Answer:
τ = 15.76 N-m
Explanation:
For rigid bodies, the net external torque acting on a rotating body, is equal to the product of the moment of inertia about its rotation axis, times the angular acceleration of the body, as follows:[tex]\tau = I * \alpha (1)[/tex]
Since I is a given of the question, we need to find out the angular acceleration.By definition, the angular acceleration is the rate of change of the angular velocity with respect to time:α = Δω/Δt = (ωf-ω₀) / (tfi-t₀)Since ω₀ = 0, and choosing t₀ =0, as tfi is a given, we need just to find out ωf.We have this value in rev/min, so we need first to convert it to rad/sec, as follows:[tex]537 rev/min * (1min/60 sec)*(2*\pi *rad/rev) = 17.9 * \pi rad/sec (2)[/tex]
Replacing in the definition of α, we have:α = 17.9*π rad/sec / 6.78 s = 8.29 rad/sec²Replacing I and α in (1), we finally get:τ = 15.76 N-m.
The chart shows Daniela’s run through her race.
A graph with horizontal axis time (seconds) and vertical axis position in (meters). The line runs straight from 0 seconds 5 meters to 10 seconds 30 meters.
What is her velocity?
1.5 m/s
2 m/s
2.5 m/s
5 m/s
Answer:
Explanation:
velocity is the rate of change of displacement with respect to time.
Velocity = displacement/time
If a line runs straight from 0 seconds 5 meters to 10 seconds 30 meters.
The velocity will be expressed as;
v = d2-d2/t2-t1
v = 30-5/10-0
v = 25/10
v = 2.5m/s
Hence her velocity is 2.5m/s
Daniela's velocity is 2.5 m/s.
Velocity is the ratio of displacement to time taken. It is given by:
Velocity = displacement / time
The velocity of Daniela can be gotten by calculating the slope of the graph. Considering the points (0, 5) and (4, 15).
[tex]Velocity(slope)=\frac{15-5}{4-0}=2.5\ m/s[/tex]
Therefore Daniela's velocity is 2.5 m/s.
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A wire of radius 0.8 cm carries a current of 106 A that is uniformly distributed over its cross-sectional area. Find the magnetic field B at a distance of 0.07 cm from the center of the wire.
Answer:
The magnetic field is [tex]B = 2.319 *10^{-3} \ T[/tex]
Explanation:
From the question we are told that
The radius of the wire is [tex]r = 0.8 \ cm = 0.008 \ m[/tex]
The current is [tex]I = 106 \ A[/tex]
The position considered is d = 0.07 cm = 0.0007 m
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * I}{2\pi * \frac{r^2}{d} }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex] 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]B = \frac{ 4\pi * 10^{-7} * 106 }{2 * 3.142 * \frac{0.008^2}{0.0007} }[/tex]
=> [tex]B = 2.319 *10^{-3} \ T[/tex]
A net force F acts on a mass m and produces an acceleration a. What acceleration results if a net force 4F acts on a mass 6m?
Answer: The acceleration results if a net force of 4F acts on a mass of 6m is 2/3a.
Explanation:
Force exerted on an object can be defined as a pull or a push on an object which leads to it's displacement. Force is taken to be a vector quantity because it has both magnitude and direction. The formula which can be used to determine force exerted on an object in physics is:
F= mass( kg) × acceleration( m/ s²)
Acceleration is defined as the rate at which the velocity of an object changes. From the formula of force given above it can be determined by making it the subject of formula. Therefore acceleration= Force/ mass.
From the question,
Force= 4F
Mass= 6m
Therefore acceleration= F/m
= 4/6
Acceleration= 2/3a
The required magnitude of acceleration when force is 4F and mass is 6m is 2/3a.
Given data:
The magnitude of net force is, 4F.
The value of mass is, 6m.
Apply the Newton's second law which says that force exerted on an object can be defined as a pull or a push on an object which leads to it's displacement. Force is taken to be a vector quantity because it has both magnitude and direction. The formula which can be used to determine force exerted on an object in physics is
F = ma
a = F/m ..........................................(1)
here, a is the acceleration.
Solving as when the force becomes 4F and mass becomes 6m.
[tex](4F) = (6m) \times a'\\\\a '= \dfrac{4F}{6m}\\\\a '= \dfrac{2}{3}a[/tex]
Thus, the required magnitude of acceleration when force is 4F and mass is 6m is 2/3a.
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If a ball rolls across a table to the left with constant speed, which of the following is true about the force(s) on the ball? (3a1)
Question 10 options:
There cannot be any forces applied to the ball.
There must be exactly one force applied to the ball.
The net force applied to the ball is zero.
The net force applied to the ball is directed to the right.
Answer:
C. The net force applied to the ball is zero.
Explanation:
From Newton's second law of motion;
F = ma
Where F is the force on an object, m is its mass and a is its acceleration.
Therefore, the force on an object is a product of its mass and acceleration as it travel from one point to another.
Since acceleration relates to the rate of change in velocity to time. Then when the object moves at uniform velocity (especially along a straight path), its acceleration is zero.
So that;
F = m x 0
= 0
No force is applied on the object.
Therefore for the ball in the given question, the net force applied to the ball is zero because it rolls with constant speed along a straight path.
gravities limit is under which sphere as the perimeter?
What percentage of an iron anchor’s weight will be supported by buoyant force when submerged in salt water?
Answer:
0.87
Explanation:
To solve this, we use the principle of Archimedes. Archimedes Principal of flotation states that "the buoyant force of an object is equal to the total weight of the fluid it displaces."
In the attachment, I stated the mathemacal formula, of which
F(B) = The buoyant force
w(fl) = The weight of the salt water displaced
p(iron) = density of iron
p(salt) = density of the salt water = 1025 kg/m³
F' = weight of the iron in air
F = weight of the iron in salt water
p(man) = density of man = 7680 kg/m³
The rest are the easy calculations done by substituting the values
match the variables to its definition
Answer:
See connections below
Explanation:
1 [tex]\Rightarrow[/tex] b
2 [tex]\Rightarrow[/tex] a
3 [tex]\Rightarrow[/tex] d
4 [tex]\Rightarrow[/tex] i
5 [tex]\Rightarrow[/tex] g
6 [tex]\Rightarrow[/tex] h
7 [tex]\Rightarrow[/tex] c
8 [tex]\Rightarrow[/tex] e
9 [tex]\Rightarrow[/tex] f
A seated musician plays a C4 note at 262 Hz . How much time Δ does it take for 346 air pressure maxima to pass a stationary listener?
Answer:
t = 1.32 s
Explanation:
We are given;. Frequency of C4 note; F_c = 262 Hz
In conversions, we know that 1 Hz = 1 cycle/s
Thus, F_c = 262 cycles/s
Now, we want to find out how much time it takes for 346 air pressure maxima to pass a stationary listener.
346 air pressure maxima denotes that the air pressure maxima is 346 cycles.
Thus, time will be;
t = 346 cycles/262 cycles/s
t = 1.32 s
The time taken for the musical note to pass the stationary listener is 1.32 s.
The given parameters:
frequency of the C4 note, f = 262 Hzair pressure maximum, n = 346The frequency of a sound wave is defined as the number of cycles completed per second by the wave.
[tex]F = \frac{n}{t}[/tex]
where;
t is the time to compete the maximum cycleThe time taken for the musical note to pass the stationary listener is calculated as follows;
[tex]262 = \frac{n}{t} \\\\t = \frac{n}{262} \\\\t = \frac{346}{262} \\\\t = 1.32 \ s[/tex]
Thus, the time taken for the musical note to pass the stationary listener is 1.32 s.
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1. When asteroids collided some of the broken materials fall into Earth's orbit. What do
astronomers call the debris when it hits planet Earth?
(2 Points)
meteors
meteoroids
meteorites
metabots
Before they meet Earth -- meteoroids
While they're falling -- meteors
After they hit the ground -- meteorites
Microwave ovens have a plate that rotates at a rate of about 7.0 rev/min. What is this in revolutions per second?
Answer:
The value is [tex]w = 0.1167 \ rev/second[/tex]
Explanation:
From the question we are told that
The rate at which the plate rotates is [tex]w =7.0 \ rev/min[/tex]
Generally the revolution per second is mathematically represented as
[tex]w = \frac{7.0}{60}[/tex]
=> [tex]w = 0.1167 \ rev/second[/tex]
Which of these should you always do at the end of a calculation
Answer:
Reverse check the answer
Explanation:
I believe it is very important that once someone is done with any calculation, the person ought to go over the calculations again. And even, recheck the answer in inverted form.
This is so because while doing the calculations, we can possibly make errors that we won't notice until after submission. Knowing 2 * 3 = 6, but writing 2 * 3 = 5 in the course of calculations can happen to anybody. So therefore, cross checking and reverse checking is needed
An unladen swallow that weighs 0.03 kg flies straight northeast a distance of 125 km in 4.0 hours. With the x x direction due east and the y y direction due north, what is the average momentum of the bird (in unit vector notation)?
Answer:
The average momentum of the bird is 0.26 kgm/s
Explanation:
The formula to be used here is that of momentum which is
momentum (in kgm/s) = mass (in kg) × velocity (in m/s)
The velocity of the bird is
velocity (in m/s) = distance (in meter) ÷ time (in seconds)
distance in meters = 125km × 1000 = 125,000 m
time in seconds = 4 hrs × 60 × 60 = 14,400 secs
velocity = 125000/14400
velocity = 8.68 m/s
momentum (p) = 0.03 × 8.68
p = 0.26 kgm/s
The average momentum of the bird is 0.26 kgm/s
The average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.
Total displacementSince the unladen swallow that weighs 0.03 kg flies straight northeast (that is at a bearing of 45°) a distance of 125 km in 4.0 hours.
Its position vector after 4.0 hours is d = (125kmcos45)i + (125kmsin45)j = (125000 × 1/√2)i + (125000 × 1/√2)j
= (62500√2)i + (62500√2)j.
If the initial position of the swallow is d' = 0i + 0j, then its total displacement after 4 hours is, D = d - d'
= (62500√2)i + (625000√2)j - (0i + 0j)
= (62500√2)i + (62500√2)j m
Average velocityThe unladen swallow's average velocity, v = D/t where
D = total displacement = (62500√2)i + (62500√2)j m and t = time = 4.0 hours = 4 × 60 min/hr × 60 s/min = 14400 sSo, v = [(62500√2)i + (62500√2)j m]/14400 s = (88388.35)i/14400 + (88388.35)j /1440
= 6.14i + 6.14j m/s
Average momentumThe average momentum of the unladen swallow is p = mv where
m = mass of unladen swallow = 0.03 kg and v = average velocity = 6.14i + 6.14j m/sSo, p = mv
p = 0.03 kg × (6.14i + 6.14j m/s)
p = (0.1842i + 0.1842j) kgm/s
So, the average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.
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Image formed by a plane mirror is
Answer: A plane mirror always forms a virtual image (behind the mirror). The image and object are the same distance from a flat mirror, the image size is the same as the object size, and the image is upright.
Explanation:
What is the force on a 5-kilogram ball that is falling freely due to the pull of gravity?
Answer:
Force = 49N
Explanation:
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force measured in Newton.
m represents the mass of an object measured in kilograms.
a represents acceleration measured in meter per seconds square.
Given the following data
Mass = 5kg
We know that acceleration due to gravity = 9.8m/s²
To find the Force:
[tex] F = 5*9.8 [/tex]
Force = 49N
What two methods are the best choices to factor this expression 18x2-8
Answer:
Please check the explanation
Explanation:
The best two methods will be:
Factor by groupingFactor out the GCFFactor by grouping
Factor by grouping deals with establishing a smaller groups from each term.
[tex]18x^2=\:\left(2\cdot 3\cdot 3\cdot x^2\right)[/tex]
[tex]8\:=\:\:2\cdot 2\cdot 2[/tex]
Therefore, the expression becomes
[tex]18x^2=\:\left(2\cdot 3\cdot 3\cdot x^2\right)-\left(2\cdot \:2\cdot \:2\right)[/tex]
Now factor out the greatest common factor (GCF) which is 2
[tex]=\:2\left(3\cdot \:\:3x^2-\left(2\right)\left(2\right)\right)[/tex]
[tex]=2\left(9x^2-2\cdot \:2\right)[/tex]
[tex]=2\left(9x^2-4\right)[/tex]
Factor out the GCF
Given the expression
[tex]18x^2-8\:\:\:[/tex]
factor out common term 2
[tex]=2\left(9x^2-4\right)[/tex]
[tex]=2\left(3x+2\right)\left(3x-2\right)[/tex] ∵ [tex]Factors\:\:\left(9x^2-4\right)=\left(3x+2\right)\left(3x-2\right)[/tex]
Elevator is accelerating upward 3.5 M/S2 and has a mass of 300 KG. The force of gravity is 2940 N. What is the tension force pulling elevator up?
Answer:
T = 3990 N
Explanation:
The free body diagram for the elevator consists of a tension force pointing up, and its weight pointing down. So the elevator's net force is:
F = T - 2940N
ad at the same time, using Newton's second law, we have that this net force should equal the elevator's mass (300 kg) times its acceleration (a):
T - 2940N = 300kg (3.5m/s^2)
then
T = 2940 N + 1050 N
T = 3990 N
In your new job, you are the technical advisor for the writers of a gangster movie about Bonnie and Clyde. In one scene Bonnie and Clyde are being pursued by a police car. They are 750m from a level railroad crossing travelling at 100 km/hr. A train is 500 meters from the crossing travelling at 130 km/hr. The level crossing is on the state line, and so if they can beat the train, they could evade capture, at least for a while until they become Federal fugitives. They accelerate at a constant rate of 4 m/s2 toward the crossing. The writers want to know if Bonnie and Clyde make it across the crossing before the train.
Answer:
Explanation:
Velocity of train 130 km /hr
distance of crossing s = 500 m = .5 km
Time taken by train to reach crossing = .5 / 130 = .003846 hr = 13.85 s
Time taken by gangster to reach the crossing = t
initial velocity u = 100km/h = 27.77 m /s
distance of crossing s = 750 m
acceleration a = 4 m /s
s = ut + .5 a t²
750 = 27.77 t + .5 x 4 x t²
2 t² + 27.77 t - 750 = 0
t = - 27.77 ± √(27.77² + 4 x 2 x 750) / 2 x 2
= - 27.77 ±√ ( 771.17 + 6000) / 4
= - 27.77 ±82.28 / 4
= 13.62 s
So this time is less than time taken by train so it will be able to cross the crossing before train arrives .
Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the direction in which such pressure fluctuations travel?A) The direction of motion of pressure fluctuations is independent of the direction of motion of the sound wave.B) Pressure fluctuations travel perpendicularly to the direction of propagation of the sound wave.C) Pressure fluctuations travel along the direction of propagation of the sound wave.D) Propagation of energy that passes through empty spaces between the particles that comprise the mediumDoes air play a role in the propagation of the human voice from one end of a lecture hall to the other?a) yesb) no
Answer:
None of them: the direction of the pressure fluctuations is parallel to the direction of motion of the wave
Explanation:
There are two identical, positively charged conducting spheres fixed in space. The spheres are 40.4 cm apart (center to center) and repel each other with an electrostatic force of F1=0.0720 N . A thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed, the spheres still repel, but with a force of F2=0.115 N . The Coulomb force constant is k=1/(4π????0)=8.99×109 N⋅m2/C2 . Using this information, find the initial charge on each sphere, q1 and q2 , if q1 is initially less than q2 .
Answer:
[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex]
Explanation:
[tex]F_1=0.072\ \text{N}[/tex]
[tex]F_2=0.115\ \text{N}[/tex]
r = Distance between shells = 40.4 cm
[tex]q_1[/tex] and [tex]q_2[/tex] are the charges
[tex]k[/tex] = Coulomb constant = [tex]8.99\times10^{9}\ \text{Nm}^2/\text{C}^2[/tex]
Force is given by
[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q_1q_2=\dfrac{F_1r^2}{k}\\\Rightarrow q_1q_2=\dfrac{0.072\times 0.404^2}{8.99\times 10^{9}}\\\Rightarrow q_1q_2=1.307\times 10^{-12}\\\Rightarrow q_1=\dfrac{1.307\times 10^{-12}}{q_2}[/tex]
[tex]F_2=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{F_2r^2}{k}}\\\Rightarrow q=\sqrt{\dfrac{0.115\times 0.404^2}{8.99\times 10^{9}}}\\\Rightarrow q=1.44\times 10^{-6}\ \text{C}[/tex]
[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2\times1.44\times 10^{-6}\\\Rightarrow q_1+q_2=2.88\times 10^{-6}[/tex]
Substituting the above value of [tex]q_1[/tex] we get
[tex]\dfrac{1.307\times 10^{-12}}{q_2}+q_2=2.88\times 10^{-6}\\\Rightarrow q_2^2-2.88\times 10^{-6}q_2+1.307\times 10^{-12}=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4\times \:1\times \:1.307\times 10^{-12}}}{2\times \:1}\\\Rightarrow q_2=2.32\times 10^{-6}, 5.64\times 10^{-7}[/tex]
[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{2.32\times 10^{-6}}\\\Rightarrow q_1=5.63\times 10^{-7}[/tex]
[tex]q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{5.64\times 10^{-7}}\\\Rightarrow q_1=2.32\times 10^{-6}[/tex]
Since we know [tex]q_1<q_2[/tex]
[tex]q_1=5.64\times 10^{-7}\ \text{C}[/tex] and [tex]q_2=2.32\times 10^{-6}\ \text{C}[/tex].
The speed of a space shuttle is 10 / express this in /�
Answer:
268.22m/s
Explanation:
Given;
10mile/min to m/s
We need to convert between the two units;
Using the dimensions;
1 mile = 1609.34m
60s = 1min
Now;
10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]
= 268.22m/s
In a control system, an accelerometer consists of a 4.63-g object sliding on a calibrated horizontal rail. A low-mass spring attaches the object to a flange at one end of the rail. Grease on the rail makes static friction negligible, but rapidly damps out vibrations of the sliding object. When subject to a steady acceleration of 0.832g, the object should be at a location 0.450 cm away from its equilibrium position.
Required:
Find the force constant of the spring required for the calibration to be correct.
Answer:
8.4 N/m
Explanation:
m = Mass of block = 4.63 gm
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
x = Displacement of spring = 0.45 cm
a = Acceleration of subject = 0.832g
k = Spring constant
Force is given by
[tex]F=ma[/tex]
From Hooke's law
[tex]F=kx[/tex]
So
[tex]ma=kx\\\Rightarrow k=\dfrac{ma}{x}\\\Rightarrow k=\dfrac{4.63\times 10^{-3}\times 0.832\times 9.81}{0.45\times 10^{-2}}\\\Rightarrow k=8.4\ \text{N/m}[/tex]
The force constant of the spring is 8.4 N/m.
Jolie is on the weightlifting team at her school. She must lift as much weight as possible from the ground to a standing straight position. How much work will Jolie do if she uses a force of 5N to lift 150 pounds to a height of 1.5m? 0 2.3) O 6.5) 7.5) 0 3.3)
Answer:
=7.5J
Explanation:
Step one:
given data
the applied force F=5N
the distance = 1.5m
Step two:
Required
work done by Jolie
Now by definition, the work done is the applied times the distance which the force is applied
Wd= F*D
Wd= 5*1.5
Wd=7.5J
If she uses a force of 5N to lift 150 pounds to a height of 1.5m, the work done will be 7.5J
what is gathering and analyzing information about an object without physical contact with the object
Answer:
Remote Sensing
Explanation:
A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 50 m. If he completes the 200 m dash in 29.6 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration (in m/s2) as he runs the curved portion of the track?
Answer:
The centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s²
Explanation:
Given;
distance traveled in the given time = 200 m
time to cover the distance, t = 29.6 s
speed of the runner, v = d / t
v = 200 / 29.6
v = 6.757 m/s
The centripetal acceleration of the runner is given by;
[tex]a_c = \frac{V^2}{r}[/tex]
where;
r is the radius of the circular arc, given as 50 m
Substitute the givens;
[tex]a_c = \frac{V^2}{r}\\\\a_c = \frac{(6.757)^2}{50}\\\\a_c = 0.91 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s².
What is the electric force between two point charges when 91 = -4e, 92 = +3 e,
and r= 0.05 m?
e = 1.6 x 10-1C, k = 9.00 x 10°Nom/C2)
Kouch
A. -1.1 * 10-24 N
B. 1.1 * 10-24 N
C. 5,5 x 10-25 N
D. -5.5 x 10-25 N
Answer:option B
Explanation:
You've just put a new wood floor in your house. An object will dent the flooring if the stress--the force divided by the area--exerted by the object is great enough.
A) Find the stress produced by a 50 kg woman in high-heeled shoes (assume a circular heel pad 0.50 cm in diameter) with all of her weight on one heel.
Express your answer with the appropriate units.
B) Find the stress produced by a 5000 kg African elephant (assume a circular contact area of 40 cm in diameter for one foot) standing on all four feet.
Express your answer with the appropriate units.
Answer:
The correct answer will be:
(A) 24955495.07 N/m²
(B) 97482.40 N/m²
Explanation:
(A)
The given values are:
mass,
m = 50 kg
diameter
= 0.50 cm
then,
radius,
r = 0.25 cm
= 0.0025 m
As we know,
⇒ [tex]A = area \ of \ cross \ section[/tex]
[tex]=\pi r^2[/tex]
and,
⇒ [tex]stress = \frac{mg}{A}[/tex]
On substituting the values, we get
[tex]=\frac{(50)(9.8)}{\pi (0.0025^2)}[/tex]
[tex]=24955495.07 \ N/m^2[/tex]
(B)
mass,
m = 5000 kg
radius,
r = 20 cm
= 0.2 m
Now,
⇒ [tex]stress=\frac{\frac{5000}{4} (9.8)}{\pi (0.2^2)}[/tex]
[tex]=97482.40 \ N/m^2[/tex]
If I ride my bike at 10 mph and traveled 5 miles, how long did I ride in both hours and
minutes?
What is the difference between the two graphs
Answer:
one of the graph is postion-time graph while the other one is velocity-time graph
A sample contains 20 kg of radioactive material. The decay constant of the material is 0.179 per second. If the amount of time that has passed is 300 seconds, how much of the of the original material is still radioactive? Show all work
Answer:
6,000 kg
Explanation:
What physical property does the symbol I_enclosed in problem 5 represent? a. The current along the path in the same direction as the magnetic field b. The current in the path in the opposite direction from the magnetic field c. The total current passing through the loop in either direction d. The net current through the loop
Answer:
C
Explanation:
Current passing through the loop in either direction