Answer:
Angular acceleration = 5 rad /s ^2
Kinetic energy = 0.391 J
Work done = 0.391 J
P =6.25 W
Explanation:
The torque is given as moment of inertia × angular acceleration
angular acceleration = torque/ moment of inertia
= 10/2= 5 rad/ s^2
The kinetic energy is = 1/2 Iw^2
w = angular acceleration/time
=5/8= 0.625 rad /s
1/2 × 2× 0.625^2
=0.391 J
The work done is equal to the kinetic energy of the motor at this time
W= 0.391 J
The average power is = torque × angular speed
= 10× 0.625
P = 6.25 W
A. The angular acceleration is 5 rad/s². B. The kinetic energy of the shaft after 8 seconds is 16000 J. C. The work done by the motor in this time is 1600 J. D. The average power delivered by the motor over this time interval is 200 W.
Explanation:A. Angular acceleration can be found using the rotational analog to Newton's second law a = net τ/I. The moment of inertia I is given and the torque τ can be found from the given torque value. So, τ = 10 Nm and I = 2.0 kgm2. Substituting these values, we get a = 10 Nm / 2.0 kgm2 = 5 rad/s².
B. Using the equation K = (1/2) I ω², where I is the moment of inertia and ω is the angular velocity, we can calculate the initial angular velocity ω₁ as 0 rad/s. Then, the final angular velocity ω can be found using the relationship ω = ω₁ + αt. So, ω = 0 rad/s + (5 rad/s²)(8 s) = 40 rad/s. Finally, substituting these values into the kinetic energy equation, we have K = (1/2)(2.0 kgm2)(40 rad/s)² = 16000 J.
C. The work done by the motor can be found using the equation W = τθ, where τ is the torque and θ is the angular displacement. In this case, θ = ω₁t + (1/2)αt² = 0 rad/s(8 s) + (1/2)(5 rad/s²)(8 s)² = 160 rad. Therefore, W = (10 Nm)(160 rad) = 1600 J.
D. The average power can be calculated using the equation P = W/t, where W is the work done and t is the time interval. Substituting the values we found in the previous calculations, we have P = 1600 J / 8 s = 200 W.
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A person walks in the following pattern: 2.9 km north, then 2.8 km west, and finally 4.4 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point
Answer:
(a) Magnitude =3.18 km
(a) Angle =28.2°
Explanation:
(a) To find magnitude of this vector recognize
[tex]R_{x}=-2.8 km\\R_{y}=-1.5km[/tex]
Use Pythagorean theorem
[tex]R=\sqrt{(R_{x})^{2}+(R_{y})^{2} }\\ R=\sqrt{(-2.8)^{2}+(-1.5)^{2} }\\ R=3.18km[/tex]
(b)To find the angle use the trigonometric property
[tex]tan\alpha =\frac{opp}{adj}\\\ tan\alpha =\frac{R_{y} }{R_{x}}\\\alpha =tan^{-1}(\frac{(-1.5)}{(-2.8)})\\\alpha =28.2^{o}[/tex]
A square steel bar has a length of 9.7 ft and a 2.9 in by 2.9 in cross section and is subjected to axial tension. The final length is 9.70710 ft . The final side length is 2.89933 in . What is Poisson's ratio for the material? Express your answer to three significant figures.
The Poisson's ratio definition is given as the change in lateral deformation over longitudinal deformation. Mathematically it could be expressed like this,
[tex]\upsilon = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}}[/tex]
[tex]\upsilon = \frac{(\delta a/a)}{(\delta l/l)}[/tex]
Replacing with our values we would have to,
[tex]\upsilon = \frac{(2.89933-2.9/2.9)}{(9.70710-9.7/97)}[/tex]
[tex]\upsilon = 0.1977[/tex]
Therefore Poisson's ratio is 0.1977.
A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 slater 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.
(a) What is the magnitude of the average velocity of the sled during each of the 2.00-s intervals after passing the 14.4-m point?
(b) What is the acceleration of the sled?
(c) What is the speed of the sled when it passes the 14.4-m point?
(d) How much time did it take to go from the top to the 14.4-m point?
(e) How far did the sled go during the first second after passing the 14.4-m point?
Answer:
(a) 5.6 m/s, 7.2 m/s, and 8.8 m/s, respectively.
(b) 0.8 m/s^2
(c) 4.8 m/s
(d) 6 s
(e) 5.2 m
Explanation:
(a) The average velocity is equal to the total displacement divided by total time.
For the first 2s. interval:
[tex]V_{\rm avg} = \frac{\Delta x }{\Delta t} = \frac{25.6 - 14.4}{2} = 5.6~{\rm m/s}[/tex]
For the second 2s. interval:
[tex]V_{\rm avg} = \frac{40 - 25.6}{2} = 7.2~{\rm m/s}[/tex]
For the third 2s. interval:
[tex]V_{\rm avg} = \frac{57.6 - 40}{2} = 8.8~{\rm m/s}[/tex]
(b) Every 2 s. the velocity increases 1.6 m/s. Therefore, for each second the velocity increases 0.8 m/s. So, the acceleration is 0.8 m/s2.
(c) The sled starts from rest with an acceleration of 0.8 m/s2.
[tex]v^2 = v_0^2 + 2ax\\v^2 = 0 + 2(0.8)(14.4)\\v = 4.8~{\rm m/s}[/tex]
(d) The following kinematics equation will yield the time:
[tex]\Delta x = v_0 t + \frac{1}{2}at^2\\14.4 = 0 + \frac{1}{2}(0.8)t^2\\t = 6~{\rm s}[/tex]
(e) The same kinematics equation will yield the displacement:
[tex]\Delta x = v_0t + \frac{1}{2}at^2\\\Delta x = (4.8)(1) + \frac{1}{2}(0.8)1^2\\\Delta x = 5.2~{\rm m}[/tex]
A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery and the plate separation is increased.
What happens to the potential difference between the plates?
A) More information is needed to answer this question
B) The potential difference between the plates stays the same.
C) The potential difference between the plates decreases.
D) The potential difference between the plates increases.
Answer:
D) The potential difference between the plates increases.
Explanation:
The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.
Where ϵ0 is the permittivity of free space.
A capacitor with increased distance, will have a new capacitance C1=ϵ0kA/d1
Where d1 = nd
since d1 > d
therefore n >1
n is a factor derived as a result of the increased distance
Therefore the new capacitance becomes:
C1=ϵ0A/d1
C1= ϵ0A/nd
C1= C/n -------1
Where C1 is the capacitance with increased distance.
This implies that the charge storing capacity of the capacitor with increased plate separation decreases by a factor of (1/n) compared to that of the capacitor with original distance.
Given points
The charge stored in the original capacitor Q=CV
The charge stored in the original capacitor after inserting dielectric Q1=C1V1
The law of conservation of energy states that the energy stored is constant:
i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.
Charge before plate separation increase same as after plate separation increase
Q = Q1
CV = C1V1
CV = C1V1 -------2
We derived C1=C/n in equation 1. Inserting this into equation 2
CV = (CV1)/n
V1 = n(CV)/C
= n V
Since n > 1 as a result of the derived new distance, the new voltage will increase
"An elevator is moving upward with a speed of 11 m/s. Three seconds later, the elevator is still moving upward, but its speed has been reduced to 5.0 m/s. What is the average acceleration of the elevator during the 3.0 s interval?
Answer:
Average acceleration = - 2 m/s^2
Explanation:
Given data:
Initial velocity = 11 m/s
Final velocity = 5.0 m/s
duration of change in velocity = 3 sec
Average acceleration [tex]= \frac{v - u}{\Delta t}[/tex]
Average acceleration [tex]= \frac{5 - 11}{3} = -2 m/s^2[/tex]
Average acceleration = - 2 m/s^2
here negative sign indicate that acceleration is proceed in downward direction.
A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC moves from the point x= 0.170 m , y= 0 to the point x= 0.250 m , y= 0.250 m .
How much work W is done by the electric force on the moving point charge?
Express your answer in joules. Use k = 8.99×109 N*m^2/ C^2 for Coulomb's constant: k=1/(4*pi*epsilon0)
The work done by the electric force on the moving point charge is approximately -5.09 × 10^-5 J.
Explanation:Work done by the electric force is given by the equation W = q1 * q2 * (1/r1 - 1/r2), where q1 and q2 are the charges, r1 is the initial distance, and r2 is the final distance.
In this case, q1 = 2.30 μC, q2 = -5.00 μC, r1 = 0.170 m, and r2 = 0.250 m. Plugging these values into the equation and solving for W, we get:
W = (2.30 μC) * (-5.00 μC) * [1/√(0.170^2) - 1/√(0.250^2 + 0.250^2)]
After simplifying, the work done is approximately -5.09 × 10^-5 J.
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What is the sign and magnitude of a point charge that produces a potential of −2.2 V at a distance of 1 mm?
The sign of the point charge that produces a potential of -2.00 V at a distance of 1.00 mm is negative, and its magnitude is calculated to be approximately -2.22×10-13 C using Coulomb's law.
Explanation:
The question pertains to determining the sign and magnitude of a point charge based on the electric potential it produces at a specific distance. The electric potential (V) at a distance (r) from a point charge (q) is given by the equation V = k * q / r, where k is Coulomb's constant (k = 8.988×109 N·m2/C2). Since the potential is negative (-2.00 V), the point charge must have a negative sign. To find the magnitude, we rearrange the formula to solve for q: q = V * r / k.
Plugging in the values gives q = (-2.00 V * 1.00×10-3 m) / 8.988×109 N·m2/C2, which calculates to a charge magnitude of approximately -2.22×10-13 C.
The probable question is in the image attached.
A 0.23 kg mass on a spring vibrates with amplitude 25 cm and frequency 1.7 Hz. Calculate (b) the speed at which the mass passes through equilibrium and (b) the total energy of the oscillation. (Answers: 0.82 J, 2.7 m/s).
Answer:
a) 2.67 m/s
b) 0.82 J
Explanation:
Amplitude A = 25 cm = 0.25 m
The period of the motion is the inverse of the frequency
[tex]T = \frac{1}{f} = \frac{1}{1.7} = 0.588 s[/tex]
So the angular frequency
[tex]\omega = \frac{2\pi}{T} = \frac{2\pi}{0.588} = 10.68 rad/s[/tex]
The speed at the equilibrium point is the maximum speed, at
[tex]v = \omega A = 10.68 * 0.25 = 2.67 m/s[/tex]
The spring constant can be calculated using the following
[tex]\omega^2 = \frac{k}{m} = \frac{k}{0.23}[/tex]
[tex]k = 0.23\omega^2 = 0.23*10.68^2 = 26.24 N/m[/tex]
The total energy of the oscillation is
[tex]E = kA^2 / 2 = 26.24*0.25^2 / 2 = 0.82 J[/tex]
A metallic sphere has a charge of +3.1 nC. A negatively charged rod has a charge of −4.0 nC. When the rod touches the sphere, 9.2×109 electrons are transferred. What are the charges of the sphere and the rod now?
Answer:
Q'sphere=2.7*10^-9 C
Q'rod=-4.7*10^-9 C
Explanation:
given data:
charge on metallic sphere Qsphere=3.1*10^-9 C ∴1n=10^-9
charge on rod Qrod =-4*10^-9 C
no of electron n= 9.2×10^9 electrons
To find:
we are asked to find the charges Q'sphere on the sphere and Q'rod on the rod after the rod touches the sphere.
solution:
the total charge transferred when the rod touches the sphere equal to the no of electrons transferred multiplied by the charge of each electron:
Q(transferred)= nq_(e)
=(9.2×10^9)(1.6×10^-19)
=-1.312×10^-9 C
because electron are negative they move from the negatively charged rod to the positively charged rod so that new charged of the sphere is:
Q'sphere =Qsphere+Q(transferred)
=(3.1*10^-9 )-(1.312×10^-9)
=2.7*10^-9 C
similarly the new charge of the rod is:
Q'rod = Qrod-Q(transferred)
= (-6*10^-9 C)-(1.312*10^-9 C)
= -4.7*10^-9 C
∴note: there maybe error in calculation but the method is correct.
Final answer:
Upon contact, a metallic sphere and a negatively charged rod share their charges until equilibrium. The total charge of -0.9 nC is equalized, with 9.2×109 electrons changing the sphere's charge to +1.628 nC and the rod's charge to -2.528 nC.
Explanation:
When two charged objects come into contact, they share their charges until equilibrium is reached. This means each object will end up with the average charge. In the case of the metallic sphere with a charge of +3.1 nC and the negatively charged rod with a charge of -4.0 nC, the total charge before contact is (-4.0 nC) + (+3.1 nC) = -0.9 nC.
Since 9.2×109 electrons are transferred, we calculate the charge transferred using the charge of one electron, which is approximately -1.6×10-19 C. Multiplying the number of electrons by the charge of one electron gives us the total charge transferred: 9.2×109 × -1.6×10-19 C/electron ≈ -1.472 nC.
This charge is added to the metallic sphere and subtracted from the rod. So, the new charge on the sphere is +3.1 nC + (-1.472 nC) = +1.628 nC, and the charge on the rod is -4.0 nC - (-1.472 nC) = -2.528 nC. Both charges are now closer in magnitude, representing the sharing of charges due to contact.
What is the electric potential V V due to the nucleus of hydrogen at a distance of 5.292 × 10 − 11 m 5.292×10−11 m ?
Answer:
27.1806500378 V
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
q = Charge = [tex]1.6\times 10^{-19}\ C[/tex]
r = Distance = [tex]5.292\times 10^{-11}\ m[/tex]
Voltage is given by
[tex]V=k\dfrac{q}{r}\\\Rightarrow V=8.99\times 10^9\dfrac{1.6\times 10^{-19}}{5.292\times 10^{-11}}\\\Rightarrow V=27.1806500378\ V[/tex]
The potential difference is 27.1806500378 V
An object is released from rest near and above Earth’s surface from a distance of 10m. After applying the appropriate kinematic equation, a student predicts that it will take 1.43s for the object to reach the ground with a speed of 14.3m/s . After performing the experiment, it is found that the object reaches the ground after a time of 3.2s. How should the student determine the actual speed of the object when it reaches the ground? Assume that the acceleration of the object is constant as it falls.
The student determine the actual speed of the object when it reaches the ground as 12.52 m/s.
Given data:
The distance from the Earth's surface is, = 10 m.
Time taken to reach the ground is, t = 1.43 s.
The speed of object is, v = 14.3 m/s.
Experimental value of time interval is, t' = 3.2 s.
Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:
[tex]h=ut+\dfrac{1}{2}a't'^{2} \\\\10=0 \times t+\dfrac{1}{2} \times a' \times 3.2^{2} \\\\a'=\dfrac{20}{3.2^{2}}\\\\a'= 1.95 \;\rm m/s^{2}[/tex]
Now, use the principle of conservation of total energy of system:
Potential energy - work done by air resistance = Kinetic energy
[tex]mgh-(ma) \times h=\dfrac{1}{2}mv^{2} \\\\gh-(a) \times h=\dfrac{1}{2}v^{2} \\\\v=\sqrt{2h(g-a)}[/tex]
Here, v is the actual speed of object while reaching the ground.
Solving as,
[tex]v=\sqrt{2 \times 10(9.8-1.95)}\\\\v=12.52 \;\rm m/s[/tex]
Thus, we can conclude that the student determine the actual speed of the object when it reaches the ground as 12.52 m/s.
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The actual speed of the object when it reaches the ground is approximately [tex]31.392 \, m/s.[/tex]
The actual speed of the object when it reaches the ground can be determined using the kinematic equation that relates initial velocity, acceleration, and time to the final velocity. Since the object is released from rest, the initial velocity [tex]\( u \)[/tex] is 0 m/s, and the acceleration [tex]\( a \)[/tex] is due to gravity, which is approximately [tex]\( 9.81 \, m/s^2 \)[/tex] near the Earth's surface.
The kinematic equation that relates these quantities to the final velocity [tex]\( v \)[/tex] is:
[tex]\[ v = u + at \][/tex]
Given that [tex]\( u = 0 \, m/s \)[/tex] and [tex]\( a = 9.81 \, m/s^2 \)[/tex], and the time [tex]\( t \)[/tex] to reach the ground is [tex]\( 3.2 \, s \)[/tex], we can substitute these values into the equation to find the actual final velocity:
[tex]\[ v = 0 + (9.81 \, m/s^2)(3.2 \, s) \] \[ v = (9.81)(3.2) \, m/s \] \[ v = 31.392 \, m/s \][/tex]
Therefore, the actual speed of the object when it reaches the ground is approximately [tex]31.392 \, m/s.[/tex]
A race car travels 765 km around a circular sprint track of radius 1.263 km. How many times did it go around the track?
Answer:
It will go 96 times around the track.
Explanation:
Given that,
Distance covered by the race car, d = 765 km
Radius of the circular sprint track, r = 1.263 km
Let n times did it go around the track. It is given by :
[tex]n=\dfrac{d}{C}[/tex]
C is the circumference of the circular path, [tex]C=2\pi r[/tex]
[tex]n=\dfrac{d}{2\pi r}[/tex]
[tex]n=\dfrac{765}{2\pi \times 1.263}[/tex]
[tex]n=96.4[/tex]
Approximately, n = 96
So, it will go 96 times around the track. Hence, this is the required solution.
With a track radius of 1.263 km, the car completes approximately 96.50 laps.
The formula for the circumference (C) of a circle is: C = 2πr, where r is the radius of the circle, and π (pi) is approximately 3.14159. Using the given radius of 1.263 km, we can calculate the circumference of the track:
C = 2π(1.263 km) ≈ 2(3.14159)(1.263 km) ≈ 7.932 km (rounded to three decimal places).
Divide the total distance traveled by the circumference of the track:
Number of laps = Total distance traveled ÷ Circumference of the track
Number of laps = 765 km ÷ 7.932 km ≈ 96.50 laps.
Therefore, the race car would have completed approximately 96.50 laps around the track.
The voltage across a conductor is increasing at a rate of 2 volts/min and the resistance is decreasing at a rate of 1 ohm/min. Use I = E/R and the Chain Rule to find the rate at which the current passing through the conductor is changing when R = 20 ohms and E = 70 volts.
Answer:
3.5 amperes
Explanation:
I = E/R
I = ?
E = 70volts
R = 20 Ohms
Therefore , I = 70/20
= 3.5 amperes
Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?
Answer:
Explanation:
Given
mass of first child [tex]m_1=20\ kg[/tex]
mass of second child [tex]m_2=30\ kg[/tex]
Distance between two children is [tex]d=3\ m[/tex]
Suppose light weight child is placed at a distance of x m from Pivot point
therefore
Torque due to heavy child [tex]T_1=m_2g\times (3-x)[/tex]
Torque due to small child [tex]T_2=m_1g\times x[/tex]
Net Torque about Pivot must be zero
Therefore [tex]T_1=T_2[/tex]
[tex]30\times g\times (3-x)=20\times g\times x[/tex]
[tex]9-3x=2x[/tex]
[tex]9=5x[/tex]
[tex]x=\frac{9}{5}[/tex]
[tex]x=1.8\ m[/tex]
In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat the athlete as a particle and let ymax be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 to the time it takes him to go from the floor to that height. Ignore air resistance.
The answer details the vertical velocity needed and the horizontal distance required for a basketball player to complete a jump.
Vertical velocity: To rise 0.750 m above the floor, the athlete needs a vertical velocity of 5.43 m/s.
Horizontal distance: The athlete should start his jump 2.27 m away from the basket to reach his maximum height at the same time as he reaches the basket.
The oscilloscope is set to measure 2 volts per division on the vertical scale. The oscilloscope display a sinusoidal voltage that vertically covers 3.6 divisions from positive to negative peak. What is the peak to peak voltage of this signal
Answer: 7.2V
Explanation:
Peak values or peak to peak voltage are calculated from RMS values, which implies VP = VRMS × √2, (assuming the source is a pure sine wave).
Since it's a sinusoidal voltage and measuring from an oscilloscope, the peak to peak voltage is gotten thus:
No of division X Volts/divisions
So, 3.6 x 2V = 7.2V
Final answer:
The peak-to-peak voltage of a sinusoidal signal covering 3.6 divisions on an oscilloscope set to 2 volts per division is 7.2 volts.
Explanation:
The question involves calculating the peak-to-peak voltage of a sinusoidal signal observed on an oscilloscope where the vertical scale is set to 2 volts per division. Given that the signal covers 3.6 divisions from positive to negative peak, we calculate the peak-to-peak voltage by multiplying the number of divisions the signal spans by the voltage per division.
To find the peak-to-peak voltage, we use the formula: Peak-to-Peak Voltage = Number of Divisions × Voltage per Division. Thus, the peak-to-peak voltage of the signal is 3.6 divisions × 2 volts/division = 7.2 volts.
Suppose the radius of the Earth is given to be 6378.01 km. Express the circumference of the Earth in m with 5 significant figures.
Round to 5 sig figs with trailing zeros --> 40074000
The mathematical description that fits to find the circumference of a circle (Approximation we will make to the earth considering it Uniform) is,
[tex]\phi = 2\pi r[/tex]
Here,
r = Radius
The radius of the earth is 6378.01 km or 6378010m
Replacing we have that the circumference of the Earth is
[tex]\phi = 2\pi (6378010m)[/tex]
[tex]\phi = 40074000 m[/tex]
[tex]\phi = 40074*10^3 m[/tex]
Therefore the circumference of the Earth in m with 5 significant figures is [tex]40074*10^3 m[/tex] and using only trailing zeros the answer will be [tex]40074000m[/tex]
The largest building in the world by volume is the boeing 747 plant in Everett, Washington. It measures approximately 632 m long, 710 yards wide, and 112 ft high.
What is the cubic volume in feet, convert your result from part a to cubic meters?
Explanation:
Given that,
The dimensions of the largest building in the world is 632 m long, 710 yards wide, and 112 ft high. It basically forms a cuboid. The volume of a cuboidal shape is given by :
Since,
1 meter = 3.28084 feet
632 m = 2073.49 feet
1 yard= 3 feet
710 yards = 2130 feet
V = lbh
[tex]V=2073.49 \ ft\times 2130\ ft\times 112\ ft[/tex]
[tex]V=494651774.4\ ft^3[/tex]
[tex]V=4.94\times 10^8\ ft^3[/tex]
Also,
[tex]V=(4.94\times 10^8\ ft^3)(\dfrac{1\ m}{3.281})^3[/tex]
[tex]V=1.39\times 10^7\ m^3[/tex]
Hence, this is the required solution.
A 30.0-g ice cube at its melting point is dropped into an aluminum calorimeter of mass 100.0 g in equilibrium at 24.0 °C with 300.0 g of an unknown liquid. The final temperature is 4.0 °C . What is the heat capacity of the liquid?
Answer:
Cu = 1453.72J/Kg°C
The heat capacity of the liquid is 1453.72J/Kg°C
Explanation:
At equilibrium, assuming no heat loss to the surrounding we can say that;
Heat gained by ice + heat gained by cold water = heat loss by hot unknown liquid (24°C) + heat loss by aluminium calorimeter.
Given;
Mass of ice = mass of cold water = mc = 30g = 0.03kg
Mass of hot unknown liquid mh= 300g = 0.3kg
Mass of aluminium calorimeter ma= 100g = 0.1 kg
change in temperature cold ∆Tc = (4-0) = 4°C
Change in temperature hot ∆Th = 24-4 = 20°C
Specific heat capacity of water Cw= 4186J/Kg°C
Specific heat capacity of aluminium Ca = 900J/kg°C
Specific heat capacity of unknown liquid Cu =?
Heat of condensation of ice Li = 334000J/Kg
So, the statement above can be written as.
mcLi + mcCw∆Tc = maCa∆Th + mhCu∆Th
Making Cu the subject of formula, we have;
Cu = [mcLi + mcCw∆Tc - maCa∆Th]/mh∆Th
Substituting the values we have;
Cu = (0.03×334000 + 0.03×4186×4 - 0.1×900×20)/(0.3×20)
Cu = 1453.72J/Kg°C
the heat capacity of the liquid is 1453.72J/Kg°C
Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of each are measured. Both stones have the same volume, 0.15 cm3. However, stone A has a mass of 0.52 g, and stone B has a mass of 0.42 g. If diamond has a density of 3.5 g/cm3, could either of the stones be real diamonds?
Answer:
stone A is diamond.
Explanation:
given,
Volume of the two stone = 0.15 cm³
Mass of stone A = 0.52 g
Mass of stone B = 0.42 g
Density of the diamond = 3.5 g/cm³
So, to find which stone is gold we have to calculate the density of both the stone.
We know,
[tex]density[/tex][tex]\density = \dfrac{mass}{volume}[/tex]
density of stone A
[tex]\rho_A = \dfrac{0.52}{0.15}[/tex]
[tex]\rho_A = 3.467\ g/cm^3[/tex]
density of stone B.
[tex]\rho_B = \dfrac{0.42}{0.15}[/tex]
[tex]\rho_B = 2.8\ g/cm^3[/tex]
Hence, the density of the stone A is the equal to Diamond then stone A is diamond.
Neither of the stones is a real diamond because their densities, calculated using mass and volume, do not match the density of a real diamond.
Explanation:The determination of whether a stone is a real diamond or a fake can be made by calculating the density of the stone. Density is calculated as the mass of an object divided by its volume. So, for stone A, the density is 0.52 g / 0.15 cm3 = 3.47 g/cm3, and for stone B, the density is 0.42 g / 0.15 cm3 = 2.8 g/cm3. The density of a real diamond is 3.5 g/cm3. Hence, neither stone A nor stone B is a real diamond as their densities are less than the density of a real diamond.
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A light bulb is connected to a 120.0-V wall socket. The current in the bulb depends on the time t according to the relation I = (0.644 A) sin [(394 rad/s)t]. (a) What is the frequency of the alternating current? (b) Determine the resistance of the bulb's filament. (c) What is the average power delivered to the light bulb?
The frequency of the alternating current is 394/2π Hz. The resistance of the bulb's filament can be determined using Ohm's Law. The average power delivered to the light bulb can be calculated using the formula P = IV.
Explanation:(a) The frequency of the alternating current can be calculated using the angular frequency formula ω = 2πf. In this case, the angular frequency is 394 rad/s. So, we can rearrange the formula to find the frequency: f = ω/2π = 394/2π Hz.
(b) The resistance of the bulb's filament can be determined using Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I). In this case, the voltage is 120.0 V and the current is given by I = (0.644 A) sin [(394 rad/s)t].
(c) The average power delivered to the light bulb can be calculated using the formula P = IV, where I is the current and V is the voltage. In this case, the voltage is 120.0 V and the current is given by I = (0.644 A) sin [(394 rad/s)t].
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A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.
Answer:
A) The expression of the electric field halfway between the plates, if the plates are in the plane Y-Z is:
[tex]\vec{E}=\displaystyle \frac{q}{LW\varepsilon_0}\vec{x}[/tex]
B) The expression for the magnitude of the electric field E₂ just in front of the plate two ends is:
[tex]|E_2|=\displaystyle \frac{q}{2LW\varepsilon_0}[/tex]
C) The charge density is:
[tex]\sigma_2=-4.2517\cdot10^{-3}C/m^2[/tex]
Completed question:
A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.
A) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an express of the electric field, E, halfway between the plates
B) Input an expression for the magnitude of the electric field, E₂. Just in front of plate two END
C) If plate two has a total charge of q =-1 mC, what is its charge density, σ in C/m2?
Explanation:
A) The expression of the field can be calculated as the sum of the field produced by each plate. Each plate can be modeled as 2 parallel infinite metallic planes. Because this is a capacitor connected by both ends to a battery, the external planes have null charge (the field outside the device has to be null by definition of capacitor). This means than the charge of each plate has to be distributed in the internal faces. Because this es a metallic surface and there is no external field, we can consider a uniform charge distribution (σ=cte). Therefore in this case for each plane:
[tex]\sigma_i=\displaystyle \frac{q_i}{LW}[/tex]
The field of an infinite uniform charged plane is:
[tex]\vec{E_i}=\displaystyle \frac{\sigma_i}{2\varepsilon_0}sgn(x-x_{0i})\vec{x} =\frac{q_i}{2LW\varepsilon_0}sgn(x-x_{0i})\vec{x}[/tex]
In this case, inside the capacitor, if the plate 1 is in the left and the plate 2 is in the right, the field for 0<x<d is:
[tex]\vec{E_1}\displaystyle=\frac{q}{2LW\varepsilon_0}sgn(x})\vec{x}=\frac{q}{2LW\varepsilon_0}\vec{x}[/tex]
[tex]\vec{E_2}\displaystyle=\frac{-q}{2LW\varepsilon_0}sgn(x-d})\vec{x}=\frac{q}{2LW\varepsilon_0}\vec{x}[/tex]
[tex]\vec{E}=\vec{E_1}+\vec{E_2}[/tex]
[tex]\vec{E}=\displaystyle \frac{q}{LW\varepsilon_0}\vec{x}[/tex]
B) we already obtain the expression of the field E₂ inside the space between the plates. Even if we are asked the expression just in front of the plate and not inside, the expression for |E₂| is still de same.
C) As seen above, we already obtain the charge density expression. Therefore we only have to replace the variables for the numerical values.
The electric field, E, halfway between the plates is (σ / ε₀). The electric field is written in terms of permittivity and surface charge density.
Given:
Length, L = 0.49 m
Width, W = 0.48 m
Distance, d = 0.1 m
Here:
E = electric field
σ = surface charge density
ε₀ = permittivity of free space
We must determine the surface charge density on each plate since the plates are wired to a battery and charged so that the first plate has a charge of q.
The area of the plate is:
Area of each plate (A) = L x W = 0.49 m x 0.48 m = 0.2352 m²
The surface charge density is given by:
σ = q / A
The electric field is computed as:
E = (σ / ε₀)
Hence, the electric field, E, halfway between the plates is (σ / ε₀). The electric field is written in terms of permittivity and surface charge density.
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#Complete question is:
A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.
A) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an express of the electric field, E, halfway between the plates
A heat engine operates at 30% of its maximum possible efficiency and needs to do 995 J of work. Its cold reservoir is at 22 ºC and its hot reservoir is at 610 ºC. (a) How much energy does it need to extract from the hot reservoir? (b) How much energy does it deposit in the cold reservoir?
Answer:
(a) The energy extracted from the hot reservoir (Qh) is 3316.67J
(b) The energy deposited in the cold reservoir (Qc) is 2321.67J
Explanation:
Part (a) The energy extracted from the hot reservoir (Qh)
e = W/Qh
where;
e is the maximum efficiency of the system = 30% = 0.3
W is the the work done on the system = 995 J
Qh is the heat absorbed from the hot reservoir
Qh = W/e
Qh = 995/0.3
Qh = 3316.67J
Part (b) The energy deposited in the cold reservoir (Qc)
e = W/Qh
W = Qh - Qc
where;
Qc is the heat deposited in the cold reservoir
e = (Qh - Qc)/Qh
Qh - Qc = e*Qh
Qc = Qh - e*Qh
Qc = 3316.67J - 0.3*3316.67J
Qc = 3316.67J - 995J
Qc = 2321.67J
A proton accelerates from rest in a uniform electric field of 680 N/C. At one later moment, its speed is 1.30 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton.
Answer:
Acceleration, [tex]a=6.51\times 10^{10}\ m/s^2[/tex]
Explanation:
Given that,
Electric field, E = 680 N/C
Speed of the proton, v = 1.3 Mm/s
We need to find the acceleration of the proton. We know that the force due to motion is balanced by the electric force as :
[tex]qE=ma[/tex]
a and m are the acceleration and mass of the proton.
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\times 680}{1.67\times 10^{-27}}[/tex]
[tex]a=6.51\times 10^{10}\ m/s^2[/tex]
So, the acceleration of the proton is [tex]a=6.51\times 10^{10}\ m/s^2[/tex]. Hence, this is the required solution.
A uniform horizontal electric field of 1.8 × 105 N/C causes a ball that is suspended from a light string to hang at an angle of 23° from the vertical. If the mass of the ball is 5.0 grams, what is the magnitude of its charge?
Answer:
[tex]1.15669\times 10^{-7}\ C[/tex]
Explanation:
[tex]\theta[/tex] = Angle with which the electric field is hung = 23°
m = Mass of ball = 5 g
E = Electric field = [tex]1.8\times 10^5\ N/C[/tex]
T = Tension
q = Charge
We have the equations
[tex]Tcos\theta=mg[/tex]
[tex]Tsin\theta=qE[/tex]
Dividing the equations
[tex]tan\theta=\dfrac{mg}{qE}\\\Rightarrow q=\dfrac{mgtan\theta}{E}\\\Rightarrow q=\dfrac{5\times 10^{-3}\times 9.81\times tan23}{1.8\times 10^5}\\\Rightarrow q=1.15669\times 10^{-7}\ C[/tex]
The magnitude of the charge is [tex]1.15669\times 10^{-7}\ C[/tex]
A soda can with a volume of 345 mL is 6.5 cm in diameter and has a mass of 20g. The can is half-filled with water, and when it is placed in a tub of water it is found to float upright. What length of the can is above the water level?
Answer:
0.0473m
Explanation:
345 ml = 0.000354 m3
6.5 cm = 0.065 m
20g = 0.02 kg
Since can is half filled with water, the water volume is 0.000354 / 2 = 0.000177 m cubed
Let water density be 1000kg/m3, the mass of this half-filled water is
1000*0.000177 = 0.177 kg
The total water-can system mass is 0.177 + 0.02 = 0.197 kg
For the system to stay balanced, this mass would be equal to the mass of the water displaced by the can submerged
The volume of water displaced, or submerged can is
0.197 / 1000 = 0.000197 m cubed
Then the volume of the can that is not submerged, aka above water level is
0.000354 - 0.000197 = 0.000157 m cubed
The base area of the can is
[tex]A = \pi r^2 = \pi (d/2)^2 = \pi (0.065)^2 = 0.003318 m squared[/tex]
The length of the can that is above water is
0.000157 / 0.003318 = 0.0473 m
The half-filled soda can displaces the equivalence of its own weight in water when placed in it. Half of the soda can's total volume will always be submerged since it is only half-filled i.e., half of the can's mass is displacing water.
Explanation:The subject of this problem is the principles of buoyancy and volume. A half-filled soda can placed in water will displace its own weight of the water. The length of the can above the water level can be calculated using an understanding of volume and displacement.
First, calculate the volume of the can using the formula for the volume of a cylinder V = πr²h, where r is radius and h is height. Given the diameter of the can is 6.5 cm, the radius is 3.25 cm. The height can be calculated by rearranging the volume formula to find h. We know that the can's complete volume is 345 mL, so h (full can height) = V / (πr²).
From this, we can calculate the height of the can that is submerged in water. Since the can is half-filled, it displaces half its full weight in water. So half of the can's total volume will always be submerged. Therefore, the length of the can above the water will be half the total height of the can.
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A 50 kg box hangs from a rope. What is the tension in the rope if: The box is at rest? The box moves up at a steady 5.0 m/s? The box has vy=5.0 m/s and is speeding up at 5.0 m/s2? The box has vy=5.0 m/s and is slowing down at 5.0 m/s2?
Answer:
(a) [tex]T_{1}=490N[/tex]
(b) [tex]T_{2}=240N[/tex]
Explanation:
For Part (a)
Given data
The box moves up at steady 5.0 m/s
The mas of box is 50 kg
As ∑Fy=T₁ - mg=0
[tex]T_{1}=mg\\T_{1}=(50kg)(9.8m/s^{2} ) \\T_{1}=490N[/tex]
For Part(b)
Given data
[tex]v_{iy}=5m/s\\ a_{y}=-5.0m/s^{2}[/tex]
As ∑Fy=T₂ - mg=ma
[tex]T_{2}=mg+ma_{y}\\T_{2}=m(g+ a_{y})\\T_{2}=50kg(9.8-5.0) \\T_{2}=240N[/tex]
The tension in the rope varies depending on whether the box is at rest, moving at a constant velocity, or accelerating. The tension equals the weight of the box when it's at rest or moving constantly, but it will be increased or decreased by the net force caused by acceleration when the box is speeding up or slowing down.
Explanation:If the box is at rest, the tension in the rope is equal to the force of gravity. We can calculate this using the formula T = mg, where m is the mass of the box and g is the acceleration due to gravity. Therefore, T = (50 kg)(9.8 m/s²) = 490 N.
When the box moves upwards with a constant velocity, the tension in the rope also equals the weight of the box (T = mg), so the tension will stay the same at 490 N.
However, when the box is speeding up, the net force is the product of mass and acceleration. In this case, acceleration = 5.0 m/s². Using the equation Fnet = ma, we find that Fnet = (50 kg)(5 m/s²) = 250 N. The total tension now includes both the tension due to the box's weight and the additional force due to the acceleration. Therefore, T = T(g) + Fnet = 490 N + 250 N = 740 N.
Lastly, when the box is slowing down at 5.0 m/s², the net force acts in the opposite direction of the initial velocity. Using the same calculations, we find Fnet = 250 N. But this force now reduces the tension originally caused by the box's weight, so the total tension in the rope becomes T = T(g) - Fnet = 490 N - 250 N = 240 N.
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A uniformly dense solid disk with a mass of 4 kg and a radius of 2 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 20 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 5 s
The angular displacement of the solid disk after 5 seconds is 0 rad.
Explanation:To determine the angular displacement of the solid disk after 5 seconds, we can use the formula:
Δθ = ΔErot / (I * ω)
where Δθ is the angular displacement, ΔErot is the change in rotational kinetic energy, I is the moment of inertia of the disk, and ω is the angular velocity of the disk.
The moment of inertia of a solid disk rotating around an axis through its center perpendicular to its plane is given by:
I = (1/2) * m * r2
where m is the mass of the disk and r is the radius.
Given that ΔErot = 20 J/s, m = 4 kg, r = 2 m, and the disk starts from rest, we can calculate the angular displacement:
Δθ = ΔErot / (I * ω) = 20 / [(1/2) * 4 * 22 * ω]
Since the disk starts from rest, the initial angular velocity ω is 0. Therefore, the angular displacement after 5 seconds is:
Δθ = 20 / [(1/2) * 4 * 22 * 0] = 0 rad
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At a lab investigating fire extinguisher foams, a heavy ball is accidentally dropped into a deep vat of foam from a crane 6.10 m above the foam. After entering the foam, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the foam. The ball reaches the bottom 3.20 s after it is released. How deep is the vat?
Answer:
22.8077659955 m deep
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 6.1+0^2}\\\Rightarrow v=10.9399268736\ m/s[/tex]
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{10.9399268736-0}{9.81}\\\Rightarrow t=1.11518112881\ s[/tex]
Time taken to fall through the foam
[tex]3.2-1.11518112881=2.08481887119\ s[/tex]
Distance is given by
[tex]s=vt\\\Rightarrow s=10.9399268736\times 2.08481887119\\\Rightarrow s=22.8077659955\ m[/tex]
The vat is 22.8077659955 m deep
The depth of the vat obtained is 44.076 m
Data obtained from the question Height of crane above the vat = 6.10 mTime to reach the bottom of the vat from the crane = 3.20 sDepth of vat =? Determination of the height from the crane to the bottom of the vat Time to reach the bottom of the vat from the crane (t) = 3.20 sAcceleration due to gravity (g) =? Height from crane to bottom of vat (H) =?H = ½gt²
H = ½ × 9.8 × 3.2²
H = 4.9 × 10.24
H = 50.176 m
How to determine the depth of the vatHeight from crane to bottom of vat (H) = 50.176 mHeight of crane above the vat (h) = 6.10 mDepth of vat =?Depth of vat = H – h
Depth of vat = 50.176 – 6.10
Depth of vat = 44.076 m
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A water balloon is launched at a speed of (15.0+A) m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located (18.0+B) m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures.
Answer:
The question is incomplete, below is the complete question,
"A water balloon is launched at a speed of (15.0+A) m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located (18.0+B) m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures. A=12, B=2"
Answe:
12.1m
Explanation:
Below are the data given
Speed, V=(15+A) = 15+12=27m/s
Angle of projection, ∝=36 degree
Distance from building = 18+B=18+2=20m
Since the motion describe by the object is a projectile motion, and recall that in projectile motion, motion along the horizontal path has zero acceleration and motion along the vertical path is under gravity,
the Velocity along the horizontal path is define as
[tex]V_{x}=Vcos\alpha \\V_{x}=27cos36\\V_{x}=21.8m/s[/tex]
the velocity along the Vertical path is
[tex]V_{y}=Vsin\alpha \\V_{y}=27sin36 \\V_{y}=15.87m/s[/tex]
Since the horizontal distance from the point of projection to the building is 20m, we determine the time it takes to cover this distance using the simple equation of motion
[tex]Velocity=\frac{distance }{time }\\ Time,t=27/21.8\\t=1.24secs[/tex]
The distance traveled along the vertical axis is given as
[tex]y=V_{y}t-\frac{1}{2}gt^{2}\\ t=1.24secs,\\V_{y}=15.87m/s\\g=9.81m/s[/tex]
if we substitute values, we arrive at
[tex]y=15.87*1.24-\frac{1}{2}9.81*1.24^{2}\\y=19.66-7.57\\y=12.118\\y=12.1m[/tex]
Hence the water balloon hit the building at an height of 12.1m