Due to the fact that a star's parallax angle and distance are inversely related, I concur with student 2 and disagree with student 1. The farther away the star is, the smaller the parallax angle is.
Student 1: That's not necessarily true. It's true that the triangle in the diagram would be pointier for a star that is farther away, but the parallax angle depends on the ratio of the distance between the Earth and the star to the distance from the star to the edge of the solar system.
Student 2: But if the star is farther away, then the distance between the Earth and the star will be larger, which means the parallax angle will be smaller.
Student 1: Not necessarily. Remember that the distance from the star to the edge of the solar system is also increasing. If the distance from the star to the edge of the solar system increases more quickly than the distance between the Earth and the star, then the ratio of these two distances will decrease, which means the parallax angle will actually increase.
Student 2: Oh, I see what you mean. So it's possible for the parallax angle to increase as the distance to a star increases.
Student 1: That's correct. It's important to consider both the distance between the Earth and the star and the distance from the star to the edge of the solar system when determining the parallax angle.
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Question- Consider the following debate between two students regarding the relationship between parallax angle and the distance we measure to a star.
Student 2: If we drew a diagram for a star that was much more than 1 parsec away from us, the triangle in the diagram would be pointier than the one we just drew in Part II. That should make the parallax angle smaller for a star farther away.
Coherent monochromatic light of wavelength l passes through a narrow slit of width a, and a diffraction pattern is observed on a screen that is a distance x from the slit. On the screen, the width w of the central diffraction maximum is twice the distance x. What is the ratio a>l of the width of the slit to the wavelength of the light
Answer:
λ = a
Explanation:
This is a diffraction exercise that is described by the expression
a sin θ = m λ
sin θ = m λ/ a
the first zero of the diffraction occurs for m = 1
sin θ = λ / a
angles are generally very small and are measured in radians
sin θ = θ = y / x
we substitute
[tex]\frac{y}{x} = \frac{\lambda}{a}[/tex]
the width of the central maximum is twice the distance to zero
w = 2y
in the exercise indicate that this width is equal to twice the distance to the screen (2x)
W = 2x
2y = 2x
we substitute
1 = λ/ a
λ = a
we see that the width of the slit is equal to the wavelength used.
What do you mean by physics?
Answer:
Explanation:
Physics is the branch of science that studies the natural world and its rules and orders. It is one of the oldest sciences as ancient people study the stars and astronomy is considered part of Physics.
The length of stereocilia actually vary from 10 to 50 micrometers. Again, assuming that they behave like simple pendula, over what frequency range of sound waves would they resonate. (The actual frequency range of human hearing is 20 Hz - 20,000 Hz, so there certainly must be other mechanisms involved in determining the frequency of these being the pendular model is rather oversimplified.)
a. About 70 Hz -160 Hz.
b. About 440 Hz - 1000 Hz.
c. About 20 Hz - 50 Hz.
d. About 0.07 to 0.16 Hz.
Answer:
the frequency range of sound waves is about 70 Hz - 160 Hz
Hence, Option a) About 70 Hz -160 Hz is the correct answer
Explanation:
Given the data in the question;
L₁ = 10 micrometers = 0.00001 m
so
T₁ = 2π√(L/g)
g = 9.8 m/s
so we substitute
T₁ = 2π√(0.00001 /9.8) = 0.006347
⇒ f₁ = 1 / T₁ = 1 / 0.006347 = 157.55 ≈ 160 Hz
L₂ = 50 micrometers = 0.00005 m
T₂ = 2π√(L/g)
g = 9.8 m/s
so we substitute
T₂ = 2π√(0.00005 /9.8) = 0.0142
f₂ = 1 / T₂ = 1 / 0.0142 = 70.422 ≈ 70 Hz
Therefore, the frequency range of sound waves is about 70 Hz - 160 Hz
Hence, Option a) About 70 Hz -160 Hz is the correct answer
Which of the following is NOT part of the grain group?
Answer:
Any food made from wheat, rice, oats, cornmeal, barley, or another cereal grain is a grain product. Anything else is not
Explanation:
It takes 20 seconds to fill a two-liter bottle with water from your kitchen faucet. What is the mass flow rate from the faucet if water has a density of 1000 fraction numerator k g over denominator m cubed end fraction?
a. 0.1kg/sec.
b. 0.01kg/sec.
c. 1g/sec.
d. 1kg/sec.
Answer:
0.1 kg/s.
Explanation:
The density of water, d = 1000 kg/m³
Volume, V = 2 L
Time, t = 20 s
We need to find the mass flow rate from the faucet. We know that the density of an object is given by :
[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\\dfrac{m}{t}=\dfrac{dV}{t}\\\\\dfrac{m}{t}=\dfrac{1000\times 0.002}{20}\\\\=0.1\ kg/s[/tex]
So, the mass flow rate is equal to 0.1 kg/s.
question is included in the picture!!! PUT REAL ANSWERS OR I WILL REPORT YOU
Answer:
Explanation:
this is like rubbing a balloon on your head to make your hair stand up. Do that to the can. The balloon is filled , ofc, and then just rub the balloon on the can. This will charge the can with static electricity. :P
I need help please someone !!!!! Would appreciate it
Answer:
Yes, it would make it back up.
Explanation:
If it has 100,000 Joules of gravitational potential energy at the top of the hill, by the time the cart gets to the bottom, it will become PE = 0, KE = 90,000 since 10% of 100,000 is 10,000. The cart only requires 80,000J to climb back up so it should easily do so.
I didn't quite understand if the 10% energy loss is total, or every time it goes up or down, but it isn't a problem because 10% of 90,000 is 9,000, which means it would have 81,000J of energy on the way back up IF it loses energy due to friction on the way back up also.
The only physical law you need to prove this is the Law of Conservation of Energy: no energy is lost, only transformed; 10% of the energy becomes heat, the rest remains mechanical energy, which is the reason why the reasoning above works.
Jovian planets have rings because:__________.
a. their thick gaseous atmospheres would disintegrate any small rock that enter them
b. there is too much material to have fit into the ball of each planet
c. tidal forces prevent the material in rings from forming into moons
d. Jovian planets rotate very rapidly, and some material near the equator of these planets was flung outward, forming the rings
e. tidal forces cause volcanic eruptions on some moons, and part of this material subsequently escaped the gravity of the moons, forming the rings.
Answer:
E
Explanation:
Planets after Mars in our solar system are called Jovian planets. Therefore, Jupiter, Saturn, Uranus and Neptune are Jovian planets. The specialty of these planets is that they mostly made of gases and have ring around them.
They have rings around them because tidal forces cause volcanic eruptions on some moons, and part of this material subsequently escaped the gravity of the moons, forming the rings.
An object A with mass 200 kg and an another object B with mass 1000 kg are moving with same speed. The ratio of kinetic energy of object A to B is
Answer:
Ratio of kinetic energy of object A to B = 1:5
Explanation:
Given:
Mass of object A = 200 kg
Mass of object B = 1,000 kg
Find:
Ratio of kinetic energy of object A to B
Computation:
Kinetic energy = (1/2)(m)(v²)
Kinetic energy of object A = (1/2)(200)(v²)
Kinetic energy of object A = (100)(v²)
Kinetic energy of object B = (1/2)(1,000)(v²)
Kinetic energy of object B = (500)(v²)
Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B
Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)
Ratio of kinetic energy of object A to B = 100 / 500
Ratio of kinetic energy of object A to B = 1/5
Ratio of kinetic energy of object A to B = 1:5
We say that the mass comes to rest if, after a certain time, the position of the mass remains within an arbitrary small distance from the equilibrium position. Will the mass ever come to rest
Answer:
No, the mass will never come to rest
Explanation:
It is so because even at arbitrarily small distance it will experience some amount of force (irrespective of how small the value of force is).
This does not allow the mass to become stationary or in a equilibrium state as it is still subject to some amount of force.
Hence, the the mass will never come to rest
What is the index of refraction of a refractive medium if the angle of incidence in air is 40 degrees and the angle of refraction is 29 degrees?
The index of refraction is 1.33.
To find the answer, we need to know about index of refraction.
What is index of refraction?
The indication of the light bending ability of a medium is the refractive index of that medium.It determines how much the path of light is bent or refracted.It's a dimensionless number.What is the mathematical expression of refractive index?
Mathematically, the refractive index(n) isn= sin∅₁ / sin∅₂
where, ∅₁= angle of incident
∅₂= angle of refraction
What is the refractive index, if the angle of incidence in air is 40 degrees and the angle of refraction is 29 degrees?
n= sin(40) /sin(24) = 1.33Thus, we can conclude that the index of refraction is 1.33.
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A 2.50 MHz sound wave travels through a pregnant woman's abdomen and is reflected from the fetal heart wall of her unborn baby. The heart wall is moving toward the sound receiver as the heart beats. The reflected sound is then mixed with the transmitted sound, and 78 beats per second are detected. The speed of sound in body tissue is 1500 m/s. Part A Calculate the speed of the fetal heart wall at the instant this measurement is made.
Answer:
the speed of the fetal heart wall at the instant is 0.0325 m/s
Explanation:
Given the data in the question;
f₀ = 2.50 MHz = 1.80 × 10⁶ Hz
f[tex]_B[/tex] = 78 beat per sec ( Hz )
V = 1500 m/s
the speed of the fetal heart wall at the instant this measurement is made = ?
now, if v is the magnitude of hart wall speed, V is speed of sound and f[tex]_h[/tex] the frequency the heart receives( and reflects), f₀ is original frequency and f, is reflected back;
so
heart is moving observer and device is stationary source
f[tex]_h[/tex] = ((V + v)/v )f₀
Heart is moving source and device is stationary observer
f' = (V/(V-v ))f[tex]_h[/tex]
Beats
f[tex]_B[/tex] = f₀ - f' = f₀ - f₀( V+v / V-v ) = f₀( 2v / V-v )
so we solve for v
v = V( f[tex]_B[/tex] / ( 2f₀ + f[tex]_B[/tex] )
so we substitute
v = 1500 ( 78 / ( (2×1.80 × 10⁶) + 78 )
v = 1500 ( 78 / ( 3,600,000 + 78 )
v = 1500 ( 78 / 3,600,078 )
v = 1500 ( 2.1666 × 10⁻⁵ )
v = 0.0325 m/s
Therefore, the speed of the fetal heart wall at the instant is 0.0325 m/s
Fluid mechanics questions and answers
Answer:
Fluid mechanics is considered one of the toughest subdisciplines within mechanical and aerospace engineering. It is unique from almost any other field an undergraduate engineer will encounter. It requires viewing physics in a new light, and that's not always an easy jump to make.
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Rank these objects in order of their resistance to change in motion (from greatest to least).
A. pyramid, cone, sphere, cube
B. cube, pyramid, cone, sphere
C. sphere, cone, pyramid, cube
D. None of the above
Answer:
pyramid, cone, sphere, cube
Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. Otherwise, we keep the experiment the same. Compared to the pucks in the video, this pair of pucks will rotate View Available Hint(s) Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. Otherwise, we keep the experiment the same. Compared to the pucks in the video, this pair of pucks will rotate four times as fast. at the same rate. one-fourth as fast. twice as fast. one-half as fast.
Answer:
w = w₀ / 2 the angular velocity is half the initial value.
Explanation:
We can analyze this exercise as if we added another disk to obtain a disk with twice the mass, for which if the system is two disks, the angular tidal wave is conserved
initial instant.
L₀ = I₀ w₀
final moment
L_f = I w
the moment is preserved
L₀ = L_f
I₀ w₀ = I w
the moment of inertia of a disk is
I = ½ m R²
we substitute
½ m R² w₀ = ½ (2m) R² w
w = w₀ / 2
for the case of a disk with twice the mass, the angular velocity is half the initial value.
How much mechanical work is required to catch a 14.715N ball traveling at a velocity of 37.5m/s?
To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.
What is work?Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.
In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that
Work is the product of the component of the force acting in the displacement's direction and its magnitude.
Weight of the ball = 14.715 N.
Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.
Velocity of the ball = 37.5 m/s
Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.
Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.
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What happens when a neutral atom gains an electrons?
Answer:
The neutral atom becomes an anion.
Explanation:
When a neutral atom gains an electron (e−), the number of protons (p+) in the nucleus remains the same, resulting in the atom becoming an anion (an ion with a net negative charge).
Explain how grality and electric charge are different
what happens during subduction
Answer:
Subduction , Latin for "carried under," is a term used for a specific type of plate interaction. It happens when one lithospheric plate meets another—that is, in convergent zones —and the denser plate sinks down into the mantle.
The steering wheel of a car has a radius of 0.19 m, and the steering wheel of a truck has a radius of 0.25 m. The same force is applied in the same direction to each steering wheel. What is the ratio of the torque produced by this force in the truck to the torque produced in the car
Answer:
[tex]\frac{T_t}{T_c} = 1.32[/tex]
Explanation:
The torque applied on an object can be calculated by the following formula:
[tex]T = Fr[/tex]
where,
T = Torque
F = Applied Force
r = radius of the wheel
For car wheel:
[tex]T_c = Fr_c\\[/tex]
For truck wheel:
[tex]T_t = Fr_t[/tex]
Dividing both:
[tex]\frac{T_t}{T_c} = \frac{Fr_t}{Fr_c}[/tex]
for the same force applied on both wheels:
[tex]\frac{T_t}{T_c} = \frac{r_t}{r_c} \\[/tex]
where,
rt = radius of the truck steering wheel = 0.25 m
rc = radius of the car steering wheel = 0.19 m
Therefore,
[tex]\frac{T_t}{T_c} = \frac{0.25\ m}{0.19\ m} \\[/tex]
[tex]\frac{T_t}{T_c} = 1.32[/tex]
How much kinetic energy does an object have that is moving at a rate of 30 m/s and has a mass of 4000 kg ?
Answer:
K = 1800 kJ
Explanation:
Given that,
The speed of the object, v = 30 m/s
Mass of the object, m = 4000 kg
We need to find the kinetic energy of the object. The formula for the kinetic energy is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 4000\times 30^2\\\\K=1800000\\\\or\\\\K=1800\ kJ[/tex]
So, the required kinetic energy is equal to 1800 kJ.
Which one of the statements below is true about mechanical waves?
They must travel in empty space.
They can travel in a vacuum.
Both sound and light are examples of mechanical waves.
They require a medium to travel through.
assuming weightless pulleys and 100% efficiency, what is the minimum input force required to lift a 120 N weight using a single fixed pulley?
A. 21 N
B. 61 N
C. 121 N
D. 241 N
A 101 kg basketball player crouches down 0.380 m while waiting to jump. After exerting a force on the floor through this 0.380 m, his feet leave the floor and his center of gravity rises 0.920 m above its normal standing erect position. (a) Using energy considerations, calculate his velocity (in m/s) when he leaves the floor. m/s (b) What average force (in N) did he exert on the floor
Answer:
[tex]4.25\ \text{m/s}[/tex]
[tex]3391.22\ \text{N}[/tex]
Explanation:
y = Height of compression = 0.38 m
m = Mass of basketball player = 101 kg
h = Height of center of gravity after jump = 0.92 m
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Energy balance of the system is given by
[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.92}\\\Rightarrow v=4.25\ \text{m/s}[/tex]
The velocity of the player when he leaves the floor is [tex]4.25\ \text{m/s}[/tex]
[tex]Fy=mgy+\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{mgy+\dfrac{1}{2}mv^2}{y}\\\Rightarrow F=\dfrac{101\times 9.81\times 0.38+\dfrac{1}{2}\times 101\times 4.25^2}{0.38}\\\Rightarrow F=3391.22\ \text{N}[/tex]
The force exerted on the floor is [tex]3391.22\ \text{N}[/tex].
A light bulb has a resistance of 360 . What is the current in the bulb when it has a potential difference of 120 V across it? 0.33 A 3 A 480 A 43,200 A
Answer:
I=R×V
360×120 V
=43,200 A
Answer:
A.)0.33 A
Explanation:
i just took the quiz 2021
a student practicing for a track meet ran 250 m in 30 seconds. What was her average speed?
Answer:
8.33 meters/sec.
time = 30 sec. 30 sec. = 8.33 meters/sec.
A charged particle (charge 1.6x10-19 C and mass 1.67x10-27 kg) is initially moving with a velocity of 2x105 m/s and then moves into a region having a magnetic field and an electric field (6x104 V/m). The direction of initial velocity is perpendicular to the electric field and magnetic field. If the charged particle keep moving in the original direction without being deflected, what is the magnitude of the magnetic field
Answer:
[tex]0.3\ \text{T}[/tex]
Explanation:
q = Charge = [tex]1.6\times 10^{-19}\ \text{C}[/tex]
m = Mass of particle = [tex]1.67\times 10^{-27}\ \text{kg}[/tex]
v = Velocity = [tex]2\times 10^5\ \text{m/s}[/tex]
E = Electric field = [tex]6\times 10^4\ \text{V/m}[/tex]
B = Magnetic field
Magnetic field is given by
[tex]B=\dfrac{E}{v}\\\Rightarrow B=\dfrac{6\times 10^4}{2\times 10^5}\\\Rightarrow B=0.3\ \text{T}[/tex]
The magnitude of magnetic field is [tex]0.3\ \text{T}[/tex]
Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. Bifocal lenses are used to correct his vision. A diverging lens is used in the top part of the lens to allow the person to clearly see distant objects.
1. What power lens (in diopters) should be used in the top half of the lens to allow her to clearly see distant objects?
2. What power lens (in diopters) should be used in the bottom half of the lens to allow him to clearly see objects 25 cm away?
Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
[tex]\frac{1}{f_1} = \frac{1}{q}[/tex]
q = f₁
2) for an object located at p = 25 cm
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]
[tex]\frac{1}{f_2}[/tex] = 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P = [tex]\frac{1}{f}[/tex]
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
A wave has a frequency of 2 Hz. Find its period
The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.3 m/s . Initially, the car is also traveling at a speed 19.3m/s and its front bumper is a distance 25.0m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.560m/s^2 , then pulls back into the truck's lane when the rear of the car is a distance 26.5m ahead of the front of the truck. The car is of length 4.50m and the truck is of length 20.7m .
Part A) How much time is required for the car to pass the truck?
Part B ) What distance does the car travel during this time?
Part C) What is the final speed of the car?
Answer:
A) t = 10.56 s, B) x = 235 m, C) v = 25.2 m / s
Explanation:
A) We can solve this problem using kinematics expressions.
The distance traveled by the truck is
x_c = v_c t
Distance traveled by the car.
The car must travel the distance that separates them from the truck x₀=25.0. Return to the lane at x₁ = 26.5 m. the length of the truck x₂=20.7m and the length of the car x₃ = 2 4.5 = 9 m, therefore the total length traveled by the car is
x_t = x₁ + x₂ + x₃
x_t = 26.5 + 20.7 +9 = 56.2 m
the distance traveled by the car when it returns to the lane is
x_c + x_t = x₀ + v₀ t + ½ a t²
when the car passes the car the distance traveled by the two vehicles is the same, we substitute
v_c t + x_t = x₀ + v₀ t + ½ a t²
½ a t² + t (v₀ -v_c) + (x₀ - x_t) = 0
we substitute the values
½ 0.560 t² + t (19.3 -19.3) + (25.0 - 56.2) =
0.28 t² -31.2 = 0
t = [tex]\sqrt{ \frac{31.2}{0.28} }[/tex]
t = 10.56 s
This is the time it takes for the car to pass the truck and back into the lane.
B) the distance traveled is
x = v₀ t + ½ a t²
x = 19.3 10.56 + ½ 0.560 10.56²
x = 235 m
C) the final velocity is
v = v₀ + a t
v = 19.3 + 0.560 10.56
v = 25.2 m / s