Answer:
Answer:196 Joules
Explanation:
Hello
Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem
the work is the product of a force applied to a body and the displacement of the body in the direction of this force
assuming that the force goes in the same direction of the displacement, that is upwards
W=F*D (work, force,displacement)
the force necessary to move the object will be
[tex]F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules[/tex]
Answer:196 Joules
I hope it helps
The maximum potential energy of a spring system (mass 15 kg, spring constant 850 N/m) is 6.5 J. a) What is the amplitude of the oscillation? b) What is the maximum speed? c) Setting φ = 0, write the equation for the potential energy as a function of time.
Answer:
a) 0.124 m
b) 0.93 ms⁻¹
c) 0.5 k A² cos ² ( ωt )
Explanation:
1) Potential energy = U = 0.5 k A² , where A is the amplitude and k = 850 N/m is the spring constant.
0.5 ( 850) (A² ) = 6.5
⇒ A = 0.124 m = Amplitude.
b) From energy conservation, 0.5 m v² = 6.5
⇒ speed = v = 0.93 ms⁻¹
c) If x = A cos ωt ,
Potential energy = 0.5 k A² = 0.5 k A² cos ² ( ωt )
An eagle is flying horizontally at a speed of 3m/s when the fish in her talons wiggles loose and falls into the lake 10m below. Calculate the velocity of the fish relative to the water when it hits the water.
Answer:
Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.
Explanation:
Vertical motion of fish:
Initial speed, u = 0
Acceleration, a = 9.81 m/s²
Displacement, s = 10 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 0² + 2 x 9.81 x 10 = 196.2
v = 14 m/s
Final vertical speed = 14 m/s
Final horizontal speed = initial horizontal speed = 3 m/s
Final velocity = 3 i - 14 j m/s
Magnitude
[tex]v=\sqrt{3^2+(-14)^2}=14.32m/s[/tex]
Direction
[tex]\theta =tan^{-1}\left ( \frac{-14}{3}\right )=-77.91^0[/tex]
Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.
Can the resistors in an "unbalanced" Wheatstone bridge circuit be treated as a combination of series and/or parallel resistors? What about a "balanced bridge?
Answer:
Explanation:
The resistors in a unbalanced wheat stone bridge cannot be treated as a combination of series and parallel combination of resistors.
In case of balanced wheat stone bridge, the resistors can be treated as the combination of series and parallel combination.
Here, In the balanced wheat stone bridge
R1 and R2 be in series and Ra and Rx is series and then their combination is in parallel combination.
Resistors in a Wheatstone bridge can be treated as combinations of series and/or parallel resistors for simplification in both balanced and unbalanced bridges. A balanced bridge allows separate treatment of two parallel branches, while unbalanced requires careful analysis. Not all resistor networks can be simplified into series or parallel models.
Explanation:In a Wheatstone bridge, the resistors can indeed be treated as combinations of series and/or parallel resistors when aiming to simplify calculations or understand the behavior of the circuit. For an unbalanced bridge, resistors are not in simple series or parallel arrangements with respect to the entire circuit due to the bridge not being in equilibrium. However, within certain parts of the bridge, resistors may appear to be in series or parallel with each other. In a balanced bridge, where the bridge is in a state of equilibrium and the central voltmeter reads zero, the two arms of the bridge can be treated separately as two parallel voltage dividers, because no current flows through the meter, effectively decoupling the two parallel branches.
Attempting to simplify a complex resistor network encountered in bridges can indeed be done by identifying and replacing series and parallel resistor combinations step by step until a single resistance value is found. However, this approach cannot always be applied to any arbitrary combination of resistors. Some configurations might contain elements that cannot be reduced to mere parallel or series connections, usually because they form more intricate networks, such as bridges or loops not separable into simpler series or parallel sections.
In conclusion, while more complex connections of resistors in circuits like the Wheatstone bridge can often be broken down into combinations of series and parallel, this is not universally the case for all resistor networks. In certain scenarios, specific techniques or theorems such as Kirchhoff's laws might be required to analyze the circuit effectively.
(c) If η = 60% and TC = 40°F, what is TH, in °F?
Part C Suppose 1 kg each of water (4.19 J/(g ⋅ ∘C)), brick (0.90 J/(g ⋅ ∘C)), iron (0.46 J/(g ⋅ ∘C)), and olive oil (1.79 J/(g ⋅ ∘C)) were held at the same initial temperature and heated for an equivalent amount of time. Indicate their relative final temperatures from lowest (left) to highest (right). Assume no heat is lost to the surroundings. You can use the Intro tab of the PhET to help visualize the temperature changes. (Assume that the mas of the water, iron, and olive oil is 1 kg each and that the mass of the brick is 0.5 kg.) Rank from lowest resulting temperature to highest resulting temperature. To rank items as equivalent, overlap them. View Available Hint(s) ResetHelp IronWaterOlive oilBrick
Answer:
Water < Olive oil < Brick < IronExplanation:
The change in temperature may be calculated from the formula:
Q = m × C × ΔT ⇒ ΔT = Q / (m × C)Where:
Q = amount of heat energy supplied (or released in case of cooling)m = massC = specific heat (a different constant, property, for every matter)ΔT = increase of temperatute (decrease when it is cooling)Then, you can make these assumptions or inferences from the conditions stated in the problem:
Q is the same for all the matters because it is supplied by a external source, taking into account that all the substances are held the same time.Initial temperature is the same for all the substances (given)The mass is the same (1 kg each sample)Then, you can state that, for those samples, ΔT = k / C, i.e. the increase in temperature is inversely related to the specific heat.
That means that the higher the specific heat the lower ΔT, and the lower the specific heat the higher ΔT.
The ranking in decrasing order of specific heat is:
Water (4.19) > Olive oil (1.79) > Brick (0.9) > Iron (0.46)Ranking in increasing order of ΔT:
Water < Olive oil < Brick < IronAnd since all of them started at the same temperature, that is the ranking in resulting temperature from lowest to highest:
Water < Olive oil < Brick < Iron ← answerThat means that the sample of water, the matter with the highest specific heat capacity (4.19 J/g°C), will reach the lowest temperature, and the sample of iron, the matter with the lowest heat capacity (0.46 J/g°C) will reach the highest temperature.
A spring has a natural length of 28 cm. If a 27-N force is required to keep it stretched to a length of 32 cm, how much work W is required to stretch it from 28 cm to 30 cm? (Round your answer to two decimal places.) W = 13.5 Incorrect: Your answer is incorrect. J
Answer:
0.14 J
Explanation:
Use the force to calculate the spring constant.
F = k Δx
27 N = k (0.32 m − 0.28 m)
k = 675 N/m
Work is the change in energy:
W = PE
W = ½ k (Δx)²
W = ½ (675 N/m) (0.30 m − 0.28 m)²
W = 0.135 Nm
W = 0.135 J
Rounding to two decimal places, W = 0.14 J.
Your answer was correct, but it was in units of Ncm, and you needed to answer in units of J.
The work done to stretch the spring from 28 cm to 30 cm is 39.00 Joules, computed using the principles of Hooke's Law and the concept of work done.
Explanation:In this problem, we are dealing with the concept of work done on a spring, which falls under Physics principles. Hooke's Law states that the force to compress or extend a spring by a distance x from its natural length is proportional to x. It can be written as F = kx, where F is the force, k is the spring constant, and x is the distance.
In this case, the force (F) is 27 N, and the length of stretch (x) is 32 cm - 28 cm = 4 cm. We can find the spring constant (k) using the formula k = F / x = 27 N / 4 cm = 6.75 N/cm.
The work done (W) to stretch the spring from 28 cm to 30 cm is the area under the force/displacement graph from '28 cm' to '30 cm'. Since the force is linear with displacement for a spring, this area can be found using the formula for the area of a trapezoid: W = ½ (F1 + F2) x d. F1 is the initial force (k*28 cm), F2 is the final force (k*30 cm), and d is the displacement (30 cm - 28 cm). Substituting the values, W = ½ [(6.75 N/cm*28 cm)+(6.75 N/cm*30 cm)]*(2 cm) = 39.00 J.
Learn more about Work Done here:https://brainly.com/question/35917320
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The space shuttle travels at a speed of about 7.6times10^3 m/s. The blink of an astronaut's eye lasts about 110 ms. How many football fields (length=91.4 m) does the shuttle cover in the blink of an eye? The three-toed sloth is the slowest-moving land mammal. On the ground, the sloth moves at an average speed of 0.033 m/s, considerably slower than the giant tortoise have gone relative to the sloth? Do not put units in your answer.
Answer:
It covers distance of 9.15 football fields in the said time.
Explanation:
We know that
[tex]Distance=Speed\times Time[/tex]
Thus distance covered in blinking of eye =
[tex]Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters[/tex]
Thus no of football fields=[tex]\frac{936}{91.4}=9.15Fields[/tex]
Name a common product produced by blow molding.
Parts made from blow molding are plastic, hollow, and thin-walled, such as bottles and containers that are available in a variety of shapes and sizes. Small products may include bottles for water, liquid soap, shampoo, motor oil, and milk, while larger containers include plastic drums, tubs, and storage tanks.
A person with mass of 68kg rides in an elevator that has an upward acceleration of 1.5 m/s2. Draw 1 body diagram of the person and determine the magnitude of the force of the elevator floor ree on the person?
Answer:
N = 768.4 N
Explanation:
As per given FBD we can see that the person inside the elevator have two forces on it
1) Normal force upwards
2) weight downwards
Now from Newton's law of motion we can say
[tex]F_{net} = ma[/tex]
[tex]N - mg = ma[/tex]
[tex]N = mg + ma[/tex]
now plug in all values in it
[tex]N = 68(1.5) + 68(9.8)[/tex]
[tex]N = 768.4 N[/tex]
Starting from rest, a 1.8 x 10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts an average upward force of 0.42 N on it. This force does +2.7 x 10^-4 J of work on the flea. (a) What is the flea's speed when it leaves the ground? (b) How far upward does the flea move while it is pushing off? Ignore both air resistance and the flea's weight.
Answer:
a)
1.73 m/s
b)
6.43 x 10⁻⁴ m
Explanation:
m = mass of the flea = 1.8 x 10⁻⁴ kg
v₀ = initial speed of the flea = 0 m/s
v = final speed of the flea
W = work done by the force on the flea = 2.7 x 10⁻⁴ J
Using work-change in kinetic energy, Work done is given as
W = (0.5) m (v² - v₀²)
Inserting the values
2.7 x 10⁻⁴ = (0.5) (1.8 x 10⁻⁴) (v² - 0²)
v = 1.73 m/s
b)
d = distance moved by the flea while pushing off
F = Upward force applied on the flea by ground = 0.42 N
Work done is also given as
W = F d
2.7 x 10⁻⁴ = (0.42) d
d = 6.43 x 10⁻⁴ m
Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.
Answer:
[tex]strain = 1.4 \times 10^{-3} [/tex]
Explanation:
As we know by the formula of elasticity that
[tex]E = \frac{stress}{strain}[/tex]
now we have
[tex]E = 110 GPA[/tex]
[tex]E = 110 \times 10^9 Pa[/tex]
Area = 15.2 mm x 19.1 mm
[tex]A = 290.3 \times 10^{-6}[/tex]
now we also know that force is given as
[tex]F = 44500 N[/tex]
here we have
stress = Force / Area
[tex]stress = \frac{44500}{290.3 \times 10^{-6}}[/tex]
[tex]stress = 1.53 \times 10^8 N/m^2[/tex]
now from above formula we have
[tex]strain = \frac{stress}{E}[/tex]
[tex]strain = \frac{1.53 \times 10^8}{110 \times 10^9}[/tex]
[tex]strain = 1.4 \times 10^{-3} [/tex]
4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with 5m/s in the same direction. They suffer an elastic collision and block B is now moving with 8m/s. What is the speed of block A after collision? 10m/s
8m/s
5m/s
13m/s
18m/s
Answer:
Option A is the correct answer.
Explanation:
Here momentum is conserved.
That is [tex]\left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}[/tex]
Substituting values
[tex]3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s[/tex]
Speed of block A after collision = 10 m/s
Option A is the correct answer.
Answer:
8m/s
Explanation:
1/32nd of a radioactive element's atoms remain in a lunar rock sample. How many half lives old is the rock?
A. 1
B. 2
C. 3
D. 4
E. 5
Answer:
E. 5
Explanation:
N₀ = initial total number of radioactive elements number
N = Number of atoms of radioactive element after "n" half lives = N₀ /32
n = number of half lives
Number of atoms of radioactive element after "n" half lives is given as
[tex]N = N_{o}\left ( \frac{1}{2} \right )^{n}[/tex]
inserting the values
[tex]\frac{N_{0}}{32} = N_{o}\left ( \frac{1}{2} \right )^{n}[/tex]
[tex]\frac{1}{32} = \left ( \frac{1}{2} \right )^{n}[/tex]
n = 5
If a 2 inch diameter rod is subjected to a centric tensile axial load of 15.71 kip, what is the average normal stress (ksi) to three significant figures?
Answer:
The average normal stress is 5 ksi.
Explanation:
Given that,
Diameter = 2 inch
Load = 15.71 kip
We need to calculate the average normal stress
Using formula of stress
Average normal stress [tex]\sigma =\dfrac{F}{A}[/tex]
Where, F = load
A = area
Put the value into the formula
[tex]\sigma=\dfrac{15.71}{\pi\times(\dfrac{2}{2})^2}[/tex]
[tex]\sigma = 5\ kip/inc^2[/tex]
[tex]\sigma=5\ ksi[/tex]
Hence, The average normal stress is 5 ksi.
In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres with masses of 1.60 kg and 16.0 g whose centers are separated by about 4.00 cm. Calculate the gravitational force between these spheres, treating each as a particle located at the center of the sphere.
Answer:
[tex]1.0672\times 10^{-9}N[/tex]
Explanation:
[tex]G[/tex] = Gravitational constant = 6.67 x 10⁻¹¹
[tex]F[/tex] = Gravitational force between these spheres
[tex]m_{1}[/tex] = mass of first sphere = 1.60 kg
[tex]m_{2}[/tex] = mass of second sphere = 16 g = 0.016 kg
[tex]r[/tex] = distance between the centers of the sphere = 4 cm = 0.04 m
Gravitational force between these spheres is given as
[tex]F = \frac{Gm_{1}m_{2}}{r^{2}}[/tex]
[tex]F = \frac{(6.67\times 10^{-11})(1.60)(0.016)}{0.04^{2}}[/tex]
[tex]F = 1.0672\times 10^{-9}N[/tex]
A solenoid having N turns and carrying a current of 2.000 A has a length of 34 00 cm. If the magnitude of the magnetic field generated at the center of the solenoid is 9.000 mT what is the value of N? (μo = 4π x10^-7 T. m/A) A) 2318 B) 1218 C) 860.0 D) 3183 E) 1591
Answer:
B) 1218
Explanation:
N = Total number of turns in the solenoid
L = length of the solenoid = 34.00 cm = 0.34 m
B = magnetic field at the center of the solenoid = 9 mT = 9 x 10⁻³ T
i = current carried by the solenoid = 2.000 A
Magnetic field at the center of the solenoid is given as
[tex]B = \frac{\mu _{o}N i}{L}[/tex]
[tex]9\times 10^{-3} = \frac{(4\pi\times 10^{-7} )N (2)}{0.34}[/tex]
N = 1218
The value of N is about B) 1218
[tex]\texttt{ }[/tex]
Further explanationLet's recall magnetic field strength from current carrying wire and from center of the solenoid as follows:
[tex]\boxed {B = \mu_o \frac{I}{2 \pi d} } [/tex]
B = magnetic field strength from current carrying wire (T)
μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)
I = current (A)
d = distance (m)
[tex]\texttt{ }[/tex]
[tex]\boxed {B = \mu_o \frac{I N}{L} } [/tex]
B = magnetic field strength at the center of the solenoid (T)
μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)
I = current (A)
N = number of turns
L = length of solenoid (m)
Let's tackle the problem now !
[tex]\texttt{ }[/tex]
Given:
Current = I = 2000 A
Length = L = 34.00 cm = 0.34 m
Magnetic field strength = B = 9000 mT = 9 T
Permeability of free space = μo = 4π × 10⁻⁷ T.m/A
Asked:
Number of turns = N = ?
Solution:
[tex]B = \mu_o \frac{I N}{L}}[/tex]
[tex]\frac{I N}{L} = B \div \mu_o[/tex]
[tex]IN = BL \div \mu_o[/tex]
[tex]N = BL \div (\mu_o I)[/tex]
[tex]N = ( 9 \times 0.34 ) \div ( 4 \pi \times 10^{-7} \times 2000 )[/tex]
[tex]\boxed {N \approx 1218}[/tex]
[tex]\texttt{ }[/tex]
Learn moreTemporary and Permanent Magnet : https://brainly.com/question/9966993The three resistors : https://brainly.com/question/9503202A series circuit : https://brainly.com/question/1518810Compare and contrast a series and parallel circuit : https://brainly.com/question/539204[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Magnetic Field
A 20 m high filled water tank develops a 0.50 cm hole in the vertical wall near the base. With what speed does the water shoot out of the hole? a) 30 m/s
b)15 m/s
c) 25m/s
d) 20 m/s
Answer:
The speed of the water shoot out of the hole is 20 m/s.
(d) is correct option.
Explanation:
Given that,
Height = 20 m
We need to calculate the velocity
Using formula Bernoulli equation
[tex]\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]
Where,
v₁= initial velocity
v₂=final velocity
h₁=total height
h₂=height of the hole from the base
Put the value into the formula
[tex]v_{1}^2=2g(h_{2}-h_{1})[/tex]
[tex]v_{1}=\sqrt{2g(h_{2}-h_{1})}[/tex]
[tex]v_{1}=\sqrt{2\times9.8\times(20-0.005)}[/tex]
[tex]v_{1}=19.7\ m/s= approximate\ 20\ m/s[/tex]
Hence, The speed of the water shoot out of the hole is 20 m/s.
The activation energy for the uncatalyzed decomposition of hydrogen peroxide at 20°C is 75.3 kJ/mol. In the presence of the enzyme catalase, the activation energy is reduced to 29.3 kJ/mol. Use the following form of the Arrhenius equation, RT ln1k1/k22 5 Ea2 2 Ea1 , to calculate how much larger the rate constant of the catalyzed reaction is.
Answer:
The rate of enzyme catalyzed reaction will increases by [tex]1.58\times 10^{8}[/tex] times.
Explanation:
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
[tex]\log K=\log A-\frac{Ea}{2.303\times RT}[/tex]
The expression used with catalyst and without catalyst is,
[tex]\log K_1=\log A-\frac{Ea_1}{2.303\times RT}[/tex]...(1)
[tex]\log K_2=\log A-\frac{Ea_2}{2.303\times RT}[/tex]...(2)
On subtracting (2) from (1)
[tex]\log \frac{K_2}{K_1}=\frac{Ea_1-Ea_2}{2.303RT}[/tex]
where,
[tex]K_2[/tex] = rate of reaction with catalyst
[tex]K_1[/tex] = rate of reaction without catalyst
[tex]Ea_2[/tex] = activation energy with catalyst = 29.3 kJ/mol = 29300 J/mol
[tex]Ea_1[/tex] = activation energy without catalyst = 75.3 kJ/mol=75300 J/mol
R = gas constant =8.314 J /mol K
T = temperature = [tex]20^oC=273+20=293K[/tex]
Now on substituting all the values in the above formula, we get
[tex]\log \frac{K_2}{K_1}=\frac{75300 kJ/mol-29300 kJ/mol}{2.303\times 8.314 J/mol K\times 293}=1.58\times 10^{8}[/tex]
The rate of enzyme catalyzed reaction will increases by [tex]1.58\times 10^{8}[/tex] times.
An electron is released from rest in a uniform electric field of 418 N/C near a particle detector. The electron arrives at the detector with a speed of 3.70 x 106 m/s (a) What was the uniform acceleration of the electron? (Enter the magnitude.) m/s (b) How long did the electron take to reach the detector? (c) What distance was traveled by the electron? cm (d) What is the kinetic energy of the electron when it reaches the detector?
Answer:
a) 7.35 x 10¹³ m/s²
b) 5.03 x 10⁻⁸ sec
c) 9.3 cm
d) 6.23 x 10⁻¹⁸ J
Explanation:
E = magnitude of electric field = 418 N/C
q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of the electron = 9.1 x 10⁻³¹ kg
a)
acceleration of the electron is given as
[tex]a = \frac{qE}{m}[/tex]
[tex]a = \frac{(1.6\times 10^{-19})(418)}{(9.1\times 10^{-31})}[/tex]
a = 7.35 x 10¹³ m/s²
b)
v = final velocity of the electron = 3.70 x 10⁶ m/s
v₀ = initial velocity of the electron = 0 m/s
t = time taken
Using the equation
v = v₀ + at
3.70 x 10⁶ = 0 + (7.35 x 10¹³) t
t = 5.03 x 10⁻⁸ sec
c)
d = distance traveled by the electron
using the equation
d = v₀ t + (0.5) at²
d = (0) (5.03 x 10⁻⁸) + (0.5) (7.35 x 10¹³) (5.03 x 10⁻⁸)²
d = 0.093 m
d = 9.3 cm
d)
Kinetic energy of the electron is given as
KE = (0.5) m v²
KE = (0.5) (9.1 x 10⁻³¹) (3.70 x 10⁶)²
KE = 6.23 x 10⁻¹⁸ J
A billiard ball moving at 6.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 5.21 m/s at an angle of 29.7° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision.
Answer:
Velocity is 3.11 m/s at an angle of -56° with respect to the original line of motion.
Explanation:
Let line of action be horizontal axis , mass of ball be m and unknown velocity be v.
Here momentum is conserved.
Initial momentum =Final momentum
Initial momentum = m x 6i + m x 0i = 6m i
Final momentum = m x (5.21cos 29.7 i + 5.21sin 29.7 j) + m x v = 4.26 m i + 2.58 m j + m v
4.26 m i + 2.58 m j + m v = 6m i
v = 1.74 i - 2.58 j
Magnitude of velocity [tex]=\sqrt{1.74^2+(-2.58)^2}=3.11m/s[/tex]
Direction,
[tex]\theta =tan^{-1}\left ( \frac{-2.58}{1.74}\right )=--56^0[/tex]
Velocity is 3.11 m/s at an angle of -56° with respect to the original line of motion.
Rank these temperatures from hottest to coldest: 32° F,32° C, and 32 K 320 F> 32° C>32 K 32°C 32° F 32 K 32° F 32 K 32° c 32° F 32° c 32 K 32° C 32° F 32 K All are the same temperature
Answer:
32 C > 32 F > 32 K
Explanation:
32 F, 32 C, 32 K
Let T1 = 32 F
T2 = 32 C
T3 = 32 K
Convert all the temperatures in degree C
The relation between F and C is given by
(F - 32) / 9 = C / 100
so, (32 - 32) / 9 = C / 100
C = 0
So, T1 = 32 F = 0 C
The relation between c and K is given by
C = K - 273 = 32 - 273 = - 241
So, T3 = 32 K = - 241 C
So, T 1 = 0 C, T2 = 32 c, T3 = - 241 C
Thus, T2 > T1 > T3
32C > 32 F > 32 K
The heat flux for a given wall is in the x-direction and given as q^n = 11 W/m^2, the walls thermal conductivity is 1.7 W/mK and the walls thickness is 1.3 cm. Assume steady state conditions exist. Determine the temperature gradient in units of K/m and degree C/m. If the temperature gradient were larger what happens to the heat flux and why? Comment on the direction of heat flux given a negative temperature gradient and again for the case of a positive temperature gradient.
Answer:
[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]
Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.
For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards
Explanation:
As we know that heat flux is given by the formula
[tex]q^n = K\frac{dT}{dx}[/tex]
here we know that
K = thermal conductivity
[tex]\frac{dT}{dx}[/tex] = temperature gradient
now we know that
[tex]q^n = 11 W/m^2[/tex]
also we know that
K = 1.7 W/mK
now we have
[tex]11 = 1.7 \frac{dT}{dx}[/tex]
so temperature gradient is given as
[tex]\frac{dT}{dx} = \frac{11}{1.7} = 6.47 K/m [/tex]
also in other unit it will be same
[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]
Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.
For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards
Two points are on a disk turning at constant angular velocity. One point is on the rim and the other halfway between the rim and the axis. Which point moves the greater distance in a given time?
Answer:
The point on the rim
Explanation:
All the points on the disk travels at the same angular speed [tex]\omega[/tex], since they cover the same angular displacement in the same time. Instead, the tangential speed of a point on the disk is given by
[tex]v=\omega r[/tex]
where
[tex]\omega[/tex] is the angular speed
r is the distance of the point from the centre of the disk
As we can see, the tangential speed is directly proportional to the distance from the centre: so the point on the rim, having a larger r than the point halway between the rim and the axis, will have a larger tangential speed, and therefore will travel a greater distance in a given time.
Why do we use the two-body problem to solve interplanetary trajectories, instead of including all of the appropriate gravitational forces that actually apply?
Answer:
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A sample of chloroform is found to contain 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen. If a second sample of chloroform is found to contain 30.0 g of carbon, what is the total mass of chloroform in the second sample?
To find the total mass of chloroform in the second sample, the mass of carbon is scaled proportionally to find the corresponding masses of hydrogen and chlorine. By summing these, the total mass of chloroform is calculated to be 298.525 grams.
Explanation:To determine the total mass of chloroform in the second sample, we must first understand that chloroform has a known molecular formula of CHCl3. Given that the first sample contains 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen, we can deduce the mass ratios of the elements within chloroform. Using the molecular mass of chloroform, which is 119.37 g/mol, we can calculate the masses of hydrogen and chlorine in the second sample based on the given mass of carbon.
For the second sample: If 12.0 g of carbon is accompanied by 1.01 g of hydrogen and 106.4 g of chlorine, then 30.0 g of carbon should be accompanied by:
Hydrogen: (1.01 g H / 12.0 g C) × 30.0 g C = 2.525 g HChlorine: (106.4 g Cl / 12.0 g C) × 30.0 g C = 266.0 g ClThus, the total mass of chloroform in the second sample would be the sum of the masses of carbon, hydrogen, and chlorine: 30.0 g C + 2.525 g H + 266.0 g Cl = 298.525 g of chloroform.
Relative to the ground, a car has a velocity of 15.3 m/s, directed due north. Relative to this car, a truck has a velocity of 22.5 m/s, directed 52.0° north of east. What is the magnitude of the truck's velocity relative to the ground
Answer:
The magnitude of the truck's velocity relative to the ground is 35.82 m/s.
Explanation:
Given that,
Velocity of car relative to ground = 15.3 m/s
Velocity of truck relative to car = 22.5 m/s
We need to calculate the magnitude of the truck's velocity relative to the ground
We need to calculate the x component of the velocity
[tex]v_{x}=22.5\cos\theta[/tex]
[tex]v_{x}=22.5\cos52^{\circ}[/tex]
[tex]v_{x}=13.852\ m/s[/tex]
We need to calculate the y component of the velocity
[tex]v_{y}=15.3+22.5\sin\theta[/tex]
[tex]v_{y}=15.3+22.5\sin52^{\circ}[/tex]
[tex]v_{y}=33.030\ m/s[/tex]
Using Pythagorean theorem
[tex]|v|=\sqrt{v_{x}^2+v_{y}^2}[/tex]
[tex]|v|=\sqrt{(13.852)^2+(33.030)^2}[/tex]
[tex]|v|=35.82\ m/s[/tex]
Hence, The magnitude of the truck's velocity relative to the ground is 35.82 m/s.
A square coil (length of side = 24 cm) of wire consisting of two turns is placed in a uniform magnetic field that makes an angle of 60° with the plane of the coil. If the magnitude of this field increases by 6.0 mT every 10 ms, what is the magnitude of the emf induced in the coil?
Explanation:
It is given that,
Length of side of a square, l = 24 cm = 0.24 m
The uniform magnetic field makes an angle of 60° with the plane of the coil.
The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :
[tex]\epsilon=N\dfrac{d\phi}{dt}[/tex]
[tex]\dfrac{d\phi}{dt}[/tex] is the rate of change if magnetic flux.
[tex]\phi=BA\ cos\theta[/tex]
[tex]\theta[/tex] is the angle between the magnetic field and the normal to area vector.
[tex]\theta=90-60=30[/tex]
[tex]\epsilon=NA\dfrac{dB}{dt}\times cos30[/tex]
[tex]\epsilon=2\times (0.24\ m)^2\times \dfrac{6\ mT}{10\ mT}\times cos(30)[/tex]
[tex]\epsilon=0.0598\ T[/tex]
[tex]\epsilon=59.8\ mT[/tex]
or
EMF = 60 mT
So, the magnitude of emf induced in the coil is 60 mT. Hence, this is the required solution.
An electron is released from rest in a uniform electric field. The electron accelerates, travelling 6.20 m in 4.50 µs after it is released. What is the magnitude of the electric field in N/C?
Answer:
E= 3.4893 N/C
Explanation:
Given s=6.20 m , t=2.50μs, m=9.11*10^-31 Kg , q= 1.6*10^-19 C
the distance traveled by the electron in time t is
s=ut+0.5at^2
here, u is the initial velocity of the electron, t is time taken and
a is acceleration.
Since the electron is initially at rest u=0
now s=0.5at^2
Therefore a=2s/t^2
also. we know that strength of electric field is
E=ma/q
[tex]E= \frac{2ma}{qt^2}[/tex]
now puting the values we get
[tex]E=\frac{9.11\times 10^-31\times 2\times 6.20}{1.6\times 10^-19\times (4.5\times 10^-6)^2}[/tex]
therefore, E= 3.4865 N/C
The magnitude of the electric field is calculated by first determining the acceleration of the electron and then using the electric force equation to find the electric field. The resulting electric field is 34.8 N/C.
To find the magnitude of the electric field, we first need to calculate the acceleration of the electron. Given that the electron travels a distance of 6.20 m in a time of 4.50 µs (4.50 × 10-6 s), we can use the equations of motion.
Initial velocity, u = 0 (since the electron is released from rest)
Time, t = 4.50 × 10-6 s
Distance, s = 6.20 m
Using the equation of motion: s = ut + 0.5at2
Substitute the values: 6.20 = 0 + 0.5a(4.50 × 10-6)2
6.20 = 0.5a(20.25 × 10-12)
a = 6.20 / (0.5 × 20.25 × 10-12)
a = 6.20 / (10.125 × 10-12)
a = 6.12 × 1011 m/s2
Now, we calculate the electric field using Newton's Second Law, F = ma, and the electric force equation, F = eE, where e is the charge of the electron (1.60 × 10-19 C) and E is the electric field.
ma = eE
(9.11 × 10-31 kg)(6.12 × 1011 m/s2) = (1.60 × 10-19 C)E
(5.57 × 10-19 N) = (1.60 × 10-19 C)E
E = 5.57 × 10-19 N / 1.60 × 10-19 C
E = 3.48 × 101 N/C
Thus, the magnitude of the electric field is 34.8 N/C.
A bullet of mass 0.093 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 2.5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block?
Explanation:
It is given that,
Mass of the bullet, m₁ = 0.093 kg
Initial speed of bullet, u₁ = 100 m/s
Mass of block, m₂ = 2.5 kg
Initial speed of block, u₂ = 0
We need to find the speed of the block after the bullet embeds itself in the block. Let it is given by V. On applying the conservation of linear momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
[tex]V=\dfrac{0.093\ kg\times 100\ m/s+0}{(0.093\ kg+2.5\ kg)}[/tex]
V = 3.58 m/s
So, the speed of the bullet is 3.58 m/s. Hence, this is the required solution.
A fox locates rodents under the snow by the slight sounds they make. The fox then leaps straight into the air and burrows its nose into the snow to catch its meal. If a fox jumps up to a height of 81 cm , calculate the speed at which the fox leaves the snow and the amount of time the fox is in the air. Ignore air resistance.
Answer:
4 m/s
0.82 s
Explanation:
h = height to which the fox jumps = 81 cm = 0.81 m
v₀ = speed at which the fox leaves the snow
v = speed of the fox at highest point = 0 m/s
a = acceleration due to gravity = - 9.8 m/s²
Using the kinematics equation
v² = v₀² + 2 a h
0² = v₀² + 2 (- 9.8) (0.81)
v₀ = 4 m/s
t = amount of time in air while going up
Using the equation
v = v₀ + a t
0 = 4 + (- 9.8) t
t = 0.41 s
T = Total time
Total time is given as
T = 2 t
T = 2 (0.41)
T = 0.82 s
The speed at which the fox leaves the snow is approximately 3.987 m/s. The fox is in the air for approximately 0.407 seconds.
Explanation:To calculate the speed at which the fox leaves the snow, we can use the concept of vertical motion and the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. Since the fox jumps straight up, the initial velocity is 0 m/s and the displacement is 81 cm (or 0.81 m). Assuming the acceleration due to gravity is 9.8 m/s^2, we can now calculate the final velocity:
v^2 = u^2 + 2as
v^2 = 0^2 + 2(9.8)(0.81)
v^2 = 15.876
v = √15.876
v ≈ 3.987 m/s
The time the fox is in the air can be calculated using the equation v = u + at, where t is the time. Again, the initial velocity is 0 m/s and the acceleration due to gravity is 9.8 m/s^2. Plugging in these values, we have:
v = u + at
3.987 = 0 + (9.8)t
3.987 = 9.8t
t = 3.987/9.8
t ≈ 0.407 s