Answer:
30 minutes
Explanation:
Energy per time is constant, so:
E₁ / t₁ = E₂ / t₂
m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂
(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t
(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t
3/min = 90 / t
t = 30 min
Two radio antennas A and B radiate in phase. Antenna B is a distance of 100 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 50.0 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.
Required:
a. What is the longest wavelength for which there will be destructive interference at point Q?
b. What is the longest wavelength for which there will be constructive interference at point Q?
Answer:
a. 200 m
b. 100 m
Explanation:
Solution:-
- We will first draw three points marked A,B and Q from left most to right most.
- We are told that the antennas at A and B radiate in phase. This means the radio-waves emitted by each antenna are synchronous in terms of ( frequency and wavelength ).
- We will denote the common wavelength of coherent sources of radio-waves ( A and B ) with λ.
- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of destructive interference is:
AQ - BQ = n*λ/2
Where,
n: The order of wavelength
AQ: The distance between antenna A and point Q
BQ: The distance between antenna B and point Q
- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,
150 - 50 = n*λ/2
- To determine the longest wavelength ( λ ) to meet destructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,
100 = λ/2
λ = 200 m .... Answer
- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of constructive interference is:
AQ - BQ = n*λ
Where,
n: The order of wavelength
AQ: The distance between antenna A and point Q
BQ: The distance between antenna B and point Q
- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,
150 - 50 = n*λ
- To determine the longest wavelength ( λ ) to meet constructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,
100 = λ
λ = 100 m .... Answer
Assuming 100% efficient energy conversion, how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery?
Complete question is;
Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes.
Answer:
Amount of water required to charge the battery = 7.35 m³
Explanation:
The formula for Potential energy of the water at that height = mgh
Where;
m = mass of the water
g = acceleration due to gravity = 9.8 m/s²
h = height of water = 50 cm = 0.5 m
We know that in density, m = ρV
Where;
ρ = density of water = 1000 kg/m³
V = volume of water
So, potential energy is now given as;
Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J
Now, formula for energy of the battery is given as;
E = qV
We are given;
q = 50 A.min = 50 × 60 = 3,000 C
V = 12 V
Thus;
qV = 3,000 × 12 = 36,000 J
E = 36,000 J
At a 100% conversion rate, the energy of the water totally powers the battery.
Thus;
(4900V) = (36,000)
4900V = 36,000
V = 36,000/4900
V = 7.35 m³
The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average rest energy is 91.19 GeV, but its short lifetime shows up as an intrinsic width of 2.5 GeV. what is the lifetime of this particle?
Answer:
The lifetime of the particle is [tex]\Delta t = 2.6*10^{-25} \ s[/tex]
Explanation:
From the question we are told that
The average rest energy is [tex]E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J[/tex]
The intrinsic width is [tex]\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J[/tex]
The lifetime is mathematically represented as
[tex]\Delta t = \frac{h}{\Delta E}[/tex]
Where h is the Planck's constant with a value of [tex]1.055*10^{-34} \ J\cdot s[/tex]
substituting values
[tex]\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}[/tex]
[tex]\Delta t = 2.6*10^{-25} \ s[/tex]
What is The mass of an electron
The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, though the nature of the prediction was not fully understood until the experimental discovery. Today, it is well accepted that all fundamental particles have antiparticles.
Suppose that an electron and a positron collide head-on. Both have kinetic energy of 3.58 MeV and rest energy of 0.511 MeV. They produce two photons, which by conservation of momentum must have equal energy and move in opposite directions. What is the energy Eloton of one of these photons?
Answer:
4.09 MeV
Explanation:
Find the given attachment
One car travels 40. meters due east in 5.0 seconds, and a second car travels 64 meters due west in 8.0 seconds. During their periods of travel, the cars definitely had the same
Answer:
They had the same speed.
Explanation:
It won't be velocity, because velocity is a vector quantity. Speed is scalar.
Velocity is the rate of change of displacement. During their periods of travel, the cars definitely had the same velocity.
What is Velocity?Velocity is the directional speed of a moving object as an indicator of its rate of change in location as perceived from a certain frame of reference and measured by a specific time standard.
Given that the first car travels 40 meters due east in 5 seconds. Therefore, we can write,
Distance = 40 meters
Time = 5 seconds
Velocity = Distance / Time = 40 meter/ 5 sec = 40 m/sec
Also, given that the second car travels 64 meters due west in 8 seconds. Therefore, we can write,
Distance = 64 meters
Time = 8 seconds
Velocity = Distance / Time = 64 meter/ 8 sec = 8 m/sec
Hence, During their periods of travel, the cars definitely had the same velocity.
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Describe how fractional distillation and cracking are used so that sufficient petrol is produced from crude oil to meet demand.
Answer:
☟
Explanation:
Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.
Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:
heated to 600-700°C
passed over a catalyst of silica or alumina
These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:
hexane → butane + ethene
C6H14 → C4H10 + C2H4
Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.
Fractional distillation of crude oil
Fractional distillation separates a mixture into a number of different parts, called fractions.
A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.
Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.
Oil fractions
The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.
As you go up the fractionating column, the hydrocarbons have:
lower boiling points
lower viscosity (they flow more easily)
higher flammability (they ignite more easily).
Other fossil fuels
Crude oil is not the only fossil fuel.
Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.
Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.
In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.
What is fractional distillation?Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.
Generally, the components have boiling points that differ by less than 25 °C from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.
The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.
Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.
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A jet plane is flying at a constant altitude. At time t1=0t 1=0, it has components of velocity vx=90m/s,vy=110m/sv x = 90m/s,v y=110m/s. At time t2=30.0st 2=30.0s, the components are vx=−170m/s,vy=40m/sv x =−170m/s,v y=40m/s.
(a) Sketch the velocity vectors at t1and t2.
How do these two vectors differ? For this time interval calculate
(b) the components of the average acceleration, and
(c) the magnitude and direction of the average acceleration.
The average acceleration [tex]\vec a_{\rm ave}[/tex] over some time interval [tex][t_1,t_2][/tex] is equal to the ratio of the change in velocity [tex]\vec v_2-\vec v_1[/tex] over the duration of the interval [tex]t_2-t_1[/tex], or
[tex]\vec a_{\rm ave}=\dfrac{\Delta\vec v}{\Delta t}=\dfrac{\vec v_2-\vec v_1}{t_2-t_1}[/tex]
which can be split into the [tex]x[/tex] and [tex]y[/tex] components as
[tex]a_{\rm{ave},x}=\dfrac{v_{2,x}-v_{1,x}}{t_2-t_1}=\dfrac{-170\frac{\rm m}{\rm s}-90\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-8.67\dfrac{\rm m}{\mathrm s^2}[/tex]
[tex]a_{\rm{ave},y}=\dfrac{v_{2,y}-v_{1,y}}{t_2-t_1}=\dfrac{40\frac{\rm m}{\rm s}-110\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-2.33\dfrac{\rm m}{\mathrm s^2}[/tex]
The magnitude of this average acceleration is
[tex]\left\|\vec a_{\rm ave}\right\|=\sqrt{{a_{\rm{ave},x}}^2+{a_{\rm{ave},y}}^2}\approx8.98\dfrac{\rm m}{\mathrm s^2}[/tex]
and its direction is [tex]\theta[/tex] such that
[tex]\tan\theta=\dfrac{a_{\rm{ave},y}}{a_{\rm{ave},x}}\implies\theta\approx-164.9^\circ[/tex]
which corresponds to a direction of about 15.1º South of West.
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying pan is only 0.400 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)? N
Answer:
normal force = 10 N
Explanation:
Given data
frictional force = 0.400 N
coefficient of kinetic friction = 0.04
Solution
we get here normal force that is express as
normal force = [tex]\frac{Frictional\ force}{coefficient\ of\ friction}[/tex] ............1
put here value and we will get value
normal force = [tex]\frac{0.400}{0.04}[/tex]
solve it we get
normal force = 10 N
Write the first equation of motion. Under what condition(s) is this equation valid?
Explanation:
The first equation of motion in kinematics is given by :
[tex]v=u+at[/tex] .....(1)
u is initial speed
a is acceleration
v is final speed
t is time
Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.
An industrial flywheel (a solid disk) of mass 10.0 kg and radius 17.3 cm is rotating at an angular speed of 22.0 rad/s. Upon being switched to a slower setting, the flywheel uniformly slows down to 13.5 rad/s after rotating through an angle of 13.8 radians. Calculate the angular acceleration of the flywheel in the process of slowing down
Answer:
Explanation:
During slowing down , initial angular velocity ω₁ = 22 rad /s
final angular velocity ω₂ = 13.5 rad /s
using the law's of motion formula for rotation
ω₂² = ω₁² + 2 αθ , α is angular acceleration and θ is angle in radian rotated during this period
13.5² = 22² - 2xα x 13.8
2xα x 13.8 = 484 - 182.25
α = 10.93 rad / s²
A roller coaster car is going over the top of a 15-m-radius circular rise. The passenger in the roller coaster has a true weight of 600 N (therefore a mass of 61.2 kg). At the top of the hill, the passengers "feel light," with an apparent weight of only 360 N. How fast is the coaster moving
Answer:
v = 7.67 m/s
Explanation:
The equation for apparent weight in the situation of weightlessness is given as:
Apparent Weight = m(g - a)
where,
Apparent Weight = 360 N
m = mass passenger = 61.2 kg
a = acceleration of roller coaster
g = acceleration due to gravity = 9.8 m/s²
Therefore,
360 N = (61.2 kg)(9.8 m/s² - a)
9.8 m/s² - a = 360 N/61.2 kg
a = 9.8 m/s² - 5.88 m/s²
a = 3.92 m/s²
Since, this acceleration is due to the change in direction of velocity on a circular path. Therefore, it can b represented by centripetal acceleration and its formula is given as:
a = v²/r
where,
a = centripetal acceleration = 3.92 m/s²
v = speed of roller coaster = ?
r = radius of circular rise = 15 m
Therefore,
3.92 m/s² = v²/15 m
v² = (3.92 m.s²)(15 m)
v = √(58.8 m²/s²)
v = 7.67 m/s
assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves the uzzel of the cannon at a highet of 200 m.( the cannon is at the edge of the cliff) A: find the horizontal distance the cannon travles. B: when does the cannon ball reach the ground? C: find the maximum highet the cannon ball reaches.
Answer:
A. xmax = 131.49 m
B. t = 8.74 s
C. ymax = 220.33 m
Explanation:
A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:
[tex]y=y_o+v_osin\theta-\frac{1}{2}gt^2[/tex] (1)
yo: height from the projectile is fired = 200m
vo: initial velocity of the projectile = 25m/s
g: gravitational acceleration = 9.8 m/s^2
θ: angle between the direction of the initial motion of the ball and the horizontal = 53°
t: time
You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.
When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:
[tex]0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2[/tex] (2)
You use the quadratic formula to obtain the value of t:
[tex]t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s[/tex]
You use the positive value because it has physical meaning.
Now, you can calculate the horizontal range of the projectile by using the following formula:
[tex]x_{max}=v_ocos\theta t[/tex]
[tex]x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m[/tex]
The cannon ball travels a horizontal distance of 131.49 m
B. The cannon ball reaches the canon for t = 8.74s
C. The maximum height is obtained by using the following formula:
[tex]y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}[/tex] (3)
By replacing in the equation (3) the values of all parameters you obtain:
[tex]y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m[/tex]
The maximum height reached by the cannon ball is 220.33m
A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Question:
A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Answer:
24 m/s
Explanation:
Given:
x=(24t - 2.0t³)m
First find velocity function v(t):
v(t) = ẋ(t) = 24 - 2*3t²
v(t) = ẋ(t) = 24 - 6t²
Find the acceleration function a(t):
a(t) = Ẍ(t) = V(t) = -6*2t
a(t) = Ẍ(t) = V(t) = -12t
At acceleration = 0, take time as T in velocity function.
0 =v(T) = 24 - 6T²
Solve for T
[tex] T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2 [/tex]
Substitute -2 for t in acceleration function:
a(t) = a(T) = a(-2) = -12(-2) = 24 m/s
Acceleration = 24m/s
The “turning effect of a force” (T = F * r) is:
(a) determined as the product of force and the moment of inertia.
(b) generated by concentric forces.
(c) equivalent to the angular momentum.
(d) determined as a product of torque and moment arm.
(e) called “moment” or “torque”.
Answer:
b and e
Explanation:
r x F is the formula for torque.
The "turning effect" or torque happens when concentric forces rotate an object along said center.
a) False because T = Fr = Ia (a = angular acceleration)
b) True
c) False. L = Iw (w = angular velocity), which does not equal Ia
d) False. It is torque, not the product of torque and something else
e) True.
A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible. Part A What is the speed of the rock when the string passes through the vertical position
Answer:
v = 3.33 m/s
Explanation:
In the position of 45 degrees, all the energy of the rock is gravitational, then we have:
E = m*g*L*cos(angle)
and in the vertical position of the string, all the energy is kinetic, so we have:
E = m*v^2/2
If there is no dissipation, both energies are equal, so we have:
m*g*L*cos(45) = m*v^2/2
9.81 * 0.8 * 0.7071 * 2 = v^2
v^2 = 11.0986
v = 3.33 m/s
A truck has a bed that is 4.50 metres long,and 2.50 metres wide, and 1.50 metres high. What is maximum volume of sand can the truck carry within this dimensions?
Answer:
since the bed is a cuboid, we find the volume by L×W×H
4.50 × 2.50 × 1.50 = 16.875m³
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Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up from left to right and a force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 2.0 m/s2. What is the magnitude of the force exerted on the middle cube by the left cube in this case
Answer:
24 Newtons
Explanation:
The force exerted in the middle cube needs to be enough to move the middle cube and the right cube with an acceleration of 2 m/s2.
The mass of those two cubes combined is 6 + 6 = 12 kg
So, using the following equation, we can find the force:
Force = mass * acceleration
Force = 12 * 2
Force = 24 Newtons
EASY HELP
As a space shuttle climbs, _____.
its mass increases
its mass decreases
its weight increases
its weight decreases
Answer: it's weight decreases
Explanation:
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is [tex]Q =2.094 C[/tex]
Explanation:
From the question we are told that
The diameter of the wire is [tex]d = 0.205cm = 0.00205 \ m[/tex]
The radius of the wire is [tex]r = \frac{0.00205}{2} = 0.001025 \ m[/tex]
The resistivity of aluminum is [tex]2.75*10^{-8} \ ohm-meters.[/tex]
The electric field change is mathematically defied as
[tex]E (t) = 0.0004t^2 - 0.0001 +0.0004[/tex]
Generally the charge is mathematically represented as
[tex]Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt[/tex]
Where A is the area which is mathematically represented as
[tex]A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2[/tex]
So
[tex]\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega[/tex]
Therefore
[tex]Q = 120 \int\limits^{t}_{0} { E(t) } \, dt[/tex]
substituting values
[tex]Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt[/tex]
[tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.[/tex]
From the question we are told that t = 5 sec
[tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.[/tex]
[tex]Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }[/tex]
[tex]Q =2.094 C[/tex]
The charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds is 2.094 Coulomb.
Given the following data:
Diameter of wire = 0.205 centimeters.Resistivity of aluminum = [tex]2.75\times 10^{-8}[/tex] Ohm-meters.[tex]E(t)=0.0004t^2-0.0001t+0.0004[/tex] Newton per coulomb.Conversion:
Diameter of wire = 0.205 cm to m = 0.00205 meter.
Radius = [tex]\frac{Diameter}{2} =\frac{0.00205}{2} =0.001025\;meter[/tex]
To determine the charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds, we would apply Gauss's law in an electric field for a surface charge:
First of all, we would find the area of the wire.
[tex]Area = \pi r^2\\\\Area = 3.142 \times 0.001025^2\\\\Area = 3.3 \times 10^{-6}\;m^2[/tex]
Mathematically, Gauss's law in an electric field for a surface charge is given by the formula:
[tex]Q = \int\limits^t_0 {\frac{A}{\rho } E(t)} \, dt[/tex]
Where:
A is the area of a conductor.[tex]\rho[/tex] is the resistivity of a conductor.t is the time.E is the electric field.Substituting the given parameters into the formula, we have;
[tex]Q= \int\limits^t_0 {\frac{3.3 \times 10^{-6}}{2.75\times 10^{-8} } (0.0004t^2-0.0001t+0.0004)} \, dt\\\\Q=120\int\limits^t_0 1{ (0.0004t^2-0.0001t+0.0004)} \, dt[/tex]
[tex]Q=120(\frac{0.0004t^3}{3} -\frac{0.0001t^2}{2} +0.0004t |\left{5} \atop {0} \right[/tex]
When t = 5 seconds:
[tex]Q=120(\frac{0.0004[5]^3}{3} -\frac{0.0001[5]^2}{2} +0.0004[5])\\\\Q=120(\frac{0.03}{3} -\frac{0.0025}{2} +0.002)\\\\Q=120(0.0167-0.00125+0.002)\\\\Q=120(0.01745)[/tex]
Q = 2.094 Coulomb.
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Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.653 kJ of heat. It shrinks on cooling, and the atmosphere does 389 J of work on the balloon. Express your answer to three significant figures and include the appropriate units.
Answer:
ΔE = ‒0.271 kJ
Explanation:
Let's begin by listing out the given variables:
q = -0.653 kJ, w = 0.389 kJ
Using the formula ΔE = q + w
ΔE = -0.653 + 0.388
ΔE = (‒0.655 + 0.382) kJ
ΔE = ‒0.271 kJ
Therefore, the change in internal energy is -271 J or -0.271 kJ which implies that the system is exothermic
A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the crate moves at constant speed, find (a) the work done by the 150-N force and (b) the coefficient of kinetic friction between the crate and surface.
Answer:
a. 900 J
b. 0.383
Explanation:
According to the question, the given data is as follows
Horizontal force = 150 N
Packing crate = 40.0 kg
Distance = 6.00 m
Based on the above information
a. The work done by the 150-N force is
[tex]W = F x = \mu N x = \mu\ m\ g\ x[/tex]
[tex]W = 150 \times 6[/tex]
= 900 J
b. Now the coefficient of kinetic friction between the crate and surface is
[tex]\mu = \frac {F}{m\timesg}[/tex]
[tex]= \frac{150}{40\times 9.8}[/tex]
= .383
We simply applied the above formulas so that each one part could calculate
We want to find the work and kinetic friction for the given situation. The solutions are:
a) W = 900 N*mb) μ = 0.38Here we have a horizontal force of 150N pushing a 40.0 kg packing crate a distance of 6.00m at a constant speed.
a) First we want to find the work, it is given by the force applied times the distance moved, so the work is just:
W = 150N*6.00m = 900 N*m
b) Now we want to find the coefficient of kinetic friction, it must be such that the kinetic friction force is equal to the pushing force, in this way there is no net force, and then there is no acceleration.
Remember that the friction force is:
F = m*g*μ
Where:
m = mass of the box = 40 kgg = gravitational acceleration = 9.8m/s^2μ = coefficient of kinetic friction.Then we must solve:
150N = 40kg*(9.8 m/s^2)*μ = 392N*μ
150N/392N = 0.38
So the coefficient of kinetic friction between the crate and the surface is 0.38
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A tank with a constant volume of 3.72 m3 contains 22.1 moles of a monatomic ideal gas. The gas is initially at a temperature of 300 K. An electric heater is used to transfer 4.5 × 104 J of energy into the gas. It may help you to recall that CV = 12.47 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
a) What is the temperature of the gas after the energy is added?___K
b) What is the change in pressure of the gas?____Pa
c) How much work was done by the gas during this process?____J
Answer:
a) 463.29 K
b) 8065.65 Pa
c) 0 J
Explanation:
The parameters given are;
Volume of the tank, V = 3.72 m³
Number of moles of gas present in the tank, n = 22.1 moles
Temperature of the gas before heating, T₁ = 300 k
Heat added to the gas, ΔQ = 4.5 × 10⁴ J
Specific heat capacity at constant volume, [tex]c_v[/tex], for monatomic gas = 12.47 J/K/mole
Avogadro's number = 6.022 × 10²³ particles per mole
a) ΔQ = n × [tex]c_v[/tex] × ΔT
Where:
ΔT = T₂ - T₁
T₂ = Final temperature of the gas
Hence, by plugging in the values, we have;
4.5 × 10⁴ = 22.1 × 12.47 × (T₂ - 300)
[tex]T_{2} - 300 = \frac{4.5\times 10^{4}}{22.1\times 12.47}[/tex]
T₂ = 300 + 163.29 = 463.29 K
b) The pressure of the gas is found from the relation;
P×V = n×R×T
[tex]P = \dfrac{n \times R \times T}{V}[/tex]
Where:
P = Pressure of the gas
R = Universal gas constant = 8.3145 J/(mol·K)
T = Temperature of the gas
V = Volume of the gas = 3.72 ³ (constant)
n = Number of moles of gas present = 22.1 moles (constant)
Hence the change in pressure is given by the relation;
[tex]\Delta P = \dfrac{n \times R \times (T_2 - T_1)}{V} = \dfrac{n \times R \times \Delta T}{V}[/tex]
Plugging in the values, we have;
[tex]\Delta P = \dfrac{22.1 \times 8.3145 \times 163.29}{3.72} = 8065.65 \, Pa[/tex]
c) Work done, W, by the gas is given by the area under the pressure to volume graph which gives;
W = f(P) × ΔV
The volume given in the question is constant
∴ ΔV = 0
Hence, W = f(P) × 0 = 0 J
No work done by the gas during the process.
Which symbol is used to show vector quantities
Answer: arrows
Explanation:
A vector quantity is usually represented by an arrow, with the direction of the vector being the direction in which the arrow points and the length of the arrow representing the vector's magnitude.
What is the vector quantity unit?
The meter is the only fundamental SI unit that is a vector. The rest are all scalars. Derived quantities might be scalar or vector, but all vector quantities require meters as part of their definition and measurement.
The term "vector quantities" refers to physical quantities whose magnitude and direction are well specified.
Arrows are used to depict vectors. An arrow has a direction and a magnitude (how long it is) (the direction in which it points).
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An LC circuit has a 6.00 mH inductor. The current has its maximum value of 0.570 A at t =0s. A short time later the capacitor reaches its maximum potential difference of 66.0 V. What is the value of the capacitance?
Answer:
C = 44.75 x 10⁻⁸ F
Explanation:
Assuming no loss of energy between capacitor and inductor
energy in inductor initially = 1/2 Li₀² where L is inductance and i₀ is peak current .
= .5 x 6 x 10⁻³ x .57²
= .97 x 10⁻³ J .
This energy is transferred to capacitor .
energy of capacitor = 1/2 CV²
= .5 x C x 66²
= 2178 C
2178C = .97 x 10⁻³
C = 44.75 x 10⁻⁸ F .
The magnetic energy stored in the inductor is transformed into electrical energy stored in the capacitor. The value of capacitance for the given circuit is 44.75 x 10⁻⁸ F
Finding the capacitance:According to the law of conservation of energy, the magnetic energy stored in the inductor will be gradually lost and this energy will be stored in the capacitor as electrical energy.
Initially, the energy in the inductor is:
E = 1/2 Li₀²
where L is inductance
and i₀ is peak current.
E = 0.5 × 6 × 10⁻³ × (0.57)²
E = 0.97 × 10⁻³J
This energy is transformed into electrical energy stored in the capacitor.
So the capacitor energy is:
E = 1/2 CV²
where C is the capacitance
E = 0.5 × C × 66²
E = 2178 C
0.97 x 10⁻³ = 2178 C
C = 44.75 x 10⁻⁸ F
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A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of it as a cord) between the locomotive and the first car (FT1) to that between the first car and the second car (FT2), for any nonzero acceleration of the train
Answer:
The ratio is [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
Here we are assume the acceleration of the train is a
which makes the acceleration of each car a
From the question we are told that
Considering the second car
The force causing it s movement is mathematically represented as
[tex]F_{T2} = ma[/tex]
Considering the first car
The force causing it s movement is mathematically represented as
[tex]F = F_{T1} -F_{T2} = ma[/tex]
=> [tex]F_{T1} -ma = ma[/tex]
=> [tex]F_{T1} = 2 ma[/tex]
=> [tex]\frac{F_{T1}}{ma} = 2[/tex]
=> [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]
Question 7 of 10
The coefficient of kinetic friction between a couch and the floor is 0.4. If the
couch has a mass of 35 kg and you push it with a force of 200 N. what is the
net force on the couch as it slides?
O A. 337 N
B. 143 N
O C. 343 N
O D. 63 N
Answer:
D
Explanation:
Now the net force is the applied force minus the frictional force; this is expressed mathematically as:
Fnet= Fappplied - Ffrictional
Now the frictional force is given as ;
Coefficient of friction × normal reaction
Normal reaction is the weight of the human acting in opposite direction.
Normal reaction of the human is ;
35 × 9.8 = 343N { note that weight = m× g and g= 9.8m/S2, a known standard }
Hence the Frictional force =343×0.4 =137.20N
Hence Fnet = 200-137.20 = 62.8N
Fnet = 63N to the nearest whole
The net force on the couch as it slides is 63N.
What is frictional force?
When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.
The friction force is given by
f = coefficient of friction x Normal force
Given, the coefficient of kinetic friction between a couch and the floor is 0.4. If the couch has a mass of 35 kg and you push it with a force of 200 N.
Normal reaction is the weight of the human acting in opposite direction.
Normal reaction N =35 × 9.8 = 343N
Frictional force f =0.4 x 343
f =137.20N
The net force will be
Fnet= Fappplied - Ffrictional
Fnet = 200-137.20 = 62.8N
Fnet = 63N
Thus, the net force on the couch as it slides is 63N.
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A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.7 m from his television set. A reporter at the press conference is located 5.5 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.
Answer:
Explanation:
Time taken by sound waves to cover distance between politician and reporter = time taken by em waves to travel distance between politician and the television viewer.
5.5 / 343 = d / 3 x 10⁸ + 2.7 / 343
d is distance between politician and television set + time taken by sound to travel distance between television and its viewer.
.0160349 = d / 3 x 10⁸ + .0078717
d / 3 x 10⁸ = .0081632
d = 2448960 m
= 2448.96 km
= 2449 km .
The battery on your car has a rating stated in ampere minute which permit you to estimate the length of time a fully charged battery could deliver any particular current before discharge. Approximately how much energy is stored by a 50 ampere minute 12 volt battery
Answer:
Thus, the energy stored by a 50 Ampere minute battery is found to be 36 KJ.
Explanation:
The power delivered by a battery is given by the formula:
P = VI
where,
P = Power Delivered by battery in 1 second
V = Voltage of battery = 12 volt
I = Current stored in battery
But, if we multiply both sides of equation by time (t), then:
Pt = VIt
where,
Pt = Power x Time = E = Energy Stored = ?
It = Rating of Battery = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
E = 36000 J = 36 KJ
Thus, the energy stored by a 50 Ampere minute battery is found to be 36 KJ.
Which of the following is analogous to the pipes in an electrical circuit?
A. capacitors storing the incoming charge from the battery
B. large resistors causing restrictions to the flow of charge
C. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery electric current being forced uphill by the battery
D. electric current being forced uphill by the battery back to the positive terminal
The correct answer is D. electric current being forced uphill by the battery back to the positive terminal.
What is Electric Current?
Electric current is the flow of electric charge through a conducting medium, such as a wire, due to the movement of electrons or ions. The flow of charge is typically caused by the presence of an electric field that creates a potential difference (voltage) between two points in a circuit
In an electrical circuit, pipes are analogous to wires or conductive paths that allow the flow of electric current. The flow of electric current is from the positive terminal of the battery to the negative terminal, which is opposite to the direction of conventional current flow. Therefore, option C is incorrect.
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