Answer:
estrutura cristalina / cloreto de sódio
Explicação:
O cloreto de sódio forma a estrutura cristalina da treliça. Nesta estrutura, um átomo de sódio é cercado por 6 íons de carga oposta. Os íons opostos são negativos porque o sódio perde elétrons, o que o torna ião carregado positivamente chamado cátion, de modo que 6 íons negativos estão presentes ao redor do sódio. A estrutura cristalina também é conhecida como cúbico simples, o que significa que eles são distribuídos em três dimensões com igual distância entre eles e formando um ângulo de 90 graus.
A single slit 1.4 mmmm wide is illuminated by 460-nmnm light. Part A What is the width of the central maximum (in cmcm ) in the diffraction pattern on a screen 5.0 mm away
Answer:
1.643*10⁻⁴cmExplanation:
In a single slit experiment, the distance on a screen from the centre point is expressed as y = [tex]\frac{\delta m \lambda d}{a}[/tex] where;
[tex]\delta m[/tex] is the first two diffraction minima = 1
[tex]\lambda[/tex] is light wavelength
d is the distance of diffraction pattern from the screen
a is the width of the slit
Given [tex]\lambda[/tex] = 460-nm = 460*10⁻⁹m
d = 5.0mm = 5*10⁻³m
a = 1.4mm = 1.4*10⁻³m
Substituting this values into the formula above to get width of the central maximum y;
y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³
y = 2300*10⁻¹²/1.4*10⁻³
y = 1642.86*10⁻⁹
y = 1.643*10⁻⁶m
Converting the final value to cm,
since 100cm = 1m
x = 1.643*10⁻⁶m
x = 1.643*10⁻⁶ * 100
x = 1.643*10⁻⁴cm
Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is 1.643*10⁻⁴cm
A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 20.5 meters?
Answer:
The final speed of the crate after being pulled these 20.5 meters is 13.82 m/s
Explanation:
I'll assume that the correct question is
A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 22.5 meters?
mass of box = 51 kg
for the first 12 m, it is pulled with a constant force of 240 N
The acceleration of the box for this first 12 m will be
from F = ma
a = F/m
where F is the pulling force
m is the mass of the box
a is the acceleration of the box
a = 240/51 = 4.71 m/s^2
Since the body started from rest, the initial velocity u = 0
applying Newton's equation of motion to find the final velocity at the end of the first 12 m, we have
[tex]v^{2}= u^{2}+2as[/tex]
where v is the final velocity
u is the initial velocity which is zero
a is the acceleration of 4.71 m/s^2
s is the distance covered which is 12 m
substituting value, we have
[tex]v^{2}[/tex] = 0 + 2(4.71 x 12)
[tex]v^{2}[/tex] = 113.04
[tex]v = \sqrt{113.04}[/tex] = 10.63 m/s
For the final 10.5 m, coefficient of friction is 0.21
from f = μF
where f is the frictional force,
μ is the coefficient of friction = 0.21
and F is the pulling force of the box 240 N
f = 0.21 x 240 = 50.4 N
Net force on the box = 240 - 50.4 = 189.6 N
acceleration = F/m = 189.6/51 = 3.72 m/s^2
Applying newton's equation of motion
[tex]v^{2}= u^{2}+2as[/tex]
u is initial velocity, which in this case = 10.63 m/s
a = 3.72 m/s^2
s = 10.5 m
v = ?
substituting values, we have
[tex]v^{2}[/tex] = [tex]10.63^{2}[/tex] + 2(3.72 x 10.5)
[tex]v^{2}[/tex] = 112.9 + 78.12
v = [tex]\sqrt{191.02}[/tex] = 13.82 m/s
Resistance and Resistivity: The length of a certain wire is doubled while its radius is kept constant. What is the new resistance of this wire?
Answer:
Explanation:
The formula for calculating the resistance of a material in terms of its resistivity is expressed as [tex]R = \rho L/A[/tex] where;
R is the resistance of the material
[tex]\rho[/tex] is the resistivity of the material
L is the length of the wire
A is the area = πr² with r being the radius
[tex]R = \rho L/\pi r^{2}[/tex]
If the length of a certain wire is doubled while its radius is kept constant, then the new length of the wire L₁ = 2L
The new resistance of the wire R₁ will be expressed as [tex]R_1 = \frac{\rho L_1}{A_1}[/tex]
since the radius is constant, the area will also be the same i.e A = A₁ and the resistivity also will be constant. The new resistance will become
[tex]R_1 = \frac{\rho(2L)}{A}[/tex]
[tex]R_1 = \frac{2\rho L}{\pi r^2}[/tex]
Taking the ratio of both resistances, we will have;
[tex]\frac{R_1}{R} = \frac{2\rho L/\pi r^2}{\rho L/ \pi r^2} \\\\\frac{R_1}{R} = \frac{2\rho L}{\pi r^2} * \frac{\pi r^2}{ \rho L} \\\\\frac{R_1}{R} = \frac{2}{1}\\\\R_1 = 2R[/tex]
This shoes that the new resistance of the wire will be twice that of the original wire
What sentence best supports the statement that hormones are involved in the regulation of homeostasis? A. The hormone cortisol suppresses the immune system and is produced when the body is under stress. B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low. C. The hormone melatonin induces sleep and its production is slowed by exposure to light. D. The hormone oxytocin promotes labor contractions of the uterus during childbirth.
B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low.
What would you estimate for the length of a bass clarinet, assuming that it is modeled as a closed tube and that the lowest note that it can play is a D b whose frequency is 69 Hz
Answer:
1.24m
Explanation:
See attached file
You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph from due north.
a) What is your airspeed?
b) What angle (direction) are you flying?
c) The wind increases to 14 mph from the north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?
Answer:
a) 17.05 mph
b) 54.7° northeast direction
c) 10.71 mph
The direction is -22.58° relative to the east.
To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.
Explanation:
The question is a little confusing but, I guess the correct question should be;
You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.
a) What is your airspeed?
b) What angle (direction) are you flying?
c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?
NB: The difference in the question and my suggestion is highlighted boldly.
Your speed = 14 mph
direction is 45° northeast
Th wind speed = 4 mph
direction is north
We resolve the your speed and the wind speed into the horizontal and vertical components
For vertical the component component
[tex]V_{y}[/tex] = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph
For the horizontal speed component
[tex]V_{x}[/tex] = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph
Resultant speed = [tex]\sqrt{V^{2} _{y}+V^{2} _{x} }[/tex]
==> [tex]\sqrt{13.89^{2} +9.89^{2} }[/tex] = 17.05 mph This is your airspeed
b) To get your direction, we use
tan ∅ = [tex]V_{y}[/tex] /[tex]V_{x}[/tex]
tan ∅ = 13.89/9.89 = 1.413
∅ = [tex]tan^{-1}[/tex](1.413) = 54.7° northeast direction
c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .
In this case,
[tex]V_{y}[/tex] = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph
Resultant speed = [tex]\sqrt{V^{2} _{y}+V^{2} _{x} }[/tex]
==> [tex]\sqrt{4.11^{2} +9.89^{2} }[/tex] = 10.71 mph This is your airspeed
Your direction will be,
tan ∅ = [tex]V_{y}[/tex] /[tex]V_{x}[/tex]
tan ∅ = -4.11/9.89 = -0.416
∅ = [tex]tan^{-1}[/tex](-0.416) = -22.58° this is the angle you'll travel relative to the east.
To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.
Force and distance are used to calculate work. Work is measured in which unit? joules watts newtons meters
Answer:
The unit of work is joules
Force and displacement are used to calculate the work done by an object. This work is measured in the units of Joules. Thus, the correct option is A.
What is Work?Work can be defined as the force that is applied on an object which shows some displacement. Examples of work done include lifting an object against the Earth's gravitational force, and driving a car up on a hill. Work is a form of energy. It is a vector quantity as it has both the direction as well as the magnitude. The standard unit of work done is the joule (J). This unit is equivalent to a newton-meter (N·m).
The nature of work done by an object can be categorized into three different classes. These classes are positive work, negative work and zero work. The nature of work done depends on the angle between the force and displacement of the object. Positive work is done if the applied force displaces the object in its direction, then the work done is known as positive work. Negative work is opposite of positive work as in this work, the applied force and displacement of the object are in opposite directions to each other and zero work is done when there is no displacement.
Therefore, the correct option is A.
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The resistance of a 0.29 m long piece of wire is measured to be 0.31 Ohms. The wire has a cross-sectional area of 0.003 m2. What is the resistivity of the wire?
Answer:
3.21×10⁻³ Ωm
Explanation:
Applying,
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.
Make ρ the subject of the equation
ρ = RA/L................... Equation 2
Given: R = 0.31 Ohms, A = 0.003 m², L = 0.29 m
Substitute into equation 2.
ρ = 0.31(0.003)/0.29
ρ = 3.21×10⁻³ Ωm
A 1.8 kg microphone is connected to a spring and is oscillating in simple harmonic motion up and down with a period of 3s. Below the microphone is 1.8 hz, calculate the spring constant
Answer:
230N/m
Explanation:
Pls see attached file
A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250 m/s and leaves at 140 m/s
With what angular speed is the stick spinning after the collision?
Answer:
63.44 rad/s
Explanation:
mass of bullet = 3.3 g = 0.0033 kg
initial velocity of bullet [tex]v_{1}[/tex] = 250 m/s
final velocity of bullet [tex]v_{2}[/tex] = 140 m/s
loss of kinetic energy of the bullet = [tex]\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})[/tex]
==> [tex]\frac{1}{2}*0.0033*(250^{2} - 140^{2} )[/tex] = 70.785 J
this energy is given to the stick
The stick has mass = 250 g =0.25 kg
its kinetic energy = 70.785 J
from
KE = [tex]\frac{1}{2} mv^{2}[/tex]
70.785 = [tex]\frac{1}{2}*0.25*v^{2}[/tex]
566.28 = [tex]v^{2}[/tex]
[tex]v= \sqrt{566.28}[/tex] = 23.79 m/s
the stick is 1.5 m long
this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m
The angular speed will be
Ω = v/r = 23.79/0.375 = 63.44 rad/s
In a fluorescent tube of diameter 3 cm, 3 1018 electrons and 0.75 1018 positive ions (with a charge of e) flow through a cross-sectional area each second. What is the current in the tube
Answer:
The current in the tube is 0.601 A
Explanation:
Given;
diameter of the fluorescent, d = 3 cm
negative charge flowing in the fluorescent tube, -e = 3 x 10¹⁸ electrons/second
positive charge flowing in the fluorescent tube, +e = 0.75 x 10¹⁸ electrons/ second
The current in the fluorescent tube is due to presence of positive and negative charges to create neutrality in the conductor (fluorescent tube).
Q = It
I = Q/t
where;
I is current in Ampere (A)
Q is charge in Coulombs (C)
t is time is seconds (s)
1 e = 1.602 x 10⁻¹⁹ C
3 x 10¹⁸ e/ s = ?
= (3 x 10¹⁸ e/s x 1.602 x 10⁻¹⁹ C) / 1e
= 0.4806 C/s
negative charge per second (Q/t) = 0.4806 C/s
positive charge per second (Q/t) = (0.75 x 10¹⁸ e/s x 1.602 x 10⁻¹⁹ C) / 1e
positive charge per second (Q/t) = 0.12015 C/s
Total charge per second in the tube, Q / t = (0.4806 C/s + 0.12015 C/s)
I = 0.601 A
Therefore, the current in the tube is 0.601 A
When a central dark fringe is observed in reflection in a circular interference pattern, waves reflected from the upper and lower surfaces of the medium must have a phase difference, in radians, of
Explanation:
Let the first wave is :
[tex]y_1=A\sin\omega t[/tex]
And another wave is :
[tex]y_2=A\sin (\omega t+\phi)[/tex]
[tex]\phi[/tex] is phase difference between waves
Let y is the resultant of these two waves. So,
[tex]y =y_1+y_2[/tex]
The waves reflected from the upper and lower surfaces of the medium, it means that the resultant to be zero. So,
[tex]\cos(\dfrac{\phi}{2})=0\\\\\cos(\dfrac{\phi}{2})=\cos(\dfrac{\pi}{2})\\\\\phi=\pi[/tex]
So, the phase difference between the two waves is [tex]\pi[/tex].
A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maximum is Io, what is the intensity 10° from the center?
Answer:
The intensity at 10° from the center is 3.06 × 10⁻⁴I₀
Explanation:
The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ
I₀ = maximum intensity of light
a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m
θ = angle at intensity point = 10°
λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m
α = πasinθ/λ
= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m
= 1.0911/650 × 10³
= 0.001679 × 10³
= 1.679
Now, the intensity I is
I = I₀(sinα/α)²
= I₀(sin1.679/1.679)²
= I₀(0.0293/1.679)²
= 0.0175²I₀
= 0.0003063I₀
= 3.06 × 10⁻⁴I₀
So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀
The diameter of a 12-gauge copper wire is 0.081 in. The maximum safe current it can 17) carry (in order to prevent fire danger in building construction) is 20 A. At this current, what is the drift velocity of the electrons?
Answer:
0.44m/s
Explanation:
drift velocity=I/nAq
diameter 12 gauge
wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*
V = 0.44m/s
The drift velocity of the electrons should be 0.44 mm/s.
Calculation of the drift velocity:Since the diameter is 0.081 in
So, the radius = r = 0.081 inch/2
= 0.0405 inch
Now the conversion of inches to cm should be
= 0.0405*2.54
= 0.10287 cm
Now
area = π r2
= 3.14 * (0.10287)2
= 0.0332
Now the velocity should be
v = i/(nqA)
= 20/(8.5e22*1.60e-19*0.0332)
= 0.044 cm/s
= 0.44 mm/s
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The force required to compress a spring with elastic constant 1500N / m, with a distance of 30 cm is
Explanation:
F = kx
F = (1500 N/m) (0.30 m)
F = 450 N
A tiger leaps horizontally out of a tree that is 3.30 m high. He lands 5.30 m from the base of the tree. (Neglect any effects due to air resistance.)
Calculate the initial speed. (Express your answer to three significant figures.)
m/s Submit
Answer:
The initial velocity is [tex]v_h = 8.66 \ m/s[/tex]
Explanation:
From the question we are told that
The height of the tree is [tex]h = 3.30\ m[/tex]
The distance of the position of landing from base is [tex]d = 5.30 \ m[/tex]
According to the second equation of motion
[tex]h = u_o * t + \frac{1}{2} at^2[/tex]
[tex]Where\ u_o[/tex] is the initial velocity in the vertical axis
a is equivalent to acceleration due to gravity which is positive because the tiger is downward
So
[tex]3 = 0 + 0.5 * 9.8 *t^2[/tex]
=> [tex]t = \frac{3 }{9.8 * 0.5}[/tex]
[tex]t = 0.6122\ s[/tex]
Now the initial velocity in the horizontal direction is mathematically evaluated as
[tex]v_h = \frac{5.30}{0.6122}[/tex]
[tex]v_h = 8.66 \ m/s[/tex]
A metal rod of length 2.0 m is moved at 6.0 m/s in a direction perpendicular to its length. A 5.0 mT magnetic field is perpendicular to both the rod and its velocity. What is the potential difference between the ends of the rod? 30 mV 15 mV 0 mV 12 mV 60 mV
Answer:
The potential difference is 60mVExplanation:
This problem bothers on application of the expression for motion emf which is expressed as
[tex]E= Blv[/tex]
where B= magnetic field in Tesla
l= length of the conductor
v= speed of conductor
Given data
l= 2 meters
v= 6 m/s
B= 5 Tesla
Applying the formula we have
[tex]E=5*2*6= 60mV[/tex]
What explains why a prism separates white light into a light spectrum?
A. The white light, on encountering the prism, undergoes both reflection and refraction; some of the reflected rays re-enter the prism merging with refracted rays changing their frequencies.
B. The white light, on entering a prism, undergoes several internal reflections, forming different colors.
C. The different colors that make up a white light have different refractive indexes in glass.
D. The different colors that make up a white light are wavelengths that are invisible to the human eye until they pass through the prism.
E. The different rays of white light interfere in the prism, forming various colors.
Answer:
I think the answer probably be B
What explains why a prism separates white light into a light spectrum ?
C. The different colors that make up a white light have different refractive indexes in glass.
✔ Indeed, depending on the radiation (and therefore colors), which each have different wavelengths, the refraction index varies: the larger the wavelength (red) the less the reflection index is important and vice versa (purple).
✔ That's why purple is more deflected so is lower than red radiation.
Part (a) Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)? 94.2 Attempts Remain 33%
Part (b) If all the blood that flows through the aorta then branches into the major arteries, what is the velocity of blood in the major arteries? Give your answer in cm/s X Attempts Remain v4.71 A 33%
Part (c) The blood flowing in the major arteries then branches into the capillaries. If the velocity of blood in the capillaries is measured to be 0.04 cm/s, what is the cross sectional area of the capillary system in cm2? Grade Summary Deductions Potential cm2 A = 0% 100% cos( tan acos Submissions sin ( 7 8 9 НОME л Attempts remaining: 10 (4% per attempt) detailed view cotan0 asin acotan A E 4 5 6 atan cosh sinh0 1 2 3 tanh) ODegrees cotanh 0 + END Radians BACKSPACE CLEAR DEL
The aorta (the main blood vessel coming out of the heart) has a radius of about 1.0 cm and the total cross section of the major arteries is about 20 cm2. The density of blood is about the same as water, 1 g/cm3.
Answer:
a) 94.26 g/s
b) 4.713 cm/s
c) 2356.5 cm^2
Explanation:
a) velocity of blood through the aorta = 30 cm/s
radius of aorta = 1 cm
density of blood = 1 g/cm^3
Area of the aorta = [tex]\pi r^{2}[/tex] = 3.142 x [tex]1^{2}[/tex] = 3.142 cm^2
Flow rate through the aorta Q = AV
where A is the area of aorta
V is the velocity of blood through the aorta
Q = 3.142 x 30 = 94.26 cm^3/s
Current of blood through aorta [tex]I[/tex] = Qρ
where ρ is the density of blood
[tex]I[/tex] = 94.26 x 1 = 94.26 g/s
b) Velocity of blood in the major aorta = 30 cm/s
Area of the aorta = 3.142 cm^2
Velocity of blood in the major arteries = ?
Area of major arteries = 20 cm^2
From continuity equation
[tex]A_{ao} V_{ao} = A_{ar} V_{ar}[/tex]
where
[tex]V_{ao}[/tex] = velocity of blood in the major arteries
[tex]A_{ao}[/tex] = Area of the aorta
[tex]V_{ar}[/tex] = velocity of blood in the major arteries
[tex]A_{ar}[/tex] = Area of major arteries
substituting values, we have
3.142 x 30 = 20[tex]V_{ar}[/tex]
94.26 = 20[tex]V_{ar}[/tex]
[tex]V_{ar}[/tex] = 94.26/20 = 4.713 cm/s
c) From continuity equation
[tex]A_{ar} V_{ar} = A_{c} V_{c}[/tex]
where
[tex]A_{ar}[/tex] = Area of major arteries = 20 cm/s
[tex]V_{ar}[/tex] = velocity of blood in the major arteries = 4.713 cm/s
[tex]A_{c}[/tex] = Area of the capillary system = ?
[tex]V_{c}[/tex] = velocity of blood in the capillary system = 0.04 cm/s
substituting values, we have
20 x 4.713 = [tex]A_{c}[/tex] x 0.04
94.26 = 0.04[tex]A_{c}[/tex]
[tex]A_{c}[/tex] = 94.26/0.04 = 2356.5 cm^2
This question involves the concepts of volumetric flow rate, continuity equation, and flow velocity.
a) Total current of the blood passing through the aorta is "94.2 g/s".
b) The velocity of blood in major arteries is "4.71 cm/s".
c) The cross-sectional area of the capillary system is "2356.2 cm²".
a)
First, we will find the volumetric flow rate of the blood, using the continuity equation's formula:
[tex]Q=Av[/tex]
where,
Q = volumetric flow rate = ?
A = cross-sectional area of aorta
A = [tex]\pi(r)^2=\pi(1\ cm)^2= 3.14\ cm^2[/tex]
v = flow velocity = 30 cm/s
Therefore,
[tex]Q=(3.14\ cm^2)(30\ cm/s)[/tex]
Q = 94.25 cm³/s
Now, the blood current will be given as:
I = Qρ
where,
I = current = ?
ρ = blood density = 1 g/cm³
Therefore,
I = (94.2 cm³/s)(1 g/cm³)
I = 94.2 g/s
b)
Now, this volumetric flow rate will be constant in major arteries:
[tex]Q = A_r v_r\\\\v_r=\frac{Q}{A_r}[/tex]
where,
Ar = cross-section area of major arteries = 20 cm²
vr = flow velocity of blood in major arteries = ?
Therefore,
[tex]v_r=\frac{94.25\ cm^3/s}{20\ cm^2}[/tex]
vr = 4.71 cm/s
c)
Now, this volumetric flow rate will be constant in capillaries:
[tex]Q = A_c v_c\\\\A_c=\frac{Q}{v_c}[/tex]
where,
Ac = cross-section area of capillaries = ?
vc = flow velocity of blood in capillaries = 0.04 cm/s
Therefore,
[tex]A_c=\frac{94.25\ cm^3/s}{0.04\ cm/s}[/tex]
Ac = 2356.2 cm²
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Si se deja caer una piedra desde un helicóptero en reposo, entonces al cabo de 20 s cual será la rapidez y la distancia recorrida por la piedra
Answer:
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
Explanation:
Si se excluye los efectos del arrastre por la viscosidad del aire, la piedra experimenta un movimiento de caída libre, es decir, que la piedra es acelerada por la gravedad terrestre. La distancia recorrida y la rapidez final de la piedra pueden obtenerse con la ayuda de las siguientes ecuaciones cinemáticas:
[tex]v = v_{o} + g\cdot t[/tex]
[tex]y - y_{o} = v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex]
Donde:
[tex]v[/tex], [tex]v_{o}[/tex] - Rapideces final e inicial de la piedra, medidas en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo al cuadrado.
[tex]y[/tex]. [tex]y_{o}[/tex] - Posiciones final e inicial de la piedra, medidos en metros.
Si [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 0\,m[/tex], entonces:
[tex]v = 0\,\frac{m}{s} +\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)[/tex]
[tex]v = -196.14\,\frac{m}{s}[/tex]
[tex]y-y_{o} = \left(0\,\frac{m}{s} \right)\cdot (20\,s) + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)^{2}[/tex]
[tex]y-y_{o} = -1961.4\,m[/tex]
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
Your friend just challenged you to a race. You know in order to beat him, you must run 15 meters within 20 seconds in a northern direction. What does your average velocity need to be to win the race? .5 meters per second, north .75 meters per second, north 1.3 meters per second, north 300 meters per second, north
The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric field is in the y direction, the electric and magnetic fields are given below. This wave is linearly polarized in the y direction.
E= EoSin(kx - ωt)y
B= Bosin(kx- ωt)z
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0 and B0 are the________ of the electric and magnetic fields.
a. maximas
b. wavelenghts
c. amplitudes.
d. velocities
Answer:
amplitudes
Explanation:
In everyday physics we define the amplitude of a wave as the maximum (this can also be called the highest)displacement or distance moved by a point on a given vibrating body or wave as measured from its equilibrium position. The key idea in defining the amplitude of a wave motion is the idea of a 'maximum displacement from the position of equilibrium'.
Given the equations;
E= EoSin(kx - ωt)y
B= Bosin(kx- ωt)z
Both Eo and Bo refer to the maximum displacement of the electric and magnetic field components of the electromagnetic wave. This maximum displacement is known as the amplitude of the electric and magnetic components of the electromagnetic wave.
A person takes a trip, driving with a constant speed of 98.5 km/h, except for a 20.0-min rest stop. The person's average speed is 68.8 km/h. (a) How much time is spent on the trip? h (b) How far does the person travel? km
Answer:
Total time taken(T) = 1.1 hour
Distance = 75.68 km
Explanation:
Given:
Average speed = 68.8 km/h
Constant speed = 98.5 km/h
Rest time = 20 min = 20 / 60 = 0.3333 hour
Find:
Total time taken(T)
Total distance (D)
Computation:
Distance = speed × time
D = 68.8 × t.........Eq1
and
D = 98.5 × [t-0.33]
D = 98.5 t - 32.8333.........Eq2
From Eq1 and Eq2
68.8 t = 98.5 t - 32.83333
29.7 t = 32.83333
t = 1.1
Total time taken(T) = 1.1 hour
Distance = speed × time
Distance = 68.8 × 1.1
Distance = 75.68 km
"What is the energy density (energy per cubic meter) carried by the magnetic field vector in a small region of space in a EM wave at an instant of time when the electric vector is a maximum of 3500V/m
Answer:
The energy density is [tex]Z = 5.4 2 *10^{-5 } \ J/m^3[/tex]
Explanation:
From the question we are told that
The electric vector is [tex]E = 3500 \ V/m[/tex]
Generally the energy vector is mathematically represented as
[tex]Z = 0.5 * \epsilon_o * E^2[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with the value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]Z = 0.5 * 8.85 *10^{-12} * 3500^2[/tex]
[tex]Z = 5.4 2 *10^{-5 } \ J/m^3[/tex]
A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid cons
Complete question:
A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid consists of 2100 turns of wire.
Answer:
The magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.
Explanation:
Given;
length of solenoid, L = 2.4 m
radius of solenoid, R = 1.7 cm = 0.017 m
current in the solenoid, I = 0.19 A
number of turns of the solenoid, N = 2100 turns
The magnitude of the magnetic field inside the solenoid is given by;
B = μnI
Where;
μ is permeability of free space = 4π x 10⁻⁷ m/A
n is number of turns per length = N/L
I is current in the solenoid
B = μnI = μ(N/L)I
B = 4π x 10⁻⁷(2100 / 2.4)0.19
B = 4π x 10⁻⁷ (875) 0.19
B = 2.089 x 10⁻⁴ T
Therefore, the magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.
The planets how and block are near each other in the Dorgon system. the Dorgons have very advanced technology, and a Dorgon scientist wants to increase the pull of gravity between the two planets. Which proposals would the scientist make to accomplish this goal? check all that apply.
Answer:
Decreasing the distance between Hox and Blox, increasing the mass of Hox, or increasing the mass of Hox and Blox.
Explanation:
The gravity force is directly proportional to the mass of the bodies and inversely proportional to the square of the distance that separates them.
Or
If we decrease the distance between both planets (Hox and Blox), the gravitational pull between them will increase.
On the other hand, if we keep the distance between Hox and Blox, but we increase the mass of one of them, or increase the mass of both, the gravitational pull between them will also increase.
A screen is placed 43 cm from a single slit which is illuminated with 636 nm light. If the distance from the central maximum to the first minimum of the diffraction pattern is 3.8 mm, how wide is the slit in micrometer
Answer: The width is 1.25692 μm
Explanation:
The data that we have here is:
Distance between the single slit to the screen = L = 43cm
λ = 636 nm
Distance from the central maximum to the first minimum = Z = 3.8mm
We know that the angle for the destructive diffraction is:
θ = pλ/a
where p is the order of the minimum, for the first minimum we have p = 1, and a is the width of the slit,
then we have:
θ = (636nm/a)
And we also know that we can construct a triangle rectangle, where the adjacent cathetus to this angle is the distance between the slit and the screen, and the opposite cathetus is the distance between the first maximum and the first minimum:
Tg(θ) = Z/L
Tan(636nm/a) = 3.8cm/43cm
First, we need to use the same units in the right side:
3.8mm = 0.38cm
Tg(636nm/a) = 0.38cm/43cm
636nm/a = Atg( 0.38cm/43cm ) = 0.506
a = 636nm/0.506 = 1,256.92 nm
1 μm = 1000nm
then:
a = 1,256.92 nm = (1,256.92/1000) μm = 1.25692 μm
5) A coil of wire consists of 20 turns, each of which has an area of 0.0015 m2. A magnetic field is perpendicular to the surface with a magnitude of B = 4.91 T/s t – 5.42 T/s2 t2. What is the magnitude of the induced emf in the coil?
Answer:
1.5x10^-1 V
Explanation:
See attached file
Answer:
The magnitude of the induced emf in the coil is 15.3 mV
Explanation:
Given;
number of turns, N = 20 turns
Area of each coil, A = 0.0015 m²
initial magnitude of magnetic field at t₁, B₁ = 4.91 T/s
final magnitude of magnetic field at t₂, B₂ = 5.42 T/s
The magnitude of the induced emf in the coil is given by;
[tex]E = -N\frac{\delta \phi}{\delta t} \\\\E =-N (\frac{\delta B}{\delta t} )A\\\\E = -NA(\frac{B_1-B_2)}{\delta t} \\\\E = NA(\frac{B_2-B_1)}{\delta t} \\\\E = 20(0.0015)(5.42-4.91)\\\\E = 0.0153 \ V\\\\E = 15.3 \ mV[/tex]
Therefore, the magnitude of the induced emf in the coil is 15.3 mV
A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?
Answer:
1.3515x10^5pa
Explanation:
Plss see attached file
Which of the following explains why metallic bonding only occurs between
metallic atoms?
A. Metallic atoms are less likely to give their electrons to nonmetallic
atoms
B. Electrical conductivity is higher in metallic atoms, which means
they are more likely to attract free electrons.
C. Metallic atoms are highly reactive and do not tend to form bonds
with other atoms.
D. Metallic atoms have valence shells that are mostly empty, which
means these atoms are more likely to give up electrons and allow
them to move freely.
Answer:
D. Metallic atoms have valence shells that are mostly empty, which
means these atoms are more likely to give up electrons and allow
them to move freely.
Explanation:
Metals usually contain very few electrons in their valence shells hence they easily give up these few valence electrons to yield metal cations.
In the metallic bond, metal cations are held together by electrostatic attraction between the metal ions and a sea of mobile electrons.
Since metals give up their electrons easily, it is very easy for them to participate in metallic bonding. They give up their electrons easily because their valence shells are mostly empty, metal valence shells usually contain only a few electrons.