Answer:
False.
Explanation:
Master production schedule (MPS) is nothing but plan for the individual commodities to be produced in a factory, during to a time period. MPS includes Planning, production, staffing , inventory, etc. It preferably used in places where it is know that when and how each product is demanded. It has nothing to deal with decision and breaking down of aggregate planning.
The definition of MPS given in question is wrong. There the given statement is false.
Yellow light travels through a certain glass block at a speed of 1.97 × 108 m/s. The wavelength of the light in this particular type of glass is 3.81 × 10−7 m (381 nm). What is the frequency of the yellow light in the glass block? Answer in units of H
The frequency of the yellow light is [tex]5.17\cdot 10^{14}Hz[/tex]
Explanation:
The relationship between wavelength, frequency and speed of a wave is given by the wave equation:
[tex]v = f \lambda[/tex]
where
v is the speed of the wave
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the yellow light in this problem, we have:
[tex]v=1.97\cdot 10^8 m/s[/tex] is the speed
[tex]\lambda=3.81\cdot 10^{-7} m[/tex] is the wavelength
Solving for f, we find its frequency:
[tex]f=\frac{v}{\lambda}=\frac{1.97\cdot 10^8}{3.81\cdot 10^{-7}}=5.17\cdot 10^{14}Hz[/tex]
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The frequency of the yellow light in the glass block is [tex]5.17 \times 10^{14}Hz[/tex]
Given information:Yellow light travels through a certain glass block at a speed of 1.97 × 108 m/s. The wavelength of the light in this particular type of glass is 3.81 × 10−7 m (381 nm).
Calculation of the frequency:[tex]= 1.97 \times 10^8 \div 3.81 \times 10^{-7}\\\\= 5.17 \times 10^{14}Hz[/tex]
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Light with a wavelength of 587.5 nm illuminates a single slit 0.750 mm in width.
(a.) At what distance from the slit should a screen be located if the first minimum in the diffraction pattern is to be 0.850 mm from the center of the screen?
b.) What is the width of the central maximum?
(c.) Sketch the intensity distribution for the diffraction pattern observed on the viewing screen.
Answer:
a) The screen should be located at 1.08 meters
b) The width of the central maximum is 1.7 mm
c) See figure below.
Explanation:
a) This is a single slit diffraction problem, the equation that describes this kind of phenomenon is:
[tex]a\sin\theta=m\lambda [/tex] (1)
Because we’re interested in a minimum near the center of the screen, we can use the approximation [tex] \sin\theta\approx\tan\theta=\frac{y}{x} [/tex]
So equation (1) is now:
[tex]a\frac{y}{x}=m\lambda [/tex] (2)
Solving (2) for x:
[tex] x=\frac{ay}{m\lambda}=\frac{(0.75\times10^{-3})(0.85\times10^{-3})}{1(587.5\times10^{-9})}\approx1.08m [/tex]
b) As you can see on the figure below a maximum is approximately between the two adjacent minimums, because the diffraction pattern is approximately symmetric respect the center of the screen the width of the central maximum is 2*0.850mm = 1.7 mm.
A spring with k = 53 N/m hangs vertically next to a ruler. The end of the spring is next to the 18 cm mark on the ruler. If a 2.4 kg mass is now attached to the end of the spring, and the mass is allowed to fall, where will the end of the spring line up with on the ruler marks when the mass is at its lowest position?
Answer:
1.07 m
Explanation:
x = Compression of the spring
k = Spring constant = 53 N/m
Initial length = 18 cm
P = Kinetic energy
K = Kinetic energy
At the lowest point of the mass the energy conservation is as follows
[tex]P_{ig}+P_{is}+K_i=P_{fg}+P_{fs}+K_f\\\Rightarrow mgx+0+0=mgx+\frac{1}{2}kx^2\\\Rightarrow x=\frac{2mg}{k}\\\Rightarrow x=\frac{2\times 2.4\times 9.81}{53}\\\Rightarrow x=0.89\ m[/tex]
At its lowest position the mark on the ruler will be
[tex]x_f=0.18+0.89\\\Rightarrow x_f=1.07\ m[/tex]
The spring line will end up at 1.07 m
In this scenario, when a 2.4 kg mass is attached to a spring, the spring gets displaced due to the force exerted by the mass due to gravity. The displacement can be calculated using Hooke's Law (F = kx), resulting in a mark of 62.4 cm on the ruler when the mass is at its lowest position.
Explanation:
The phenomenon described in this question is related to Hooke's Law of Physics, which states that the force required to extend or compress a spring by a distance is proportional to that distance. The force exerted by the spring is measured in Newtons (N) and is given by the equation F = kx, where k is the spring constant and x is the distance the spring is stretched or compressed.
In this case, when the 2.4 kg mass is attached to the spring, it will exert a force due to gravity which is equal to the mass times the acceleration due to gravity (g = 9.8 m/s²), therefore F = m*g = 2.4kg * 9.8m/s² = 23.52 N. The spring reacts to this force and gets displaced, which can be calculated using x = F/k = 23.52N / 53N/m = 0.444 m or 44.4 cm. Therefore, with the mass attached, the bottom of the spring would fall to 18cm + 44.4 cm = 62.4 cm on the ruler.
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Stars of spectral type A and F are considered ________.
a. reasonably likely to have Earth-like planets with complex plant- and animal-like life
b. reasonably likely to have habitable planets but much less likely to have planets with complex plant- or animal-like life
c. unlikely to have planets of any kind
d. unlikely to have habitable planets
Answer:
B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.
Explanation:
The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:
• They live at least a few billion years, allowing life a chance to evolve. More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.
• They emit enough high-frequency ultraviolet radiation to trigger important atmospheric dynamics such as ozone formation, but not so much that ionisation destroys incipient life.
• They emit sufficient radiation at wavelengths conducive to photosynthesis.
• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.
Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.
A turntable is spinning on a frictionless axle when a ball of putty is dropped onto it from above. The putty sticks to the turntable when it lands. Is angular momentum conserved for the turntable and putty (the system) during this process?
Answer:
Yes
Explanation:
Gravity and forces form the axle (external forces) do not cause forces that tend to cause rotation around the turntable's axle.
People fighting forest fires carry emergency tents that haveshiny aluminum outer surfaces. If there is trouble, a fire fightercan lie under the tent to block the heat from burning treesoverhead. The tent helps because
A) radiation carries heat downward toward the fire fighter and thealuminum tent reflects most of that radiation.
B) convection carries heat downward toward the fire fighter and thealuminum tent blocks most of the heat carried by convection.
C) both conduction and radiation carry heat downward toward thefire fighter and the aluminum tent blocks most of that heat.
D) conduction carries heat downward toward the fire fighter and thealuminum tent conducts that heat harmlessly into the ground.
Answer:
option A
Explanation:
The correct answer is option A
People fighting forest fire carry emergency tent which has a shiny outer surface because the radiation of the heat which is downward toward firefighter the aluminium tents reflects the heat from the fire.
Aluminium Reflects 95 % of the infrared heat that hit the shiny surface of aluminium i.e. it is used by the firefighter to prevent from the heat radiation.
Final answer:
Emergency tents with shiny aluminum outer surfaces protect firefighters from heat primarily by reflecting infrared radiation. Convection plays a lesser role as hot air rises, and conduction is minimal. Radiative heat, the main form of heat transfer from fire, is what the tents are designed to protect against.
Explanation:
The reason emergency tents with shiny aluminum outer surfaces help firefighters when they lie under them to block heat from burning trees overhead is due to the tent's ability to reflect infrared radiation. Heat from fires is primarily transferred through radiation, particularly infrared radiation, which the shiny aluminum surface is effective at reflecting away. This reflection helps to protect the firefighter by reducing the amount of heat that reaches them. Convection plays a smaller role in transferring heat downward in this scenario, since hot air tends to rise, and conduction is negligibly slow here, especially from the fire to the tent. Therefore, the tents are designed to reflect the intense radiative heat that a fire emits, which is the main mechanism at play in this scenario for heat transfer.
At one point in time, the price of gold was about $1400 per ounce, while that of silver was about $20 an ounce. The "ounce" in this case is the troy ounce, which is equal to 31.1035 g. (The more familiar avoirdupois ounce is equal to 28.35 g.) The density of gold is 19.3 g/cm3 and that of silver is 10.5 g/cm3.
A. If you found a spherical gold nugget worth $5 million, what would be its diameter?
ANSWER: ______________ cm
B. How much would a silver nugget of that size be worth?
ANSWER: $ ____________
Answer:
Gold nugget diameter 22,24 cm
Silver nugget price 38860 $
Explanation:
Gold price 1400 $ /ou 1 ounce = 31.1035 grs
so 1400 /31.1035
Gold price is 45,01 $/grs.
If a nugget worth 5000000 $ then 5000000/ 45.01
a nugget mass : 111086,43 grs
Now gold density is d = 19.3 grs/cm³
And she volume is V = 4/*π*r³
d = m/V V = m/d V = 111086.43/19.3 cm³
V = 5755,77 cm³
Now V of the sphere is V = 5755,77 = 4/3*π*r³
r³ =3*5755,77 / 4π r³ = 1374,79
r = 11.12 cm 2r = 22,24 = Φ (sphere diameter)
B) d (silver) = m/V m = d*V
V = 5755,77 cm³ The same size the same volume
m = 10,5 * 5755,77 [grs/cm³ * cm³] m = 60435,59 grs
Silver nugget worth :
20 $ /ou 20/31.1035 = 0.643 $ /grs
Price 0,643 * 60435,59 = 38860 $
An angstrom is about the size of an atom and a fermi is about the size of the nucleus of an atom. Given that 1 angstrom unit = 10⁻¹⁰ m and 1 fermi = 10⁻¹⁵ m, what is the relationship between these units?
Answer:
See explanation below
Explanation:
To do this, we need the relation between a fermi and angstrom. We know the relation between the angstrom and meters, and fermi and meters, so, we can actually solve this by doing the conversion of meters.
1 A = 1x10^-10 m
1 m = 1 A / 1x10^-10 m
1 m = 1x10^10 A
Now if we do the same thing with the fermi:
1 f = 1x10^-15 m
1 m = 1 f / 1x10^-15 m
1 m = 1x10^15 f
then:
1x10^10 A = 1x10^15 f
A/f = 1x10^10 / 1x10^15
A/f = 1x10^-5
A 70.0-kg person throws a 0.0420-kg snowball forward with a ground speed of 35.0 m/s. A second person, with a mass of 57.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.20 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged?
Answer:
so throwers new velocity = 2.18032m/s
so catchers new velocity = 0.02577m/s
Explanation:
Directly by conservation of momentum we can write
[tex]m_1u_1+m_2u_2= m_1v_1+m_2v_2[/tex]
let x be the thrower's new velocity
(70+0.042)×2.2 + 57×0 = 70× x +0.042×35 +57×0
x = 2.18032m/s
so the velocity of 70 kg man = 2.18032m/s
so throwers new velocity = 2.18032m/s
now again by conservation of momentum
0.042×35 = (57+0.042) ×y
y = 0.02577m/s
so catchers new velocity = 0.02577m/s
On the moon the surface temperature ranges from 379 K during the day to 1.04 x 102 K at night. Convert these temperatures to the Celsius and Fahrenheit scales.
(a) 379 K in degrees Celsius;
(b) 379 K in degrees Fahrenheit;
(c) 1.04 x 102 K in degrees Celsius;
(d) 1.04 x 102 K in degrees Fahrenheit.
Answer:
Explanation:
Relation between Celsius and Kelvin is
K = C + 273
Relation between Kelvin and Fahrenheit is
[tex]\frac{K-273}{100}=\frac{F -32}{180}[/tex]
(a) 379 K
Relation between Celsius and Kelvin is
K = C + 273
So, C = K - 273 = 379 - 273 = 106
Thus, 379 K = 106°C.
(b) 379 K
Relation between Kelvin and Fahrenheit is
[tex]\frac{K-273}{100}=\frac{F -32}{180}[/tex]
[tex]\frac{379-273}{100}=\frac{F -32}{180}[/tex]
F = 222.8°F
Thus, 379 K = 222.8°F
(c) 1.04 x 10^2 K = 104 K
Relation between Celsius and Kelvin is
K = C + 273
So, C = K - 273 = 104 - 273 = -169
Thus, 104 K = - 169°C.
(d) 1.04 x 10^2 K = 104 K
Relation between Kelvin and Fahrenheit is
[tex]\frac{K-273}{100}=\frac{F -32}{180}[/tex]
[tex]\frac{104-273}{100}=\frac{F -32}{180}[/tex]
F =- 272.2°F
Thus, 104 K = - 272.2°F
An airplane takes hours to travel a distance of kilometers against the wind. The return trip takes hours with the wind. What is the rate of the plane in still air and what is the rate of the wind?
Answer:
Rate of plane in still air = P = W (t1 +t2)/ (t1-t2)
Rate of wind in still air = W = P (t1 - t2)/(t1 + t2)
Explanation:
Assuming speed of plane are the same on both trips
Rate (D/t1) = (P-W).... EQU 1 going from city a to b
Rate (D/t2) = (P +W)...Equ2 going back to city a
Where t1 is not equal to t2
Where D=distance between two cities
P &W are the speed of plane and wind
t1 &t2 = time taken for travel
Equ 1 & equ 2 becomes
D = ( P - W ) t1.. equ3
D = ( P + W ) t2 equa4
Equating equ 3 and 4
Pt1 - Wt1 = Pt2 +Wt2
P ( t1 - t2) = W ( t1 + t2)
Rate of plane in still air = P = W (t1 +t2)/ (t1- t2)
Rate of wind in still air = W = P (t1 - t2)/(t1 + t2)
A small car traveling with 5,000 kgm/s hits a light truck that was at rest at a red light. After the collision the small car bounces backwards with a momentum of 3,000 kgm/s. What would the momentum of the truck be after the collision?
Answer:
8000
Explanation:
cause in the problem we dont have any ext Forces we know that the momentum of system is constant.
Fext=0
P1 + P2 = P1' + P2'
P2 = 0
5000 = -3000 + P2'
P2' = 5000+3000 = 8000
A wheel rotates with a constant angular speed ω. Which one of the following is true concerning the angularacceleration α of the wheel, the tangential accelerationaT of a point on the rim of the wheel, and thecentripetal acceleration ac of a point on the rim?
a)α= 0rad/s2, aT= 0m/s2, ac= 0 m/s2
b)α=0 aT0 ac= 0
c) α0 aT=0 ac=0
d)α=0 aT=0 ac0
e)α 0 aT0 ac0
Answer:
None of the given options
Explanation:
If the angular speed is constant, there is no angular acceleration:
α=0
If there is no angular acceleration, tangential acceleration will also be 0.
Centripetal acceleration, on the other hand, will be:
[tex]ac =\omega^2*R[/tex] (Different to 0)
A piano tuner hears three beats per second when a tuning fork and a note are sounded together and six beats per second after the string is tightened. What should the tuner do next, tighten or loosen the string? Explain.
Final answer:
The piano tuner should loosen the piano string after hearing an increase from three beats per second to six beats per second upon tightening the string, as this indicates the string's frequency was adjusted away from the tuning fork's frequency.
Explanation:
When a piano tuner hears beats, it indicates that there is a frequency difference between a tuning fork and the piano string that is being compared. Initially, the tuner hears three beats per second, which means the frequency of the piano string is either slightly higher or lower than that of the tuning fork. After tightening the string, the number of beats per second increases to six. This indicates that the frequency of the string has moved further away from the frequency of the tuning fork.
The fact that the beats increased after tightening the string implies that the tuner has adjusted the frequency of the piano string in the wrong direction. Since the goal is to eliminate the beats entirely by matching frequencies, the tuner should loosen the string to reduce the frequency instead of tightening it further.
Which is the best aerobic exercise plan?
Answer:
Running or Jogging
Running and jogging are both great options for aerobic conditioning. Whether you run at the gym or outside, you are in control of setting the intensity of your workout. When aiming to build muscle mass, you can add more resistance or jog at an incline, along with increasing your speed.
Explanation:
Answer:
Running
Jogging
Yoga
Explanation:
Running Jogging and yoga best describes aerobic exercise plan. Aerobic exercises are those where cardiovascular conditioning happen it can be swimming, cycling ,running, jogging etc. aerobic exercise is good for health as it reduces risk of:- obesity, high blood pressure type 2, diabetes, metabolic syndrome , strokes and some type of cancer too can be avoided by doing aerobic exercise.
Humans can see several thousand shades of color but have cone photoreceptors that are sensitive to only three (perhaps four) wavelengths of light. What is the best explanation for why we see so many colors?
Select one:
a.Color perception is dependent on the millions of rods as well as cone photoreceptors.
b.Color perception is achieved by activation of various combinations between the three cone types.
c.Shades of color are purely psychological and learned by association with age, infants only seeing in black and white. Incorrect
d. Colors are added and enhanced in the primary visual cortex of the brain.
Answer:
B. Color perception is achieved by activation of various combinations between the tree cone types.
Explanation:
Human eye have only three photo receptor ( three wavelength) but they can see several thousands of shades of color because color perception is gained by combination of these three wave length cone types ( RED, BLUE AND GREEN) into various combinations. Our retina have two types of cell:- cone cells and cylindrical cells. Cone cells are responsible for the identification of color.
A space telescope travels about Earth in a circular orbit at a distance of 380 miles from Earth's surface. It makes one orbit every 95 minutes. Find its linear velocity in miles per hour. (The radius of Earth is approximately 3960 miles.) Round to the nearest tenth place.
Final answer:
The linear velocity of the space telescope is about 17,193.2 miles per hour when rounded to the nearest tenth. This is calculated using the orbit radius of 4340 miles and the orbital period of 1.5833 hours.
Explanation:
To calculate the linear velocity of a space telescope traveling around Earth, we need to use the formula v = 2πr / T, where v is the linear velocity, r is the radius of the orbit, and T is the orbital period. To find the orbit radius, we add Earth's radius and the altitude of the telescope above Earth's surface: r = 3960 miles + 380 miles = 4340 miles.
Next, we convert the orbital period from minutes to hours. There are 60 minutes in an hour, so T = 95 minutes / 60 minutes/hour = 1.5833 hours. We can then plug the radius and period into the formula to find the linear velocity: v = 2 x π x 4340 miles / 1.5833 hours = 17,193 miles/hour. When rounded to the nearest tenth, the linear velocity is approximately 17,193.2 miles per hour.
A surface receiving sound is moved from its original position to a position three times farther away from the source of the sound. The intensity of the received sound thus becomes A. Nine times higher. B. Nine times lower. C. Three times higher. D. Three times lower.
The intensity of the sound will be B. Nine times lower
Explanation:
The intensity of a sound follows an inverse square law, which means that it is inversely proportional to the square of the distance from the source:
[tex]I\propto \frac{1}{r^2}[/tex]
where
I is the intensity
r is the distance from the source
In this problem, the siund has an intensity of I when the receiver is placed at a distance r from the source.
Later, the receiver is placed three times farther away, so the new distance is
r' = 3r
Therefore, the new intensity of the sound will be:
[tex]I'\propto \frac{1}{r'^2}=\frac{1}{(3r)^2}= \frac{1}{9} (\frac{1}{r^2})= \frac{1}{9}I[/tex]
Therefore, the intensity of the sound received will be nine times lower.
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Answer: B. Nine times lower
Explanation:
A mountain biker encounters a jump on a race course that sends him into the air at 35,2° to the horizontal. If he lands at a horizontal distance of 30,1 m and 14,7 m below his launch point, what is his initial speed?
The initial speed of the mountain biker can be determined using principles of projectile motion, by first calculating the time of flight from the vertical displacement and then substitifying this into the equation for horizontal displacement.
Explanation:This problem can be solved using principles of projectile motion, where motion is analysed separately along the vertical and horizontal axes.
The time of flight can be determined by considering the vertical displacement and using the equation y = V0y*t - 0.5*g*t^2, where y is the vertical displacement, V0y is the initial vertical velocity, t is the total time, and g is the acceleration due to gravity. Solving for t, we can substitute this into the equation for horizontal displacement x = V0x*t, where x is the horizontal displacement and V0x is the initial horizontal velocity. This allows us to solve for the initial speed.
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The mountain biker's initial speed is found using the principles of projectile motion by resolving into horizontal and vertical components and solving a quadratic equation. The initial speed needed to achieve the jump is approximately 13.65 m/s.
Determining the Initial Speed of the Mountain Biker
To find the initial speed of the mountain biker, we'll use the principles of projectile motion. We'll perform a step-by-step analysis considering horizontal and vertical components separately.
Given data:
Angle of projection: 35.2°Horizontal distance: 30.1 mVertical distance: -14.7 m (below launch point)Step-by-Step Solution
Step 1: Separate the initial speed into horizontal (Vx) and vertical (Vy) components:
[tex]V_x = V_0 \cos(35.2^o) \\V_y = V_0 \sin(35.2^o)[/tex]
Step 2: Use the horizontal motion formula to express time (t):
[tex]x=V_xt[/tex]
[tex]$\begin{equation}t = \frac{x}{Vx} = \frac{30.1}{V_0 \cos(35.2^o)}\end{equation}$[/tex]
Step 3: Use the vertical motion formula considering the displacement:
[tex]$\begin{equation}y = V_y t + \frac{1}{2} a t^2\end{equation}$[/tex]
Substituting known values and simplifying, we have:
[tex]$\begin{equation} -14.7 = (V_0 \sin(35.2^o)) \times \frac{30.1}{V_0 \cos(35.2^o)} + \frac{1}{2}(-9.8) \left( \frac{30.1}{V_0 \cos(35.2^o)}\right)^2\end{equation}$[/tex]
[tex]or, -14.7=V_0\times0.57\times \frac{30.1}{V_0\times 0.817} -4.9\times (\frac{30.1}{V_0\times 0.817})^2 \\or,-14.7=20.99-4.9\times\frac{906.01}{V_0^2\times0.66} \\or, 6650.9/V_0^2=20.99+14.7\\or,V_0^2=6650.9/35.69\\or, V_0=13.65m/sec[/tex]
The initial speed is 13.65 m/sec.
A 2 kg object moves in a circle of radius 4 m at a constant speed of 3 m/s. What is the angular momentum of the object with respect to an axis perpendicular to the circle and through its center? 7. (A) 9 Nm/kg.
(B) 12 m2/s.
(C) 13.5 kg m2/s2.
(D) 18 N m /kg.
(E) 24 kg m2/s
Answer:
The angular momentum in this case is [tex] \mathbf{24\,\frac{kg\,m^{2}}{s}} [/tex]
Explanation:
The angular momentum of a point mass moving around an axis of rotation is the cross product between the distance of the object to the axis (r) of rotation and the linear momentum (p) of the particle:
[tex] \overrightarrow{L}=\overrightarrow{r}\times\overrightarrow{p} [/tex] (1)
But linear momentum is defined as mv, so (1) is:
[tex]\overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v} [/tex](2)
And its magnitude is:
[tex]L=rmv*\sin\theta=(4)(2)(3)\sin(90)=\mathbf{24\,\frac{kg\,m^{2}}{s}} [/tex] (3)
It is important to note that [tex] \theta [/tex] is the angle between the radius vector and the velocity vector, because the axis of rotation is perpendicular to the circle and through its center this angle is equal to 90° and [tex] \sin(90) = 1 [/tex]
1.67 points When comparing equal volumes of gases at the same pressure and temperature, different gases have different densities. Which property or properties of gas particles contribute to different gases having different macroscopic densities
Answer:
Molar mass due to different atomic masses
Explanation:
Two gases that have the same volume, pressure and gases have the same number of moles, as we can deduce from the ideal gas equation. So the gases will have the same number of moles of gases but they can have densities due to the fact that they have different molar masses due to the fact that they have different atomic masses. So one gas will be heavier than the other, for the same volume.
What condition is required for cloud formation in the atmosphere?
Answer: To form a cloud, the air that rises must cool to the point where some of the water vapor molecules "clump together" at a faster pace than they are pulled apart by their thermal energy. These molecules then condense to form the clouds and water droplets.
We can see here that Cloud formation in the atmosphere requires two main conditions: the presence of water vapor and the cooling of air.
What is atmosphere?The atmosphere is a layer of gases that surrounds a planet or celestial body, held in place by gravity. On Earth, the atmosphere is the layer of gases that envelops the planet and sustains life. It plays a crucial role in protecting and supporting life on Earth by providing essential elements, regulating temperature, and facilitating various atmospheric processes.
Clouds play a crucial role in the Earth's climate system by reflecting sunlight back into space (resulting in cooling) and trapping heat radiated from the Earth's surface (resulting in warming).
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Technician A says that for best results, loosen the fastener immediately after spraying it with penetrating oil. Technician B says that rusted fasteners may be loosened by tapping with a hammer. Which technician is correct?
a. technician A only
b. technician B only
c. both technicians A and B
d. neither technician A nor B
Answer:
C. Both Technicians A and B
Explanation:
To loosen a fastener immediately especially if it is a rusted one we can follow these steps.
1. Tap the sides of the fastener so that the rust particle fall off.
2. These rust particles can be brushed off.
3. Spray the fastener with the penetrating oil.
4. Again tap the sides with a hammer to let the oil penetrate the inter-locked parts (threads in case of a nut and bolt).
5. Loosen the fastener after oil has penetrated.
Note: It should be kept in consideration that spraying alone won't do the job, tapping is essential to let the oil penetrate deep in to free the rusted parts.
A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the loaded sled with its rider has a mass of 210 kg. (a) Calculate the magnitude of the acceleration starting from rest if each dog exerts an average force of 185 N backward on the snow. (b) What is the magnitude of the acceleration once the sled starts to move
Answer
given,
number of dog = 8
mass of each dog= 19 Kg
mass of sled = 210 Kg
average force = 185 Nss
a) writing all the horizontal force
force acting by dog - friction force = (M + 8m) a
8 F_d - μ m g = (M + 8m) a
assuming coefficient of friction of snow be μ = 0.14
8 x 185 - 0.14 x 210 x 9.8 = (210 + 8 x 19 ) x a
a = 3.29 m/s²
b) the kinetic friction of coefficient is less than static friction
hence, we can suggest that acceleration of the sled will increase once the sled start to move.
a > 3.29 m/s²
A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exerts an 8.0 N thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the launch point. At what horizontal distance left of the hoop should you launch?
Answer:
x = 7.62 m
Explanation:
First we need to calculate the weight of the rocket:
W = mg
we will use the gravity as 9.8 m/s². We have the mass (500 g or 0.5 kg) so the weight is:
W = 0.5 * 9.8 = 4.9 N
We know that the rocket exerts a force of 8 N. And from that force, we also know that the Weight is exerting a force of 4.9. From here, we can calculate the acceleration of the rocket:
F - W = m*a
a = F - W/m
Solving for a:
a = (8 - 4.9) / 0.5
a = 6.2 m/s²
As the rocket is accelerating in an upward direction, we can calculate the distance it reached, assuming that the innitial speed of the rocket is 0. so, using the following expression we will calculate the time which the rocket took to blast off:
y = vo*t + 1/2 at²
y = 1/2at²
Solving for t:
t = √2y/a
t = √2 * 20 / 6.2
t = √6.45 = 2.54 s
Now that we have the time, we can calculate the horizontal distance:
x = V*t
Solving for x:
x = 3 * 2.54 = 7.62 m
Duplain st. is 300m long and runs from west to east between Baron and Burkey. If keith is strolling east from Baron at an average velocity of 3km/hr, and Sue is power-walking west from Burkey at an average velocity of 6km/hr, how long will it take them to meet?
A. 1 minute
B. 2 minutes
C. 3 minutes
D. 6 minutes
Answer:
Time, t = 2 minutes
Explanation:
Given that,
Length of Duplain st. d = 300 m = 0.3 km
If keith is strolling east from Baron at an average velocity of, v = 3 km/hr
Sue is power-walking west from Burkey at an average velocity of, v' = 6 km/hr
To find,
How long will it take them to meet ?
Solution,
When both objects are travelling in opposite direction, then the total speed is given by :
V = v + v'
V = 3 km/hr + 6 km/hr
V = 9 km/hr
Let t is the time taken will it take them to meet. It can be calculated as :
[tex]t=\dfrac{d}{V}[/tex]
[tex]t=\dfrac{0.3\ km}{9\ km/hr}[/tex]
t = 0.033 hour
or
t = 1.98 minutes
i.e. t = 2 minutes
So, they will take 2 minutes to meet. Hence, this is the required solution.
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support. To what distance x can a person who weighs 522 N walk on the overhanging part of the plank before it just begins to tip?
Answer:
x = 0.6034 m
Explanation:
Given
L = 5 m
Wplank = 225 N
Wman = 522 N
d = 1.1 m
x = ?
We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.
∑τ = 0 ⇒ τ₁ + τ₂ = 0
Torque come from the weight of the plank = τ₁
Torque come from the weight of the man = τ₂
⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)
⇒ τ₂ = Wman*x = 522 N*x (clockwise)
then
τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0
⇒ x = 0.6034 m
Which type of bonding is found in all molecular substances
Answer:
Covalent bonding
Explanation:
Covalent bonding is the type of bonding found in all molecular substances much as water, carbon dioxide or methane. Unlike ionic bonding which is found in ionic substances, covalent bonding involves sharing, not transfer, of electrons between the bonding atoms to form molecules.
A tennis ball bouncing on a hard surface compresses and then rebounds. The details of the rebound are specified in tennis regulations. Tennis balls, to be acceptable for tournament play, must have a mass of 57.5 g. When dropped from a height of 2.5 m onto a concrete surface, a ball must rebound to a height of 1.4 m. During impact, the ball compresses by approximately 6 mm.
How fast is the ball moving when it hits the concrete surface? (Ignore air resistance.)
The velocity of the ball is 7 m/s
Explanation:
The motion of the ball is a free fall motion, so it means that the ball falls down under the effect of the force of gravity only. Therefore, it has a constant acceleration (acceleration of gravity, g), and we can use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
For the ball in this problem, we have:
u = 0 (initial velocity, the ball is dropped from rest)
[tex]a=g=9.8 m/s^2[/tex] (acceleration of gravity)
s = 2.5 m (vertical displacement)
Solving for v, we find the velocity at which the ball hits the concrete surface:
[tex]v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(2.5)}=7 m/s[/tex]
Learn more about free fall motion:
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Refrigerant-134a enters the condenser of a residential heat pump at 800 kPA and 35oC at a rate of 0.018 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor consumes 1.2 kW of power, determine
(a) the COP of the heat pump
(b) the rate of heat absorbtion from the outside air.
Answer:
(A) COP = 2.64
(B) rate of heat absorption= 1.9637 kW
Explanation:
mass flow rate (m) = 0.018 kg/s
work input (Win) = 1.2kW
inlet pressure (P1) = 800kPa
inlet temperature (T1) = 35 degree Celsius
h1 = 271.24 KJ/Kg
outlet pressure (P2) = 800 kPa
outlet temperature (T2) = ?
entalphy (h2) = 95.48 KJ/Kg
The entalphies are gotten from tables for refrigerant 134a at the temperatures and pressures above
(A) COP = Qh ÷ Win
where Qh = m(h1 -h2) from the energy balance equation
Qh = 0.018 ( 271.24 - 95.48 ) = 3.1637 kW
COP = 3.1637 ÷ 1.2 = 2.64
(B) rate of heat absorption = Qh - Win
= 3.1637 - 1.2 = 1.9637 kW
The COP of a heat pump is calculated by using the formula COP = Q_H/W, where Q_H is the heat delivered to the house and W is the work done. Utilizing the specific enthalpies and flow rate, Q_H can be calculated. Subsequently, the rate of heat absorption from the outside can be deduced using Q_L = Q_H - W.
Explanation:The thermal efficiency or COP (Coefficient of Performance) of the heat pump can be calculated using the equation: COP = Q_H/W where Q_H is the heat delivered to the house and W is the work input. In this case, since the fluid leaves the condenser as a saturated liquid, Q_H = m_dot*(h_1 - h_2), where h_1 and h_2 are the specific enthalpies at the state 1 and state 2 and m_dot is the mass flow rate of refrigerant.
From the saturated liquid tables, we know that the specific enthalpy h_2 = 257 kJ/kg for Refrigerant-134a at 800 kPa. Thus, Q_H can be calculated and this value (in kW) can be used to calculate COP.
Then, to answer part (b), the rate of heat absorption from the outside air can be calculated using the equation Q_L = Q_H - W. Q_L is the rate of heat absorption, Q_H is the heat delivered to the house and W is the power absorbed by the compressor.
Learn more about Heat Pump Efficiency here:https://brainly.com/question/31560607
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