Answer:
D the development of an ozone layer
Explanation:
the rest don't relate to the atmosphere
describe free fall and weightlessness ?
weightlessness is the complete or near complete absense of the sensation of weight.
when in freefall,the only force acting upon your body is the force of gravity a noun contact force.since the gravity cannot be felt without any other opposing forces,you would have no sensation of it.you would feel weightless when in a state of freefall.
please help asap!!!!
Answer:
D. 100% Chance of having a child with dimple
A 75.0kg diver falls from a 3.00m board into the water. She goes 2.40m into the water before stopping. What is the average force the water exerted on the diver?
Answer:
The potential energy of the diver at the top is equal to the net work needed to be done to stop the diver.
PE = W
mg(h+d) = Fd
Explanation:
What do both types of bone marrow do that is found in our long bones.
(answer asappp)
Answer:
There are two types of bone marrow – red bone marrow (myeloid tissue) and yellow bone marrow (fatty tissue). Red bone marrow is involved in production of blood cells, while yellow marrow is important for fat storage. As you age, yellow bone marrow replaces red bone marrow.
Explanation:
pls mark brainliest :D
Water at a pressure of 3.00 × 105 Pa flows through a horizontal pipe at a speed of 1.00 m/s. The pipe narrows to one-fourth its original diameter. What is the speed of the flow in the narrow section?
Answer:
The answer is "[tex]16 \ \frac{m}{s}[/tex]"
Explanation:
[tex]Pressure \ (P) = 3.00 \times 10^{5}\ Pa\\\\Velocity\ V_1 = 1.00 \ \frac{m}{s}[/tex]
Using the formula for equating the continuity equation:
[tex]\to V_1 \times A_1 = V_2 \times A_2\\\\\to A_2 = \frac{A_1}{16} \\\\\therefore \\\\\to V_2 = V_1 \times \frac{A_1}{A_2} \\\\\to V_2 = 1 \times 16 = 16 \ \frac{m}{s}[/tex]
The speed of the flow in the narrow section is [tex]16m/s[/tex].
To find the speed of the flow in the narrow section, We use equation of continuity.
[tex]V_{1}*A_{1}=V_{2}*A_{2}[/tex]
Given that , initial speed [tex]V_{1}=1m/s[/tex]
Let us consider that, initial cross-section area is[tex]A_{1}=A[/tex].
When the pipe narrows to one-fourth its original diameter.
Then, Area [tex]A_{2}=\frac{A}{4^{2} }=\frac{A}{16}[/tex]
Substituting values in above equation.
[tex]1*A=V_{2}*\frac{A}{16}\\\\V_{2}=16m/s[/tex]
Hence, the speed of the flow in the narrow section is [tex]16m/s[/tex].
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Una cantidad de aire se lleva del estado a al estado b siguiendo una trayectoria recta en una grafica PV como es mostrado en la figura. Suponga que la expansión es desde un volumen de 0.06 metros cúbicos a un volumen de 0.13 metros cúbicos y que la presión aumenta desde 125,432 pascales a 168,793 pascales.¿Cuánto trabajo efectúa el gas en este proceso?
Answer:
El trabajo de frontera ejercido por el gas es 10,297.875 joules.
Explanation:
Supongamos que el gas se comporta idealmente y que el proceso es isotérmico y adiabático. El trabajo efectuado por el gas en el proceso equivale al área bajo la curva de la gráfica PV. Puesto que es una línea recta y que tanto la presión como el volumen son variables positivas, tenemos que esa área es la suma del rectángulo inferior (límite inferior de la presión) y el triángulo superior (entre los límites inferior y superior de la presión). Es decir:
[tex]W = P_{min}\cdot (V_{2}-V_{1})+\frac{1}{2}\cdot (P_{max}-P_{min})\cdot (V_{2}-V_{1})[/tex] (1)
Donde:
[tex]W[/tex] - Trabajo de frontera ejercido por el gas, medida en joules.
[tex]V_{1}[/tex], [tex]V_{2}[/tex] - Volúmenes del gas antes y después de la expansión, medidas en metros cúbicos.
[tex]P_{min}[/tex] - Límite inferior de la presión del gas, medida en pascales.
[tex]P_{max}[/tex] - Límite superior de la presión del gas, medida en pascales.
Si sabemos que [tex]V_{1} = 0.06\,m^{3}[/tex], [tex]V_{2} = 0.13\,m^{3}[/tex], [tex]P_{min} = 125,432\,Pa[/tex] y [tex]P_{max} = 168,793\,Pa[/tex], entonces el trabajo de frontera ejercido por el gas es:
[tex]W = (125,432\,Pa)\cdot (0.13\,m^{3}-0.06\,m^{3})+\frac{1}{2}\cdot (168,793\,m-125,432\,m) \cdot (0.13\,m^{3}-0.06\,m^{3})[/tex]
[tex]W = 10,297.875\,J[/tex]
El trabajo de frontera ejercido por el gas es 10,297.875 joules.
One long wire carries current 30.0 A to the left along the x axis. A second long wire carries current 50.0 A to the right along the line (y = 0.280m, z = 0). (b) A particle with a charge of -2.00μC is moving with a velocity of 150i Mm/s along the line (y = 0.100m , z=0 ). Calculate the vector magnetic force acting on the particle.
The vector magnetic force on the particle is [tex]3.47 \times 10^{-2}{~}\hat{j}{~} N[/tex].
A stream of charged particles, such as electrons or ions, traveling through an electrical conductor or a vacuum is known as an electric current.
One long wire carries a current I = 30.0 A to the left along the x-axis. I' = 50.0 A of current is being carried by a second lengthy wire to the right along the line (y = 0.280 m, z = 0).
Now,
The charge of a particle is -2.00μC which is moving with a velocity of v = [tex]150\hat{i} {~}Mm/s[/tex].
At y = 0.1 m the total magnetic field will be:
B = [ μ₀ (I') / 2πr ] + [μ₀ (I) / 2πr]
[tex]B = [\frac{\mu_{0} }{2 \pi}] \times( \frac{50}{0.28-0.1} (-\hat{k})] + \frac{30}{0.1}(-\hat{k})[/tex]
B = 1.16 × 10⁻⁴ T
Then the force on the particle is:
F =q(v × B)
[tex]F = (-2 \times 10^{-6} ) ( 150 \times 10^{6})\hat{i} \times (1.16 \times 10^{-4})(-\hat{k})[/tex]
F = 3.47 × 10⁻² N
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a cube of silver has a mass of 21 kg. calculate the length of the cube's side in inches. the density of silver is 10.49 g/cm3.
the length of the cube's side in inches is 4.96 inches
Density- It is a measurement of that how material is packed together tightly.
Its formula is ρ= m/v
where ρ is the density of material
m is the mass
v is the volume
here we have to find the length of cube's side
a cube of silver has a mass = 21 kg. = 21×10³ [∵1kg = 1000g ]
density of silver = 10.49g/cm³
the length of the cube's side = ?
density is given by
ρ= m/v
to find the volume we rearrange the above equation
v=m/ρ
volume of a cube = (length)³ = l³
using values we get
l³ = 21×10³ /10.49 ≈ 2×10³ cm³
l =∛(2×10³)cm³
l = 1.26 ×10 = 12.6 cm
l = 4.96 inches [∵ 1 cm = 0.394 inch ]
the length of the cube's side in inches is 4.96 inches
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Which two statements correctly describe transmutation?
Answ. B and C
Explanation:
The two statements which correctly describes transmutation include the following:
The number of atom is not conserved.The parent and daughter elements are different.What is Transmutation?This is the process of conversion of a chemical element into another with the atomic nuclei being changed.
Transmutation has true statements which are mentioned above thereby making option B and C the most appropriate choice.
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Is question II correct?
Answer:
it most likly right I'm not 100% sure
A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced.
Which free body diagram shows the ball falling at terminal velocity?
The free body that shows the ball falling at terminal velocity is B. A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.
How to illustrate the information?It should be noted that at terminal velocity, the ball will stop accelerating and the net force on the ball will be zero.
In this case, for the net force to be zero, equal, their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂
Since F₁ = 20 N, then F₂ = -F₁ = -20 N
Therefore, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.
Therefore, the correct option is B.
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A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced. Which free body diagram shows the ball falling at terminal velocity? A free body diagram with one force pointing downward labeled F Subscript g Baseline 20 N. A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N. A free body diagram with 2 forces. The first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 5 N. A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 30 N.
Question 14 (3 points)
What part of volcano carries magma from the magma chamber to the vent? (3 points)
Vent
Magma Chamber
Lava Flow
Conduit
d
you may check it online
a electron with an initial speed of 500,000 m/s is brought to rest by an electric field. what was the potential difference that stopped electron? m
The potential difference that stopped electron is 1.422v
potential difference in between two points is the amount of work done is moving a unit charge from one point to another point.
The mass of the electron,
m = 9.109*10⁻³¹
Va (initial speed) = 500,000 m/s
Vb (final speed) = 0 m/s
Electron has been stopped because the electrons move from higher potential to lower potential.
by the work energy theorem
W = ½ mv² (1 equation)
The Potential difference is
W = eV
V = W/e
put the value of W form equation 1.
[tex]V = \frac{\frac{1}{2} (V_{b} ^{2} - V_{a} ^{2} )}{e}[/tex]
e = 9.109*10⁻³¹
V = 1/2 [(500000m/s²) - 0m/s² ] / 9.109*10⁻³¹
V = 1.422v
The potential difference that stopped electron is 1.422v
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Develop a definition of chemical change using the words atoms and molecules in your definition.
Answer:
can you be more espisific
Explanation:
Answer:
a chemical change is a substance change in to a new substance
Explanation:
define direct current
Answer:
DC (direct current) is the unidirectional flow or movement of electric charge carriers (which are usually electrons). The intensity of the current can vary with time, but the general direction of movement stays the same at all times.
Direct current is produced by rectifiers, batteries, generators with commutators, and fuel cells. For example, the current flowing in appliances running on batteries or in a flashlight is a direct current.
A car has a speed of 12m/s. The mass of the car and its passengers is 1250 kg. What is the total momentum of the car/passengers? (assuming the car moves in the positive direction)
p= kgm/s
Answer:
15000kgm/s
Explanation:
Given parameters:
Speed of the car = 12m/s
Mass of car and passengers = 1250Kg
Unknown:
Momentum of the car = ?
Solution:
Momentum is the quantity of motion a body posses;
Momentum = mass x velocity
Now insert the given parameters and solve;
Momentum = 12 x 1250 = 15000kgm/s
Which statement best explains the relationship between the electric force between two charged objects and the distance between them? A. As the distance increases by a factor, the electric force increases by the square of that factor. B. As the distance increases by a factor, the electric force increases by twice that factor. C. As the distance increases by a factor, the electric force decreases by twice that factor. D. As the distance increases by a factor, the electric force decreases by the same factor. E. As the distance increases by a factor, the electric force decreases by the square of that factor.
Answer: I think it’s C but I’m not completely sure about it please correct me if I’m wrong
a ball is thrown upward at 25 m/s from the ground. what distance has the ball travelled after 5 seconds?
Answer:
125m
Explanation:
25m/s x 5 = 125
PLEASE HELP AND HURRY IM FOING A TEST Right now
Gamma rays and X-rays are both high-energy forms of radiation that can penetrate the
skin. X-rays, however, are the chosen form of radiation used to take pictures inside our
bodies. Which of the following is a reasonable explanation for the use of X-rays over
gamma rays when taking images?
A The energy and intensity of X-rays can be controlled.|
B X-rays have higher energy than gamma rays.
C The energy and intensity of gamma rays can be controlled.
D Gamma rays have higher energy than X-rays.
Answer: A The energy and intensity of X-rays can be controlled.|
Explanation:
Compare and contrast types of radioactive decay involving alpha particles, beta particles, and gamma radiation.
In alpha decay, the nucleus loses two protons, in beta decay, the nucleus either loses or gains a proton whereas in gamma decay, no change in proton number occurs only energy is released.
How do the types of radioactive decay compare?Alpha decay is a type of radioactive decay in which the nucleus of an atom emits an alpha particle which leads to transforming into a new atomic nucleus by losing two protons. During beta decay, the proton in the nucleus is transformed into a neutron and and neutron is transformed into proton while on the other hand, gamma decay is a type of decay in which nucleus changes from a higher energy state to a lower energy state due to the release of electromagnetic radiation.
So we can conclude that In alpha decay, the nucleus loses two protons, in beta decay, the nucleus either loses or gains a proton whereas in gamma decay, no change in proton number occurs only energy is released.
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Which is the work required to move a charge?
O electric field
O electric field line
O electric potential
O electric potential energy
Answer:
O electric potential
Explanation:
Answer:
[tex]electric \: potential[/tex]
Explanation:
[tex]hope \: helps \\ and \: brainliest \: if \: helped[/tex]
Which forms of heat transfer do not require matter(particles) to happen? (Select all that apply)
Conduction
Convection
Radiation
None of these
Answer:
Radiation
Explanation:
The form of heat that does not require matter or particles to happen is known as radiation.
Radiation is a form of heat that is carried by electromagnetic spectrum. It does not depend on particles of the medium. Convection and conduction depends on the particles in the medium for heat transfer. Conduction and convection cannot occur in a vacuum. Radiation can transfer heat in a vacuum.Suppose you fill two rubber balloons with air, suspend both of them from the same point, and let them hang down on strings of equal length. You then rub each with wool or on your hair so that the balloons hang apart with a noticeable separation between them. Make order-of-magnitude estimates of (c) the field each creates at the center of the other
The electric field each creates at the centre of the other is 1.7×10^4N/C.
Consider two balloons of diameter 0.200m each with a mass 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical
E = ke q /r^2
= [(8.99×10^9N⋅m^2C^2)(1.2×10^−7C)] / (0.25m)^2
= 1.7×10^4N/C
Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the electric field.
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A 90.5-kg cannon is loaded with a 104-gram ball. The cannon is fired and propels the ball forward. A photogate measurement determines that the cannon recoiled backwards with a speed of 20.8 m/s. Determine the speed of the ball.
Answer:
18100 m/s.
Explanation:
conservation of momentum can be applied in solving this question, by using below expresion
mv = MV
where M = mass of recoil of canon =90.5kg
V =velocity of recoil of canon = 20.8 m/s.
m= mass of tennis ball=204g=104x 10⁻³kg
v = velocity of tennis ball
104 x 10⁻³ x v = 90.5 x 20.8
We can make V subject of formula
v = (90.5 x 20.8) / 104 x 10⁻³
v = 18100 m/s.
Hence, the speed of the ball is 18100 m/s.
Why d/dt(a+bt^2)=2bt
Because a is not a function of t .
So d/dt of a is zero.
Using the formula, Force = mass'acceleration. (F = m xa)(stated in Newtons (N)). Solve: A water-filled balloon with an overall mass of 1 kg undergoes an acceleration of 2 m/s2. How much force is being applied to the balloon?
Answer:
2N
Explanation:
Given parameters:
Mass of the balloon = 1kg
Acceleration = 2m/s²
Unknown:
Applied force = ?
Solution
According to Newton's second law of motion:
Force = mass x acceleration
Now insert the given parameters and solve;
Force = 1 x 2 = 2N
QC A string fixed at both ends and having a mass of 4.80g , a length of 2.00 m , and a tension of 48.0 N vibrates in its second (n=2) normal mode. (b) What is the ratio of the wavelength in air of the sound emitted by this vibrating string and the wavelength of the wave on the string?
The ratio of the wavelength in air of the sound emitted by this vibrating string and the wavelength of the wave on the string is 2:1.
Frequency of the wavef = n/2L(√T/μ)
where;
n is number of nodesL is length of the stringT is the tension of the springμ is mass per unit lengthμ = 0.0048 kg / 2 m = 0.0024 kg/m
f = 2/(2 x 2) (√48/0.0024)
f = (¹/₂)(√20,000)
f = (¹/₂)(141.42)
f = 70.71 Hz
Speed of sound in stringv = √T/μ
v = (√48/0.0024)
v = 141.42 m/s
wavelength of the wave on the stringλ = v/f
λ = (141.42) / (70.71)
λ = 2 m
Wavelength in air of the sound emittedSpeed of sound in air, v = 332 m/s
λ = v/f
λ = 332 / 70.71
λ = 4.7 m
Ratio of the wavelengths = 4.7 m / 2m = 2.3/1 ≈ 2 : 1
Thus, the ratio of the wavelength in air of the sound emitted by this vibrating string and the wavelength of the wave on the string is 2:1.
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An engine has 180,000 W of power and can accelerate from rest to a top speed in 9 s. How much work did the engine do?
Answer:
1620000J
Explanation:
Given parameters:
Power = 180000W
Time = 9s
Unknown:
Work done by the engine = ?
Solution:
Power is the rate at which work is being done.
Now,
Power = [tex]\frac{Work done }{time}[/tex]
Work done = Power x time taken
Work done = 180000 x 9 = 1620000J
Give one example which describes Newton's first law of motion.
Answer:
a moving car
Explanation:
a body remains in its constant state unless un extenal force is applied.
A model car company produces battery-powered model cars. At its maximum
acceleration, their latest model accelerates from rest to its top speed in 3.5 s.
a) While the car is travelling at its top speed, it covers 180 m in 9.0 s.
Calculate the car's maximum acceleration.
b) Calculate how fast the car would be moving if it accelerated from rest for 1.5 seconds
with its maximum acceleration.
Answer:
a) 5.714 m/s²
b) 13.33 m/s²
Explanation:
, [tex]\text{The final velocity, }\text{\ensuremath{v_{f}}}\text{ of the car is }\text{\ensuremath{\dfrac{\Delta v}{\Delta t}=}}\dfrac{{180m}}{9.0s}=20\;m/s[/tex]
The car goes from an initial velocity of 0 m/s to this final velocity in t=3.5s
Therefore the acceleration of the car is given by
[tex]\displaystyle a=\boldsymbol{\dfrac{{v}_{f}-{v}_{0}}{t}}[/tex]
where [tex]v_0[/tex] is the initial velocity
[tex]a=\dfrac{20-0}{3.5}=\dfrac{20}{3.5}=5.714\mathrm{m/s^{2}}\[/tex]
Part a
[tex]\text{Maximum\;Acceleration}}}=5.714 \;m/s^{2}[/tex]
Part b)
If the car accelerated from rest and reached final speed of 20 m/s in 1.5s instead of 3.5s then
[tex]a=\dfrac{\boldsymbol{20}}{\boldsymbol{1.5}}={\displaystyle \textbf{13.33}\;m/s^{2}}[/tex]