The container has the greatest total force acting on its base is the conatainer with the greatest total weight.
How to find which container has the greatest total force acting on its base?For each container, there is a pressure, P acting on its base given by
P = F/A where
F = force on base and A = area of baseNow, since the force on the base is the weight of water acting on the base, W, we have that
P = W/A where w = total weight of water acting on base
Now, for each container, the area of base is the same. So, we have that
P ∝ W
So, we see that the pressure is directly proportional to the weight on the container.
Since pressure is directly proportional to force, we see that the force on the base is directly proportional to the weight of water acting on the base.
So, for the greatest force acting on the base, we would have the greatest total weight of water acting on the base.
So, the container has the greatest total force acting on its base is the conatainer with the greatest total weight.
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define limitations in the operation conditions of a pn junction
Answer:
Such limitations are given below.
Explanation:
Each pn junction provides limited measurements of maximum forwarding current, highest possible inversion voltage as well as the maximum output level.If controlled within certain adsorption conditions, the pn junction could very well offer satisfying performance. In connector operation, the maximum inversion voltage seems to be of significant importance.Complete the following statement:
Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different:_____
a. volume.
b. specific heat capacities.
c. expansion coefficients.
d. mass.
e. length.
Answer: c. expansion coefficients.
Explanation: Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different expansion coefficients.
I found the answer on Quizlet. :)
Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different expansion coefficients. The correct option is c.
What is the expansion coefficient?The coefficient of thermal expansion (CTE) is the rate at which a material expands as its temperature rises. This coefficient is determined at constant pressure and without a phase change, i.e. the material is expected to remain solid or fluid.
Bimetallic strips, which are utilized as adjustable switches in electric appliances, are made up of metallic strips with differing expansion coefficients. The coefficient of thermal expansion indicates how the size of an object varies as temperature changes.
Therefore, the correct option is c. expansion coefficients.
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Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The power created is [tex]P_{avg} = F * v_{avg}[/tex]
Explanation:
From the question we are told that
The that the average power is mathematically represented as
[tex]P_{avg} = \frac{W }{\Delta t }[/tex]
Where W is is the Workdone which is mathematically represented as
[tex]W = F * s[/tex]
Where F is the applies force and s is the displacement due to the force
So
[tex]P_{avg} = \frac{F *s }{\Delta t }[/tex]
Now this displacement can be represented mathematically as
[tex]s = v_{avg} * \Delta t[/tex]
Where [tex]v_{avg }[/tex] is the average velocity and [tex]\Delta t[/tex] is the time taken
So
[tex]P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }[/tex]
=> [tex]P_{avg} = F * v_{avg}[/tex]
Answer:
Pavg = Fvavg
Explanation:
Since the P (power) done by the F (force) is:
P = Fs/t
and we are looking for the velocity, so then it would be:
P = Fv
with the average velocity the answer is:
Pavg = Favg
If an object is moving at a constant speed, and the force F is also constant, this formula can be used to find the average power. If v is changing, the formula can be used to find the instantaneous power at any given moment (with the quantity v in this case meaning the instantaneous velocity, of course).
An ultrasound machine uses 1.64 × 105 watts of power. If it draws 12.0 amps of current, what is the resistance?
Answer:
R = 1138.9 Ω
Explanation:
Hello,
In this case, for the given power (P) and current (I), we can compute the resistance (R) via:
R = P / I²
Thus, we obtain:
R = 1.64x10⁵ W / (12.0 A)²
R = 1138.9 Ω
Best regards.
A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the drift velocity of the electrons.
Complete Question
A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s
Answer:
The drift velocity is [tex]v = 0.0002808 \ m/s[/tex]
Explanation:
From the question we are told that
The current on the copper is [tex]I = 20 \ A[/tex]
The cross-sectional area is [tex]A = 5.261 \ mm^2 = 5.261 *10^{-6} \ m^2[/tex]
The number of copper atom in the wire is mathematically evaluated
[tex]n = \frac{\rho * N_a}Z}[/tex]
Where [tex]\rho[/tex] is the density of copper with a value [tex]\rho = 8.93 \ g/m^3[/tex]
[tex]N_a[/tex] is the Avogadro's number with a value [tex]N_a = 6.02 *10^{23}\ atom/mol[/tex]
Z is the molar mass of copper with a value [tex]Z = 63.55 \ g/mol[/tex]
So
[tex]n = \frac{8.93 * 6.02 *10^{23}}{63.55}[/tex]
[tex]n = 8.46 * 10^{28} \ atoms /m^3[/tex]
Given the 1 atom is equivalent to 1 free electron then the number of free electron is
[tex]N = 8.46 * 10^{28} \ electrons[/tex]
The current through the wire is mathematically represented as
[tex]I = N * e * v * A[/tex]
substituting values
[tex]20 = 8.46 *10^{28} * (1.60*10^{-19}) * v * 5.261 *10^{-6}[/tex]
=> [tex]v = 0.0002808 \ m/s[/tex]
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displaced from its equilibrium position and undergoes simple harmonic oscillations. What is the period of the oscillations
Answer:
The period is [tex]T = 0.700 \ s[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 0.350 \ kg[/tex]
The extension of the spring is [tex]x = 12.0 \ cm = 0.12 \ m[/tex]
The spring constant for this is mathematically represented as
[tex]k = \frac{F}{x}[/tex]
Where F is the force on the spring which is mathematically evaluated as
[tex]F = mg = 0.350 * 9.8[/tex]
[tex]F =3.43 \ N[/tex]
So
[tex]k = \frac{3.43 }{ 0.12}[/tex]
[tex]k = 28.583 \ N/m[/tex]
The period of oscillation is mathematically evaluated as
[tex]T = 2 \pi \sqrt{\frac{m}{k} }[/tex]
substituting values
[tex]T = 2 * 3.142* \sqrt{\frac{0.35 }{28.583} }[/tex]
[tex]T = 0.700 \ s[/tex]
Classify the bending of light as exhibited by the ray diagrams. According to your data, is light refracted away from or toward the normal as it passes at an angle into a medium with a higher index of refraction?
Answer:
the ray of light should approach normal
Explanation:
When light passes through two means of different refractive index, it fulfills the equation
n₁ sin θ₁ = n₂ sin θ₂
where index 1 and 2 refer to each medium
In this problem, they tell us that light passes to a medium with a higher index, which is why
n₁ <n₂
let's look for the angle in the second half
sinθ₂ = n₁ /n₂ sin θ₁
θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)
let's examine the angle argument the quantity n₁ /n₂ <1 therefore the argument decreases, therefore the sine and the angle decreases
Consequently the ray of light should approach normal
What happens when you increase the number of slits per millimeter (decrease the spacing between slits)?
Answer:
Increasing the number of slits not only makes the diffraction maximum sharper, but also much more intense. If a 1 mm diameter laser beam strikes a 600 line/mm grating, then it covers 600 slits and the resulting line intensity is 90,000 x that of a double slit. Such a multiple-slit is called a diffraction grating.
An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the speed remained the same, by what factor would its centripetal acceleration change?
Answer:
The centripetal acceleration changed by a factor of 0.5
Explanation:
Given;
first radius of the horizontal circle, r₁ = 500 m
speed of the airplane, v = 150 m/s
second radius of the airplane, r₂ = 1000 m
Centripetal acceleration is given as;
[tex]a = \frac{v^2}{r}[/tex]
At constant speed, we will have;
[tex]v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1[/tex]
a₂ = 0.5a₁
Therefore, the centripetal acceleration changed by a factor of 0.5
what dangers higher voltages have for humans whose resistance changes depending on hand lotion, rubber gloves, other types of gloves etc.
Answer:
Higher voltages pose huge dangers to humans with changes in resistance however, the use of hand lotion and rubber gloves help to reduce the amount of current flowing through the human. Notwithstanding higher currents can lead to death by stopping blood flow to the heart causing a heart attack. Best way to survive voltages above 120 volts is to use a rubber glove and avoid touching bare wires.
Explanation:
Other classes of gloves that can be used include: Class 1 gloves which can be used for current up to 7,500 volts of AC, Class 2 up to 17,000 volts AC, Class 3 up to 26,500 volts AC, and Class 4 up to 36,000 volts AC. However, cotton gloves can be used inside to absorb perspiration and to improve the comfort of the user.
A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h.
a) If they are released from rest and roll without slipping, determine the velocity vring of the ring when it reaches the bottom.
b) Verify your answer by calculating their speeds when they reach the bottom in terms of h.
Explanation:
velocity of disc [tex]=\sqrt((gh)/0.75)[/tex]
lets call (h) 1 m to make it simple.
= 3.614 m/s
[tex]\sqrt((4/3) x 1 x 9.8) = 3.614[/tex] m/s pointing towards this:
[tex]4×V_d=\sqrt(4/3hg)[/tex]
[tex]V_h=\sqrt(hg)[/tex]
velocity of hoop=[tex]\sqrt(gh)[/tex]
lets call (h) 1m to make it simple again.
[tex]\sqrt(9.8 x 1) = 3.13[/tex] m/s
[tex]\sqrt(gh) = sqrt(hg)
so [tex]4×V_d= \sqrt(4/3hg)V_h=\sqrt(hg)[/tex]
The disc is the fastest.
While i'm on this subject i'll show you this:
Solid ball [tex]=0.7v^2= gh[/tex]
solid disc [tex]= 0.75v^2 = gh[/tex]
hoop [tex]=v^2=gh[/tex]
The above is simplified from linear KE + rotational KE, the radius or mass makes no difference to the above formula.
The solid ball will be the faster of the 3, like above i'll show you.
solid ball: velocity [tex]=\sqrt((gh)/0.7)[/tex]
let (h) be 1m again to compare.
[tex]\sqrt((9.8 x 1)/0.7) = 3.741[/tex] m/s
solid disk speed [tex]=\sqrt((gh)/0.75)[/tex]
uniform hoop speed [tex]=\sqrt(gh)[/tex]
solid sphere speed [tex]=\sqrt((gh)/0.7)[/tex]
During a snowball fight, two snowball with masses of 0.30 kg and 0.70 kg, respectively, are thrown in such a manner that they meet head-on (traveling opposite directions) and combine to form a single mass. The magnitude of initial velocity for each is 10.4 m/s. What is the speed of the 1.0 kg mass immediately after the collision
Answer:
10.4 m/s
Explanation:
Given that
mass of the first snowball, m1 = 0.3 kg
mass of the second snowball, m2 = 0.7 kg
Magnitude of initial velocity for both masses, u = 10.4 m/s
To start with, we use the formula of conservation of linear momentum which states that
magnitude of initial momentum is equal to magnitude of final momentum.
m1u1 + m2u2 = V(m1 + m2)
0.3 * 10.4 + 0.7 * 10.4 = V(0.3 + 0.7)
3.12 + 7.28 = V(1)
10.4 = V
The 1 kg mass is an addition Of the 0.3 mass & 0.7 kg mass.
Thus, the speed of the 1 kg mass is 10.4 m/s
a person lifts 4.5kg block up a vertical distance of 1.2meters and then carries it horizontally for 7.3meters. Calculate the Total Work done by the person and the block.
Answer:
The total work done by the person is given as = m g h
= 4.5kg x 9.8m/s²x1.2m
= 52.92J
This is the work done in moving the block in a vertical distance
However there is no work done when the block is moved in a horizontal direction since ko work is done against gravity.
Explanation:
6a. A special lamp can produce UV radiation. Which two statements
describe the electromagnetic waves emitted by a UV lamp? *
They have a higher frequency than X-rays.
They have the same wave speed as visible light
They have a longer wavelength than microwaves.
They have a lower frequency than gamma rays.
They have a greater wave speed than radio waves.
Answer:
The correct options are:
B) They have the same wave speed as visible light
D) They have a lower frequency than gamma rays.
Explanation:
B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s
D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).
We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.
At what minimum speed must a roller coaster be traveling when upside down at the top of a 7.4 m radius loop-the-loop circle so the passengers will not fall out?
Answer:
v = 8.5 m/s
Explanation:
In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,
Weight = Centripetal Force
but,
Weight = mg
Centripetal Force = mv²/r
Therefore,
mg = mv²/r
g = v²/r
v² = gr
v = √gr
where,
v = minimum speed required = ?
g = 9.8 m/s²
r = radius = 7.4 m
Therefore,
v = √(9.8 m/s²)(7.4 m)
v = 8.5 m/s
Minimum speed for a roller coaster while travelling upside down so that the person will not fall out = 8.5 m/s
For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.
In the given condition the weight of the person must be balanced by the centrifugal force.
and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person
According to the Newton's Second Law of motion we can write force balance
[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]
Given Radius of loop = r = 7.4 m
Putting the value of r = 7.4 m in equation (1) we get
[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]
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A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost by the system as a result of this collision.
Answer:
20J
Explanation:
In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;
m₁u₁ + m₂u₂ = (m₁ + m₂)v ---------------(i)
Where;
m₁ and m₂ are the masses of first and second objects respectively
u₁ and u₂ are the initial velocities of the first and second objects respectively
v is the final velocity of the two objects after collision;
From the question;
m₁ = 2.0kg
m₂ = 8.0kg
u₁ = 5.0m/s
u₂ = 0 (since the object is initially at rest)
Substitute these values into equation (i) as follows;
(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v
(10.0) + (0) = (10.0)v
10.0 = 10.0v
v = 1m/s
The two bodies stick together and move off with a velocity of 1m/s after collision.
The kinetic energy(KE₁) of the objects before collision is given by
KE₁ = [tex]\frac{1}{2}[/tex]m₁u₁² + [tex]\frac{1}{2}[/tex]m₂u₂² ---------------(ii)
Substitute the appropriate values into equation (ii)
KE₁ = ([tex]\frac{1}{2}[/tex] x 2.0 x 5.0²) + ([tex]\frac{1}{2}[/tex] x 8.0 x 0²)
KE₁ = 25.0J
Also, the kinetic energy(KE₂) of the objects after collision is given by
KE₂ = [tex]\frac{1}{2}[/tex](m₁ + m₂)v² ---------------(iii)
Substitute the appropriate values into equation (iii)
KE₂ = [tex]\frac{1}{2}[/tex] ( 2.0 + 8.0) x 1²
KE₂ = 5J
The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision
K = KE₂ - KE₁
K = 5 - 25
K = -20J
The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J
How much electromagnetic energy is contained in each cubic meter near the Earth's surface if the intensity of sunlight under clear skies is 1000 W/m2
Answer:
344.8 x10^-8J/m³
Explanation:
Using=> energy intensity/ speed oflight
= 1000/2.9x10^8
= 344.8 x10^-8J/m³
The electromagnetic energy is 344.8 x10⁻⁸J/m³
We have to use the formula which says
Electromagnetic energy = energy intensity/ speed of light
We are given intensity as 1000 W/m²
Electromagnetic energy = 1000/2.9 x 10⁸
= 344.8 x 10⁻⁸J/m³
Therefore the electromagnetic energy is contained in each cubic meter will be 344.8 x 10⁻⁸J/m³
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According to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of
Answer:
Order of 10^(-35) m.
Explanation:
The string theory is a theoretical concept whereby the very small particles of particle physics are replaced by one dimensional objects which are called strings. This theory is also applicable to black hole physics, nuclear physics, cosmology, etc.
Now, according to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of 10^(-35) m.
This is because the length of the scale is assumed to be on the order of the Planck length, or 10^(−35) meters which is the scale at which the effects of quantum gravity are usually believed to become very significant.
A cylinder of mass 8.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 13.0 m/s.
(a) Determine the translational kinetic energy of its center of mass.
J
(b) Determine the rotational kinetic energy about its center of mass.
J
(c) Determine its total energy.
J
Answer:
b
Explanation:
determine the rotational kinetic energy about it's center of mass
A fluid is forced through a pipe of changing cross section as shown. In which section would the pressure of the fluid be a minimum?
Answer:
straight
Explanation:
because it is good in that direction
A slender rod of length L has a varying mass-per-unit-length from the left end (x=0) according to dm/dx=Cx where C has units kg/m2. Find the total mass in terms of C and L, and then calculate the moment of inertia of the rod for an axis at the left end note: you need the total mass in order to get the answer in terms of ML^2
Answer:
ML²/6
Explanation:
Pls see attached file
The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,
Moment of inertia:The length of the rod is L. It has a non-uniform distribution of mass given by:
dm/dx = Cx
where C has units kg/m²
dm = Cxdx
the total mass M of the rod can be calculated by integrating the above relation over the length:
[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]
Thus,
C = 2M/L²
Now, the moment of inertia of the small element dx of the rod is given by:
dI = dm.x²
dI = Cx.x²dx
[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]
I = ML²/2
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In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos 5t + π 8 where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos (5t + π / 8) where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm
Answer:(a) 7.392cm
(b) -15.32 cm/s
(c) -184cm/s²
(d) 0.4πs and 8.00cm
Explanation:The general equation of a simple harmonic motion (SHM) is given by;
x(t) = A cos (wt + Φ) --------------(i)
Where;
x(t) = position of the body at a given time t
A = amplitude or maximum displacement during oscillation
w = angular velocity
t = time
Φ = phase constant.
Given from question:
x(t) = 8.00 cos (5t + π / 8) ---------------(ii)
(a) At time t = 0;
The position, x(t), of the body (piston) is given by substituting the value of t = 0 into equation (ii) as follows;
x(0) = 8.00 cos (5(0) + π / 8)
x(0) = 8.00 cos (π /8)
x(0) = 8.00 x 0.924
x(0) = 7.392 cm
Therefore, the position of the piston at time t = 0 is 7.392cm
(b) To get the velocity, v(t), of the piston at t = 0, first differentiate equation (ii) with respect to t as follows;
v(t) = [tex]\frac{dx(t)}{dt}[/tex]
v(t) = [tex]\frac{d(8.00cos(5t + \pi / 8 ))}{dt}[/tex]
v(t) = 8 (-5 sin (5t + π / 8))
v(t) = -40sin(5t + π / 8) --------------------(iii)
Now, substitute t=0 into the equation as follows;
v(0) = -40 sin(5(0) + π / 8)
v(0) = -40 sin(π / 8)
v(0) = -40 x 0.383
v(0) = -15.32 cm/s
Therefore, the velocity of the piston at time t = 0 is -15.32 cm/s
(c) To find the acceleration a(t) of the piston at t = 0, first differentiate equation (iii), which is the velocity equation, with respect to t as follows;
a(t) = [tex]\frac{dv(t)}{dt}[/tex]
a(t) = [tex]\frac{d(-40sin (5t + \pi /8))}{dt}[/tex]
a(t) = -200 cos (5t + π / 8)
Now, substitute t = 0 into the equation as follows;
a(0) = -200 cos (5(0) + π / 8)
a(0) = -200 cos (π / 8)
a(0) = -200 x 0.924
a(0) = -184.8 cm/s²
Therefore, the acceleration of the piston at time t = 0 is -184cm/s²
(d) To find the period, T, first, let's compare equations (i) and (ii) as follows;
x(t) = A cos (wt + Φ) --------------(i)
x(t) = 8.00 cos (5t + π / 8) ---------------(ii)
From these equations it can be deduced that;
Amplitude, A = 8.00cm
Angular velocity, w = 5 rads/s
But;
w = [tex]\frac{2\pi }{T}[/tex] [Where T = period of oscillation]
=> T = [tex]\frac{2\pi }{w}[/tex]
=> T = [tex]\frac{2\pi }{5}[/tex]
=> T = 0.4π s
Therefore, the period and amplitude of the piston's motion are respectively 0.4πs and 8.00cm
A small merry-go round is spinning about its center in a clockwise direction at 1.6 r a d s and with rotational kinetic energy of 580 J. Calculate the moment of inertia of the merry-go round.
Answer:
The moment of inertia of the merry-go round is 453.125 kg.m²
Explanation:
Given;'
angular velocity of the merry-go round, ω = 1.6 rad/s
rotational kinetic energy, K = 580 J
Rotational kinetic energy is given as;
K = ¹/₂Iω²
Where;
I is the moment of inertia of the merry-go round
[tex]I = \frac{2K}{\omega^2} \\\\I = \frac{2*580}{1.6^2} \\\\I = 453.125 \ kg.m^2[/tex]
Therefore, the moment of inertia of the merry-go round is 453.125 kg.m²
Since the small merry-go round is spinning about its center in a clockwise direction, its moment of inertia is equal to 453.13 [tex]Kgm^2[/tex]
Given the following data:
Angular velocity = 1.6 rad/sRotational kinetic energy = 580 Joules.To calculate the moment of inertia of the small merry-go round:
Mathematically, the rotational kinetic energy of an object is giving by the formula:
[tex]E_{rotational} = \frac{1}{2} Iw^2[/tex]
Where:
I is the moment of inertia.w is the angular speed.Making moment of inertia (I) the subject of formula, we have:
[tex]I = \frac{2E_{rotational}}{w^2}[/tex]
Substituting the given parameters into the formula, we have;
[tex]I = \frac{2(580)}{1.6^2}\\\\I = \frac{1160}{2.56}[/tex]
Moment of inertia (I) = 453.13 [tex]Kgm^2[/tex]
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A trough is filled with a liquid of density 810 kg/m3. The ends of the trough are equilateral triangles with sides 8 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough. (Use 9.8 m/s2 for the acceleration due to gravity.)
Answer:
The hydrostatic force on one end of the trough is 54994.464 N
Explanation:
Given;
liquid density, ρ = 810 kg/m³
side of the equilateral triangle, L = 8m
acceleration due to gravity, g = 9.8 m/s²
Hydrostatic force is given as;
H = ρgh
where;
h is the vertical height of the equilateral triangle
Draw a line to bisect upper end of the trough, to the vertex at the bottom, this line is the height of the equilateral triangle.
let the half side of the triangle = x
x = ⁸/₂ = 4m
The half section of the triangle forms a right angled triangle
h² = 8² - 4²
h² = 48
h = √48
h = 6.928m
F = ρgh
F = 810 x 9.8 x 6.928
F = 54994.464 N
Therefore, the hydrostatic force on one end of the trough is 54994.464 N
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.5, what minimum force magnitude is required from the rope to start the crate moving? (b) If µk= 0.35, what is the magnitude of the initial acceleration of the crate?
Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ball is also thrown at the same speed horizontally at the wall, but sticks to it upon hitting. [THE CLAY BALL / THE RUBBER BALL /NEITHER] (circle one) exerts a greater magnitude of impulse on the wall. Briefly explain (either in words or calculations)
Answer:
THE RUBBER BALL
Explanation:
From the question we are told that
The mass of the rubber ball is [tex]m_r = 2 \ kg[/tex]
The initial speed of the rubber ball is [tex]u = 3 \ m/s[/tex]
The final speed at which it bounces bank [tex]v - 3 \ m/s[/tex]
The mass of the clay ball is [tex]m_c = 2 \ kg[/tex]
The initial speed of the clay ball is [tex]u = 3 \ m/s[/tex]
The final speed of the clay ball is [tex]v = 0 \ m/s[/tex]
Generally Impulse is mathematically represented as
[tex]I = \Delta p[/tex]
where [tex]\Delta p[/tex] is the change in the linear momentum so
[tex]I = m(v-u)[/tex]
For the rubber is
[tex]I_r = 2(-3 -3)[/tex]
[tex]I_r = -12\ kg \cdot m/s[/tex]
=> [tex]|I_r| = 12\ kg \cdot m/s[/tex]
For the clay ball
[tex]I_c = 2(0-3)[/tex]
[tex]I_c = -6 \ kg\cdot \ m/s[/tex]
=> [tex]| I_c| = 6 \ kg\cdot \ m/s[/tex]
So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5 × 10−4 mm and a crack length of 2.5 × 10−2 mm when a tensile stress of 170 MPa is applied (in MPa)
Answer:
2404 MPa
Explanation:
See attachment for solution
The maximum stress that exists at the tip of the internal crack is 3,400 Mpa.
The given parameters;
radius of the internal crack, r = 2.5 x 10⁻⁴ mmcrack length, l = 2.5 x 10⁻²tensile stress, σ = 170 MPa = 170 x 10⁶ N/m²The maximum stress that exists at the tip of the internal crack is calculated as follows;
[tex]\sigma _{max} = 2\sigma \times \sqrt{(\frac{l}{r} )} \\\\\sigma _{max} = 2 \times 170 \times 10^6 \times \sqrt{(\frac{2.5\times 10^{-2}}{2.5 \times 10^{-4}})} \\\\\sigma _{max} = 3.4 \times 10^{9} \ Pa\\\\\sigma _{max} = 3,400\ Mpa[/tex]
Thus, the maximum stress that exists at the tip of the internal crack is 3,400 Mpa.
Learn more here:https://brainly.com/question/19264500
what is the maximum torque on a coil 5 cm x 12 cm, composed of 600 turns, when it is carrying a current of 10^-5 A in a uniform field of .1 T
Answer:
3.6×10⁻⁶ Nm
Explanation:
From the question,
The expression for maximum torque is given as
τ = BANI.................Equation 1
where τ = maximum torque, B = magnetic field, A = Area of the coil, N = number of turns, I = current carried by the coil.
Given: B = 0.1 T, A = (5×12) = 60 cm² = 0.006 m², N = 600 turns, I = 10⁻⁵ A.
Substitute these values into equation 1
τ = 0.1(0.006)(600)(10⁻⁵)
τ = 3.6×10⁻⁶ Nm
A straw has an outer diameter of 14 mm, inner diameter of 11 mm, and length of 21 cm. It is made of glass with a Young's modulus of 68 GPa and tensile strength of 7 MPa. a. What is the effective spring constant of the straw with respect to elongation in N/m? b. When you hold either end you can stretch the straw with up to 90 N. How much does the straw elongate in mm? c. By how much can you extend the length of the straw before it breaks?
Given that,
Outer diameter = 14 mm
Inner diameter = 11 mm
Length = 21 cm
Young's modulus = 68 GPa
Tensile strength = 7 Mpa
(a). We need to calculate the effective spring constant of the straw with respect to elongation
Using formula of effective spring constant
[tex]\dfrac{Y}{\Delta l}=\dfrac{YA}{l}[/tex]
[tex]k=\dfrac{YA}{l}[/tex]
Where, k = effective spring constant
Y= Young's modulus
A = area
l = length
Put the value into the formula
[tex]k=\dfrac{68\times10^{9}\times\dfrac{\pi}{4}(0.014^2-0.011^2)}{21\times10^{-2}}[/tex]
[tex]k=1.90\times10^{7}\ N/m[/tex]
(b). Force = 90 N
We need to calculate the stretch the straw
Using formula of stretch
[tex]\Delta l=\dfrac{F}{k}[/tex]
Where, F = force
k = effective spring constant
Put the value into the formula
[tex]\Delta l=\dfrac{90}{1.90\times10^{7}}[/tex]
[tex]\Delta l=0.0000047\ m[/tex]
[tex]\Delta l=4.7\times10^{-6}\ m[/tex]
(c). We need to calculate the extend the length of the straw before it breaks
Using formula of extend length
[tex]E_{max}=\dfrac{Y\Delta l}{l}[/tex]
[tex]\Delta l=\dfrac{E_{max}l}{Y}[/tex]
Put the value into the formula
[tex]\Delta l=\dfrac{7\times10^6\times21\times10^{-2}}{68\times10^{9}}[/tex]
[tex]\Delta l=0.0000216\ m[/tex]
[tex]\Delta l=0.0216\times10^{-3}\ m[/tex]
[tex]\Delta l=0.0216\ mm[/tex]
Hence, (a). The effective spring constant of the straw with respect to elongation is [tex]1.90\times10^{7}\ N/m[/tex]
(b). The stretch the straw is [tex]4.7\times10^{-6}\ m[/tex]
(c). The extend length of the straw before it breaks is 0.0216 mm.
A total electric charge of 2.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm . The potential is zero at a point at infinity.
a) Find the value of the potential at 45.0 cm from the center of the sphere.
b) Find the value of the potential at 26.0 cm from the center of the sphere.
c) Find the value of the potential at 16.0 cm from the center of the sphere.
Answer:
a) 40 V
b) 69.23 V
c) 69.23 V
Explanation:
See attachment for solution