Answer:
E = 12.25 x 10³ N/C = 12.25 KN/C
Explanation:
In order to balance the weight of the object the electrostatic force due to the electric field must be equal to the weight of the body or charge. Therefore,
Electrostatic Force = Weight
E q = mg
where,
E = Electric Field = ?
m = Mass of the Charge = 5 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
q = magnitude of charge = 4 μC = 4 x 10⁻⁶ C
Therefore,
E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)
E = 0.049 N/4 x 10⁻⁶ C
E = 12.25 x 10³ N/C = 12.25 KN/C
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 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.
Answer:
1.6nT [in the negative z direction]
Explanation:
The magnetic field, B, due to a distance of finite value b, is given by;
B = (μ₀IL) / (4πb[tex]\sqrt{b^2 + L^2}[/tex]) -----------(i)
Where;
I = current on the wire
L = length of the wire
μ₀ = magnetic constant = 4π × 10⁻⁷ H/m
From the question,
I = 20A
L = 2.0cm = 0.02m
b = 5.0m
Substitute the necessary values into equation (i)
B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 [tex]\sqrt{5.0^2 + 0.02^2}[/tex])
B = (10⁻⁷ x 20 x 0.02) / (5.0 [tex]\sqrt{5.0^2 + 0.02^2}[/tex])
B = (10⁻⁷ x 20 x 0.02) / (5.0 [tex]\sqrt{25.0004}[/tex])
B = (10⁻⁷ x 20 x 0.02) / (25.0)
B = 1.6 x 10⁻⁹T
B = 1.6nT
Therefore, the magnetic field at the point x = 5.0m on the x-axis is 1.6nT.
PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.
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°.
If radio waves were used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, Earth would receive their signals at a speed of
Answer:
Explanation:
speed of alien spaceship = .1 c
We shall apply formula of relativistic mechanics to solve the problem
relative velocity =
[tex]\frac{v+v_1}{1 -\frac{v\times v }{c^2} }[/tex]
Here v = v₁ = .1 c
relative velocity = .1c + .1 c / 1 - .1²
= .2 c / .99
= .202 c
The earth would receive the signal at the speed of .202 c .
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
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]
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
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.
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
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
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.
Learn more about Work here:
https://brainly.com/question/18094932
#SPJ6
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.
A flat loop of wire consisting of a single turn of cross-sectional area 8.60 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.40 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.80
Answer:
The induced current is [tex]I = 5.72*10^{-4 } \ A[/tex]
Explanation:
From the question we are told that
The cross-sectional area is [tex]A = 8.60 \ cm^2 = \frac{8.60 }{10000} = 8.60 *10^{-4} \ m[/tex]
The initial value of magnetic field is [tex]B_1 = 0.500 \ T[/tex]
The value of magnetic field at time t is [tex]B_f = 2.40 \ T[/tex]
The number of turns is N = 1
The time taken is [tex]dt[/tex]= 1.02 \ s
The resistance of the loop is [tex]R = 2.80\ \Omega[/tex]
Generally the induced emf is mathematically represented as
[tex]e = - \frac{d \phi}{dt }[/tex]
Where [tex]d \phi[/tex] is the change n the magnetic flux which is mathematically represented as
[tex]d \phi = N *A * d B[/tex]
Where [tex]dB[/tex] is the change in magnetic field which is mathematically represented as
[tex]d B = B_f - B_i[/tex]
substituting values
[tex]d B = 2.40 - 0.500[/tex]
[tex]d B = 1.9 \ T[/tex]
So
[tex]d \phi = 1 * 1.9 * 8.60 *10^{-4}[/tex]
[tex]d \phi = 1.63*10^{-3} \ T[/tex]
So
[tex]e = - \frac{1.63 *10^{-3}}{ 1.02 }[/tex]
[tex]e = - 1.60*10^{-3} \ V[/tex]
Here the negative only indicates that the emf is acting in opposite direction of the motion producing it so the magnitude of the emf is
[tex]e = 1.60*10^{-3} \ V[/tex]
Now the induced current is evaluated as follows
[tex]I = \frac{e}{R }[/tex]
substituting values
[tex]I = \frac{1.60 *10^{-3}}{2.80 }[/tex]
[tex]I = 5.72*10^{-4 } \ A[/tex]
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
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 primary of an ideal transformer has 100 turns and its secondary has 200 turns. If the input current at the primary is 100 A, we can expect the output current at the secondary to be
Answer:
Explanation:
For current in ideal transformer the formula is
I₁ / I₂ = N₂ / N₁
I₁ and I₂ are current in primary and secondary coil respectively and N₁ and N₂ are no of turns in primary and secondary coil .
Putting the given values
100 / I₂ = 200 / 100 = 2
I₂ = 50 A .
output current = 50 A .
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
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
shows a mixing tank initially containing 2000 lb of liquid water. The tank is fitted with two inlet pipes, one delivering hot water at a mass flow rate of .8 lb/s and the other delivering cold water at a mass flow rate of 1.2 lb/s. Water exits through a single exit pipe at a mass flow rate of 2.5 lb/s. Determine the amount of water, in lb, in the tank after one hour
Answer:
the water that remain in the tank in one hour will be 200 lb
Explanation:
Initial mass of water in the tank = 2000 lb
hot water is delivered through the first inlet pipe at a rate of = 0.8 lb/s
cold water is delivered through the second inlet pipe at a rate of = 1.2 lb/s
exit pipe flow rate = 2.5 lb/s
amount of water in the tank after one hour = ?
In one hour, there are 60 x 60 seconds = 3600 sec, therefore
the water through the first inlet pipe in one hour = 0.8 x 3600 = 2880 lb
the water through the second inlet pipe in one hour = 1.2 x 3600 = 4320 lb
the water through the exit in one hour = 2.5 x 3600 = 9000 lb
The total amount of water in the tank = 2000 + 2880 + 4320 = 9200 lb
The total amount of water that leaves the tank = 9000 lb
therefore, in one hour, the water that remain in the tank will be
==> 9200 lb - 9000 lb = 200 lb
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
Learn more about velocity here: https://brainly.com/question/19279241
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
Two uniform solid balls are rolling without slipping at a constant speed. Ball 1 has twice the diameter, half the mass, and one-third the speed of ball 2. The kinetic energy of ball 2 is 37.0 J.
Part A What is the kinetic energy of ball 1?
Express your answer with the appropriate units.
K7 = Value Units
Answer:
The kinetic energy of the ball 1 is 2.06 J
Explanation:
The kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I is its rotational inertia, ω its angular speed, m its mass and v = its velocity of center of mass.
Let m₁, I₁, v₁, d₁ represent the mass, rotational inertia, speed and diameter of solid ball 1. and Let m₂, I₂, v₂, d₂ represent the mass, rotational inertia, speed and diameter of solid ball 2.
Since both objects are spheres, I =2/5mr²
Let r₁ = radius of ball 1 and r₂ = radius of ball 2. Since d₂ = 2d₁
⇒ 2r₂ = 4r₁ ⇒ r₂ = 2r₁
Now, the the kinetic energy of sphere 1 is
K₁ = 1/2I₁ω₁² + 1/2m₁v₁² ω₁ = v₁/r₁ which is the angular speed of solid ball 1.
K₁ = 1/2(2/5mr²)v₁²/r₁² + 1/2m₁v₁²
K₁ = 1/5m₁v₁² + 1/2m₁v₁²
K₁ = 7/10m₁v₁²
Also, the the kinetic energy of sphere 2 is
K₂ = 1/2I₂ω₂² + 1/2m₂v₂² ω₂ = v₂/r₂ which is the angular speed of solid ball 2.
K₂ = 1/2(2/5m₂r₂²)v₂²/r₂² + 1/2m₂v₂²
K₂ = 1/5m₂v₂² + 1/2m₂v₂²
K₂ = 7/10m₂v₂²
Now, m₁ = m₂/2 and v₁ = v₂/3
Substituting these into K₁, we have
K₁ = 7/10(m₂/2)(v₂/3)²
K₁ = 7/10 × 1/18m₂v₂²
K₁ = (1/18)(7/10m₂v₂²)
K₁ = K₂/18
K₂ = 37.0 J/18
K₂ = 2.06 J
So, the kinetic energy of the ball 1 is 2.06 J
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]
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
Learn more about the continuity equation here:
https://brainly.com/question/24905814?referrer=searchResults
0.25-kg block oscillates on the end of a spring with a spring constant of 200 N/m. If the oscillations is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is A :
Answer:
5.2m/s
Explanation:
Plss see attached file
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.
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.
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