of linear expansion for the material of the rod can be calculated using the formula ΔL/L = α ΔT to be α = 2 x 10⁻⁵ °C⁻¹.
Explanation:The subject of this question is the calculation of the coefficient of linear expansion for a certain material. The relevant formula here is: ΔL/L = α ΔT, where ΔL is change in length, L is original length, α is coefficient of linear expansion and ΔT is change in temperature.
Firstly, we calculate the change in length of the rod (ΔL) which is the final length minus initial length, 20.11 cm - 20.05 cm = 0.06 cm. Then, the change in temperature (ΔT) is the final temperature minus the initial one: 270°C - 20°C = 250°C.
After substitution into the formula, we get 0.06 cm / 20.05 cm = α × 250°C. Solving for α, we get the coefficient of linear expansion for the material of the rod to be α = 2 x 10-5 °C⁻¹.
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The coefficient of linear expansion for the material of the rod is approximately 0.011974 cm/°C.
Explanation:Solution:To find the coefficient of linear expansion for the material of which the rod is made, we can use the equation for linear thermal expansion:
ΔL = αLΔT
where ΔL is the change in length, α is the coefficient of linear expansion, L is the initial length of the rod, and ΔT is the change in temperature.
Given that the initial length of the rod is 20.05 cm, the final length is 20.11 cm, and the change in temperature is 270 °C - 20 °C = 250 °C, we can substitute these values into the equation to solve for α:
ΔL = αLΔT20.11 cm - 20.05 cm = (α)(20.05 cm)(250 °C)0.06 cm = 5.0125 α °Cα = 0.06 cm / 5.0125 °Cα = 0.011974 cm/°CHence, the coefficient of linear expansion for the material of which the rod is made is approximately 0.011974 cm/°C.
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Which moon has a thick atmosphere made mostly of nitrogen?
Titan, Saturn's moon, has a thick atmosphere primarily composed of nitrogen. Its unique chemistry and thick atmosphere make it an interesting place for scientific study.
Explanation:Titan, Saturn's moon, has a thick atmosphere made mostly of nitrogen. The atmosphere on Titan is thicker than Earth's and is primarily composed of nitrogen, with about 5% methane. The thick atmosphere and unique chemistry of Titan make it an interesting place for scientific study, as it may provide clues about the early chemistry of the solar system and the potential for life.
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Order the electromagnetic waves from lowest
frequency (1) to highest (4).
Gamma:
Infrared:
X-ray:
Visible light:
The electromagnetic waves from the lowest frequency to the highest frequency are gamma rays, X-ray, visible light, and infrared.
What are electromagnetic waves?Electromagnetic waves are the waves which are formed when an electric field couples with a magnetic field. Magnetic and the electric fields of an electromagnetic wave are perpendicular to each other and to the direction of the wave propagation. Electromagnetic waves are used widely in the food processing for destroying microbes.
Therefore, the correct order of electromagnetic waves from the lowest frequency to the highest frequency are gamma rays, X-ray, visible light, and infrared.
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An archer tests various arrowheads by shooting arrows at a pumpkin
that is suspended from a tree branch by a rope, as shown to the right.
When struck head‐on by the arrow, the pumpkin swings upward on the
rope. The maximum angle θ that the rope makes with the vertical is
different for each arrowhead that the archer tests. Each arrow, including
its arrowhead, has the same mass m and is shot with the same velocity
vo toward the right.
The arrowheads are made of different materials, however, and each
behaves differently when it strikes the pumpkin, as described below.
Embedded arrow: Strikes the pumpkin and remains embedded,
while the pumpkin swings to angle θemb
Pass arrow: Passes all the way through the pumpkin and continues traveling away from the archer, while the
pumpkin swings to angle θpass
Bounce arrow: Bounces off the pumpkin back toward the archer, while the pumpkin swings to angle θbounce
(a) Rank the three angles θemb, θpass, θbounce from greatest to least in the spaces indicated below. Use “1” for the greatest
angle, “2” for the next greatest, and so on. If any two or all three angles are the same, use the same number for their
ranking.
(b) In a clear, coherent, paragraph‐length response that may also contain figures and/or equations, justify your ranking
a. The greatest angle will be bouncing, then embedded and finally the least is passing angle.
b. The Momentum is greatest when the arrow bounces and least when the arrow passes. This justify the obtained ranking of angles.
The given problem is based on the concept of momentum and angle made at various instance. Momentum is defined as an impact caused due to applied force. It is expressed as the product of mass and velocity of an object, with respect to the direction. Taking forward direction of arrow as positive then backward direction will be negative.
From the law of conservation of linear momentum, the sum of initial and final momentum should be equal.
When the arrow moves forward then bounces, it means the pumpkin will have positive momentum while arrow during bouncing will have negative energy. When the arrow is embedded in the pumpkin, both pumpkin and arrow will have same momentum. Finally, when the arrow passes through the pumpkin, it means it will have positive momentum hence pumpkin momentum is very little.
Thus we can conclude that by ranking the angles, consider that the maximum angle that the pumpkin reaches depends on the speed of the pumpkin. The greatest angle will be bouncing, then embedded and finally the least is passing angle.
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Your system engineer has started making negative comments during your weekly team meeting. He has had a heated argument with the marketing manager, and you have heard from various team members that he has become difficult to work with. What is the best course of action for you to take?
A. You should write a memo to the system engineer's functional manager and request a replacement as soon as possible.
B. The system engineer is critical to the project, so you should give him some slack and wait to see whether the behavior stops.
C. You should confront the system engineer openly at the next team meeting. Let him know that his behavior is unacceptable and that he will be replaced if there is not an immediate change.
D. You should schedule an individual meeting with the system engineer to determine whether he has issues with the project that need to be resolved. Get his perspective on how the project is progressing and how he feels about his role.
Answer: D. To address the issue, you need to understand what is behind the system engineer's current behavior. He may have been given additional work that you are not aware of, or he may misunderstand the project goals, to name just a couple of possibilities. The situation cannot be ignored, no matter how valuable the person is, and it should be handled in private.
Answer :
D. You should schedule an individual meeting with the system engineer to determine whether he has issues with the project that need to be resolved. Get his perspective on how the project is progressing and how he feels about his role.
What are the wavelengths of electromagnetic wave in free space that have the following frequencies?
(a) 4.00 1019 Hz
(b) 5.50 109 Hz
Explanation:
Given that,
(a) Frequency, [tex]f_1=4\times 10^{19}\ Hz[/tex]
All electromagnetic wave moves with the speed of light. It is given by :
[tex]c=f\lambda[/tex]
[tex]\lambda_1=\dfrac{c}{f_1}\\\\\lambda_1=\dfrac{3\times 10^8}{4\times 10^{19}}\\\\\lambda_1=7.5\times 10^{-12}[/tex]
(b) Frequency, [tex]f_2=5.5\times 10^{1=9}\ Hz[/tex]
All electromagnetic wave moves with the speed of light. It is given by :
[tex]c=f\lambda[/tex]
[tex]\lambda_2=\dfrac{c}{f_2}\\\\\lambda_2=\dfrac{3\times 10^8}{5.5\times 10^{9}}\\\\\lambda_2=0.054\ m[/tex]
Hence, this is the required solution.
If a gas is compressed isothermally, which of the following statements is true? Energy is transferred into the gas by heat. No work is done on the gas. The temperature of the gas increases. The internal energy of the gas remains constant. None of those statements are true.
Answer:
The internal energy of the gas remains constant,
Explanation:
The isothermal process can be defined as a thermodynamic process in which the temperature of the system remains constant.In isothemal process the process of transfer of heat energy from the surrounding to the system or to the system from the surrounding takes place at constant temperature.
in isothermal process internal energy of the system remains unchanged or constant.
For ideal gas internal energy of the system depends on the temperature.
For ideal gas when it is compressed the work is done by the surrounding on the gas is positive and heat added to the system is negative .When the heat energy is absent,both temperature and internal energy increases. As it is a isothermal process and temperature remains constant the gas must transfer the positive amount of heat to the system.
Hence internal energy of the system of gas remains constant.
Final answer:
During an isothermal compression of a gas, energy is transferred into the gas by heat to maintain its constant temperature, while work is done on the gas and its internal energy remains constant.
Explanation:
When a gas is compressed isothermally, specific events related to energy take place. Isothermal processes occur at a constant temperature, implying that the thermal energy of the gas does not change. If we refer to the first law of thermodynamics, stating that the change in internal energy (\(\Delta E_{int}\)) is equal to heat added to the system (Q) minus the work done by the system (W), we can understand these events. Since the temperature remains constant during an isothermal compression, the internal energy of an ideal gas, which depends only on temperature, also remains constant (\(\Delta E_{int} = 0\)). Work is done on the gas during compression, meaning that there is a transfer of energy into the gas by way of work (W > 0).
To maintain the constant temperature, heat must be removed from the gas equal to the work done on it (Q = W), so there is effective energy transfer into the gas by heat to maintain isothermality. Therefore, the correct statement among the given options is: Energy is transferred into the gas by heat.
An inverted image of an object is viewed on a screen from the side facing a converging lens. An opaque card is then introduced covering only the upper half of the lens. What happens to the image on the screen?
Answer:
d) The image would be dimmer
Explanation:
When inverted image of an object is viewed on a screen from the side facing a converging lens, the
lens focuses the diverging, and blurred light rays from the distance object through refraction of the rays two times.Then the rays is converge by the double bending cl at a focal point behind the lens inorder for a sharper image to be be observed.
But when an opaque card is then introduced to cover only the upper half of the lens then the image becomes dimmer because of the reduction in the light intensity on the screen by 50%
When the opaque card should be introduced so here the image should be considered as the dimmer.
Inverted image:At the time when the inverted image should be viewed on the screen that faced the converging lens so here the lens should be focused on the diverging also there should be blurred light rays from the distance object via the refraction of the rays.
However, at the time there is the opaque card so here the image should be dimmer since there is the decrease in the light intensity by 50%/
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Which explains earthquakes and volcanic eruptions
Answer: plate tectonics
Explanation: plate tectonics move constantly movement in narrow zones along plate boundaries causes earthquakes and plate tectonics come together and create pressure upon the magma deposit underground forcing it up into a volcanic eruption
A small sphere of mass m is launched horizontally over a body of water from a height h above the water and with a launch speed v0. Determine expressions for the following in terms of m, v0, h, and g. Air resistance is negligibly small.
(a) W is the amount work done by the force of gravity on the projectile during its flight.
W =
−mgh
For a conservative force, how does the work done by the force compare to the change in potential energy associated with the force?
Answer:
a) W = m g h , b) W / U = 1
Explanation:
a) work is defined by
W = F. dy
in this case the force of gravity goes down and the displacement of the particle is down, so the two are parallel and the scalar product is reduced to the algebraic product
W = F y
strength is the weight of the body
F = m g
W = m g h
where h is the distance the body descends
b) the only force acting on the body is the weight of the work and we calculated them in part a, the potential energy is
U = m g h
to compare the two magnitudes of let's find their relationship
W / U = mgh / mgh
W / U = 1