What is the rate of the entropy change of the universe as heat leaks out a window, consisting of a single pane of glass that is 0.5 cm thick and 1.0 m2 in area, where the indoor temperature is 25°C and the outdoor temperature is -10°C?
Answer:
The change in entropy is 1.6 W/K.
Explanation:
Thickness, d = 0.5 cm
Area, A = 1 m^2
T = 25°C
T' = - 10°C
Coefficient of thermal conductivity of glass, K = 0.8 W/mK
The change in entropy is given by
S = Q/T
Here,
[tex]S =\frac{Q}{T}\\\\S = \frac{K A (T - T')}{d(T - T')}\\\\S = \frac{0.8\times 1}{0.5} = 1.6 W/K[/tex]
A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positive direction, what is the truck's resultant displacement
Answer:
140m east
Explanation:
If East is positive then lets rephrase the problem into integers
A truck moves +70 m, then moves -120m, and finally moves +90m.
So totally Displacement = +70-120+90= +140m
Since east is positive, the trucks resultant displacement is 140 m east of origin
Physics is killing me. Any help?
Answer:
The simplified expression is 3.833 x 10⁷ g
Explanation:
Given expression;
3.88 x 10⁷ g - 4.701 x 10⁵ g
The expression above is simplified as follows;
= (3.88 x 100 x 10⁵ )g - ( 4.701 x 10⁵) g
= (388 x 10⁵ )g - ( 4.701 x 10⁵) g
= (388 - 4.701 ) x (10⁵ )g
= 383.299 x 10⁵ g
In standard form, the simplified expression can be expressed as;
= (3.83299 x 100 x 10⁵) g
= 3.83299 x 10⁷ g
= 3.833 x 10⁷ g
Therefore, the simplified expression is 3.833 x 10⁷ g
a. Give an example of the conversion of light energy to electrical energy.
b. Give an example of chemical energy converting to heat energy.
c. Give an example of mechanical energy converting to heat energy.
Explanation:
a) photovoltaic cell is a semiconductor device and it converts light energy to electrical energy
b) burning of coal converts chemical energy to heat energy
c) rubbing of both hands against each other converts mechanical to heat energy
Answer:
a. solar cells
b.coal,wood,petroleum
c.rubbing ours palms
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 26.5 m/s2 with a beam of length 5.89 m , what rotation frequency is required
Answer:
The angular acceleration is 4.5 rad/s^2.
Explanation:
Acceleration, a = 26.5 m/s2
length, L = 5.89 m
The angular acceleration is
[tex]\alpha =\frac{a}{L}\\\\\alpha = \frac{26.5}{5.89}=4.5 rad/s^2[/tex]
(b) Name the devices used to measure the volume of liquid.
Answer:
Liquid volume is usually measured using either a graduated cylinder or a buret. As the name implies, a graduated cylinder is a cylindrical glass or plastic tube sealed at one end, with a calibrated scale etched (or marked) on the outside wall.
A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes
Answer:
T1 = 499.9N, T2 = 865.8N, T3 = 1000N
Explanation:
To find the tensions we need to find the vertical and horizontal components of T1 and T2
T1x = T1 cos60⁰, T1y = T1 sin60⁰
Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰
For the forces to be in equilibrium,
the sum of vertical forces must be zero and the sum of horizontal forces must also be zero
Sum of Fx = 0
That is, T1x - T2x=0
NB: T2x is being subtracted because T1x and T2x are in opposite directions
T1 cos60⁰ - T2 cos30⁰ = 0
0.866T1 - 0.5T2 = 0 ............ (1)
Sum of Fy = 0
T1y + T2y - 1000 = 0
T1 sin60⁰ + T2 sin30⁰ - 1000 = 0
NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.
0.5T1 - 0.866T2 - 1000 = 0 ........(2)
From (1)
make T1 the subject
T1 = 0.5T2/0.866
Substitute T1 into (2)
0.5 (0.5T2/0.866) - 0.866T2 = 1000
(0.25/0.866)T2 - 0.866T2 = 1000
0.289T2 - 0.866T2 = 1000
1.155T2 = 1000
T2 = 865.8N
Then T1 = 0.5 x 865.8 / 0.866
T1 = 499.9N
T3 = 1000N
NB: The weight of the bag is the Tension above the rope, which is T3
If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm, what is the natural length (in cm) of the spring
Answer:
the natural length of the spring is 9 cm
Explanation:
let the natural length of the spring = L
For each of the work done, we set up an integral equation;
[tex]5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2} \ \ \ -----(1)[/tex]
The second equation of work done is set up as follows;
[tex]9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)[/tex]
solve equation (1) and equation (2) together;
[tex]\frac{2(9)}{(27-l)^2 - (21-l)^2} = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}\\\\\frac{2(9)}{2(5.4)} = \frac{(27-l)^2 - (21-l)^2}{(21-l)^2 - (15-l)^2}\\\\\frac{9}{5.4} = \frac{(729 - 54l+ l^2) - (441-42l+ l^2)}{(441-42l+ l^2) - (225 -30l+ l^2)} \\\\\frac{9}{5.4 } = \frac{288-12l}{216-12l} \\\\\frac{9}{5.4 } =\frac{12}{12} (\frac{24-l}{18 -l})\\\\\frac{9}{5.4 } = \frac{24-l}{18 -l}\\\\9(18-l) = 5.4(24-l)\\\\162-9l = 129.6-5.4l\\\\162-129.6 = 9l - 5.4 l\\\\32.4 = 3.6 l\\\\l = \frac{32.4}{3.6} \\\\[/tex]
[tex]l = 9 \ cm[/tex]
Therefore, the natural length of the spring is 9 cm
if one branch of a 120-v power lines is protected by a 20-A fuse, will the fuse carry an 8-Ώ load
Answer:
No I won't.
It will carry 6ohm load.
Explanation:
It obeys ohms law therefore V=IR
120=20R
R=120/20
R= 6 ohms
A caris initially at rest starts moving with a constant acceleration of 0.5 m/s2 and travels a distance of 5 m. Find
(i) Final velocity
(ii)The time taken
Answer:
(I)
[tex] { \bf{ {v}^{2} = {u}^{2} - 2as }} \\ {v}^{2} = {0}^{2} - (2 \times 0.5 \times 5) \\ {v}^{2} = 5 \\ { \tt{final \: velocity = 2.24 \: {ms}^{ - 1} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}[/tex]
4. Consider a 1 kg block is on a 45° slope of ice. It is connected to a 0.4 kg block by a cable
and pulley. Does the 1 kg block move or down the slope? What is the net force on it and
its acceleration? (8 pts)
If an icy surface means no friction, then Newton's second law tells us the net forces on either block are
• m = 1 kg:
∑ F (parallel) = mg sin(45°) - T = ma … … … [1]
∑ F (perpendicular) = n - mg cos(45°) = 0
Notice that we're taking down-the-slope to be positive direction parallel to the surface.
• m = 0.4 kg:
∑ F (vertical) = T - mg = ma … … … [2]
Adding equations [1] and [2] eliminates T, so that
((1 kg) g sin(45°) - T ) + (T - (0.4 kg) g) = (1 kg + 0.4 kg) a
(1 kg) g sin(45°) - (0.4 kg) g = (1.4 kg) a
==> a ≈ 2.15 m/s²
The fact that a is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is a ≈ 2.15 m/s², which means the net force on the block would be ∑ F = ma ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.
When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.
Answer:
angular acceleration.
Explanation:
Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.
Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.
if the tin is made of a metal which has a density of 7800 kg per metre cubic calculate the volume of the metal used to make tin and lead
Answer:
XL sleep usual Addison officer at home and ear is not a short time to be be free and ear is a short time to make a short time
Explanation:
so that I can take the class on Monday and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to be be free and ear is not a short time to time for a day or night and ear buds is Anshu and duster and duster fgor a day or night is Anshu and duster for a day or not a week of computer science from your computer and I am in the same as I am a short of ti and you can be the first time I will be be
Every object around you is attracted to you. In fact, every object in the galaxy is attracted to every other object in the galaxy.
a. True
b. False
Answer:
True
Explanation:
With the gravitational pull that our planets have, we are able to remain in orbit. This demonstrates how every object in the galaxy is attracted to every other object. Every object in the universe that has mass exerts a gravitational pull on every other mass. We as humans do it too, but since our force isn't strong, we don't have much of an effect. I hope this helped and please don't hesitate to reach out with more questions!
20. How much charge will flow through a 2002 galvanometer
connected to a 40092 circular coil of 1000 turns on a wooden
stick 2 cm in diameter? If a magnetic field B=0.011 T parallel to
the axis of the stick is decreased suddenly to zero?
Answer:
5.76 μC
Explanation:
The induce emf, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = NAΔB where N = number of turns of coil = 1000, A = cross-sectional area of coil = πd²/4 where d = diameter of coil = 2 cm = 2 × 10⁻² m and ΔB = change in magnetic field strength = B' - B where B' = final magnetic field = 0 T and B = initial magnetic field strength = 0.011 T. So, ΔB = 0 T - 0.011 T = -0.011 T
So, ε = -ΔΦ/Δt
ε = -NAΔB/Δt
ε = -NAΔB/Δt
Also ε = iR where i = current and R = combined resistance of circular coil and galvanometer = 200 Ω + 400 Ω = 600 Ω (since they are in series)
So, iR = -NAΔB/Δt
iΔt = -NAΔB/R
Δq = -NAΔB/R where Δq = charge = iΔt
substituting the values of the variables into the equation, we have
Δq = -1000 × π(2 × 10⁻² m)²/4 × -0.011 T/600 Ω
Δq = -1000 × 4π × 10⁻⁴ m²/4 × -0.011 T/600 Ω
Δq = 0.011π × 10⁻¹ m²T/600 Ω
Δq = 0.03456 × 10⁻¹ m²T/600 Ω
Δq = 5.76 × 10⁻⁶ C
Δq = 5.76 μC
A block of mass M is connected by a string and pulley to a hanging mass m.The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. Find the acceleration of the system and tensions on the string.
The free body diagram for the block of mass M consists of four forces:
• the block's weight, Mg, pointing downward
• the normal force of the table pushing upward on the block, also with magnitude Mg
• kinetic friction with magnitude µMg = 0.2 Mg, pointing to the left
• tension of magnitude T pulling the block to the right
For the block of mass m, there are only two forces:
• its weight, mg, pulling downward
• tension T pulling upward
The m-block will pull the M-block toward the edge of the table, so we take the right direction to be positive for the M-block, and downward to be positive for the m-block.
Newton's second law gives us
T - 0.2Mg = Ma
mg - T = ma
where a is the acceleration of either block/the system. Adding these equations together eliminates T and we can solve for a :
mg - 0.2 Mg = (m + M) a
a = (m - 0.2M) / (m + M) g
a = 1.96 m/s²
Then the tension in the string is
T = m (g - a)
T = 78.4 N
what is conservation energy?
Explanation:
Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant
hope it is helpful to you
What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes
The question is incomplete. The complete question is :
A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s^2. Its maximum cruising speed is 90 mi/h. What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?
Solution :
Given :
Speed of the bullet train, v = 90 mi/h
= [tex]$90 \times \frac{5280}{3600}$[/tex]
= 132 ft/s
Time = 15 minutes
= 15 x 60
= 900 s
Acceleration from rest,
[tex]$a(t) = 4 \ ft/s^2$[/tex]
[tex]$v(t) = 4t + C$[/tex]
Since, v(0) = 0, then C = 0, so velocity is
v(t) = 4t ft/s
Then find the position function,
[tex]$s(t) = \frac{4}{2}t^2 + C$[/tex]
[tex]$=2t^2+C$[/tex]
It is at position 0 when t = 0, so C = 0, and the final position function for only the time it is accelerating is :
[tex]$s(t) = 2t^2$[/tex]
Time to get maximum cruising speed is :
4t = 132
t = 33 s
Distance travelled (at cruising speed) by speed to get the remaining distance travelled.
[tex]$900 \ s \times 132 \ \frac{ft}{s} = 118800 \ ft$[/tex]
Total distance travelled, converting back to miles,
[tex]$2178 + 118800 = 120978\ ft . \ \frac{mi}{5280 \ ft}$[/tex]
= 22.9125 mi
Therefore, the distance travelled is 22.9125 miles
You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area of 245 SF. The berm is 300 ft long and is assumed to taper evenly between the two cross-sectional areas, what is the calculated volume of the berm in cubic feet
The outer surface of a spacecraft in space has an emissivity of 0.44 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.
Answer:
[tex]T=326.928K[/tex]
Explanation:
From the question we are told that:
Emissivity [tex]e=0.44[/tex]
Absorptivity [tex]\alpha =0.3[/tex]
Rate of solar Radiation [tex]R=0.3[/tex]
Generally the equation for Surface absorbed energy is mathematically given by
[tex]E=\alpha R[/tex]
[tex]E=0.3*950[/tex]
[tex]E=285W/m^2[/tex]
Generally the equation for Emitted Radiation is mathematically given by
[tex]\mu=e(\sigmaT^4)[/tex]
Where
T=Temperature
[tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]
Therefore
[tex]\alpha*E=e \sigma T^4[/tex]
[tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]
[tex]T=326.928K[/tex]
The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 6. 7 m, measured from its tip to the center of the circle. Find the ratio of the centripetal acceleration at the end of the blade to that which exists at a point located 3.0 m from the center of the circle.
Answer:
[tex]\frac{a_{c1}}{a_{c2}} = 2.23[/tex]
Explanation:
The centripetal acceleration is given as follows:
[tex]a_c = \frac{v^2}{r}\\[/tex]
where,
ac = centripetal acceleration
v = linear speed = rω
r = radius
ω = angular speed
Therefore,
[tex]a_c = \frac{(r\omega)^2}{r}\\\\a_c = r\omega^2[/tex]
Therefore, the ratio will be:
[tex]\frac{a_{c1}}{a_{c2}} = \frac{r_1\omega^2}{r_2\omega^2}\\\\\frac{a_{c1}}{a_{c2}} = \frac{r_1}{r_2}\\\\[/tex]
where,
r₁ = 6.7 m
r₂ = 3 m
Therefore,
[tex]\frac{a_{c1}}{a_{c2}} = \frac{6.7\ m}{3\ m}\\\\[/tex]
[tex]\frac{a_{c1}}{a_{c2}} = 2.23[/tex]
Pure water is an example of alan
A. Insulator
B. Metalloid
C. Conductor
D. Nonmetal
Answer: I think its A or C I'm not sure though sorry.
In a
DC source, which has more cuwent?
(i)R L Circuit
(ii)RC Circuit (series)
(iii)LC Cirenit (series)
(iv)RLC Circuit (series)
Answer:
Answer is LC Cirenit (seres)
A mass-spring system oscillates with an amplitude of 4.20 cm. If the spring constant is 262 N/m and the mass is 560 g, determine the mechanical energy of the system.
Answer:
[tex]M.E=41J[/tex]
Explanation:
From the question we are told that:
Amplitude [tex]a=4.20cm[/tex]
Spring Constant [tex]K=262N/m[/tex]
Mass [tex]m=560g[/tex]
Generally the equation for mechanical energy is mathematically given by
[tex]M.E=\frac{1}{2}km^2[/tex]
[tex]M.E=0.5*262*0.56^2[/tex]
[tex]M.E=41J[/tex]
trình bày nguyên lý Đa lăm be
A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of the point charge and the total charge on shell were, respectively:
Complete question is;
A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of point charge and charge of the shell were respectively:
+5q; 0
-6q; +2q
+2q; -3q
-4q; +12q
Rank the results of experiments according to the charge on the inner surface of the shell, most positive first:
a. 2, 3, 1, 4
b. 1, 2, 3, 4
c. 2, 4, 3, 1
d. 1, 3, 4, 2
Answer:
c. 2, 4, 3, 1
Explanation:
In this question, we can say that;
q_in = q_b
Where;
q_in is the charge on the inner surface of the shell
q_b is the point charge on the shell.
Thus q_in = -q_b was written because, as the shell is conducting, it means that the electric field would have a value of zero and thus the radius inside will be zero.
Thus;
- For +5q; 0:
q_in = -(+5q)
q_in = -5q
- For -6q; +2q :
q_in = - (-6q)
q_in = +6q
- For +2q; -3q :
q_in = -(+2q)
q_in = -2q
- For -4q; +12q:
q_in = -(-4q)
q_in = +4q
Ranking the most positive to the least positive ones, we have;
+6q, +4q, -2q, -5q
This corresponds to options;
2, 4, 3, 1
In the early 1900's ____
began leading the automobile exploration in the US automotive industry.
-Karl Benz
-Henry Ford
-Gottlieb Daimler
-None of the above
Answer:
Henry Ford
Explanation:
he built the first ford
A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill?
A) 3.57 m.
B) 4.28 m.
C) 3.14 m.
D) 2.68 m.
Answer:
A(3.56m)
Explanation:
We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.
We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.
Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2
Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).
Ef= -mgh+3/4mv^2
Since Ef=Ei=0
Mgh=3/4mv^2
gh=3/4v^2
h=0.75v^2/g
plug in givens to get h= 3.57m
How much work is required to stretch an ideal spring of spring constant (force constant) 40 N/m from x
Answer:
The work done will be "0.45 J".
Explanation:
Given:
K = 40 N/m
x₁ = 0.20 m
x₂ = 0.25 m
Now,
The required work done will be:
= [tex]\frac{1}{2}k[x_2^2-x_1][/tex]
By putting the values, we get
= [tex]\frac{40}{2}[(0.25)^2-(0.20)^2][/tex]
= [tex]20\times 0.0225[/tex]
= [tex]0.45 \ J[/tex]
Three 30 g metal balls, one of aluminum, copper and lead, are placed in a large beaker of hot water for a few minutes. [The specific heats of aluminum, copper, and lead are 903, 385, and 130 J / (kg ° C), respectively].
to. Which of the balls, if any, will reach the highest temperature? Explain.
b. Which of the balls, if any, will have the most heat energy? Explain.
Answer:
The answer is below
Explanation:
Specific heat capacity is an intensive property of a material. The specific heat of a material is the amount of energy required to raise the temperature of one unit mass m of material by one unit of temperature.
a) Temperature is inversely proportional to specific heat capacity. If the same amount of heat is applied to all three balls, the ball that will reach the highest temperature is the ball with the least specific heat capacity.
Hence lead will have the highest temperature since it has the least specific heat capacity.
b) The quantity of heat is directly proportional to the specific heat capacity. Hence if all balls experience the same temperature change, the ball that have the most energy will be that with the highest specific heat capacity.
Hence aluminum will have the most heat since it has the highest specific heat capacity.
