Complete Question:
Which one of the following is a
true statement?
Group of answer choices.
A) The second law of thermodynamics is a consequence of the first law of thermodynamics.
B) It is possible for heat to flow spontaneously from a hot body to a cold one or from a cold one
to a hot one, depending on whether or not the process is reversible or irreversible.
C) It is not possible to convert work entirely into heat.
D) It is impossible to transfer heat from a cooler to a hotter body.
E) All of these statements are false.
Answer:
E) All of these statements are false.
Explanation:
All of the following statements are false;
I. The second law of thermodynamics is a consequence of the first law of thermodynamics. The second law of thermodynamics is about a physical quantity known as entropy while the first law of thermodynamics states that energy cannot be created or destroyed but can only be transformed from one form to another.
II. It is possible for heat to flow spontaneously from a hot body to a cold one or from a cold one
to a hot one, depending on whether or not the process is reversible or irreversible. According to the second law of thermodynamics this is false due to the entropy.
III. It is not possible to convert work entirely into heat. It is possible to completely convert work into heat because it involves transferring energy.
IV. It is impossible to transfer heat from a cooler to a hotter body.
what kind of mathematical relationship do you predict wll exist between velocity and number of filters
Answer:
[tex]PK_T = constant[/tex]
Explanation:
[tex]PK_T[/tex] = constant
terminal velocity is the maximum const. velocity to achieved by the partical when particle is free fall with the effect of gravity
so that here
eq of motion is
Fnet = mg - kv ................1
here a = 0
v = terminal velocity
0 = mg - kV
v = [tex]\frac{mg}{k}[/tex] m/s
What is the distance of separation between objects of masses 5.6 x 10 5 kg and 8.8 x 10 6 kg if the force of gravity between them is 440 N
Answer:
2.87m
Explanation:
Using the law of gravitation to solve this question
F = GMm/r²
G is the gravitational constant
M and m are the masses
r is the distance between the masses
Substitute the given values
G = 6.67×10^-11 m³/kgs²
M =8.8 x 10^6 kg
m = 5.6 x 10^5 kg
F =440N
400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²
400r² = 328.698×10
400r² = 3286.98
r² = 3286.98/400
r² = 8.21745
r = √8.21745
r = 2.87m
Hence the distance of separation is 2.87m
The distance of separation will be "0.86 m".
Given:
Mass,
[tex]m_1 = 5.6\times 10^5 \ kg[/tex][tex]m_2 = 8.8\times 10^6 \ kg[/tex]Force,
[tex]F = 440 \ N[/tex]As we know,
→ [tex]F = \frac{Gm_1 m_2}{r^2}[/tex]
By putting the values, we get
→ [tex]440=\frac{6.67\times 10^{-11}\times 5.6\times 10^5\times 8.8\times 10^6}{r^2}[/tex]
→ [tex]r^2 = \frac{6.67\times 5.6\times 8.8}{440}[/tex]
→ [tex]r^2 = 0.74704[/tex]
→ [tex]r = \sqrt{0.74704}[/tex]
→ [tex]= 0.86 \ m[/tex]
Thus above approach is correct.
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what can you conclude about the force of gravity?
Answer:
All objects, irrespective of their mass, experience the same acceleration g when falling freely under the influence of gravity at the same point on the Earth. If gravity is the only force acting on an object, the sum of kinetic energy and gravitational energy is constant. ...
Answer:
Gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter. ... On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth's mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects.The Earth's gravitational force accelerates objects when they fall. It constantly pulls, and the objects constantly speed up.Gravity is the force of attraction between two objects with mass and is dependent on the distance between these objects. Gravity is the force that repels two objects that have opposite charges. It is dependent upon the charges of the objects.
Explanation:
A low rider truck accelerates from 0 m/s to 30.0 m/s with a displacement of 50m. What is the vehicle's acceleration?
a. 2.00m/s
b. 5.00m/s
c. 9.0m/s
d. 10.0m/s
Answer:
a = 9 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}^{2}=v_{o}^{2} +2*a*x[/tex]
where:
Vf = final velocity = 30 [m/s]
Vo = initial velocity = 0
a = acceleration [m/s²]
x = displacement = 50 [m]
Now replacing:
[tex]30^{2}=0 +2*a*50\\100*a=900\\a=9[m/s^{2} ][/tex]
please help me this is timed .
Find x if a = 3.0 m/s^2
Answer:
x = 50 N
Explanation:
Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.
Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.
_______
F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a
m (mass which we are given) = 150 kg
a (acceleration which we are given) = 3.0m/s
________
So F = m × a → F2 - F1 = m × a →
500 - F1 = 150 × 3.0 → 500 - F1 = 450 →
-F1 = -50 → F1 = 50
If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
Complete Question
The speed of a transverse wave on a string of length L and mass m under T is given by the formula
[tex]v=\sqrt{\frac{T}{(m/l)}}[/tex]
If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
Answer:
[tex](m/l)=\frac{10}{V^2}[/tex]
Explanation:
From the question we are told that
Speed of a transverse wave given by
[tex]v=\sqrt{\frac{T}{(m/l)}}[/tex]
Maximum Tension is [tex]T=10.0N[/tex]
Generally making [tex](m/l)[/tex] subject from the equation mathematically we have
[tex]v=\sqrt{\frac{T}{(m/l)}}[/tex]
[tex]v^2=\frac{T}{(m/l)}[/tex]
[tex](m/l)=\frac{T}{V^2}[/tex]
[tex](m/l)=\frac{10}{V^2}[/tex]
Therefore the Linear mass in terms of Velocity is given by
[tex](m/l)=\frac{10}{V^2}[/tex]
What do you think causes the changes in sunrise and sunset times over the course of the year?
Answer:
The combination of Earth's elliptical orbit and the tilt of its axis results in the Sun taking different paths across the sky at slightly different speeds each day. This gives us different sunrise and sunset times each day.
Explanation:
If force c is 30n and force d us 20 N what is the size of the horizontal force
(Are these statements about the spring constant true or
false?
a. The spring constant is a measure of the stiffness of
the spring.
b. The spring constant tells you how many newtons of
force it takes to stretch the spring one meter.
c. If a spring stretches easily, it will have a high spring
constant.
d. The spring constant of a spring varies with the
amount of stretch or compression of the spring.
Answer:
true
true
false
true
pls follow me and thank you
A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head-on collision with a 0.03 kg shooter marble moving to the left at 0.18 m/s. After the collision, the smaller marble moves to the left at 0.315 m/s. What is the velocity of the 0.03 kg marble after the collision
Answer:
0.135m/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = m1v1 +m2v2
m1 and m2 are the masses
u1 and u2 are the initial velocities
v1 and v2 are thee final velocities
Substitute the given parameters and get v2
0.015(0.225)+0.03(0.18) = 0.015(0.315)+0.03v2
0.003375+0.0054 = 0.004725 + 0.03v2
0.008775 - 0.004725 = 0.03v2
0.00405 = 0.03v2
v2 = 0.00405/0.03
v2 = 0.135m/s
Hence the final velocity of the 0.03kg marble after collision is 0.135m/s
A constant force of 20. newtons applied to a box causes it to move at a constant speed of 4.0 meters per second. How much work is done in the boxin 6.0 seconds?
1) 240 joules
2) 120 joules
3) 480 joules
4) 80. joules
The work done is 480 Joules.
Option 3 is correct.
The work done is computed as multiplication of force and displacement.
Given that, Force = 20 N
Displacement = 4 meter per second
For 6 second,
Displacement [tex]=4*6=24m[/tex]
[tex]Workdone=Force*displacement\\\\Workdone=20*24=480J[/tex]
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explain why work is not considered a form of energy.
The large window air conditioner in Anita Breeze's room draws 11 amps of current. The unit runs for 8.0 hours during the course of a day. Determine the quantity of charge that passes through Anita's window AC during these 8.0 hours.
Answer:
quantity of charge that passes through Anita's window AC, Q = 316800 C
Explanation:
given data
current I = 11 amps
runs t = 8.0 hours
solution
we get here quantity of charge that passes through Anita's window AC
the quantity of charge passing Q = I × t .................1
Q = 11 A × ( 8 hr )× ( 3600 s/hr)
so
quantity of charge that passes through Anita's window AC, Q = 316800 C
The quantity of charge that passes through Anita's window AC, will be
Q = 316800 C
What is Current?
The current is defined as the quantity of charge flowing from the wire per unit time. Also defined as the rate of flow of the charge through any circuit. The formula will be given as:
[tex]\rm I=\dfrac{Q}{t}[/tex]
Now it is given that:
current I = 11 amps
runs t = 8.0 hours
The quantity of charge that passes through Anita's window AC during these 8.0 hours. Will be calculated as:
[tex]Q=I\times t[/tex]
[tex]Q=11\times 8\times 3600[/tex]
[tex]Q=316800 \ C[/tex]
Thus the quantity of charge that passes through Anita's window AC, Q = 316800 C
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A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight
Answer:
There is no work done.
Explanation:
Given the following data;
Mass = 10 kg
To find the work done?
In Physics, work done can be defined as the amount of energy transfered when an object or body is moved over a distance due to the action of an external force.
Mathematically, work done is given by the formula;
Work done = force * distance
[tex] W = F * d[/tex]
Where,
W is the work doneF represents the force acting on a body. d represents the distance covered by the body.However, the weight suspended in the air by a strong cable does not move or experience any form of displacement. Therefore, there is no work done.
2- A car on a straight highway goes in the positive direction for 8 km and then backs up for 3.6 km. What are the distance and displacement covered by the train?
Answer:
11.6km
4.4km in the negative direction
Explanation:
Distance is the total length of path covered and traveled by a body.
So, for this car on a straight line;
Total distance = 8km + 3.6km = 11.6km
Displacement is the distance traveled along a path and the direction it takes.
It is a vector quantity with magnitude and directional attributes.
For this journey;
Displacement = 8km - 3.6km = 4.4km in the negative direction.
Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is released
Answer:
171.5m
Explanation:
The velocity of sound in water = 343m/s
Time taken = 1.00secs
using the formula to calculate the distance
2x = vt
x is the distance
v is the speed of sound
t is the time
x = vt/2
x = 343(1)/2
x = 171.5m
hence their separation 1.00 s after the second object is released is 171.5m
Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation is also tripled, what happens to the magnitude of the electrostatic force that each charge exerts on the other
Answer:
The force remains the same.
Explanation:
Let the magnitude of the forces in each case be F1 and F2 respectively.
The charges are Q1 and Q2.
The distance of separation between them is R.
Hence, for F1;
F1 = KQ1Q2/R^2
For F2:
F2 = K * 3Q1 * 3Q2/(3R)^2
F2 = 9KQ1Q2/9R^2
F2 =KQ1Q2/R^2
Hence F1=F2
The force is the same in both cases!
How much extra water does a 147-lb concrete canoe displace compared to an ultra lightweight 36-lb Kevlar canoe of the same size carrying the same load
Answer:
1.7791 ft^3
Explanation:
The computation of the extra water is as follows;
Here we use the following formula
Volume of the water = ÷ Weight of concrete canoe ÷ value of the weight of the specific water
= 147 lb ÷ 62.4 lb /ft^3
= 2.356 ft^3
Now the volume of water occupied in ultra lightweight
= 36 lb ÷ 62.4 lb /ft^3
= 0.5769 ft^3
Now the water volume displaced is
= 2.356 - 0.5769
= 1.7791 ft^3
15 points! Answer FAST!
When heat is transferred, it flows_____.
A. From the warmer object to the cooler object only if the cooler object is an insulator
B. From the cooler object to the warmer object regardless of the materials
C. From the warmer object to the cooler object regardless of the materials
D. From the warmer object to the cooler object only if the warmer object is a conductor.
Answer:heat brings it up then down
Explanation:
In a chemical reaction, the mass of the reactants is 13,3 grams. Which of the
following could represent the mass of the products?
Answer:
my assuytyyhyyyyyy gftdrrdtiifyb tvhyvth rv b yy
Which is greater, the energy of one photon of orange light or the energy of one quantum ofradiation having a wavelength of 3.36 * 10^-9
The question is incomplete, here is the complete question:
Which is greater, the energy of one photon of orange light or the energy of one quantum of radiation having a wavelength of [tex]3.36\times 10^{-9}m[/tex]
Answer: The energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
Explanation:
To calculate the energy of one photon, we use the Planck's equation:
[tex]E=\frac{N_Ahc}{\lambda}[/tex]
where,
E = energy of radiation
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}mol^{-1}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8 m/s[/tex]
[tex]\lambda}[/tex] = wavelength of radiation
For orange light:For 1 photon, the term [tex]N_A[/tex] does not appear
[tex]\lambda}[/tex] = 620 nm = [tex]620\times 10^{-9}m[/tex] (Conversion factor: [tex]1nm=10^{-9}m[/tex] )
Putting values in above equation, we get:
[tex]E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{620\times 10^{-9}m}\\\\E=3.206\times 10^{-19}J[/tex]
For one quantum of radiation:[tex]\lambda}[/tex] = [tex]3.36\times 10^{-9}m[/tex]
Putting values in above equation, we get:
[tex]E=\frac{6.022\times 10^{23}mol^{-1}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.36\times 10^{-9}m}\\\\E=3.56\times 10^{7}J/mol[/tex]
Hence, the energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
A medicine ball has a mass of 8.0 kg and is thrown with a speed of 3.0 m/s. What is its kinetic energy?
O A 12 J
O
B. 24 J
C. 36 J
0 D. 192 J
Answer:
C. 36J
Explanation:
KE = ½ m V²
= ½ × 8 × 3²
= 4 × 9
= 36 J
A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h; The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s2. What is the shortest time in which the driver of the car can complete the passing operation? (assuming that the driver of the car does not pay any attention to the speed limit
Answer:
The shortest time in which the driver can complete the operation is approximately 6.32456 seconds
Explanation:
The given parameters are;
The speed of the truck and the car = 35 mi/h ≈ 51.33 ft./s
Let "t" represent the time it takes the car to pass the truck by 40 ft., we have;
The distance covered by the truck = 35 mi/h × t
The distance covered by the car = 35 mi/h × t + 80 ft = 51.33
Let the distance over which the car accelerates = d
We have;
d = 51.33×t₁ + 1/2×5×t₁²
51.33·t + 80 - d = (51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)²
51.33 = 51.33 + 5·t₁ - 20·(t - t₁)
∴ 5·t₁ = 20·(t - t₁)
5·t₁ = 20·t - 20·t₁
25·t₁ = 20·t
t₁ = 4·t/5
(51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)² + 51.33×t₁ + 1/2×5×t₁² = 51.33·t + 80
We get;
(51.33 + 5·(4·t/5))·(t - (4·t/5)) - 1/2·20·(t - (4·t/5))² + 51.33×(4·t/5) + 1/2×5×(4·t/5)² = 51.33·t + 80
(51.33 + 4·t)·(t/5) - 2·t²/5 + 51.33 × (4·t/5) + 8·t²/5 = 51.33·t + 80
51.33×t/5 + 51.33×4·t/5 + 4·t²/5 - 2·t²/5 + 8·t²/5 = 51.33·t + 80
51.33·t + 10·t²/5 = 51.33·t + 80
2·t² = 51.33·t + 80 - 51.33·t = 80
t² = 80/2 = 40
t = √40 = 2·√10 ≈ 6.32456
The shortest time in which the driver can complete the operation is "t" ≈ 6.32456 seconds
The resistance of a wire depends on its length i and on its cross sectional area A the resistance is
Answer:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area
Explanation:
h. If the mass of an object is 390 g and its volume is 50 cm?, then its density is ....... g/cm' or ....... kg/m”.
- The relative density of the object in part "h" is
Edit: nvm I'm an idiot I remembered this but thanks anyway
Answer: See explanation
Explanation:
The density would like be calculated by dividing the mass by volume. This will be:
Density = Mass / Volume
Density = 390 / 50
Density = 7.8g/cm
All of the following are ways in which sports psychologists help athletes except __________.
A.
staying motivated
B.
managing fear of failure
C.
improving performance
D.
enhancing memory
Please select the best answer from the choices provided
A
B
C
D
Answer:
D-Enhancing memory
Explanation:
What are the three longest wavelengths for standing waves on a270-cm-long string that is fixed at both ends
Answer:
The answer is below
Explanation:
a) What are the three longest wavelengths for standing waves on a 270-cm-long string that is fixed at both ends? b. If the frequency of the second-largest wavelength is 50.0 Hz, what is the frequency of the third-longest wave length?
Solution:
a) The wavelengths (λ) for standing waves is given by the formula:
[tex]\lambda_m=\frac{2*length\ of\ string}{m}\\\\Where\ m=1,2,3,.\ .\ .\\\\Given\ that\ length\ of\ string = 270\ cm=2.7\ m,\ m=1,2,3(three\ longest\ wavelengths)\\\\Hence:\\\\\lambda_1=\frac{2(2.7)}{1}=5.4\ m\\\\\lambda_2=\frac{2(2.7)}{2}=2.7\ m \\\\\lambda_3=\frac{2(2.7)}{3}=1.8\ m[/tex]
b) The frequency (f) and wavelength (λ) is given by:
fλ = constant
Hence:
[tex]f_2\lambda_2=f_3\lambda_3\\\\f_2=50\ Hz\\\\2.7*50=f_3(1.8)\\\\f_3=\frac{2.7*50}{1.8} \\\\f_3=75\ Hz[/tex]
The three longest wavelengths for the standing waves on a 270-cm long string that is fixed at both ends are:
1. 5.4 meters.
2. 2.7 meters.
3. 1.8 meters.
Given the following data:
Length of string = 270 cm to m = [tex]\frac{270}{100} =2.7\;m[/tex]To determine the three (3) longest wavelengths for these standing waves:
Mathematically, the wavelength for standing waves is given by the formula:
[tex]\lambda_n = \frac{2L}{n}[/tex]
Where:
[tex]\lambda_n[/tex] is the wavelength for standing waves.L is the length of string.Note: n = 1, 2, and 3.
When n = 1:
[tex]\lambda_1 = \frac{2\times 2.7}{1} \\\\\lambda_1 = 5.4 \;meters[/tex]
When n = 2:
[tex]\lambda_2 = \frac{2\times 2.7}{2} \\\\\lambda_2 = 2.7 \;meters[/tex]
When n = 3:
[tex]\lambda_3 = \frac{2\times 2.7}{3} \\\\\lambda_3 =\frac{5.4}{3} \\\\\lambda_3 = 1.8 \;meters[/tex]
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1. Sodium-24 has a half-life of 15 hours. How much sodium-24 will remain in an
26.0 g sample after 60 hours?
Answer:
1.625g
Explanation:
The half - life of a radioactive substance is the time taken for half of it to decay to half of its original composition.
For the given specie, the original composition is 26g and the half life is 15hrs:
At time 0hrs Mass = 26g
For the first half life at time 15hrs Mass = 13g
30hrs Mass = 6.5g
45hrs Mass = 3.25g
60hrs Mass = 1.625g
A helicopter flies with an air speed of 175 km/h, heading south. The wind is blowing at 85 km/h to the east relative to the ground. Calculate the speed and direction of the helicopter.
Answer:
154° at 195 km/h
Explanation:
The helicopter is moving south at 175 km/h, relative to the wind.
But the wind is moving east at 85 km/h, relative to the ground.
This means that the helicopter is moving south east relative to the ground.
Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.
This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.
Refer to the triangle b1.
The distance traveled by the helicopter in 1 hour is denoted by d.
d is the hypotenuse of the right triangle.
Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)
Hence the helicopter is traveling at 195 km/h relative to the ground.
To calculate the direction we use,
tan (x) = opposite/adjacent = 85/175
So the angle x is,
[tex]x = arctan (\frac{85}{175} )[/tex] = 25.9°
Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)
Which three statements describe mechanical waves?
A. The waves can travel through empty space.
B. The waves transfer energy by causing particles of matter to
move.
C. The waves need matter to transfer energy.
D. The waves can transfer energy through solids, liquids, and gases.
The three statements that are correct about the mechanical waves are options B, C, and D.
Mechanical waves:Mechanical waves are produced due to disturbances in any matter. Which clearly indicates that the mechanical waves cannot travel through empty space.
The mechanical waves require matter to transfer energy through oscillation in the matter. So it is true that the waves transfer energy by causing particles of matter to move.
Also, the mechanical waves need matter to transfer energy, the waves can transfer energy through solids, liquids, and gases.
An example of a mechanical wave is the sound produced by the guitar strings.
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