Matter is composed of elementary particles i.e. quarks and leptons.
Matter is composed of elementary particles i.e. quarks and leptons.Matter is composed of elementary particles which is called quarks and leptons. Quarks consist of protons, neutrons and electrons. All observable matter is made up of up quarks, down quarks and electrons.
Matter is composed of elementary particles i.e. quarks and leptons.Matter is composed of elementary particles which is called quarks and leptons. Quarks consist of protons, neutrons and electrons. All observable matter is made up of up quarks, down quarks and electrons.Lepton is an elementary particle consist of half-integer spin that does not undergo strong interactions. Leptons exist on two main classes i.e. charged leptons, and neutral leptons. Electron, electron neutrino, muon, muon neutrino, tau and tau neutrino are the six types of leptons.
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...
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
What is the value of the magnitude of the difference of vectors A and B, |A - B|?
Answer:
Let, θ be the angle between vector a and vector b.
So, the angle between vector a and vector (-b) must be (180° - θ).
Magnitude of the difference of vectors a and b
= |a - b|
= Magnitude of the resultant of the vector sum of the vectors a and (-b)
= |{√|a|^2 + |b|^2 + 2 * |a| * |b| * cos (180° - θ)}|
= |{√|a|^2 + |b|^2 - 2 * |a| * |b| * (cos θ)}|.
Explanation:
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
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:
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.
A girl sitting on a beach counts six waves passing a buoy in 3.0 s. She measured the distance between the wave crests to be 1.5 meters. What is the speed of the waves the girl observed?
[tex]\\ \sf\longmapsto Frequency=\dfrac{1}{Time}=\dfrac{1}{3}=0.33s^{-1}[/tex]
Velocity be v
[tex]\\ \sf\longmapsto v=f\lambda[/tex]
[tex]\\ \sf\longmapsto V=0.33(1.5)[/tex]
[tex]\\ \sf\longmapsto V\approx 0.5ms^{-1}[/tex]
Given,
Wavelength = 1.5 metres
T = 3 seconds OR 3.0 seconds
We know that the formula of Frequency–
1/Time
According to the question we have seen that Time is 3 seconds
[tex] \sf \longmapsto \: Frequency = \frac{1}{3 .0 \: seconds} [/tex]
[tex] \sf \longmapsto \: Frequency = 0.33 seconds^{-1} [/tex]
We Know,
Velocity = Frequency
Solving Further[tex]\sf \longmapsto Velocity =0.33 \times 1.5 [/tex]
[tex]\sf \longmapsto \: 0.5 \: m/s ^{-1}[/tex]
Therefore the speed is[tex]\sf \: 0.5 \: m/s ^ {-1}[/tex]Imagine you and a friend are trying to rearrange the furniture in your classroom. You push on a desk with a force of 50 N to the right. Your friend pushes on the same desk with a force of 50 N to the left. What is the net force on the desk?
Answer:
The net force on the desk is zero.
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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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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what makes the morion mask expensive ?
Answer:
a lot of the masks are hand made a long time ago its really an art
Explanation:
Because it is a piece of art that is hard to craft
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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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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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:
A car equipped with a massless spring in front collides with a stationary car as shown in the diagram. The mass of the incoming car is three times the mass of the target car. While the spring is being compressed, the cars are moving closer together, and while it is expanding they are moving farther apart, but at the instant that the spring is fully compressed, they have no relative motion, which means they are moving at the same speed. At this instant, the cars are moving with a speed of 2.4 m/s.
(a) Find the speed of the incoming car before the collision.
(b) Find the fraction of the system’s energy that is stored in the spring when
it is fully compressed.
(c) Find the final speed and direfction for both cars after the spring expands and the cars separate.
Answer:
Explanation:
Let m be target car mass
a)
conservation of momentum
(3m)u + m(0) = (3m + m)(2.4)
3u = 4(2.4)
u = 3.2 m/s
b)
system energy is originally in the striking car as kinetic energy
KE = ½(3m)3.2² = 15.36m J
at full compression, kinetic energy is
½(3m + m)2.4² = 11.52m J
so spring potential energy fraction of total energy is
(15.36 - 11.52)/15.36 = 0.25 or 25%
c)
The center of mass (CoM) is moving at 2.4 m/s
The CoM sees the larger mass car approach at
3.2 - 2.4 = 0.8 m/s and will see it depart at - 0.8 m/s
so a ground based observer would see it moving at
v₁ = 2.4 + -0.8 = 1.6 m/s in its original direction
The CoM sees the smaller mass car approach at
0 - 2.4 = -2.4 m/s and will see it depart at + 2.4 m/s
so a ground based observer would see it moving at
v₂ = 2.4 + 2.4 = 4.8 m/s in its original direction of the larger car
so the relative velocity of approach at 3.2 - 0 = 3.2 m/s
equals the relative velocity of departure at 4.8 - 1.6 = 3.2 m/s
checks.
what is the distance between two corresponding points of adjacent waves?
Answer:Wavelength
Explanation: The wavelength of a wave is the distance between any two corresponding points on adjacent waves.
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.
Please explain what the passage mean in your own words.
Remember the Sabbath day, to keep it holy. Six days shalt thou labor, and do all thy work: But the seventh day is the Sabbath of the Lord thy God: in it thou shalt not do any work, thou, nor thy son, nor thy daughter, thy manservant, nor thy maidservant, nor thy cattle, nor thy stranger that is within thy gates: For in six days the Lord made heaven and earth, the sea, and all that in them is, and rested the seventh day: wherefore the Lord blessed the Sabbath day, and hallowed it.
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
Helium gas is confined within a chamber that has a moveable piston. The mass of the piston is 8.7 kg; and its radius is 0.013 m. If the system is in equilibrium, what is the pressure exerted on the piston by the gas
Helium gas is in a chamber with a moveable piston, whose mass is 8.7 kg and its radius is 0.013 m. The pressure exerted on the piston by the gas is 1.6 × 10⁵ Pa.
Helium is confined in a chamber with a moveable 8.7-kg piston (m). We can calculate the weight of the piston (w) using Newton's second law of motion.
[tex]w = m \times g = 8.7kg \times 9.8 m/s^{2} = 85 N[/tex]
The radius of the piston (r) is 0.013 m. We can calculate the area of the piston (A) using the following expression.
[tex]A = \pi \times r^{2} = \pi \times (0.013)^{2} = 5.3 \times 10^{-4} m^{2}[/tex]
We can calculate the pressure exerted by the piston (P) using the following expression.
[tex]P = \frac{w}{A} = \frac{85N}{5.3 \times 10^{-4}m^{2} } = 1.6 \times 10^{5} Pa[/tex]
Since the system is in equilibrium, the pressure exerted on the piston by the gas is also 1.6 × 10⁵ Pa.
Helium gas is in a chamber with a moveable piston, whose mass is 8.7 kg and its radius is 0.013 m. The pressure exerted on the piston by the gas is 1.6 × 10⁵ Pa.
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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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#SPJ2
100 POINTS PLEASE answer ASAP
Answer:
uhmmm
Explanation:
uhmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
name a suitable thermometer for a winter temperature at noth pole?
following the new year and it will not have a great day and service offering the best way to our attention that you are interested please send the following URL Adobe Flash player to by the following link unsubscribe the new year and the new Sprint network from the new one for the new one and the following URL to the new year is available for remote playback is most convenient time ago the new one is available for the new one is available using to our u to the followings to our to our attention the information contained IN the product and the product and it will take a look the following URL Adobe Flash memory of a great day and it was great to our to by the the following link to the new year and the following link unsubscribe if possible I would like a few minutes late and I have been working in lowest terms if I can be in touch and it was great meeting with the new year is available for remote playback not allowed because the new year is available using this email and the following URL Adobe
jjjjjjjjjjjoooooooooooooooooo
Answer:
.
Explanation:
g The human ear can perceive sounds with frequencies ranging from 20.0 Hz to 20000 Hz. If the speed of sound is 330 m/s in air and 1500 m/s under water, what is the longest and shortest wavelength that can be heard in air and in water
Answer: lamda max= 16,5m lamda min= 0,0165m (air)
lamda max= 75m, lamda min=0,075m (water)
Explanatin
A worker pushes 1500N crate with a horizontal force of 345N a distance of 24m.
How much work is done by the Floor on a crate?
Answer:
8280J
Explanation:
W = f x d
345 x 24 = 8280J
If a car has a suspension system with a force constant of 5.00x104 N/m, how much energy must the car's shocks remove to dampen an oscillation starting with a maximum displacement of 0.0750 m
Answer: 140.625
Explanation: Because energy gained due to damped motion and needs to be absorbed
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
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:
Complete the following sentence: If the car is 100% efficient, it could travel a distance of 900 meters on a single battery charge.
W = Fd
1800 = 2d
d = 900
Assuming 100 % efficient means turning battery power into motion.
A car totally 100% efficient would have zero friction and could travel an infinite distance on a full battery charge.
Please help on question e and F
E- what will be the total charge of the compound formed?
F what type of band will form?
Answer:
E-MgO(Magnesium Oxide)
