Answer:
Soil erosion is the act of removing the top layer of the soil either by rain wind or human activities
Answer:
l erosion is a gradual process that occurs when the impact of water or wind detaches and removes soil particles, causing the soil to deteriorate. ... The impact of soil erosion on water quality becomes significant, particularly as soil surface runoff. Sediment production and soil erosion are closely related.
Explanation:
If a small child swallowed a safety pin, why
would an X-ray photograph clearly show the
location of the pin?
Answer:
yes
Explanation:
it is in the body system
Answer:
it would show clearly because it is a metal piece in the body.
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
A) 8.18 x 10^-14 J
B) 2.73 x 10^-22 J
C) 1.5053 x 10^-10 J
D) 1.5032 x 10^-10 J
Answer:
djfjci3jsjdjdjdjdjddndn
ds
Airplane lift is achieved when air pressure on the bottom of its wings is
A) greater than pressure on top.
B) less than pressure on top.
C) the same as pressure on top.
Answer:
C the same as pressure on top
how can the starch be removed from the leaves of potted plants
Answer:
Explanation:
There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.
pls help! George pushes a wheelbarrow for a distance of 12 meters at a constant speed for 35 seconds by applying a force of 20 newtons. What is the
power applied to push this wheelbarrow?
A. 1.2 watts
B. 3.4 watts
C. 6.9 watts
D. 13 watts
Answer:
C. 6.9 watts
Explanation:
Power = work/time
if work = force×distance...
Then... power= (force×distance)/time
Power = (20×12)/35
= 6.9 watts
2 coplas o pregones inventadas por ti relacionadas con la región caribe.
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 550 lines/mm , and the light is observed on a screen 1.7 m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.
Answer:
Δd = 7.22 10⁻² m
Explanation:
For this exercise we must use the dispersion relationship of a diffraction grating
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the angles are small
tant θ = sinθ /cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
let's use direct ruler rule to find the distance between two slits
If there are 500 lines in 1 me, what distance is there between two lines
d = 2/500
d = 0.004 me = 4 10⁻⁶ m
diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1
let's calculate for each wavelength
λ = 656 nm = 656 10⁻⁹ m
d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶
d₁ = 2.788 10⁻¹ m
λ = 486 nm = 486 10⁻⁹ m
d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶
d₂ = 2.066 10⁻¹ m
the distance between the two lines is
Δd = d1 -d2
Δd = (2,788 - 2,066) 10⁻¹
Δd = 7.22 10⁻² m
An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where there is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?
a. 1min
b. 2min
c. 3min
d. 4min
e. 5min
f. 6min
Answer:
T = 188.5 s, correct is C
Explanation:
This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved
initial instant. Before the crash
L₀ = r m v₀ + I₀ w₀
the angular speed of the fan is zero w₀ = 0
final instant. After the crash
L_f = I₀ w + m r v
L₀ = L_f
m r v₀ = I₀ w + m r v
angular and linear velocity are related
v = r w
w = v / r
m r v₀ = I₀ v / r + m r v
m r v₀ = (I₀ / r + mr) v
v = [tex]\frac{m}{\frac{I_o}{r} +mr} \ r v_o[/tex]
let's calculate
v = [tex]\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4[/tex]
v = [tex]\frac{0.020}{2.345} \ 2.4[/tex]
v = 0.02 m / s
To calculate the time of a complete revolution we can use the kinematics relations of uniform motion
v = x / T
T = x / v
the distance of a circle with radius r = 0.6 m
x = 2π r
we substitute
T = 2π r / v
let's calculate
T = 2π 0.6/0.02
T = 188.5 s
reduce
t = 188.5 s ( 1 min/60 s) = 3.13 min
correct is C
2. Our solar system is made up of the Sun, 8 planets, and other bodies such as
asteroids orbiting the Sun. The solar system is very large compared to anything we see on
Earth. The distance between planets is measured in astronomical units (AU). One AU is
equal to 149.6 million kilometers, the average distance between the Sun and Earth. Scale
models are useful for helping us understand the size of the solar system.
Mr. Wilson’s science class made a scale model of the solar system. They went out to the
school’s football field, and they used the chart shown below to mark out the scale distance
from the Sun to each planet
in which states of matter will a substance have a fixed volume
Answer:
Solid is the state in which Matter maintains a fixed volume
Answer:
The state of matter that has a fixed volume is Solid.
Explanation:
Solid substances will maintain a fixed volume and shape.
Find the ratio of speeds of a proton and an alpha particle accelerated through the same voltage, assuming nonrelativistic final speeds. Take the mass of the alpha particle to be 6.64 ✕ 10−27 kg.
Answer:
The required ratio is 1.99.
Explanation:
We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.
We know that,
[tex]eV=\dfrac{1}{2}mv^2[/tex]
The LHS for both proton and an alpha particle is the same.
So,
[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]
So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.
A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d. A battery fully charges the capacitor and is then disconnected. The plate separation is now increased to a distance of 2d. What would be the change, if any, of the voltage across the capacitor, the electric field between the plates, and the energy stored in the capacitor?
Answer:
The answer is "Option D".
Explanation:
Please find the complete question in the attached file.
As plate separation increased to 2d the capacitance get halred but the change remain same
[tex]\therefore V=\frac{Q}{C}[/tex]
The voltage doubles are now electric field remain same because both the distance and voltage get doubled.
[tex]\to E=\frac{v}{d}\ = \frac{2v}{2d}\\\[/tex]
So,
[tex]energy=\frac{1}{2}\ \frac{Q^2}{C}\\\\c'=\frac{C}{2}\\\\E'=2E[/tex]
PLEASE HELP ME WITH THIS ONE QUESTION
Given the atomic mass of Boron-9 is 9.0133288 u, what is the nuclear binding energy of Boron-9? (Mproton = 1.0078251, Mneutron = 1.0086649, c^2 = 931.5 eV/u)
A) 59 eV
B) 58 eV
C) 57 eV
D) 56 eV
Answer:
a. 59 ev. helpful answer
A thin, uniform rod has length L and mass M. A small uniform sphere of mass m is placed a distance x from one end of the rod, along the axis of the rod.
a) Calculate the gravitational potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. (Hint: Use the power series expansion for ln(1+x).)
b) Use Fx=−dU/dx to find the magnitude of the gravitational force exerted on the sphere by the rod.
Answer:
Explanation:
From the given information:
a)
Let's have an imaginary view of the rod located at a given distance r from he the mass (m) of the sphere.
Then the equation for the potential energy as related to the small area of the dr of the rod can be computed as:
[tex]dU = -\dfrac{GMm}{L}*\dfrac{dr}{r}[/tex]
where,
G = gravitational constant
[tex]U = - \int^{x+L}_{x}Gm\dfrac{M}{L}*\dfrac{dr}{r}[/tex]
[tex]U = - \dfrac{GMm}{L}\int^{x+L}_{x}\dfrac{dr}{r}[/tex]
By taking the integral within the limit
[tex]U = - \dfrac{GMm}{L} \Big[In \ r\Big]^{x+L}_{x}[/tex]
[tex]\mathbf{\implies - \dfrac{GmM}{L} In \Big(\dfrac{{x+L}}{{x}}\Big)}[/tex]
b)
By using [tex]F= -\dfrac{dU}{dx}[/tex], the magnitude of the gravitational force can be determined as follows:
Here, we have:
[tex]F = -\dfrac{d}{dx} \Big [\dfrac{-GmM}{L}In(\dfrac{x+l}{x}) \Big ] \\ \\ = \dfrac{GmM}{L}\times \dfrac{x}{x+L}\times (0-\dfrac{L}{x^2}) \\ \\ By \ solving \\ \\ \mathbf{ =-\dfrac{GmM}{x(x+L)}}[/tex]
From above, the negative sign indicates an attractive force
Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?
a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rim
b. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.
c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.
d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.
Answer:
Option (a).
Explanation:
Let the angular velocity is w.
The centripetal acceleration is given by
[tex]a = r w^2[/tex]
where, r is the distance between the axle and the spoke.
So, more is the distance more is the centripetal acceleration.
(a) For the points on the spoke closer to the hub than for points closer to the rim is larger distance, so the centripetal force is more.
The statement is true.
(b) The direction of centripetal acceleration is always towards the center, so the statement is false.
(c) It is false.
(d) It is false.
Option (a) is correct.
A circular ice rink is 20 m in diameter and is to be temporarily enclosed in a hemispherical dome of the same diameter. The ice is maintained at 270 K. On a particular day the inner surface of the dome is measured to be 290 K. Estimate the radiant heat transfer from the dome to the rink if both surfaces can be taken as blackbody.
570k or 1050k
270k+290k= 570k
270k•290k = 1050k
Need in hurry important please
Answer:
I don't see anything on your question?
I NEED THE ANSWER QUICK PLEASEE
A farmhand pushes a 23 kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 91 N on the hay, how much work has she done?
Answer:
W = 354.9 J
Explanation:
Given that,
The mass of a bale of hay, m = 23 kg
The displacement, d = 3.9 m
The horizontal force exerted on the hay, F = 91 N
We need to find the work done. We know that,
We know that,
Work done, W = Fd
So,
W = 91 N × 3.9 m
W = 354.9 J
So, the required work done is 354.9 J.
define emperical formula and what is the dimensional formula of force and energy
Answer:
An empirical formula represents the simplest whole number ratio of various atoms present in a compound.The dimensional formula of force is [[tex]MLT^{-2}[/tex]]The dimensional formula of energy is [[tex]ML^{2} T^{-2}[/tex]]You should extending your throwing hand straight up to the sky to follow-through.
O True
O False
False
It's not straight up
what does Newton's third law ? Describe.
Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A. In other words, forces result from interactions.
Hope it helps
Answer:
Newton's third law state that every action there is equal but opposite reactions
hope this will help you more
A charge Q is transferred from an initially uncharged plastic ball to an identical ball 24 cm away.The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other?
Answer:
The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
Explanation:
Given;
magnitude of the attractive force, F = 17 mN = 0.017 N
distance between the two objects, r = 24 cm = 0.24 m
The attractive force is given by Coulomb's law;
[tex]F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C[/tex]
The charge of 1 electron = 1.602 x 10⁻¹⁹ C
n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷
[tex]n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons[/tex]
Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
A kangaroo kicks downward with a 1000N force. According to Newton's Law the kangaroo is propelled into the air by:
A) gravitational force
B) his muscles
C) The earth
D) wallabies
Explanation:
Specifically his leg muscles. As the leg muscles expand, they push down on the ground. Newton's 3rd law says that for any action, there's an opposite and equal reaction. That means a downward push into the ground will have the ground push back, more or less, and that's why the kangaroo will jump. The ground (and the earth entirely) being much more massive compared to the animal means that the ground doesn't move while the kangaroo does move. Perhaps on a very microscopic tiny level the ground/earth does move but it's so small that we practically consider it 0.
This experiment can be done with a wall as well. Go up to a wall and lean against it with your hands. Then do a pushup to move further away from the wall, but you don't necessarily need to lose contact with the wall's surface. As you push against the wall, the wall pushes back, and that causes you to move backward. If the wall was something flimsy like cardboard, then you could easily push the wall over and you wouldn't move back very much. It all depends how much mass is in the object you're pushing on.
P5. A bullet with an initial velocity of 280 m/s in the x-direction penetrates an initially stationary block of mass 11 kg and emerges on the other side with a final velocity of 70 m/s in the x-direction. The velocity of the block after the collision is 0.2 m/s, also in the x-direction. Assume the block slides on a horizontal frictionless surface. What is the mass of the bullet
Answer:
the mass of the bullet is 10.5 g
Explanation:
Given;
initial velocity, u₁ = 280 m/s
final velocity of the bullet, v₁ = 70 m/s
final velocity of the block, v₂ = 0.2 m/s
mass of the block, m₂ = 11 kg
initial velocity of the block, u₂ = 0
let the mass of the bullet = m₁
Apply the principle of conservation of linear momentum for elastic collision to calculate the mass of the bullet.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
280m₁ + 11(0) = 70m₁ + 11 x 0.2
280m₁ = 70m₁ + 2.2
280m₁ - 70m₁ = 2.2
210m₁ = 2.2
m₁ = 2.2/210
m₁ = 0.0105 kg
m₁ = 10.5 g
Therefore, the mass of the bullet is 10.5 g
Explain why energy cannot escape from the room by conduction
Answer: Heat energy transfer by conduction, convection and radiation
Heat energy is a very difficult energy to store as it can transfer in three different ways from warm surroundings to cooler surroundings. The three processes are conduction, convection or radiation.
Understanding energy, how it is transferred and how the amount of energy that is usefully transferred can be used to measure the efficiency is very important to physics and to the world.
A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, carrying a 30 A current. What acceleration (in g's) does this interaction give the airplane?
Answer:
[tex]a=0.2*10^{-5}g[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=180=>0.18kg[/tex]
Charge [tex]Q=18mC=18*10^-^3C[/tex]
Velocity [tex]v=2.2m/s[/tex]
Length of Wire [tex]L=8.6cm=>0.086[/tex]
Current [tex]I=30A[/tex]
Generally the equation for Magnetic Field of Wire B is mathematically given by
[tex]B=\frac{\mu_0*I}{2\pi*l}[/tex]
[tex]B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}[/tex]
[tex]B=6.978*10^{-5}T[/tex]
Generally the equation for Force on the plane F is mathematically given by
[tex]F=qvB[/tex]
Therefore
[tex]ma=qvB[/tex]
[tex]a=\frac{qvB}{m}[/tex]
[tex]a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}[/tex]
[tex]a=2.37*10^{-5}[/tex]
Therefore in Terms of g's
[tex]a=\frac{2.37*10^{-5}}{9.8}[/tex]
[tex]a=0.2*10^{-5}g[/tex]
Two microwave signals of nearly equal wavelengths can generate a beat frequency if both are directed onto the same microwave detector. In an experiment, the beat frequency is 170MHz . One microwave generator is set to emit microwaves with a wavelength of 1.410cm.
If the second generator emits the longer wavelength, what is that wavelength?
Answer:
The wavelength of the wave is 1.419 cm.
Explanation:
wavelength emitted by the one micro wave = 1.41 cm
beat frequency = 170 M Hz
Let the frequency of the wave is f.
[tex]f=\frac{c}{\lambda }\\\\f =\frac{3\times 10^8}{0.0141}\\\\f = 2.13\times 10^{10} Hz[/tex]
Let the frequency of the other wave is f'.
[tex]f' = 2.13\times 10^{10}-170\times10^6\\\\f' = 21130\times 10^6 Hz[/tex]
The wavelength is given by
[tex]f' = 21130\times 10^6 Hz\\\lambda = \frac{3\times 10^8}{21130\times 10^6}\\\\\lambda = 0.01419 m = 1.419 cm[/tex]
A toddler weighs 10 kg and raises herself onto tiptoe (on both feet). Her feet are 8 cm long with each ankle joint being located 4.5 cm from the point at which her feet contact the floor. While standing on tip toe:
(a) what is the upward normal force exerted by the floor at the point at which one of the toddler's feet contacts the floor?
(b) what is the tension force in one of her Achilles tendons? (c) what is the downward force exerted on one of the toddler's
ankle joints?
Answer:
a.49 n
b. 63 n
c. 112 n
Explanation:
a.10 times 9.8 from gravity/2 = 49 n
b. 49n times 4.5/8-4.5 = 63 n
c 49n + 63 n = 112 n
Coherent light with wavelength 597 nm passes through two very narrow slits, and theinterference pattern is observed on a screen a distance of 3.00{\rm m} from the slits. The first-order bright fringe is adistance of 4.84 {\rm mm} from the center of the central bright fringe.
For what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?
Answer:
The required wavelength is 1.19 μm
Explanation:
In the double-slit study, the formula below determines the position of light fringes [tex]y_m[/tex] on-screen.
[tex]y_m = \dfrac{m \lambda D}{d}[/tex]
where;
m = fringe order
d = slit separation
λ = wavelength
D = distance between screen to the source
For the first bright fringe, m = 1, and we make (d) the subject, we have:
[tex]d = \dfrac{(1) \lambda D}{y_1}[/tex]
[tex]d = \dfrac{ \lambda D}{y_1}[/tex]
replacing the value from the given question, we get:
[tex]d = \dfrac{ (597 \ nm )\times (3.00 \ m)}{4.84 \ mm} \\ \\ d = \dfrac{ (597 \ nm \times (\dfrac{1 \ m}{10^9\ nm}) )\times (3.00 \ m)}{4.84 \ mm(\dfrac{1 \ m}{1000 \ mm })} \\ \\ d = 3.7 \times 10^{-4} \ m[/tex]
In the double-slit study, the formula which illustrates the position of dark fringes [tex]y_m[/tex] on-screen can be illustrated as:
[tex]y_m = (m+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
The value of m in the dark fringe first order = 0
∴
[tex]y_0 = (0+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
[tex]y_0 = (\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
making λ the subject of the formula, we have:
[tex]\lambda = \dfrac{2y_o d}{D} \\ \\ \lambda = \dfrac{2(4.84 \ mm) \times \dfrac{1 \ m}{1000 \ mm} (3.7 \times 10^{-4} \ m) }{3.00 \ m}[/tex]
[tex]\lambda = 1.19 \times 10^{-6} \ m ( \dfrac{10^6 \mu m }{1\ m}) \\ \\ \lambda = 1.19 \mu m[/tex]