Answer:
transgression
Explanation:
Multiple-Concept Example 10 provides one model for solving this type of problem. Two wheels have the same mass and radius of 4.0 kg and 0.35 m, respectively. One has the shape of a hoop and the other the shape of a solid disk. The wheels start from rest and have a constant angular acceleration with respect to a rotational axis that is perpendicular to the plane of the wheel at its center. Each turns through an angle of 13 rad in 8.0 s. Find the net external torque that acts on each wheel.
Newton's second law for rotational motion allows finding the net torque for each body is:
Ring torque is: τ = 0.199 Nm Solid disc the torque is: τ = 0.995 N m
Newton's second law for rotational motion establishes a relationship between the torque, the moment of inertia, and the angular acceleration of the body.
τ = I α
Where τ is the torque, I the moment of inertia and α the angular acceleration.
They indicate that the rotated angle is θ = 13 rad in a time of 8.0 s, let's use the rotational kinematics relations.
θ = w₀ t + ½ α t²
The body starts from rest, therefore its inertial velocity is zero.
θ = ½ α t²
[tex]\alpha =\frac{2 \theta }{t^2}[/tex]
Let's calculate
α = [tex]\frac{2 \ 13^2}{8.0^2 }[/tex]
α = 0.406 rad / s²
The moments of inertia of symmetrical bodies are tabulated:
Ring I = m R² Solid disc I = ½ m R²
Let's look for every torque.
Ring
τ = m R² α
τ = 4.0 0.35² 0.406
τ = 0.199 N m
τ = ½ m R² α
τ = ½ 4.0 0.35² 0.406
τ = 0.0995 N m
In conclusion using Newton's second law for rotational motion we can find the net torque for each body is:
Ring torque is: τ = 0.199 Nm Solid disc the torque is: τ = 0.995 N m
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Saturn orbits the Sun at a radius of about 1.4×109 km. At this distance the force of gravity on Saturn due to the Sun is 3.7×1022 N. Assuming Saturn's orbit to be perfectly circular, how much work grav does the Sun's gravity do on Saturn in one year?
Explanation:Saturn is the second largest planet of the solar system in mass and size and the sixth nearest planet in distance to the Sun.
In the night sky Saturn is easily visible to the unaided eye as a non twinkling point of light.
Scoring a bogey on a hole means that you were:
1. One over par
2. One under par
3. Right on par
4. Nowhere near par
Answer: One over par
Answer:
Dear questioner;
Paddington's Answer is One over par that's he see in his book and Aunt Lucy help him
100 POINTS PLEASE answer ASAP
Answer:
uhmmm
Explanation:
uhmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
a parking lot is going to be 50m wide and 150m long. which dimensions could be used for a scale model of the lot
Answer:
5m and 15m
Explanation:
A remote-control car has batteries that store 1.8 kJ of energy. When the car is
driven across a sports field there is friction dragging on the car, producing a force
of 2 N.
Complete the following sentence: If the car is 100% efficient , it could travel a
distance of ……….metres on a single battery charge.
Answer:
Explanation:
If the car is 100% efficient , it could travel a
distance of 900 metres on a single battery charge.
1800 J = 2 N•d m
d = 900 m
The surface of the sun has a temperature of about 5800K and consists largely of hydrogen atoms. Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is 1.67×10−27kg.) The escape speed for a particle to leave the gravitational influence of the sun is given by (2GM/R)1/2, where M is the sun's mass, R its radius, and G the gravitational constant. The sun`s mass is M=1.99×1030kg, its radius R=6.96×108m and G=6.673×10−11N⋅m2/kg2. Calculate the escape speed for the sun.
This question involves the concepts of kinetic energy, escape velocity, and rms speed.
a) The rms speed of hydrogen atom at 5800 K is "11991 m/s".
b) The escape velocity for the Sun is "6.18 x 10⁵ m/s".
a)
We will use the formula of the average kinetic energy of gas molecules to find out the rms speed of the hydrogen atom. rms speed is the root mean square speed of an atom:
[tex]K.E = \frac{3}{2}KT=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{3KT}{m}}[/tex]
where,
v =rms speed = ?
K = Boltzman's Constant = 1.38 x 10⁻²³ J/k
T = absolute temperature = 5800 k
m = mass of hydrogen atom = 1.67 x 10⁻²⁷ kg
Therefore,
[tex]v=\sqrt{\frac{3(1.38\ x\ 10^{-23}\ J/k)(5800\ k)}{1.67\ x\ 10^{-27}\ kg}}[/tex]
v = 11991 m/s
b)
The escape velocity of the Sun is given by the following formula:
[tex]v_e=\sqrt{\frac{2GM}{R}}[/tex]
where,
ve = escape velocity = ?
G = Gravitational Constant = 6.673 x 10⁻¹¹ N.m²/kg²
M = Mass of Sun = 1.99 x 10³⁰ kg
R = radius of Sun = 6.96 x 10⁸ m
Therefore,
[tex]v_e=\sqrt{\frac{2(6.673\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{6.96\ x\ 10^8\ m}}[/tex]
[tex]v_e = 6.18\ x\ 10^5\ m/s[/tex]
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Three identical clay balls collide as in the diagram below. Two of the clay balls come in at equal angles of 35 degrees and equal (unknown) speeds, and strike the third (stationary) ball simultaneously. All three stick together and move off together without rotating. There are no forces on the balls of clay other than those that they exert on each other. Find the fraction of the original energy of the system that becomes thermal?
Answer:
Explanation:
Let the initial velocity of the two moving balls be u
Let the mass of one ball be m
Let the velocity after collision be v
Initial kinetic energy is
2(½mu²) = mu²
Assuming the 35° is measured above and below a common axis,
conservation on momentum.
the vertical portion of that velocity and momentum will be equal and will cancel out.
the horizontal momentum will be
m(ucos35) + m(ucos-35) + m(0) = 3mv
v = (2/3)ucos35 = 0.5461u
final kinetic energy is
½(3m)(0.5461u²) = 0.44734mu²
so the fraction of kinetic energy converted to heat energy is
(mu² - 0.44734mu²) / mu² = 0.5526599... = 0.55
a car stopped at a red light, not moving?
a) 1st Newton's law
b) 2nd Newton's law
c) 3rd Newton's law
I think it's a) 1st Newton's law... so sorry if it's wrong...
Hãy giải thích các khái niệm sau và lấy ví dụ cụ thể để minh họa:
Áp suất chân không tuyệt đối, áp suất tuyệt đối, áp suất khí trời, áp suất dư và áp suất chân
không?
Hãy giải thích các khái niệm sau và lấy ví dụ cụ thể để minh họa:
1) The unit of power is
A. Watt
B. Joules
C. Newton
D. Meter
2) How long does Samah need to do 100J of work with 50W?
A. 2s
B. 0.5s
C. 5000s
D. None of the above
Answer:
1) the unit of power is A.Watt
2)100J=50W
100=50s
divide both sides by 50s
S=2s(A)
On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 94.0 cm and diameter 2.75 cm from a storage room to a machinist. Calculate the weight of the rod, w. Assume the free-fall acceleration is g
Answer:
Explanation:
Wt = ρVg
Wt = ρ(πD²/4)hg
Wt = 7800(π0.0275²/4)0.94(9.81)
Wt = 42.72152359...
Wt = 42.7 N
Draw an energy pie chart above each of the cart’s four positions.
i. Choose your reference point for determining height.
ii. Create a pie chart that accurately represents the ratio of the three forms of energy and label the sections of the pie Eg for gravitational potential energy, Ek for kinetic energy, and Eth for thermal energy.
Answer:
Attached below! Ignore the name please. Thanks!
Explanation:
A physics class pushes a small wagon along a flat horizontal parking lot with a force of 81.17 N. Starting from rest the wagon travels 11.6 meters in 9.8 seconds. What is the wagon's acceleration?
Answer:
Explanation:
s = ½at²
a = 2s/t²
a = 2(11.6) / 9.8²
a = 0.2415660...
a = 0.24 m/s²
Acceleration of the wagon is 0.241 m/s²
Force exerted on the wagon, F = 81.17 N
Displacement of the wagon, S = 11.6 m
Time taken by the wagon to travel this distance, t = 9.8 s
Using the second equation of motion,
S = ut + 1/2 at²
where u is the initial velocity and a is the acceleration of the wagon.
Since the wagon was initially at rest, u = 0.
So,
S = 1/2 at²
Therefore acceleration of the wagon,
a = 2S/t²
a = 2 x 11.6/(9.8)²
Acceleration, a = 0.241 m/s²
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what is the angular speed, w, of the snowball after it has traveled a distance D down the slope of the roof
Newton's second law and the kinematics of rotation allow us to find the angular velocity of the snowball as it rolls across the roof is:
w = [tex]\frac{2g}{R}[/tex]
The kinematics of rotational motion studies the rotational motion of bodies.
w = w₀ + 2 α θ
Where w and w₀ are the current and initial angular velocities, α the angular acceleration and θ the angle traveled.
Newton's second law establishes a relationship between the force, mass, and acceleration of the body.
The linear and rotational moviments are related.
a = α R
Where a and α are the linear and rotational accelerations, respectively, and R is the radius of the body.
Let's find the linear acceleration of the body, in the attached we see a diagram of the forces, let's use trigonometry to decompose the weight.
[tex]sin \theta = \frac{W_x}{W}[/tex]
Wₓ = W sin θ
Wₓ = m a
mg sin θ = m a
a = g sin θ
Now we can find the angular acceleration.
α = a / R
α = [tex]\frac{g}{ R \ sin \theta }[/tex]
The body is released therefore its initial velocity is zero, we substitute in the kinematics expression.
[tex]w = 2 ( \frac{g}{R \ sin \theta }) \ \theta[/tex]
in rotational motion the angles are measured in radians We use trigonometry to find the relationship between the angle and the distance traveled
[tex]\theta = \frac{h}{D}[/tex]
Where h is the height of the ceiling and D is the distance traveled. Let's substitute.
[tex]w = 2 \frac{g}{R sin \theta } \frac{h}{D}[/tex]
Let's tirgonmetry.
sin θ = [tex]\frac{h}{D}[/tex]
w = [tex]\frac{2g}{R}[/tex]
In conclusion, using Newton's second law and the kinematics of rotation, we can find the angular velocity of the snowball when rolling on the roof is:
w = [tex]\frac{2g}{R}[/tex]
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A jumbo jet needs to reach a speed of 364 km/hr on the runway for takeoff. Assuming a constant acceleration and a
runway 1.5 km long, what minimum acceleration (in units of m/s2 ) from rest is required?
Answer:
Diaplacement s=1.80km
Final velocity v=360km/h
It starts from rest. So, initial velocity u=0
We need to find the constant acceleration (a).
using the equation of motion v
2
=u
2
+2as
360
2
=0
2
+2×a×1.8
a=
3.6
360×360
a=36000km/h
2
Could a car drive on a frictionless surface? Explain using the terms action
force and reaction force. *
Answer:
No, it cannot. The car needs the friction of the surface to drive because the car pushes the surface backwards, and the surfaces makes a reaction force pushing the car forward, and that works because of the friction. In a frictionless surface the tires would rotate in the same place
Two blocks, which can be modeled as point masses, are connected by a massless string which passes through a hole in a frictionless table. A tube extends out of the hole in the table so that the portion of the string between the hole and M1 remains parallel to the top of the table. The blocks have masses M1 = 1.9 kg and M2 = 2.8 kg. Block 1 is a distance r = 0.95 m from the center of the frictionless surface. Block 2 hangs vertically underneath. find the speed of m1 assume m2 does not move relative to the table.
The speed of the block m1 on the frictionless table is 1.34 m/s.
The given parameters;
mass of the first block, m1 = 1.9 kgmass of the second block, m2 = 2.8 kgdistance of block m1, R = 0.95 mThe net torque on both blocks is calculated as;
[tex]\tau _{net} = I \alpha[/tex]
[tex]T_2R- T_1 R_1 = I \alpha \\\\[/tex]
where;
T₁ is the tension on first blockI is the moment of inertia of point massα is the angular acceleration[tex]T_1 = m_1 g + m_1 a\\\\T_2 = m_2 g - m_2 a[/tex]
The acceleration of both blocks is calculated as follows;
[tex]R(T_ 2- T_1) = MR^2 \times (\frac{a}{R} )\\\\R(T_2 -T_1) = MRa\\\\T_2 - T_1 = Ma\\\\(m_2g - m_2 a) - (m_1 g + m_1 a) = Ma\\\\m_2 g - m_1 g - m_2 a - m_1 a = Ma\\\\g(m_2 - m_1) = Ma + m_2a+ m_1a\\\\g(m_2 - m_1) = a(M+ m_2 + m_1)\\\\where;\\\\M \ is \ mass \ of \ string = 0 \\\\g(m_2 - m_1) = a (0+ m_2 + m_1)\\\\g(m_2 - m_1) = a(m_1 + m_2)\\\\a = \frac{g(m_2 - m_1)}{m_1 + m_2} \\\\a = 1.88 \ m/s^2[/tex]
The speed of the block m1 is calculated as follows;
[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{a r} \\\\v = \sqrt{1.88 \times 0.95} \\\\v = 1.34 \ m/s[/tex]
Thus, the speed of the block m1 on the frictionless table is 1.34 m/s.
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Displacement (m) 0 20 40 40 80 80 60 400 Time (s) 10 10 20 30 40 50 60 70 80 a) Explain the different sections of the graph in as much detail as you can. b) Use the graph to determine the maximum velocity. c) Find the average velocity after 45 s. d) Find the instantaneous velocity at 45 s.
Answer:
because I dont know
Explanation:
first you add the multiply
The instantaneous velocity at 60 m is zero, at - 40 m is - 1 m / s, at 15 s is zero and at 25 s is 0.8 m / s if Displacement (m) 0 20 40 40 80 80 60 400 Time (s) 10 10 20 30 40 50 60 70 80.
What is Instantaneous velocity?Instantaneous velocity = Position with respect to time / Time
1 ) At d = 60 m, the instantaneous velocity is zero because the car is at rest.
2 ) At d = - 40 m, t = 40 s
Instantaneous velocity = - 40 / 40
Instantaneous velocity = - 1 m / s
3 ) At t = 15 s, d = 60 m
At 60 m, the car is at rest, so the Instantaneous velocity is zero
4 ) At t = 25 s, d = 20 m
Instantaneous velocity = 20 / 25
Instantaneous velocity = 0.8 m / s
Instantaneous velocity of a given curve in a position-time graph can be found by drawing a tangent to the curve and finding the slope of the tangent. Instantaneous velocity = 0
Instantaneous velocity = - 1 m / s
Instantaneous velocity = 0
Instantaneous velocity = 0.8 m / s
Therefore, The instantaneous velocity at 60 m is zero, at - 40 m is - 1 m / s, at 15 s is zero and at 25 s is 0.8 m / s if Displacement (m) 0 20 40 40 80 80 60 400 Time (s) 10 10 20 30 40 50 60 70 80.
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A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters. Which of the following best represents the final speed of the race car?
Answer:
Explanation:
Givens
vi = 10 m/s
a = 1.5 m/s^2
d = 600 m
vf = ?
Formula
vf^2 = vi^2 + 2*a*d
Solution
vf^2 = 10^2 + 2*1.5 * 600
vf^2 = 100 + 1800
vf^2 = 1900
sqrt(vf^2) = sqrt(1900)
vf = 43.59 m/s
How high is the image of a 1.5 mm high object that is magnified by a 20X lens system?
Answer:
Explanation:
1.5(20) = 30 mm
A solid sphere with a radius of 5cm initially at rest rolls 270cm down a slope 37° above the horizontal. What is its velocity (rad/s) at that point? [PS: it has not reached the foot of the slope]
This question involves the concepts of the law of conservation of energy, kinetic energy, potential energy, and moment of inertia.
The angular velocity of the solid sphere at this point is "95.44 rad/s".
According to the law of conservation of energy:
Loss in Potential Energy = Gain in Rotational Kinetic Energy + Gain in Translational Kinetic Energy
[tex]mgh = \frac{1}{2}I\omega^2+\frac{1}{2}mv^2[/tex]
where,
m = mass of sphere
g = acceleration due to gravity = 9.81 m/s²
h = loss in height = (270 cm) Sin 37° = 162.5 cm = 1.625 m
r = radius of sphere = 5 cm = 0.05 m
ω = angular speed of sphere = ?
v = linear speed of sphere = rω
I = moment of inertia of sphere = [tex]\frac{2}{5}mr^2[/tex]
Therefore,
[tex]mgh=\frac{1}{2}\frac{2}{5}mr^2\omega+\frac{1}{2}m(r\omega)^2\\\\gh=\frac{1}{5}r^2\omega^2+\frac{1}{2}r^2\omega^2\\\\(9.81\ m/s^2)(1.625\ m)=(\frac{1}{5}+\frac{1}{2})(0.05\ m)^2\omega^2\\\\\omega=\sqrt{\frac{15.941\ m^2/s^2}{0.00175\ m^2}}[/tex]
ω = 95.44 rad/s
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The attached picture explains the law of conservation of energy.
Analog signals and storage are used in most modern electronics.
True
False
Analog signals and storage are used in most modern electronics is a false statement.
Analog signals and storage are not used in most modern electronics because analog signals are used in old instruments and technologies while on the other hand, in modern electronics digital signals and storage system are used that have higher speed, compatibility and space for storage.
Analog signals and storage were used in old electronics and can't be used in modern electronics due to non-compatibility so we can conclude that digital signals and storage are used in modern electronics.
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A car driving at 32.6 miles/hours slowed to a stop in 2.4 seconds
The driver's acceleration was
miles/hour second. Record your answer to two places past the decimal point
Explanation:
using v=u+at
0=32.6 + (a×(2.4/3600))
-32.6=(a×(2.4/3600))
a=-32.6×3600/2.4
a= -48900miles/hr^2
Answer each of the following questions regarding energy and heat transfer: 1. Explain why it is important to understand conductors. 2. Give a real world example where thermal conductors are used. 3. Explain why it is important to understand insulators. 4. Give a real world example where thermal insulators are used. 5. Suppose you do not have electricity or gas for heating up food and you are hungry. Explain what would you do to maximize thermal energy transfer in this situation to safely cook you something to eat.
si Ana camina sobre una cuerda que la sostiene dos edificios separados 10m, la soga tiene un angulo de 10grados, La masa de Ana es de 50kg ¿cual es la tension de la cuerda?
As a passenger balloon rises, its gas bag tends to A. Become smaller B. Leak C. Distort D. Expand E. Remain unchanged
Answer:
I think it expands
Explanation:
Answer:
E. remain unchanged
Explanation:
A rotating furnace has been developed recently to give a rough para-
bolic curve to molten glass. This makes the manufacture of very large
telescope mirrors easier and more economical than ever before. The
largest mirror made in this furnace to date has a radius of 4.2 m. While
in the furnace, the centripetal acceleration of the molten glass for this
mirror was 2.13 m/s?. What was the tangential speed at the edge of the
molten glass?
The definition of centripetal acceleration allows to find the result for the linear velocity of the body is:
v = 2.99 m / s
Kinematics studies the movement of bodies, the centripetal acceleration is the acceleration that a body has where all the energy is used to change the direction of the speed, leaving its modulus constant.
[tex]a= \frac{v^2}{R}[/tex]
Where a is the acceleration ccentripette, v the modulus of the velocity and R the radius of gyration.
They indicate that the radius is R = 4.2 m and the centripetal acceleration is a = 2.13 m / s, let's find the modulus of the velocity.
[tex]v= \sqrt{a R}[/tex]
Let's calculate
v = [tex]\sqrt{2.13 \ 4.2}[/tex]
v = 2.99 m / s
In conclusion using the definition of centripetal acceleration we can find the result for the tangential velocity of the body is:
v = 2.99 m / s
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A 2 kg block is in equilibrium on a 36 degree incline. What is the normal force acting on the block?
15.9 N
Explanation:
Let's assume that the downward direction on the inclined is the (+)-direction. Since the block is in equilibrium, the x-component of its weight is pointing in the +x-direction and the frictional force [tex]f_s[/tex] is pointing up the incline. So the net force acting parallel to the incline can be written as
[tex]mg\sin36 - f_s = 0 \Rightarrow \mu N = mg\sin36[/tex]
where N is the normal force.
The net force perpendicular to the incline can be written as
[tex]N - mg\cos36 = 0 \Rightarrow N = mg\cos36[/tex]
or
[tex]N = (2\:\text{kg})(9.8\:\text{m/s}^2)\cos36 = 15.9\:\text{Newtons}[/tex]
At what speed does the squid eject the water?
Answer:
3.5 m/s
Explanation:
