Answer:
[tex] v_s =7.74\ m/s[/tex]
Explanation:
given,
Speed of sound = 343 m/s
frequency of horn = 260 Hz
the friend is approaching, the frequency is increased by the Doppler Effect. The frequency is 266 Hz
using formula
[tex]f' = \dfrac{v}{v-v_s}f_0[/tex]
[tex]266= \dfrac{343}{343 - v_s}(260)[/tex]
[tex]1.023= \dfrac{343}{343 - v_s}[/tex]
[tex]343 - v_s = 335.26[/tex]
[tex] v_s =7.74\ m/s[/tex]
the speed of friends approaching is equal to [tex] v_s =7.74\ m/s[/tex]
A box slides with uniform acceleration up an incline. The box has an initial speed of 9.0 m/s and rises vertically 2.60 m before coming to rest. If the angle of the incline is 30°, what is the coefficient of kinetic friction between the box and the incline?
Answer: 0.58
Explanation:
First we need to get the acceleration of the body using equation of motion
v²=u²-2as
v is the final velocity
u is the initial velocity
a is acceleration
s is the distance moved
0²=9²-2a(2.6)
-81=-5.2a
a=81/5.2
a= 15.6m/s²
Angle of inclination =30°
To get the coefficient of friction, we use the formula
Ff =nR
Ff is frictional force
n is coefficient of friction
R is normal reaction
n = Ff/R = Wsin30°/Wcos30°
n = tan30°
n = 0.58
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 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:
Given the atomic radius of argon, 0.97 Å, and knowing that a sphere has a volume of 4πr3/3, calculate the fraction of space that Ar atoms occupy in a sample of argon at STP. Express your answer using two significant figures.
Answer:
1.0x10^-4
Explanation:
First, in order to do this, we need to calculate the volume of 1 simple atom of Ar. Using the formula of the volume of a sphere we have the following
Converting A to cm:
0.97 * 1x10^-8 = 9.7x10^-9 cm
Now the volume:
V = 4/3π(9.7x10^-9)³
V = 3.82x10^-24 cm³
We know that 1 cm³ is 1 mL, and 1 L is 1000 mL so:
V = 3.82x10^-24 mL / 1000 = 3.82x10^-27 L
Now, using avogadro's number, we should get the total volume of all atoms of Ar so:
3.82x10^-27 * 6.02x10^23 = 2.3x10^-3 L
Finally, at STP the volume of an ideal gas is 22.4 L so:
2.3x10^-3 / 22.4 = 1.03x10^-4
With two significant figure, it would be 1.0x10^-4
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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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.
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)
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.
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.
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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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.
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 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of μs = 0.70 and μk = 0.50. The turntable very slowly speeds up.
Answer:
Angular speed will reach 6.833rad/s before the coin starts slipping
Explanation:
There is no question but I'll asume the common one: Calculate the speed of the turntable before the coin starts slipping.
With a sum of forces:
[tex]Ff = m*a[/tex]
[tex]Ff=m*V^2/R[/tex]
At this point, friction force is maximum, so:
[tex]\mu*N=m*V^2/R[/tex]
[tex]\mu*m*g=m*V^2/R[/tex]
Solving for V:
[tex]V=\sqrt{\mu*g*R}[/tex]
V=1.025 m/s
The angular speed of the turntable will be:
ω = V/R = 6.833 rad/s This is the maximum speed it can reach before the coin starts slipping.
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]
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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.
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.
During fabrication processes, polymeric materials are generally subjected to which of the following conditions? The ambient temperature. Elevated temperatures. The ambient pressure. Elevated pressures.
Answer:
- Elevated temperatures.
- Elevated pressures.
Explanation:
In the manufacture of plastic products are used:
Raw material (pellets) which are the monomers that promote the chemical reaction.
To these are added the charges, in order to reduce the cost of the final product and improve some properties. These charges can be fiberglass, paper, metal structures.
Additives are also added whose mission is to improve or achieve certain properties, such as reducing friction, reducing chemical degradation, increasing electrical conductivity, coloring the product, and all this happens in the presence of a catalyst that is responsible for initiating and accelerating the chemical reaction process.
There are different methods of production of plastics, one of the most frequent is by injection.
Injection is a process that is carried out on machines similar to extrusion machines, in which the spindle, in addition to rotating, has an axial displacement.
In the injection, once the mold is filled, it is separated from the nozzle of the machine, breaking the feed channel. After a certain time, the piece already cooled is demoulded.
High pressures and temperatures are necessary, but parts of good finish and at high production speeds are obtained
The correct answer is: Elevated temperatures.
During fabrication processes, polymeric materials are often subjected to elevated temperatures. This is because many polymers are thermoplastic, meaning they soften and can be reshaped when heated. The elevated temperatures allow the polymer chains to move more freely, enabling the material to be molded, extruded, or otherwise formed into the desired shape or structure.
While ambient temperature and pressure can also affect polymeric materials, the key condition for shaping and fabricating these materials is typically elevated temperature. Ambient temperature is usually not sufficient to induce the necessary molecular mobility for fabrication. Ambient pressure is generally constant and does not play a significant role in the fabrication process unless specific pressure conditions are required for certain molding techniques, such as injection molding or compression molding, where controlled pressure is applied to ensure the polymer fills the mold properly.
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 $
A physics student looks into a microscope and observes that small particles suspended in water are moving about in an irregular motion. Which of the following statements is the best explanation for this observation?
a. water molecules strike the particles giving them the same average kinetic energy as the water
b. the particles are carried by convection currents in the water
c. the small particles may be considered a fluid; and thus, move about randomly
d. the actual motion is regular, but the speeds of particles are too large to observe the regular motion
e. the particles are moving to be uniformly distributed throughout the volume of the water
Answer:
d. the actual motion is regular, but the speeds of particles are too large to observe the regular motion
Explanation:
The speeds of the particles are very large and comparatively the average free path is very small . Therefore time taken in covering the free path ( path between two consecutive collision with medium particles ) is very small . Hence the st line path covered by particles between two collision is less likely to be visible. Hence motion appears irregular or zig-zag.
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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Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the heat of vaporization = 2256 J/g, and the specific heat capacities of ice = 2.06 J/gK and liquid water = 4.184 J/gK
Answer:
[tex]Q=35011.6\ J[/tex]
Explanation:
Given:
mass of ice, [tex]m=50\ g[/tex]initial temperature of ice, [tex]T_i=-30^{\circ}C[/tex]final temperature of liquid water, [tex]T_f=73^{\circ}C[/tex]heat of fusion of ice, [tex]L=333\ J.g^{-1}[/tex]specific heat capacity of ice, [tex]c_i=2.06\ J.g^{-1}[/tex]specific heat capacity of liquid water, [tex]c_w=4.184\ J.g^{-1}.K^{-1}[/tex]Now, total heat energy required get to the final state:
[tex]Q=m(c_i.\Delta T_i+L+c_w.\Delta T_w)[/tex]
where:
[tex]\Delta T_w=[/tex] change in temperature of water from 0 to 73 degree C
[tex]\Delta T_i=[/tex] change in temperature of ice from -30 to 0 degree C
[tex]\therefore Q=50(2.06\times 30+333+4.184\times 73)[/tex]
[tex]Q=35011.6\ J[/tex]
Answer:
The heat energy required is 350.1J
Explanation:
Given data
Mass of ice =50g---kg =50/1000= 0.05kg
Temperature of ice T1= - 30°c
Temperature of ice T2=0°c
Temperature of liquid T3=0°c
Temperature of water T4= 73°c
The heat of fusion = 333 J/g
heat of vaporization = 2256 J/g, and the specific heat capacities of ice = 2.06 J/gK
and liquid water = 4.184 J/gK
It will be a good idea to first understand the path this process will follow
Ice at - 30°c - - ice at 0°c
Ice at 0°c - - - - liquid at 0°c fusion
Liquid at 0°c -- - liquid at 73°c
First, you have to calculate the heat absorbed by ice going from -30 C to 0 C. Use the equation:
q = m c (T2-T1)
q= 0.5*2.06(0-(-30))
q= 30.9J
Then, calculate the heat required to melt that ice at 0C. Use the equation:
q = m *(heat of fusion)
q=0.5*333
q=166.5J
Then, calculate the heat required to raise the temperature of water from 0°C to 73°C. Use
q = m c (T4-T3)
q= 0.5*4.184(73-0)
q= 152.7J
Finally, we will sum up the heat required
Total heat energy required = 30.9+166.5+152.7= 350.1J
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 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
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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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 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 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
A beam of electrons passes through a single slit, and a beam of protons passes through a second, but identical, slit. The electrons and the protons have the same speed. Which one of the following correctly describes the beam that experiences the greatest amount of diffraction?
(a) The electrons, because they have the smaller momentum and, hence, the smaller de Broglie wavelength
(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength
(c) The protons, because they have the smaller momentum and, hence, the smaller de Broglie wavelength
(d) The protons, because they have the larger momentum and, hence, the smaller de Broglie wavelength
(e) Both beams experience the same amount of diffraction, because the electrons and protons have the same de Broglie wavelength.
Answer:
(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength
Explanation:
de Broglie wavelength λ = h / m v
Since both electrons and protons have same velocity , momentum mv will be less for electrons because mass of electron is less .
for electron , momentum is less so . Therefore de Broglie wavelength λ will be more for electrons .
Amount of diffraction that is angle of diffraction is proportional to λ
Therefore electrons having greater de Broglie wavelength will show greater diffraction.
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.