Answer:
the quantity of rotation of a body, which is the product of its moment of inertia and its angular velocity.
angular momentum : The property of rotating object given by radius times linear momentum.
L = r x p
p = linear momentum
or
L = I x ω
I = inertia
ω = angular velocity
BRAINLIETS IF CORRECT
What does Hess's law say about the enthalpy of a reaction? A. The enthalpy of a reaction does not depend on the reactant path taken. B. The enthalpy of a reaction depends on the pathway the reactants followed See SUS C. The sum of the enthalpy and entropy is the free energy of a reaction. O D. The entropy of a reaction is the sum of the enthalpies of intermediate steps.
Answer: B
Explanation: I think its B or A but mostly B
Answer:
The enthalpy of a reaction does not depend on the reactant path taken
Explanation:
i just took the test on a pex :)
After being accelerated to a speed of 1.17×105 m/s , the particle enters a uniform magnetic field of strength 0.800 T and travels in a circle of radius 35.0 cm (determined by observing where it hits the screen as shown in the figure). The results of this experiment allow one to find m/q .
Find the ratio m/q for this particle.
Express your answer numerically in kilograms per coulomb.
The ratio m/q for this particle is mathematically given as
m/q=2.39*10^{-6}
What is the ratio m/q for this particle?Question Parameters:
After being accelerated to a speed of 1.17×105 m/s
the uniform magnetic field of strength 0.800 T and travels in a circle of radius 35.0 cm
Generally, the equation for the Centrifugal force is mathematically given as
[tex]qvB=\frac{mv^2}{r}[/tex]
Therefore
m/q=(0.8*0.35)/(1.17*10^{5})
m/q=2.39*10^{-6}
In conclusion, the ratio m/q for this particle
m/q=2.39*10^{-6}
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HELPPPPPP PLEASEEEEEEE-
Answer:
One piece has a north pole only, and the other piece has à soutn pole only.
Explanation:
mark me brainliest!!
Answer: B
Explanation:
When we split a magnet in half, both halves act as magnets, with each having a north and south pole. As with the original magnet, the matching north and south poles remain on the sides as in the original magnet. This means that the 2 broken sides must be polarized in the opposite direction.
A ball has a mass of 0.5 kg. Dropped from a cliff top. the ball hits the sea below at a speed of 10 m/s. What was the ball's gravitational potential energy before it was dropped?
Answer:
25 J
Explanation:
Using Law of Conservation of Energy
Total Energy before = Total Energy After
Potential Energy1 + Kinetic Energy1 = Potential Energy2 + Kinetic Energy2
Potential Energy1 + 0 = 0 + Kinetic Energy2
Potential Energy1 = [tex]\frac{1}{2}*m*v^{2}[/tex] =0.5*0.5*10*10= 25 J
what is polygon of force
Answer:
a closed polygon whose sides taken in order represent in magnitude and direction a system of forces in equilibrium.
Explanation:
Answer: a closed polygon whose sides taken in order represent in magnitude and direction a system of forces in equilibrium.
Explanation: Hope that helps.
How many oxygen atom has protein
Answer:
1 or 2
Explanation:
Compared to RNA and DNA backbone, protein backbone has a relatively simple chemical structure - a nitrogen atom, two carbon atoms, one or two oxygen atoms, and a few hydrogens.
You drive in a straight line at 18.0 m/s m / s for 10.0 miles, then at 34.0 m/s m / s for another 10.0 miles.
Answer:
1 mile = 1609 meters
10 miles = 16090 meters
16090m/(24.0m/s) = 670s
16090m/(30.0m/s) = 536s
Average Speed = Total distance/Elapsed time
Average Speed = [16090m(2)]/(670s+536s) = 32080m/1206s = 26.7m/s
(a) The car's average speed is less than 27.0m/s
(b) The car's average speed is 26.7m/s
Explanation:
1 mile = 1609 meters
10 miles = 16090 meters
16090m/(24.0m/s) = 670s
16090m/(30.0m/s) = 536s
Average Speed = Total distance/Elapsed time
Average Speed = [16090m(2)]/(670s+536s) = 32080m/1206s = 26.7m/s
(a) The car's average speed is less than 27.0m/s
(b) The car's average speed is 26.7m/s
Using the following information, calculate the theoretical lower limit of electricity consumption for melting scrap steel in terms of [kWh/ton]. The melting temperature of steel is about 1,500 [℃] and the heat capacity of steel is around 0.466 [J/g℃]. The latent heat of the fusion of steel is approximately 276 [J/g]. How far from this limit were EAF’s by the year 2000?
The theoretical electricity consumption is 268.179 kilowatt-hours per metric tonne.
The theoretical minimum of electricity consumption is 60.340 % of the practical minimum of electricity consumption.
Comparison of theoretical electricity consumption for steel production by electric arc furnace with real consumptionThe theoretical electricity consumption is equal to the sum of sensible heat ([tex]q_{s}[/tex]) and latent heat ([tex]q_{l}[/tex]) needed to melt a metric ton of steel, in kilowatt-hours per metric tonne:
[tex]q = q_{s}+q_{l}[/tex] (1)
By definitions of sensible and latent heat, we have the following formulas:
[tex]q = c\cdot (T_{f}-T_{o}) + L_{f}[/tex] (1)
Where:
[tex]c[/tex] - Specific heat of steel, in kilowatt-hours per metric ton-degree Celsius.[tex]T_{o}[/tex] - Initial temperature, in degrees Celsius.[tex]T_{f}[/tex] - Final temperature, in degrees Celsius.[tex]L_{f}[/tex] - Latent heat of fusion of steel, in kilowatt-hours per metric ton.If we know that [tex]c = 1.294\times 10^{-1}\,\frac{kWh}{ton\cdot ^{\circ}C}[/tex], [tex]T_{o} = 20\,^{\circ}C[/tex], [tex]T_{f} = 1500\,^{\circ}C[/tex] and [tex]L_{f} = 76.667\,\frac{kWh}{ton}[/tex], then the theoretical electricity consumption is:
[tex]q = \left(1.294\times 10^{-1}\,\frac{kWh}{ton\cdot ^{\circ}C} \right)\cdot (1500\,^{\circ}C-20\,^{\circ}C)+76.667\,\frac{kWh}{ton}[/tex]
[tex]q = 268.179\,\frac{kWh}{ton}[/tex]
The theoretical electricity consumption is 268.179 kilowatt-hours per metric tonne. [tex]\blacksquare[/tex]
According to the document "Theoretical Minimum Energies to Produce Steel for Selected Conditions" done by R.J. Fruehan et al. in March 2000. The practical minimum of electricity consumption for a electric arc furnace (EAF) is:
[tex]q_{real, min} = 444.444\,\frac{kWh}{ton}[/tex]
This indicator consider common energy losses throughout steel production.
The theoretical minimum of electricity consumption is 60.340 % of the practical minimum of electricity consumption. [tex]\blacksquare[/tex]
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What is the difference between a sonic boom and a shock wave?
Answer:
When a sound source moves faster than the speed of sound, a shock wave is produced as the sound waves interfere. A sonic boom is the intense sound that occurs as the shock wave moves along the ground.
More importantly, all of the energy gets concentrated into a very small distance this is called a shock wave. In this case, the observer does not hear the approaching source at all until the shock wave hits with all of the energy in the wave. For sound waves, this can cause a very loud noise, called a sonic boom.
Sonic booms produced by aircraft flying supersonic at altitudes of less than 100 feet, creating between 20 and 144 pounds overpressure, have been experienced by humans without injury. Damage to eardrums can be expected when overpres- sures reach 720 pounds
Explanation:
marke as brainliest
If a student weighs 100 pounds on Earth, how much more or less would she weigh on Pluto than Mars?
A. She would weigh 38 pounds less on Pluto
B. She would weigh 38 pounds more on Mars
C. She would weigh 31 pounds more on Mars
D. She would weigh 31 pounds less on Mars
Answer:
c because that's the answer
She would weigh 31 pounds more on Mars than on Pluto. Therefore option C is correct.
To calculate the differences in weight using the given information.
1. Weight on Earth: 100 pounds
Now, let's calculate the weight differences for Pluto and Mars:
2. Weight on Pluto:
Pluto's gravity is about 0.06 times that of Earth's.
Weight on Pluto = Weight on Earth × Pluto's gravity
Weight on Pluto = 100 pounds × 0.06 = 6 pounds
3. Weight on Mars:
Mars' gravity is about 0.38 times that of Earth's.
Weight on Mars = Weight on Earth × Mars' gravity
Weight on Mars = 100 pounds × 0.38 = 38 pounds
Comparing the weight differences:
Weight difference between Earth and Pluto = 100 pounds - 6 pounds = 94 pounds
Weight difference between Earth and Mars = 100 pounds - 38 pounds = 62 pounds
Therefore, the correct answer is C. She would weigh 31 pounds more on Mars than on Pluto.
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what is propagation
Nira here :)
Answer ASAP pls
[tex]\large{\underline{\underline{\pmb{\frak {\color {red}{Question:}}}}}}[/tex]
[tex] \sf \red{what \: is \: propagation}[/tex]
[tex]\large{\underline{\underline{\pmb{\frak {\color {blue}{Answer:}}}}}}[/tex]
Propagation is an asèxual method of reproduction. In propagation, the DNA from the parents is taken to form another organism.
[tex]\green\star[/tex]There are two types of propagation:
Vegetative propagation Artificial propagation.1. Vegetative propagation: In vegetative propagation, new plants are obtained from the parts of old plants (like stem, roots and leaves), without the help of any reproductive organs.
2. Artificial propagation: Human grow by some man made methods are called artificial propagation.
[tex] \boxed{ \frak \red{brainlysamurai}}[/tex]
l
If iron filings were sprinkled evenly across the
entire paper circle, at which location would
the
grealest concentration of iron filings be found after
30 seconds?
While traveling home at dusk, a motorcyclist gets on the highway and increases the combined mass (400 kg
for the motorcycle and 150 kg for the motorcyclist) from 30 mi/hr to 70 mi/hr in 3 seconds. While struggling to
clean his helmet later that evening, he wonders what would have been the acceleration of one of those bugs
striking his helmet if it had a mass of 5 g.
The acceleration of one of those bugs is equal to 305mi/s.
Acceleration calculation
To calculate the insect's acceleration, the action and reaction force of the impact must be considered.
As the insect will hit the helmet, the force it hits is the same force it receives, so we can make the following expression:
[tex]m_m \times a_m = m_b \times a_b[/tex]
[tex]550 \times 0.0027 = 5 \times 10^{-3} \times a_b[/tex]
Speed has been converted to miles per second
[tex]a_b = 305 mi/s[/tex]
So, the acceleration of one of those bugs is equal to 305mi/s.
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How might having a testing site like the "Mars Yard" be helpful in mission planning? Provide an example of how this might help scientists solve a problem during a mission.
Your answer should include at least two complete sentences.
Be sure to check your grammar and spelling.
We can say that having a testing site such as the mars yard helps scientists to test previously theoretical knowledge in a realistic environment.
Why are testing sites important?Testing sites help scientists by providing a realistic place to test knowledge that may have previously been only theoretical through performing experiments on such test sites. This is of vital importance in the detection of errors in order to achieve a successful future mission.
Therefore, we can confirm that having a testing site such as the mars yard helps scientists to test previously theoretical knowledge in a realistic environment.
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Data
Table A
Record your data either in your lab notebook or in the tables below.
Cart Speed
Cart Speed
Elapsed Time
(s)
(Low fan speed) (Medium fan speed)
(cm/s)
(cm/s)
16.4
23
Cart Speed
(High fan speed)
(cm/s)
31.7
1
31.5
64.0
2
54
36
89.8
3
118.1
4
5
6
7
The speed will be calculated by dividing the distance traveled by the object by the time elapsed.
What is speed?Based on the information given, an overview will be given to you on how to calculate the speed of the cart.
Average speed is a scalar quantity and is gotten by dividing the distance by time taken. For example, if the distance traveled is 100km and the time taken is 4 hours. The speed will be:
= Distance/Time
= 100/4.
= 25km/h.
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Give a reason, why the bulb is not glowing in the given circuit A. O Closed circuit B. o Complete circuit C. O Incomplete circuit D. O Cell is not connected. Previous Next
Answer:
Closed circuit
Explanation:
See the switch is not connected with circuit.
It's still off So the bulb is not glowing
Since antiquity people have used the sling to increase the speed of a rock and send it swiftly in a specific
direction. While the rock is being spun overhead, the force that keeps the rock moving in a circle is provided by
the tensile strength of the sling material. Leather has a fairly high tensile strength, so that a strip of leather with
a cross-sectional area of 0.25 cm2can withstand a pulling force of 800 N. Assume that, for a certain sling, 8.00
10²N is the largest possible centripetal force. If the rock in the sling is 0.40 m from the center of rotation and has
a tangential speed of 6.0 m/s, what is the largest mass the rock can have?
Show Your Work
Answer:
"F = 52.93 [N]
Since the Rock is thrown upwards, there are only two forces, one is the force of the rubber (55[N]) and the other is the force of gravitational acceleration acting in the opposite direction to the movement of the rock, therefore this acceleration will be taken as negative.
F = 55 - m*g
where:
F = total force [N] (units of Newtons)
m = mass = 0.21 [kg]
g = gravity acceleration = 9.81 [m/s²]
F = 55 - (0.21*9.81)
F = 52.93 [N]"
precaution taken to ensure accuracy
A ball with a volume of 3 cm3 and a density of 9.0 g/cm3 increases its velocity from 2 m/s to 6 m/s over a 12
second period of time. What force was applied to get it moving?
Answer:
M = ρ V = 9 gm/cm^ 3 * cm^3 = 27 gm
a = (V2 - V1) / t = (6 - 2) m/s / 12 s = 1/3 m/s^2 the acceleration
F = M a = 27 gm * 1/3 m/s^2 = 9 dynes net force applied
A 1kg blob of clay moving at 8 m/s collides inelastically with a 3 kg woode a. What is the initial momentum of the system? 6 kg m/s 9 kg m/s 8 kg m/s 4 kg m/s 4 s
The momentum of the blob of clay before collision which is the same as the initial momentum of the system is 8kgm/s.
Given the data in the question;
Mass of the blob of clay; [tex]m_b = 1 kg[/tex]Velocity of the blob of clay; [tex]v_b = 8m/s[/tex]Mass of wood; [tex]m_w = 3kg[/tex]Initial momentum of the system; [tex]P = \ ?[/tex]
MomentumMomentum is simply the product of the mass of an object and its velocity.
Its is expressed as;
[tex]P = m * v[/tex]
Where m is the mass of the object and v is its velocity.
To determine the initial momentum which is the momentum of the blob of clay before collision, we substitute our given values into the expression above.
[tex]P = m * v\\\\P = 1kg * 8m/s\\\\P = 8kg.m/s[/tex]
Therefore, the momentum of the blob of clay before collision which is the same as the initial momentum of the system is 8kgm/s.
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Determine the magnitude and direction of vector V2 such that V=2V1 + V2 acts along the positive z-axis having a magnitude of 20 from which vector V = 20i- 15j + 15k.
Answer:
Explanation:
V₁ = <20, -15, 15>, V = <0, 0, 20>
V₂ = V - 2V₁ = <0, 0, 20> - <40, -30, 30> = <-40, 30, -10> = -40i + 30j - 10k
magnitude = [tex]\sqrt{40^2+30^2+10^2}}=51[/tex]
how much pressire is created when a force of 55n is applied over sn area of 8m²
how much pressure is created when a force of 55N is applied over an area of 8m²
SolutionWe have been asked to determine the pressure exerted when given magnitude of force acting on given area.
[tex]☄\bf \qquad \mathcal{Pressure} = \mathcal{\dfrac{Force}{Area}}[/tex]
Put all given value we obtain
[tex]\qquad \sf \longrightarrow \:Pressure \: = \dfrac{55}{8} \\ \\ \\ \qquad \sf \longrightarrow \:Pressure \: = \cancel\dfrac{55}{8} \\ \\ \\ \qquad \sf \longrightarrow \:Pressure \: = 6.875 \: Pa.[/tex]
Therefore,
6.875 Pascal Pressure is created .L4 What is meant by the internal energy of an object?
Answer:
Internal energy is all the energy contained in an object, including both kinetic energy and potential energy
Select the correct answer.
in which situation is no work being done?
А
a person carrying a box from one place to another
В.
a person picking up a box from the ground
C.
a person pushing a box from one place to another
D. a person pulling a box from one place to another
A 1,550 kg roller coaster cart is pulled at a constant speed of 2.3 m/s to the top of a 53.5 m tall ramp
that has a 30° incline. The coefficient of friction between the ramp and the roller coaster cart is
M = 0.075.
tow cable
53.5 m
Calculate the work done by friction on the roller coaster cart during its motion Jup the ramp.
Write your answer to three significant figures.
kJ
The work done by friction on the roller coaster cart during its motion up the ramp is -813 kJ
Conservation of energyThe total mechanical energy at the bottom of the tall ramp, E equals the total mechanical energy at the top of the tall ramp, E'
E = E'
U + K + W = U' + K' + W' where
U = initial potential energy of block = mgh , K = initial kinetic energy of block = 1/2mv², W = work done by friction at bottom of rampU = final potential energy of block = mgh', K' = final kinetic energy of block = 1/2mv'² and W' = work done by friction at top of rampSo,
mgh + 1/2mv² + W = mgh' + 1/2mv'² + W' where
m = mass of block = 1550 kg, g = acceleration due to gravity = 9.8 m/s², h = initial height of block = 0 m, v = initial velocity of block = 2.3 m/s, W = work done by friction at bottom of ramp = 0 J, h' = height of ramp = 53.5 m, v' = final velocity of block = 2.3 m/s = v (since it moves at constant speed), and W' = work done by friction at top of ramp. Work done by frictionSubstituting the values of h, W and v' into the equation, and making W' subject of the formula, we have
mgh + 1/2mv² + W = mgh' + 1/2mv'² + W'
mg(0) + 1/2mv² + 0 = mgh' + 1/2mv² + W'
0 + 1/2mv² + 0 = mgh' + 1/2mv² + W'
1/2mv² - 1/2mv² - mgh' = W'
W' = -mgh
Substituing the values of the variables into the equation, we have
W = -mgh
W = 1550 kg × 9.8 m/s² × 53.5 m
W = -812665 kgm²/s²
W = -812665 J
W = -812.665 kJ
W ≅ -813 kJ
So, the work done by friction on the roller coaster cart during its motion up the ramp is -813 kJ
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A 50.0 kg rock starting from rest free falls through a distance of 300.0 m with no air resistance. Find the momentum change of the rock caused by its fall and the resulting change in the magnitude of earth’s velocity. Earth’s mass is 6.0 × 1024 kg. Show all your work, assuming the rock-earth system is closed
Answer:
Momentum change of the rock = 3835 kgm/s
Chane in earth's velocity = 6.4 x 10⁻²² m/s
Explanation:
m = 50.0 kg, M = 6.0 x 10²⁴ kg
d = 300.0 m
final velocity of rock = v, v² = 2gd = 2 * 9.81 * 300, so v = 76.7 m/s
Momentum change of the rock = mv = 50.0 * 76.7 = 3835 kgm/s
Inital momentum of the rock+earth system = 0
final momentum = 0 (momentum conservation)
momentum of rock and momentum of earth are same magnitude and opposite directions.
mv - MV = 0
Chane in earth's velocity = mv/M = 3835/(6.0 x 10²⁴) = 6.4 x 10⁻²² m/s
Which diagram represents diffuse reflection?
Answer:
A ,D
....
....
.....
....
....
..
Figure B and Figure C represents diffuse reflection. So, the correct options are B and C
What is Diffuse reflection?Diffuse reflection is defined as the reflection of light or other waves or particles from a surface, such that a ray incident on a surface is scattered at many angles rather than just at a single angle, as is the case with specular reflection.
For example, an illuminated ideal diffuse reflective surface will have the same brightness from all directions in the hemisphere surrounding the surface, i.e. Lambertian reflectance.
If the surface is microscopically rough, the light rays will be reflected and scattered in many different directions as in figure B and C above, with rough surfaces showing diffuse reflections.
Thus, Figure B and Figure C represents diffuse reflection. So, the correct options are B and C
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A bird accelerating from rest at a constant rate , experiences a displacement of 28m in 11s .what is the final velocity after 11s?
Answer:
308m/s
Explanation:
Let the final velocity, displacent, time taken and initial velocity be V, S, T and U respectively.
Given,
U=0( The bird started from rest)
T= 11s
S=28m
then,
V=U+S×T
or, V=0+11×28
or, V= 308m/s
Running with an initial velocity of +10.2 m/s m / s , a horse has an average acceleration of -1.77 m/s2 m / s 2 . How much time does it take for the horse to decrease its velocity to +6.1 m/s m / s ?
Using first equation of kinematics
[tex]\\ \rm{:}\dashrightarrow v=u+at[/tex]
[tex]\\ \rm{:}\dashrightarrow v-u=at[/tex]
[tex]\\ \rm{:}\dashrightarrow 6.1-10.2=-1.77t[/tex]
[tex]\\ \rm{:}\dashrightarrow -4.1=-1.77t[/tex]
[tex]\\ \rm{:}\dashrightarrow t=2.32s[/tex]
Find the velocity of the watch whose radius is 5 meter and time is 2 seconds
Question :-
Find the Velocity of the Watch whose Radius is 5 meter and Time is 2 seconds .Answer :-
Velocity of the Watch is 2.5 m/s .[tex] \rule {180pt} {2pt} [/tex]
Given :-
Radius = 5 meterTime = 2 seconds
To Find :-
Velocity = ?Solution :-
As per the provided information in the given question, we have been given that the Radius of the Watch is 5 meter , Time is given 2 seconds . And, we have been asked to calculate the Velocity of the Watch .
For calculating the Velocity , we will use Formula :-
Velocity = Displacement / TimeTherefore , by Substituting the given values in the above Formula :-
⇒ Velocity = Displacement / Time
⇒ Velocity = 5 / 2
⇒ Velocity = 2.5
Hence :-
Velocity = 2.5 m/s .[tex] \underline {\rule {180pt} {4pt}} [/tex]
⇒ Velocity = Displacement / Time
⇒ Velocity = 5 / 2
⇒ Velocity = 2.5