Answer:
so yeah it is Respuesta
Explain in detail why you cannot hear in space but you can see light(explanation includes particle movement of liquid, solids, gases, and plasma)
Answer:
The reason why one cannot hear in space but light can be seen is because light and sound are different kinds of wave
Light is an electromagnetic wave that propagates by a changing magnetic field hat creates a changing electric field. Light is created by a vibrating electric charge, and is transmitted through a medium by the absorption and reemission of the light by the medium particles
Sound is a longitudinal mechanical wave that is propagated by the vibration of the particles of the medium.
Sound requires a medium to propagate and it cannot be propagated or heard through space which is a vacuum.
Sound is created by a source that is vibrating and the vibration is transmitted to particles in the medium including solids, liquids, gases and plasma that the vibrating energy is in direct contact with and the transfer of the sound energy continues between layers of adjacent particles for the medium for the sound to be propagated from place to place
Explanation:
A 8.57-m ladder with a mass of 21.4 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 258 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.63 rad/s2 about an axis passing through the bottom end of the ladder. The ladder's center of gravity lies halfway between the top and bottom ends. (a) What is the net torque acting on the ladder
Answer:
[tex]1311.5\ \text{Nm}[/tex]
Explanation:
l = Length of ladder = 8.57 m
m = Mass of ladder = 21.4 kg
F = Force on ladder = 258 N
[tex]\alpha[/tex] = Angular acceleration = [tex]1.63\ \text{rad/s}^2[/tex]
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Net torque is given by
[tex]\tau=lf-\dfrac{l}{2}mg\\\Rightarrow \tau=8.57\times 258-\dfrac{8.57}{2}\times 21.4\times 9.81\\\Rightarrow \tau=1311.5\ \text{Nm}[/tex]
The net torque acting on the ladder is [tex]1311.5\ \text{Nm}[/tex].
Kepler's second law states that a line from the sun to a planet sweeps out equal areas in equal times. 2 d coordinate system with an elliptical orbit of planet around the sun. The elliptical orbit is centered at the origin. The Sun is located on the x axis between the origin and the far left edge of the orbit. The Planet is on the orbital path in the first quadrant and has a velocity vector that is pointing counter-clockwise on the orbital path. There is a line connecting the sun and the planet. The Perihelion is marked as the point where the orbit crosses the x axis closest to the Sun. The Aphelion is marked as the point where the orbit crosses the x axis farthest away from the Sun. The aphelion is the point in the orbit where the planet is the furthest distance from the sun, The perihelion is the point at which the planet is closest to the sun. At which point must the speed of the planet be greater
Answer:
the speed is higher in the PERIHELIUM
Explanation:
As stated in your statement, Kepler's second law says that a vector from the sun to the planet also sweeps equal in equal times. This is a consequence that the sun-planet system is isolated, therefore the angular momentum is conserved.
L = r x p = m r x v
where m is the planet mass and the Sun is considered fixed
Let's analyze this expression, if the anglar momentum is a constant when r is less (perihelion) the speed must increase, so that the product remains fixed
So the speed is higher in the PERIHELIUM
.
[tex]25 {?}^{?} \times \frac{?}{?} [/tex]
Answer:
what is your exact question.
A sound wave is propagating through the air from left to right as shown in the diagram below.
Answer:
Explanation: Pules
Answer: longitudinal (compression)
Explanation:
If the mass of the cement is 15000kg. calculate the density of this cement sample in kgm-3
The density of this cement sample is equal to [tex]2727.27\;kg/m^3[/tex]
Given the following data:
Mass of cement = 15000 kgLength = 1.1 mWidth = 2 mHeight = 2.5 mTo calculate the density of this cement sample:
First of all, we would determine the volume of the rectangular block.
[tex]Volume = length \times width \times height\\\\Volume = 1.1 \times 2 \times 2.5\\\\Volume = 5.5 \;m^3[/tex]
Now, can calculate the density of this cement sample:
Mathematically, the density of a substance is given by the formula;
[tex]Desnity =\frac{Mass}{Volume}[/tex]
Substituting the values into the formula, we have;
[tex]Density=\frac{15000}{5.5} \\\\Density =2727.27\;kg/m^3[/tex]
Read more: https://brainly.com/question/18320053
Note: 1.1 m, 2 m, 2.5 m are the figures provided from a rectangular block of cement.
A + 32.2 μC charge feels a 0.544 N force from a + 12.3p μC charge. How far apart are they? (u stands for micro.) [?] m
Answer:
r = 2.55 m
Explanation:
Given that,
First charge, q₁ = 32.2 μC
Second charge, q₂ = + 12.3 μC
The force between charges, F = 0.544 N
We need to find the distance between charges. The force between two charges is given by the formula as follows :
[tex]F=k\dfrac{q_1q_2}{r^2}\\\\r=\sqrt{\dfrac{kq_1q_2}{F}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 32.2\times 10^{-6}\times 12.3\times 10^{-6}}{0.544}} \\\\r=2.55\ m[/tex]
So, the charges are 2.55 m apart.
Answer:
2.55
Explanation:
Right on Acellus
Give the other guy brainliest
You can only give brainliest if 2 people answer
please help!!
A baseball cannon fires balls at an initial speed of 29 m/s. The cannon is tilted at a 52o angle.
(a) How long could one of these balls stay in the air ?
(b) How far from the cannon, will the ball hit the ground ?
(c) What will be ball’s maximum altitude ?
Answer:
(a) The time the ball stays in the air is approximately 4.66 seconds
(b) The distance from the cannon at which the ball will hit the ground is approximately 83.18 meters
(c) The maximum altitude of the ball is approximately 26.62 m
Explanation:
The given parameters of the question are;
The initial velocity at which a baseball cannon fires balls, u = 29 m/s
The angle at which the cannon is tilted, θ = 52°
(a) The time duration the ball stays in the air is given by the time of flight of the projected ball as follows;
[tex]2 \cdot t = \dfrac{2 \cdot u \cdot sin (\theta)}{g}[/tex]
Where;
t = The time it takes the baseball to reach maximum height
2·t = The total time of flight = The time the ball stays in the air
θ = The angle at which the ball is tilted = 52°
g = The acceleration due to gravity ≈ 9.81 m/s²
u = The initial velocity = 29 m/s
Therefore, we have;
[tex]2 \cdot t = \dfrac{2 \times 29 \ m/s\times sin (52^{\circ})}{9.81 \ m/s^2} \approx 4.66 \ s[/tex]
The time the ball stays in the air, 2·t ≈ 4.66 seconds
(b) The distance from the cannon at which the ball will hit the ground is given by the horizontal range, 'R', of the projectile as follows;
[tex]Horizontal \ range, R = \dfrac{u^2 \cdot sin(2 \cdot \theta) }{g}[/tex]
∴ The distance from the cannon at which the ball will hit the ground = R
[tex]Horizontal \ range, R = \dfrac{(29 \ (m/s))^2 \cdot sin(2 \times 52^{\circ}) }{9.81 \ m/s^2} \approx 83.18 \, m[/tex]
The distance from the cannon at which the ball will hit the ground = R ≈ 83.18 meters
(c) The maximum altitude of the ball is equal to the maximum height reached by the ball, H, which is given as follows;
[tex]H = \dfrac{u^2 \cdot sin^2 (\theta)}{2 \cdot g}[/tex]
Therefore, we have;
[tex]H = \dfrac{(29 \ m/s)^2 \times sin^2 (52^{\circ})}{2 \times 9.81 \ m/s^2} \approx 26.62 \ m[/tex]
The maximum altitude of the ball ≈ 26.62 m
580 nm light shines on a double slit with d=0.000125 m. What is the angle of the third dark interference minimum (m=3)?
(Remember, nano means 10^-9.)
(Unit=deg)
Answer:
0.66 degrees
Explanation:
The computation of the angle of the third dark interference is shown below:
The condition of the minima is
Path difference = (2n +1) × [tex]\lambda[/tex]÷ 2
For third minima, n = 2
Now
xd ÷ D = (2 × 2 + 1) × [tex]\lambda[/tex]÷ 2
d tan Q_3 = 5[tex]\lambda[/tex] ÷ 2
tan Q_3 = 5[tex]\lambda[/tex] ÷ 2d
Q_3 = tan^-1 × (5[tex]\lambda[/tex] ÷2d)
= tan^-1 × (5 × 580 × 10^-9) ÷ (2 × 0.000125)
= 0.66 degrees
PLZ HELP ME ASAP!!! 50 POINTS IF CORRECT! AND BRAINLEST
Answer:
B and C is the Answer
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A fruit basket of mass 44.5 kg weighs approximately how much in newton? (Answer should have a number and a proper unit) *
Answer:
445N
Explanation:
F=mg
m=44.5kg
g=10m/s2
F=44.5 * 10
=445N
a solid sphere of mass and radius 20kg and 4m is rotating with an angular velocity of 40rad/sec is accelerated uniformly to 60rad/sec in 2sec what is the torque acting on the sphere? (I=0•4MR^2)
Answer:
1280Nm
Explanation:
torque= I×alpha(angular acceleration)
so Angular acceleration is found by the following process.
why brittles of a paint brush spread when in water and cling when taken out of water
Answer:
surface tension
Explanation:
because of surface tension in water
fuel was consumed at a certain rate of 0.05Kg\s in a rocket engine and ejected as a gas with a speed of4000m\s . Determine the thrust on the rocket
Answer:
Thrust = 200 N
Explanation:
The engine thrust can be found by using the following formula:
[tex]Thrust = mv[/tex]
where,
m = mass flow rate of the fuel = 0.05 kg/s
v = velocity of ejected gases = 4000 m/s
Therefore, using the given values in the equation, we get:
[tex]Thrust = (0.05\ kg/s)(4000\ m/s)[/tex]
Thrust = 200 N
1. Which term is the name given to the underwater mountains in the
middle of oceans? *
mid-ocean crust
mid-ocean mountas
mid-ocean ridges
mid-ocean basins
Answer:
The name given to the underwater mountains in the middle of oceans is;
Mid-ocean ridges
Explanation:
Mid-ocean ridges, also known as mid-oceanic ridge is a mountain range made by plate tectonics under the water.
The magma created at a divergent boundary where two tectonic plates meet due to the rise of convection currents in the Earth's mantle which is beneath the oceanic crust results in the uplifting of the ocean floor.
Easy Guided Online Tutorial A special electronic sensor is embedded in the seat of a car that takes riders around a circular loop-the-loop ride at an amusement park. The sensor measures the magnitude of the normal force that the seat exerts on a rider. The loop-the-loop ride is in the vertical plane and its radius is 21 m. Sitting on the seat before the ride starts, a rider is level and stationary, and the electronic sensor reads 770 N. At the top of the loop, the rider is upside down and moving, and the sensor reads 350 N. What is the speed of the rider at the top of the loop?
Answer:
v = 17.30 m / s
Explanation:
For this exercise we will use Newton's second law
at the bottom of the loop and stopped
∑ F = 0
N-W = 0
N = W
W = 770 N
the mass of the body is
W = mg
m = W / g
m = 770 / 9.8
m = 78.6 kg
on top of the loop and moving
∑ F = m a
N + W = m a
note that the three vectors go in the same vertical direction down
the centripetal acceleration is
a = v² / r
we substitute
N + W = m v² / r
v = [tex]\sqrt{(N+W) \frac{r}{m} }[/tex]
let's calculate
v = [tex]\sqrt{ (350+770) \frac{21}{78.6} }[/tex]
v = 17.30 m / s
A satellite orbits earth at constant speed in circular orbit.Which statement is correct? A The resultant force on the satellite is zero. B The resultant force on the satellite is towards the Earth. C The resultant force on the satellite is away from the Earth. D The resultant force on the satellite is in the direction of motion.
Answer:
B The resultant force on the satellite is towards the Earth.
Explanation:
In the case when the satellite orbits earth would be at constant speed in the circular orbit so hee the resultant force that on the satellite would be with regard to the earth
So as per the given situation, the option B is correct
And, the rest of the options would be wrong
And, the same would be relevant
The body needs small amounts of ________ to help enzymes break down proteins.
Answer:
I would say vitamin B-6
Explanation:
it is also known as pyridoxine. it helps ebzymes break down protein and carry the dismantled amino acids to the blood stream
Hope this helps ✌✌
Answer:
The body needs small amount of vitamin B-6 to help enzymes break down protein and carry the dismantled amino acids to the blood stream.
A wire of radius 5 x 10⁻⁴ m is needed to prepare a coil of resistance 40 Ω. The resistivity of the material of the wire is 3.14x10⁻⁷ Ωm. Calculate the length of the wire.
Answer:
100 m
Explanation:
From the question,
R = Lρ/A.................... Equation 1
Where R = resistance of the wire, L = length of the wire, ρ = resistivity of the wire, A = cross sectional area of the wire.
But,
A = πr².................... Equation 2
Where r = radius of the wire.
Substitute equation 2 into equation 1
R = Lρ/πr²
Make L the subject of the equation
L = Rπr²/ρ...................... Equation 3
Given: R = 40 Ω, r = 5×10⁻⁴ m, ρ = 3.14×10⁻⁷ Ωm
Constant: π = 3.14
Substitute these values into equation 3
L = [40×3.14×( 5×10⁻⁴)²]/ (3.14×10⁻⁷)
L = 40×3.14×25×10⁻⁸/(3.14×10⁻⁷)
L = 100 m
Hence the length of the wire is 100 m
A 1,000 kg car traveling 15.0 m/s brakes and comes to a stop after traveling 20.0 m.
a. What is the car’s initial kinetic energy?
b. What is the car’s final kinetic energy?
c. How much work does it take to stop the car?
d. How much constant force is applied in bringing the car to a stop?
A. The car’s initial kinetic energy is 112500 J
B. The car’s final kinetic energy is 0 J
C. The amount of work done to stop the car is –112500 J
D. The force applied in bringing the car to a stop is –5625 N
A. How to determine the initial kinetic energy Mass (m) = 1000 KgInitial velocity (u) = 15 m/sInitial kinetic energy (KEᵢ) =?KEᵢ = ½mu²
KEᵢ = ½ × 1000 × 15²
KEᵢ = 112500 J
B. How to determine the final kinetic energy Mass (m) = 1000 KgFinal velocity (v) = 0 m/sFinal kinetic energy (KEբ) =?KEբ = ½mv²
KEբ = ½ × 1000 × 0²
KEբ = 0 J
C. How to determine the workdone Initial kinetic energy (KEᵢ) = 112500 J Final kinetic energy (KEբ) = 0 JWorkdone (Wd) =?Wd = KEբ – KEᵢ
Wd = 0 – 112500
Wd = –112500 J
D. How to determine the forceWe'll begin by calculating the acceleration. This can be obtained as follow:
Initial velocity (u) = 15 m/sFinal velocity (v) = 0 m/sDistance (s) = 20 mAcceleration (a) =?v² = u² + 2as
0² = 15² + (2 × a × 20)
0 = 225 + 40a
Collect like terms
0 – 225 = 40a
–225 = 40a
Divide both side by 40
a = –225 / 40
a = –5.625 m/s²
Finally, we shall determine the force. This is illustrated below:
Mass (m) = 1000 KgAcceleration (a) = –5.625 m/s²Force (F) =?F = ma
F = 1000 × –5.625
F = –5625 N
Learn more about energy:
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9- When the moon comes in an orbit higher from earth, ......... solar eclipse are formed.
a) total
b) annular
c) partial. d) annular and partial
A race car has a mass of 710kg. It starts from rest and speeds up to 12m/s in 12.0s. The car is uniformly accelerated during the entire time. What net force is applied to it?
8.How long is a day? A year?
Answer:
24 hours. 365 days
Explanation:
An object of mass 1.5 kg is moving forwards along the floor against an applied force of
40.0 N [backwards]. If the coefficient of kinetic friction is 0.25, determine the
acceleration of the object.
Answer:
The acceleration of the object is -29.12 m/s².
Explanation:
The acceleration of the object can be calculated by Newton's second law:
[tex] \Sigma F = ma [/tex]
[tex] - F - F_{\mu} = ma [/tex]
[tex] - F - \mu mg = ma [/tex]
Where:
F: is the applied force = 40.0 N
μ: is the coefficient of kinetic friction = 0.25
m: is the mass of the object = 1.5 kg
g: is the acceleration due to gravity = 9.81 m/s²
a: is the acceleration =?
[tex] a = \frac{- F - \mu mg}{m} = \frac{-40.0 N - 0.25*1.5 kg*9.81 m/s^{2}}{1.5 kg} = -29.12 m/s^{2} [/tex]
The minus sign is because means that the object is decelerating due to the applied force and the friction.
Therefore, the acceleration of the object is -29.12 m/s².
I hope it helps you!
Series circuit when you had one bulb and battery voltage was at 9 volts, what was current into battery?
1) .90amps
2) .40amps
3) .30 amps
Answer:
incomplete question, resistor must be there
Explanation:
Excluding the noble gas group, how does the number of valence electrons in an element influence its chemical stability?
A. Elements with intermediate numbers of valence electrons are the last chemically stable.
B. Elements with the highest number of valence electrons are the most chemically stable.
C. Elements with intermediate number of valence electrons are the most chemically stable.
D. Elements wit the lowest number of valence electrons are the most chemically stable.
Answer:
C. Elements with intermediate number of valence electrons are the most chemically stable.
A light wave has a frequency of 5 x 108. What is the wavelength of the light wave? (Assume the speed of light to be 3 x 108.)
Answer:
Wavelength of light wave = 0.6 m
Explanation:
Given:
Frequency of wave = 5 x 10⁸
Speed of light = 3 x 10⁸ m/s
Find:
Wavelength of light wave
Computation:
The size of a wave form is its spatial duration, or the duration in which the wave's form repeats.
Wavelength of light wave = Speed of light / Frequency of wave
Wavelength of light wave = [3 x 10⁸] / [5 x 10⁸]
Wavelength of light wave = 0.6 m
Wondering if you have enough rope to rappel to the ground, you drop a rock off the top, and hear the sound of it hitting the bottom 4.2 seconds later. Find the height of the cliff ignoring the time that the sound takes to travel back to you from the bottom.
Answer:
86.5 m
Explanation:
[tex]s = ut + \frac{1}{2} at {}^{2} \\ s = (0)(4.2) + \frac{1}{2} (9.81)(4.2) {}^{2} \\ s = 86.5m [/tex]
Answer:
86.5 m
Explanation:
Can you describe how and why the molecules move from one side to the other?
The molecules move from one side to another across the concentration gradient by breaking weaker bonds among the atom into stronger bonds. This is done to decrease the overall kinetic energy to become a more stable molecule.
The kinetic strength of the molecules consequences in random movement, causing diffusion. In simple diffusion, this method proceeds without the useful resource of a transport protein. it is the random motion of the molecules that reasons them to move from a place of excessive attention to a place with decreased awareness.
The molecules in a gas, a liquid, or a strong are in consistent movement due to their kinetic electricity. Molecules are in steady movement and collide with each different. those collisions cause the molecules to move in random guidelines. over time, however, greater molecules may be propelled into the less concentrated place.
The majority of the molecules flow from better to decrease awareness, although there can be some that circulate from low to excessive. the general (or net) motion is consequently from high to low concentration.
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When electrons are shared unequally a/an
bond is formed