Answers
Answer:
49 N
Explanation:
The diagram of the bar is obtained online and attached to this solution.
The free body diagram is also attached.
Since the weight of the bar acts at the middle of the bar, the torque due to the weight of the bar is given by
τ = mgx
where x = (L/2) cos 35° = 0.45 × cos 35° = 0.3686 m
τ = (7)(9.8)(0.3686) = 25.29 Nm
The force acting on pin A = torque ÷ (length × sin 35°) = 25.29 ÷ (0.9 × sin 35°)
= 25.29 ÷ 0.5162 = 48.99 N = 49 N
Hope this Helps!!!
Answer:
The force supported by the pin at A is 69.081 N
Explanation:
The diagram is in the figure attach. The angular acceleration using the moment expression is:
[tex]-mg(\frac{Lcos\theta }{2} )=I\alpha \\\alpha =\frac{-3g}{2L} cos\theta[/tex]
Where
L = length of the bar = 900 mm = 0.9 m
[tex]\alpha =\frac{-3*9.8cos35}{2*0.9} =-13.38rad/s^{2}[/tex]
The acceleration in point G is equal to:
[tex]a_{G} =a_{A} +\alpha kr_{G/A} -w^{2} r_{G/A}[/tex]
Where
aA = acceleration at A = 0
w = angular velocity of the bar = 3 rad/s
rG/A = position vector of G respect to A = [tex]\frac{L}{2} cos\theta i-\frac{L}{2} cos\theta j[/tex]
[tex]a_{G} =(\frac{L}{2}\alpha sin\theta -\frac{w^{2}Lcos\theta }{2} )i+(\frac{L}{2}\alpha cos\theta +\frac{w^{2}Lsin\theta }{2} )j=(\frac{0.9*(-13.38)*sin35}{2} -(\frac{3^{2}*0.9*cos35 }{2} )i+(\frac{0.9*(-13.38)*cos35}{2} +\frac{3^{2} *0.9*sin35)}{2} )j=-6.76i-2.61jm/s^{2}[/tex]
The force at A in x is equal to:
[tex]A_{x} =ma_{G} =7*(-6.76)=-47.32N[/tex]
The force at A in y is:
[tex]A_{y} =ma_{G} +mg=(7*(-2.61))+(7*9.8)=50.33N[/tex]
The magnitude of force A is equal to:
[tex]A=\sqrt{A_{x}^{2}+A_{y}^{2} } =\sqrt{(-47.32^{2})+50.33^{2} } =69.081N[/tex]
Related Questions
The ultracentrifuge is an important tool for separating and analyzing proteins. Because of the enormous centripetal accelerations, the centrifuge must be carefully balanced, with each sample matched by a sample of identical mass on the opposite side. Any difference in the masses of opposing samples creates a net force on the shaft of the rotor, potentially leading to a catastrophic failure of the apparatus. Suppose a scientist makes a slight error in sample preparation and one sample has a mass 10 mg larger than the opposing sample.
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The net force is [tex]F_{net}= 6.44 *10^{-4} N[/tex]
Explanation:
Generally the net force is a force that come up due to the unequal centripetal force(A difference in centripetal force ) and it is mathematically represented as
[tex]F_{net} = \Delta F_{cen}[/tex]
and the difference in centripetal force [tex]\Delta F_{cen}[/tex] is mathematically represented as
[tex]\Delta F_{cen} = \Delta m* rw^2[/tex]
Which the difference in mass multiplied by the centripetal acceleration
Substituting 10 mg = [tex]10 *10^{-3}g[/tex] for [tex]\Delta m[/tex] , 12 cm = [tex]\frac{12}{100} = 0.12m[/tex] for radius
and 70,000 rpm = [tex]70,000 *[\frac{2 \pi rad}{1 rev}][\frac{1 min}{60s} ] = 7326.7 rad/s[/tex]
[tex]F_{net} = \Delta F_{cen} = 10*10^{-3} * 0.12 * 7326.7[/tex]
[tex]F_{net}= 6.44 *10^{-4} N[/tex]
You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a magnetic system by modeling it inthe laboratory. The safety system is a conducting bar that slides on two parallel conducting rails thatrun down the ramp (similar to the one in the previous problem). The bar is perpendicular to the railsand is in contact with them. At the bottom of the ramp, the two rails are connected together. The box slides down the rails through a uniform vertical magnetic field. The magnetic field is supposed to causethe bar to slide down the ramp at a constant velocity even when friction between the bar and the rails negligible.Before setting up the laboratory model, your task is to calculate the constant velocity of the bar (sliding down the ramp on rails in a vertical magnetic field) as a function of the mass of the bar, the strength ofthe magnetic field, the angle of the ramp from the horizontal, the length of the bar (which is the same asthe distance between the rails), and the resistance of the bar. Assume that all of the other conductors inthe system have a much smaller resistance than the bar.a) If the force due to the changing flux exactly cancells out the net force due to the combination of gravity and normal force, then the bare will cease to accelerate and instead move at a constant velocity. Please solve for this velocity algebraically.b)Write out the units for each of your variables and show (by cancellation and substitution) that the units for your veloctiy will be m/s on both left and right side of your equation.
Answers
Answer:
Note that the emf induced is
emf = B d v cos (A)
---> v = emf / [B d cos (A)]
where
B = magnetic field
d = distance of two rails
v = constant speed
A = angle of rails with respect to the horizontal
Also, note that
I = emf/R
where R = resistance of the bar
Thus,
I = B d v cos (A) / R
Thus, the bar experiences a magnetic force of
F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.
Thus, the component of this parallel to the incline is
F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R
As this is equal to the component of the weight parallel to the incline,
B^2 d^2 v cos^2 (A) / R = m g sin (A)
where m = the mass of the bar.
Solving for v,
v = [R m g sin (A) / B^2 d^2 cos^2 (A)] [ANSWER, the constant speed, PART A]
******************************
v = [R m g sin (A) / B^2 d^2 cos^2 (A)]
Plugging in the units,
m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]
Note that T = kg / (s * C), and ohm = J * s/C^2
Thus,
m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]
= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]
As J = kg*m^2/s^2, cancelling C^2,,
= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]
Cancelling kg^2,
= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]
Cancelling m^2/s^2,
= [s * m/s^2]
Cancelling s,
=m/s [DONE! WE SHOWED THE UNITS ARE CORRECT! ]
n the melt atomization process to make fine metal powders, there usually is a variation in powder size because not all atomized droplets are of the same size. In an experiment, it was found that titanium droplets of 10 μm dia reached their solidification temperature (1660 C) in 2 milliseconds when the surrounding gas temperature was 1100 C and the initial droplet temperature was 1860 C. How long would it take for droplets of 5 μm, 20 μm, 30 μm and 50 μm diameter to cool by the same amount if they all had the same initial temperature? Given: density of Ti = 4510 kg.m-3.
Answers
Answer:
Check the explanation
Explanation:
Atomization, also called the spraying technique or procedure, refers to a process in which molten metals are broken down into little or tiny drops of liquid through the use of high-speed fluids (liquid as water, gas as air or inert gas) or fluids with centrifugal force, and then solidified into a powdered state.
Kindly check the attached image below to get the step by step solution to the above question.
How much current is in a circuit that includes a 9.0-volt battery and a bulb with a resistance of 4.0 ohms? A. 0.44 amps B. 36 amps C. 2.3 amps D. 13 amps
Answers
Answer:
C. 2.3A
Explanation:
V/Ω=A
9V / 4Ω = 2.25 ≅ 2.3A
The current flowing through the circuit is 2.3 A. So, option C is correct.
What is Ohm's law ?Ohm's law states that, the voltage applied across a circuit is directly proportional to the steady current flowing through the circuit and also proportional to the resistance of the circuit.
Here,
Voltage applied in the circuit, V = 9 V
Resistance of the bulb, R = 4 Ω
According to Ohm's law,
V [tex]\alpha[/tex] I
V [tex]\alpha[/tex] R
So, V = IR
Therefore, current flowing through the circuit,
I = V/R
I = 9/4
I = 2.25 A approximately, 2.3 A
Hence,
The current flowing through the circuit is 2.3 A.
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The compressive strength of our bones is important in everyday life. Young’s modulus for bone is about 1.4 * 1010 Pa. Bone can take only about a 1.0% change in its length before fracturing. (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 cm2 ? (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point.) (b) Estimate the maximum height from which a 70 kg man could jump and not fracture his tibia. Take the time between when he first touches the floor and when he has stopped to be 0.030 s, and assume that the stress on his two legs is distributed equally.
Answers
Answer:
Explanation:
Young modulus ε = 1.4 × 10¹⁰ Pa
ΔL = 1% of the original length = 0.01 x where x is the original length
cross sectional area = 3.0 cm² =( 3 .0 / 10000) m²= 0.0003 m²
ε = Stress / strain
stress = ε × strain
stress = F /A
F force = ε × A × ( ΔL / L) = 1.4 × 10¹⁰ Pa × 0.0003 m² × 0.01 = 4.2 × 10⁴ N
b) F net = F max - mg ( weight) = 84000 - ( 70 × 9.8 m/s² ) ( F is double since the stress on the two leg is equally distributed)
f net = ma = 84000 - ( 70 × 9.8 m/s² )
a = (84000 - ( 70 × 9.8 m/s² )) = 1190.2 m/s²
v = u + at
where final velocity equal zero
- u = -at since it coming downwards
u = at = 1190.2 m/s² × 0.03s = 35.706 m/s
using conservation of energy
1/2 mv² = mgh
1/2v²/ g = h
h = 0.5 × (35.706 m/s )² / 9.8 = 65.04 m
20 POINTS TRUE OR FALSE:
An object with constant velocity will change its motion only if a balanced force acts on it.
A) TRUE
B) FALSE
Answers
Hello :)
True :D
Gülün selamı vaar <3
Cerenin de selamı vaar <3
-Your Welcome--Merve-
A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a distance of 4 m. Given that the vertical component of the pulling force is 12 N, calculate the work done by the force in moving the crate.
Answers
Answer:
[tex]W=173.48J[/tex]
Explanation:
information we know:
Total force: [tex]F=45N[/tex]
Weight: [tex]w=100N[/tex]
distance: [tex]4m[/tex]
vertical component of the force: [tex]F_{y}=12N[/tex]
-------------
In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).
horizontal component: [tex]F_{x}=Fcos\theta[/tex]
vertical component: [tex]F_{y}=Fsen\theta[/tex]
but from the given information we know that [tex]F_{y}=12N[/tex]
so, equation these two [tex]F_{y}=Fsen\theta[/tex] and [tex]F_{y}=12N[/tex]
[tex]Fsen\theta =12N[/tex]
and we know the force [tex]F=45N[/tex], thus:
[tex]45sen\theta=12[/tex]
now we clear for [tex]\theta[/tex]
[tex]sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466[/tex]
the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:
[tex]F_{x}=Fcos\theta[/tex]
[tex]F_{x}=45cos(15.466)\\F_{x}=43.37N[/tex]
whith this horizontal component we calculate the work to move the crate a distance of 4 m:
[tex]W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J[/tex]
the work done is W=173.48J
To calculate the work done by the force in moving the crate, we use the formula Work = Force x Distance x cos(theta). We are given the force, distance, and the vertical component of the force, so we can find the horizontal component using trigonometry. Substituting the values into the formula gives us the work done as 174 J.
Explanation:To calculate the work done by a force, we use the formula:
Work = Force x Distance x cos(theta)
In this case, the force is 45 N, the distance is 4 m, and the vertical component of the force is 12 N. We need to find the horizontal component of the force, which can be calculated using the given information.
Given that the vertical component of the force is 12 N, we can use the trigonometric relationship between the horizontal and vertical components of a force to find the horizontal component as follows:
sin(theta) = vertical component / total force
sin(theta) = 12 N / 45 N
sin(theta) = 0.267
theta = sin^(-1)(0.267)
theta = 15.1 degrees
Once we have the horizontal component of the force, we can calculate the work done as follows:
Work = Force x Distance x cos(theta)
Work = 45 N x 4 m x cos(15.1 degrees)
Work = 180 J x cos(15.1 degrees)
Work = 180 J x 0.966
Work = 174 J
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Two negative charges that are both - 3.8 C push each other apart with a force of 19.0 N. How far apart are the two charges?
Answers
Answer:
The separation distance between the two charges must be 82704.2925 m
Explanation:
Given:
Two negative charges that are both q = -3.8 C
Force of 19 N
Question: How far apart are the two charges, s = ?
First, you need to get the electrostatic force of this two negative charges:
[tex]F=\frac{kq}{s^{2} } \\s=\sqrt{\frac{kq}{F} }[/tex]
Here
k = electric constant of the medium = 9x10⁹N m²/C²
Substituting values:
[tex]s=\sqrt{\frac{9x10^{9}*(-3.8)^{2} }{19} } =82704.2925m[/tex]
A cyclist coasts up a 9.00° slope, traveling 12.0 m along the road to the top of the hill. If the cyclist’s initial speed is 9.00 m/s, what is the final speed? Ignore friction and air resistance.
Answers
Answer:
[tex]v_{f} \approx 6.647\,\frac{m}{s}[/tex]
Explanation:
The final speed of the cyclist is determined by applying the Principle of Energy Conservation:
[tex]\frac{1}{2}\cdot m\cdot v_{o}^{2} + m\cdot g\cdot h_{o} = \frac{1}{2}\cdot m\cdot v_{f}^{2} + m\cdot g\cdot h_{f}[/tex]
[tex]\frac{1}{2} \cdot v_{o}^{2} + g\cdot (h_{o}-h_{f}) = \frac{1}{2}\cdot v_{f}^{2}[/tex]
[tex]v_{f}^{2}=v_{o}^{2} + 2\cdot g \cdot (h_{o}-h_{f})[/tex]
[tex]v_{f} = \sqrt{v_{o}^{2}+2\cdot g \cdot (h_{o}-h_{f})}[/tex]
[tex]v_{f} = \sqrt{v_{o}^{2}-2\cdot g \cdot \Delta s \cdot \sin \theta}[/tex]
[tex]v_{f} = \sqrt{(9\,\frac{m}{s} )^{2}-2\cdot (9.807\,\frac{m}{s} )\cdot (12\,m)\cdot \sin 9^{\textdegree}}[/tex]
[tex]v_{f} \approx 6.647\,\frac{m}{s}[/tex]
Answer:
the final speed of the 10.85 m/s.
Explanation:
Given that,
Slope with respect to horizontal, [tex]\theta=9^{\circ}[/tex]
Distance travelled, d = 12 m
Initial speed of the cyclist, u = 9 m/s
We need to find the final speed of the cyclist. Let h is the height of the sloping surface such that,
[tex]h=d\times sin\thetah\\\\=12\times sin(9) \\\\h = 1.877 m[/tex]
v is the final speed of the cyclist. It can be calculated using work energy theorem as
[tex]\dfrac{1}{2}m(v^2-u^2)\\\\=mgh\dfrac{1}{2}(v^2-u^2)\\\\=gh\dfrac{1}{2}\times (v^2-(9.0)^2)\\\\=9.8\times 1.87\\\\v = 10.85 m/s[/tex]
Thus,the final speed of the 10.85 m/s.
Consider a spring with a force constant of 965 N/m. Suppose the mass of the block is 1.70 kg, but its initial speed can be varied. What initial speed is required to give a maximum spring compression of 3.70 cm.
Answers
Answer: 0.88 m/s
Explanation:
given
Force constant of the spring, k = 965 N/m
Compression of the spring, x = 3.7 cm = 0.037m
Mass of the block, m = 1.7 kg
To solve this, we would be using law of conservation of energy, where initial energy is all kinetic and elastic energy is elastic but potential.
E(f) = E(i)
1/2kx² = 1/2mv²
kx² = mv²
965 * 0.037² = 1.7 * v²
1.321 = 1.7 v²
v² = 1.321 / 1.7
v² = 0.777
v = 0.88 m/s
Thus, the initial speed of the block is v = 0.88 m/s
A disk is initially at motionless, then a motor is turned on and the disk begins to spin. If second the angular acceleration of the disk is a constant 3.5 r a d / s 2 , how many complete revolutions does the disk make in 9 seconds?
Answers
The number of revolutions is 301
Explanation:
Angular acceleration, α = 3.5 rad/s
Time, t = 9s
Number of revolutions, n = ?
We know,
[tex]\alpha = \frac{w}{t}[/tex]
where,
ω = angular velocity
t = time
It can also be written as:
[tex]\alpha = \frac{\pi n}{t X 30}[/tex]
On substituting the value, we get:
[tex]3.5 = \frac{\pi X n}{9 X 30} \\\\n = 300.8[/tex]
Therefore, the number of revolutions is 301
The answer is 22 complete revolutions.
The number of complete revolutions the disk makes in 9 seconds is given by the formula for angular displacement due to constant angular acceleration:
[tex]\[ \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 \][/tex]
where:
-[tex]\( \theta \)[/tex] is the final angular displacement in radians,
-[tex]\( \theta_0 \)[/tex] is the initial angular displacement, which is zero since the disk starts from rest,
-[tex]\( \omega_0 \)[/tex] is the initial angular velocity, which is also zero since the disk starts from rest,
- [tex]\( \alpha \)[/tex] is the angular acceleration, which is 3.5 [tex]rad/s\( ^2 \)[/tex],
- [tex]\( t \)[/tex] is the time in seconds, which is 9 seconds.
Since the disk starts from rest, [tex]\( \theta_0 = 0 \)[/tex] and [tex]\( \omega_0 = 0 \)[/tex], the equation simplifies to:
[tex]\[ \theta = \frac{1}{2} \alpha t^2 \][/tex]
Substituting the given values:
[tex]\[ \theta = \frac{1}{2} \times 3.5 \, \text{rad/s}^2 \times (9 \, \text{s})^2 \][/tex]
[tex]\[ \theta = \frac{1}{2} \times 3.5 \times 81 \][/tex]
[tex]\[ \theta = 1.75 \times 81 \][/tex]
\[ \theta = 141.75 \, \text{radians} \]
To find the number of complete revolutions, we need to convert the angular displacement from radians to revolutions. Since there are [tex]\( 2\pi \)[/tex] radians in one revolution, we divide the angular displacement by [tex]\( 2\pi \)[/tex]:
[tex]\[ \text{Number of revolutions} = \frac{\theta}{2\pi} \][/tex]
[tex]\[ \text{Number of revolutions} = \frac{141.75}{2\pi} \][/tex]
[tex]\[ \text{Number of revolutions} = \frac{141.75}{6.2831853071796} \][/tex]
[tex]\[ \text{Number of revolutions} \approx 22.57 \][/tex]
A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.827 m and 2.89 kg, respectively. When the propeller rotates at 507 rpm (revolutions per minute), what is its rotational kinetic energy K
Answers
Answer:
3241.35J
Explanation:
No. Of rods = 5
Mass = 2.89kg
Length (L) = 0.827m
W = 507rpm
Kinetic energy of rotation = ½I*ω²
For each rod, the moment of inertia (I) = ML² / 3
I = ML² / 3
I = [2.89*(0.827)²] / 3
I = 1.367 / 3 = 0.46kgm²
ω = 507 rev/min. Convert rev/min to rev/sec.
507 * 2Πrads/60s = 53.09rad/s
ω = 53.09rad/s
k.e = ½ I * ω²
K.E = ½ * 0.46 * (53.09)²
K.E = 648.27.
But there five (5) rods, so kinetic energy is equal to
K.E = 5 * 648.27 = 3241.35J
The rotational kinetic energy of a propeller with five blades, each modeled as a uniform rod, can be determined using formulas for moment of inertia and rotational kinetic energy.
Explanation:The kinetic energy of rotating objects depends on their moment of inertia and angular velocity. Given that each propeller blade is modeled as a uniform rod rotating about its end, we can calculate the moment of inertia (I) of all five blades using the formula I=5*(1/3*m*L^2), where m is the mass and L is the length of each rod. After we find the moment of inertia, we can determine the angular velocity (ω) in radian per second, given that the propeller rotates at 507 rpm, by using the conversion factor ω=(507*2*pi)/60. Finally, we calculate the rotational kinetic energy (K) using the formula K=1/2*I*ω^2.
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A uniform meter stick (with a length of 1 m) is pivoted to rotate about a horizontal axis through the 25 cm mark on the stick. The stick is released from rest in a horizontal position. The moment of inertia of a uniform rod about an axis perpendicular to the rod and through the center of mass of the rod is given by 1 12 ML2. Determine the magnitude of the initial angular acceleration of the stick
Answers
Answer:
The initial angular acceleration is [tex]16.8 s^{-2}[/tex]
Explanation:
From the parallel axis theorem the moment of inertia about 25cm mark is
[tex]I = \dfrac{1}{12}ML^2+MD^2[/tex]
since [tex]L = 1m[/tex] & [tex]D = 0.25m[/tex], we have
[tex]I = \dfrac{1}{12}M(1)^2+M(0.25m)^2[/tex]
[tex]I = \dfrac{7}{48} M[/tex]
Now, the gravitational force (equal to the weight of the object) acts as a torque [tex]\tau[/tex] on the center of mass of the rod, which induces angular acceleration [tex]\alpha[/tex]according to
[tex]\tau = I\alpha[/tex]
since [tex]\tau = Mg D[/tex]
[tex]MgD =I \alpha[/tex]
[tex]MgD = \dfrac{7}{48} M \alpha[/tex]
solving for [tex]\alpha[/tex] we get
[tex]\boxed{\alpha = \dfrac{48}{7} gD}[/tex]
putting in [tex]g= 9.8m/s^2[/tex] and [tex]D = 0.25m[/tex] we get:
[tex]\boxed{\alpha = 16.8\: s^{-2}.}[/tex]
which is the initial angular acceleration.
The initial angular acceleration of the stick is 16.812 radians per square second.
First, we calculate the resulting moment of inertia by Steiner theorem:
[tex]I_{O} = I_{g} + M\cdot r^{2}[/tex] (1)
Where:
[tex]I_{g}[/tex] - Moment of inertia with respect to center of mass, in kilogram-square meters.[tex]M[/tex] - Mass of the stick, in meters. [tex]r[/tex] - Distance between the point of rotation and the center of mass, in meters.[tex]I_{O}[/tex] - Moment of inertia with respect to the point of rotation, in meters.If we know that [tex]I_{g} = \frac{1}{12} \cdot M\cdot L^{2}[/tex], [tex]L = 1\,m[/tex] and [tex]r = 0.25\,m[/tex], then the formula for the moment of inertia is:
[tex]I_{O} = \frac{1}{12}\cdot M + \frac{1}{16}\cdot M[/tex]
[tex]I_{O} = \frac{7}{48}\cdot M[/tex] (2)
We know that initial angular acceleration ([tex]\alpha[/tex]), in radians per square second, is solely due to gravity. By the second Newton's law and the D'Alembert principle, we derive an expression for the initial angular acceleration of the stick:
[tex]\Sigma M = M\cdot g\cdot r = I_{O}\cdot \alpha[/tex] (3)
Where:
[tex]g[/tex] - Gravitational acceleration, in meters per square second.[tex]r[/tex] - Distance between the point of rotation and the center of mass, in meters.
By (2) and (3) we have the following simplified expression:
[tex]M\cdot g \cdot r = \frac{7}{48} \cdot M\cdot \alpha[/tex]
[tex]\alpha = \frac{48}{7}\cdot g \cdot r[/tex] (4)
If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]r = 0.25\,m[/tex], then the initial angular acceleration of the stick is:
[tex]\alpha = \frac{48}{7}\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.25\,m)[/tex]
[tex]\alpha = 16.812\,\frac{rad}{s^{2}}[/tex]
The initial angular acceleration of the stick is 16.812 radians per square second.
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Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of operation, and low cost. Suppose the maximum speed of a moped is normally distributed with mean value 46.6 km/h and standard deviation 1.75 km/h. Consider randomly selecting a single such moped.
Answers
Answer:
a) P ( X ≤ 49 ) = 0.9149
b) P ( X ≥ 48 ) = 0.2119
c) 0.9876
Explanation:
Solution:-
- Lets define a random variable X: the maximum speed of a moped.
- The random variable follows normal distribution with the following parameters:
X ~ Norm ( u , σ^2 )
Where,
u = Mean = 46.6 km/h
σ = Standard deviation 1.75 km/h
Hence,
X ~ Norm ( 46.6 , 1.75^2 ).
a) The probability that the maximum speed of mopeds is atmost 49 km/h?
- To evaluate the probability of P ( X ≤ 49 ). We will find the standard Z-score value using the following formula:
[tex]Z-score = \frac{x-u}{s} \\\\ Z-score = \frac{49-46.6}{1.75} \\\\Z-score = 1.37142[/tex]
- Now use the standard normal tables to determine the required probability of:
P ( Z < 1.37142 ) = 0.9149
Hence,
P ( X ≤ 49 ) = 0.9149
b) The probability that the maximum speed of mopeds is at-least 48 km/h?
- To evaluate the probability of P ( X ≥ 48 ). We will find the standard Z-score value using the following formula:
[tex]Z-score = \frac{x-u}{s} \\\\ Z-score = \frac{48-46.6}{1.75} \\\\Z-score = 0.8[/tex]
- Now use the standard normal tables to determine the required probability of:
P ( Z ≥ 0.8 ) = 0.2119
Hence,
P ( X ≥ 48 ) = 0.2119
c) The probability that the maximum speed of mopeds differs from mean by at-most 2.5 standard deviation.
- The required probability is the standard error from the mean value "u" of 2.5 standard deviation.
- We don't need to evaluate the test statistics as we are already given the standard error about mean.
- Using, the standard normal Z-score: The required probability is:
P ( -2.5 < Z < 2.5 ) = 2*P ( Z < 2.5 ) - 1
= 2*0.9938 - 1
= 0.9876
A circular saw spins at 6000 rpm , and its electronic brake is supposed to stop it in less than 2 s. As a quality-control specialist, you're testing saws with a device that counts the number of blade revolutions. A particular saw turns 75 revolutions while stopping.Does it meet its specs?
a.yes
b.no
Answers
Answer:
a. yes
Explanation:
The initial speed of the circular saw is:
[tex]\dot n_{o} = \frac{6000}{60} \,\frac{rev}{s}[/tex]
[tex]\dot n_{o} = 100\,\frac{rev}{s}[/tex]
Deceleration rate needed to stop the circular saw is:
[tex]\ddot n = -\frac{100\,\frac{rev}{s} }{2\,s}[/tex]
[tex]\ddot n = - 50\,\frac{rev}{s^{2}}[/tex]
The number of turns associated with such deceleration rate is:
[tex]\Delta n = \frac{\dot n^{2}}{2\cdot \ddot n}[/tex]
[tex]\Delta n = \frac{\left(100\,\frac{rev}{s} \right)^{2}}{2\cdot \left(50\,\frac{rev}{s^{2}} \right)}[/tex]
[tex]\Delta n = 100\,rev[/tex]
Since the measured number of revolutions is lesser than calculated number of revolution, the circular saw meets specifications.
which description best tells how the arrangement of the sun, moon, and earth affect the range of the tides during a spring tide
Answers
Answer:in a spring tide the sun , moon and the earth are in a straight line which causes regular higher tide and low tide to increase.
Explanation:
In a spring tide, the Moon, Earth, and Sun are all in allignment, which causes the greatest gravitational pull on the water of Earth. Basically, when the Moon and Sun align towards and in the same direction at the Earth, their gravitational pulls join together making the pull stronger. (see image below)
What this very strong pull does is make high tides higher than usual, and low tides lower than usual. This is because the pull on tides is very strong and more water is being pulled to form higher high tides, making there less water for there to be in low tides.
(image is from oceanservice.gov)
Once Kate's kite reaches a height of 52 ft (above her hands), it rises no higher but drifts due east in a wind blowing 6 ft divided by s. How fast is the string running through Kate's hands at the moment that she has released 105 ft of string?
Answers
Answer:
5.213ft
Explanation:
Z² = x² + y²
x = √(z² - y²)
y = 52ft, dx = 6ft, z = 105ft, dz = ?
d(z² = x² + y²)
2zdz = 2xdx
dz = xdx/z
But x = √(z² - y²)
dz = √(z² - y²)/z * dx
dz = [√(105² - 52²)/105] * 6
dz = √(8521)/ 17.5
dz = 5.213ft
What are potential impacts of pollution on a watershed? Check all that apply.
excess runoff
loss of farmland
loss of areas for tourism
contaminated drinking water
deaths of sea creatures that are used as a food source
limits to potential economic activities such as a fishing
Answers
Answer:
contaminated drinking water
deaths of sea creatures that are used as a food source
limits to potential economic activities such as a fishing
Explanation:
A watershed is a large area that comprises of drainage area of all the surrounding water bodies meeting at a common affluence point before draining into sea or ocean or any other large water body. Pollution in this area can pollute the small water streams flowing through it, thereby polluting the larger water body into which it drains.
Thus, the water extracted for drinking from such area will be contaminated. Pollution in larger water body can cause death of water creature and hence pose a threat to fishing.
Answer:
loss of areas for tourism
contaminated drinking water
deaths of sea creatures that are used as a food source
limits to potential economic activities such as fishing
4. Analyze: What can you say about the acceleration of dividers when the pressure increases
from left to right?
Answers
Velocity changes with time and this is called acceleration.
What is acceleration?The question is incomplete but I will try to explain the meaning of acceleration to you. The term acceleration refers to the change of velocity with time.
a = Δv/t
a = accelerationΔv = change in velocityt = time intervalHence, velocity changes with time and this is called acceleration.
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The increase of the pressure of the system would lead to an increase of acceleration.
What happens when the pressure equates to the force of action?As pressure equals to the force upon action. Here F is the force and the A the area. As Newton's second law F = ma We substitute P A = m a
Hence the pressure is directly proportional to acceleration.
Find out more information about acceleration.
brainly.com/question/4279093.
The speed of light in a vacuum is 300,000.0 kmls. How long does it take light to travel
150,000,000 km from the Sun to Earth?
Answers
Answer:
8 minutes
Explanation:
The time it takes for light to travel from the Sun to Earth can be found using the formula time = distance/speed. Given the speed of light is known to be 300,000 km/s and the distance is 150,000,000 km, it will take light about 500 seconds to travel from the Sun to the Earth.
Explanation:To calculate the time it takes for light to travel from the Sun to Earth, you need to use the formula for time which is distance/speed. The speed of light is 300,000 km/s and the distance from the Sun to Earth is 150,000,000 km.
So, Time = Distance / Speed = 150,000,000 km / 300,000 km/s = 500 seconds
Therefore, it takes approximately 500 seconds for light to travel from the Sun to Earth.
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how large would a coin be with diameter of 3/4 in appear through a 5x lens
Answers
Answer:
h ’= 3.75 inch
Explanation:
For this exercise we use the definition of magnification which is the height of the image over the height of the object
m = h ’/ h
h’= m h
Let's calculate
h ’= 5 3/4
h ’= 3.75 inch
Jim is driving a 2268-kg pickup truck at 30.0m/sand releases his foot from the accelerator pedal. The car eventually stops due to an effective friction force that the road, air, and other things exert on the car. The friction force has an average magnitude of 700N .Part A Determine the initial kinetic energy of the truck.Part B Determine the stopping distance of the truck.
Answers
Explanation:
Given that,
Mass of the truck, m = 2268 kg
Speed of the truck, v = 30 m/s
The friction force has an average magnitude of 700 N, f = -700 N
(a) The initial kinetic energy of the truck is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2268\times (30)^2\\\\K=1.02\times 10^6\ J[/tex]
(b) Finally, the stops due to an effective friction force that the road, final speed is 0. Let d is the stopping distance of the truck. using third equation of motion to find it as :
[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}[/tex]
Since, f = ma
[tex]d=\dfrac{-u^2m}{2f}\\\\d=\dfrac{-(30)^2\times 2268}{-2\times 700}\\\\d=1458\ m[/tex]
So, the stopping distance of the truck is 1458 meters.
A block of mass m attached to a horizontally mounted spring with spring constant k undergoes simple harmonic motion on a frictionless surface. How would the maximum speed of the block be affected if the spring constant was increased by a factor of 4 while holding the amplitude of oscillation constant? It would decrease by a factor of 14 . It would increase by a factor of 4. It would remain unchanged. It would increase by a factor of 2. It would decrease by a factor of 12 .
Answers
Question:
a. It would decrease by a factor of 14 .
b. It would increase by a factor of 4.
c. It would remain unchanged.
d. It would increase by a factor of 2.
e. It would decrease by a factor of 12 .
Answer:
The correct option is;
d. It would increase by a factor of 2
Explanation:
Here we have
[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + F_kx[/tex]
The formula for maximum mass is given by the following relation;
[tex]v_{max1} =A\sqrt{\frac{k}{m} }[/tex]
Where:
[tex]v_{max}[/tex] = Maximum speed of mass on spring
k = Spring constant
m = Mass attached to the spring
A = Amplitude of the oscillation
Therefore, when k is increased by a factor of 4 we have;
[tex]v_{max2} =A\sqrt{\frac{4 \times k}{m} } = 2 \times A\sqrt{\frac{ k}{m} }[/tex]
Therefore, [tex]v_{max2} = 2 \times v_{max1}[/tex], that is the velocity will increase by a factor of 2.
You stare at a bright red screen for so long that your red cones become saturated and no longer function. The red screen is then rapidly replaced by a white screen. Immediately afterwards, what color do you see on the screen?
Answers
Answer: you'll see cyan color on the screen
Explanation:
Saturating the red cone causes them to stop functioning, hence you can't perceive the red part of white light. White light is made up of three main colors which are blue, red and green. When one can no longer perceive the red part of light, one is left with the grean and blue part. The green and blue part of light will superimpose to give a cyan color.
Answer:
Cyan color
Explanation:
When an individual stares at a red object and immediately look at a white area afterward, there will be an image formation almost immediately that is the same size and shape as the previous one. The only difference is there will be a color change to blue-green, or cyan.
This is due to the eyes generally using the red, blue and green cone cells to perceive white light. Since the red cone cells are fatigued due to long view of it and cone cells being saturated , the blue and green color which are the remaining color cone cells use to perceive white light is viewed. This gives rise to the blue green or cyan color.
Sheet metal of two different alloys is produced by rolling to 0.1 mm thickness. You must select one of these alloys for a sheet forming operation. To make the selection, tensile samples are cut from each alloy at 0o , 45o , and 90o to the rolling direction. The initial gauge section of each tensile sample is 0.75 mm width. After subjecting all samples to the same tensile load, th
Answers
Answer: Alloy A is best chosen for the sheet forming operation
Explanation:
taking both alloy A and B into consideration;
For alloy A;
the true strain ration, R₀ = width strain / thickness strain
R₀ = in (0.65/0.75) / in (0.092/0.1) = 1.716
at 45⁰ and 90⁰, the true strain ratio becomes;
R45⁰ = in (0.63/0.75) / in (0.093/0.1) = 2.4025
R90⁰ = in (0.67/0.75) / in (0.087/0.1) = 0.8099
where R(avg) = (R₀+2Ras+Rao) / A = (1.716+2(2.4025)+0.8099) = 3.0340
R(avg) = 3.0340
For alloy B;
R₀ = in (0.7/0.75) / in (0.082/0.1) = 0.3477
R45⁰ = in (0.69/0.75) / in (0.083/0.1) = 0.4475
R90⁰ = in (0.71/0.75) / in (0.078/0.1) = 0.2206
R(avg) = (R₀+2Ras+Rao) / A = (0.3477+2(0.4475)+0.2206) = 0.365825
R(avg) = 0.365825
comparing both we have that,
R(avg) for allow A > R(avg) for allow B
∴ Alloy A is the best to be selected for sheet formation operation.
cheers i hope this helps!!!1
An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/magnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field. Your device tells you that the electric field is pointing in the negative z direction and the magnetic field is pointing in the positive y direction. In which direction does the released electromagnetic wave travel
Answers
Answer:
The answer is -x direction.
Explanation:
According to the right hand rule for electromagnetism, when the thumb points up, index finger points forward and the middle finger is perpendicular to the index finger, the thumb points in the direction of the magnetic force, the index finger in the direction of the charge movement and the middle finger in the direction of the electromagnetic lines. If the electric field is pointing in the -z direction which is into the screen and the magnetic field is in the +y direction which is upwards, then the magnetic lines are in the -x direction.
I hope this answer helps.
a 500 pound metal star is hanging on two cables which are attached to the ceiling. the left hand cable makes a 13 degree angle with the ceiling while the rigth hand cables makes a 77 degree angle with the ceiling. What is the tension on each of the cables
Answers
Answer:
T1 = 112.07[lb]
T2 = 487.3 [lb]
Explanation:
To solve this problem we must perform a static balance analysis, for this we perform a free body diagram. In this free body diagram we use the angles mentioned in the description of the problem.
Performing a sum of forces on the X-axis equal to zero, we can find an equation that relates the tension of the T1 & T2 cables.
Then we perform a summation of forces on the Y-axis, in which we can find another equation. In this new equation, we replace the previous one and we can find the tension T2.
T1 = 112.07[lb]
T2 = 487.3 [lb]
If 4.1 × 1021 electrons pass through a 40 Ω resistor in 5 min, what is the potential difference across the resistor? The fundamental charge is 1.602 × 10−19 C . Answer in units of V
Answers
Answer:
Potential difference across resistor will be 87.66 volt
Explanation:
We have given number of electrons [tex]n=4.1\times 10^{21}[/tex]
Charge on one electron [tex]e=1.6\times 10^{-19}C[/tex]
So total charge [tex]Q=4.1\times 10^{21}\times 1.6\times 10^{-19}=656C[/tex]
Time is given t = 5 min
1 minute = 60 sec
So 5 minute = 5×60 = 300 sec
So current [tex]i=\frac{Q}{t}=\frac{656}{300}=2.1866A[/tex]
Resistance is given R = 40 ohm
Sp from ohm's law potential difference across resistor v = iR = 2.1866×40 = 87.466 volt
The potential difference across the 40 Ω resistor when 4.1 × 1021 electrons have passed through it over a time period of 5 minutes is calculated using Ohm's law, which results in 87.6 Volts.
Explanation:The subject of this question is based on Ohm's law, an important concept in Physics. Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
First, calculate the total charge (Q) passed through the resistor. This can be done by multiplying the number of electrons by the fundamental charge of one electron (Q = electrons x electron charge). Thus, Q = 4.1 × 1021 electrons x 1.602 × 10-19 C = 656.82 Coulombs.
Next, use the formula to obtain the electric current (I) passed through the resistor. Since current (in Amperes) is equal to the total charge (in Coulombs) divided by the time (in seconds) and time here is given in minutes, we first need to convert minutes to seconds. So, 5 min = 5 x 60 = 300 seconds. Hence, I = Q / t = 656.82 C / 300 s = 2.19 Ampere.
Lastly, we can now find the potential difference (V) across the resistor. According to Ohm's law, V = I * R. Thus, V = 2.19 Ampere * 40 Ω = 87.6 Volts.
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A magnetic dipole with a dipole moment of magnitude 0.0243 J/T is released from rest in a uniform magnetic field of magnitude 57.5 mT. The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is 0.458 mJ. (a) What is the initial angle between the dipole moment and the magnetic field
Answers
Answer:
47.76°
Explanation:
Magnitude of dipole moment = 0.0243J/T
Magnetic Field = 57.5mT
kinetic energy = 0.458mJ
∇U = -∇K
Uf - Ui = -0.458mJ
Ui - Uf = 0.458mJ
(-μBcosθi) - (-μBcosθf) = 0.458mJ
rearranging the equation,
(μBcosθf) - (μBcosθi) = 0.458mJ
μB * (cosθf - cosθi) = 0.458mJ
θf is at 0° because the dipole moment is aligned with the magnetic field.
μB * (cos 0 - cos θi) = 0.458mJ
but cos 0 = 1
(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³
1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³
1 - cos θi = 0.3278
collect like terms
cosθi = 0.6722
θ = cos⁻ 0.6722
θ = 47.76°
Answer:
The initial angle between the dipole moment and the magnetic field is 47.76⁰
Explanation:
Given;
magnitude of dipole moment, μ = 0.0243 J/T
magnitude of magnetic field, B = 57.5 mT
change in kinetic energy, ΔKE = 0.458 mJ
ΔKE = - ΔU
ΔKE = - (U₂ -U₁)
ΔKE = U₁ - U₂
U₁ -U₂ = 0.458 mJ
[tex](-\mu Bcos \theta_i )- (-\mu Bcos \theta_f) = 0.458 mJ\\\\-\mu Bcos \theta_i + \mu Bcos \theta_f = 0.458 mJ\\\\\mu Bcos \theta_f -\mu Bcos \theta_i = 0.458 mJ\\\\\mu B(cos \theta_f - cos \theta_i ) = 0.458 mJ[/tex]
where;
θ₁ is the initial angle between the dipole moment and the magnetic field
[tex]\theta_f[/tex] is the final angle which is zero (0) since the dipole moment is aligned with the magnetic field
μB(cos0 - cosθ₁) = 0.458 mJ
Substitute the given values of μ and B
0.0243 x 0.0575 (1 - cosθ₁) = 0.000458
0.00139725 (1 - cosθ₁) = 0.000458
(1 - cosθ₁) = 0.000458 / 0.00139725
(1 - cosθ₁) = 0.327787
cosθ₁ = 1 - 0.327787
cosθ₁ = 0.672213
θ₁ = cos⁻¹ (0.672213)
θ₁ = 47.76⁰
Thus, the initial angle between the dipole moment and the magnetic field is 47.76⁰
2. On January 21 in 1918, Granville, North Dakota, had a surprising change in temperature. Within 12 hours, the temperature changed from 237 K to 283 K. What is this change in temperature in the Celsius and Fahrenheit scales?
Answers
Answer:
The temperature change in Celsius is 46°C.
The temperature change in Fahrenheit is 82.8°F.
Explanation:
A degree of Celsius scale is equal to that of kelvin scale; therefore,
[tex]\Delta C = \Delta K = 283K-237K\\\\ \boxed{\Delta C = 46^oC}[/tex]
A degree in Fahrenheit is 1.8 times the Celsius degree; therefore
[tex]\Delta F = 1.8(46^o)[/tex]
[tex]\boxed{\Delta F = 82.8^oF}[/tex]
Hence, the temperature change in Celsius is 46°C, and the temperature change in Fahrenheit is 82.8°F.
Final answer:
The temperature change of 46 degrees from Kelvin to Celsius in Granville, North Dakota, is equivalent to a change of 82.8 degrees Fahrenheit.
Explanation:
On January 21, 1918, the temperature in Granville, North Dakota, changed from 237 K to 283 K within 12 hours. To convert the change in temperature to the Celsius scale, we first recognize that 0 degrees Celsius is equivalent to 273.15 K. Therefore, the initial temperature in Celsius would have been 237 K - 273.15 K = -36.15 extdegree C, and the final temperature would have been 283 K - 273.15 K = 9.85 extdegree C. The change in the Celsius scale is then 9.85 extdegree C - (-36.15 extdegree C) = 46 extdegree C
To convert the change in temperature to the Fahrenheit scale, we start with the change in Celsius and use the conversion formula (F = C imes \frac{9}{5} + 32). Since we are looking for the change, we only need to convert the difference in Celsius to Fahrenheit. Therefore, the change in Fahrenheit is 46 extdegree C imes \frac{9}{5} = 82.8 extdegree F.