Answer:
B. microwaves, visible light, X-rays, gamma rays
The order that shows increasing frequency for gamma rays, microwaves, visible light, and X-rays is:
microwaves, visible light, X-rays, gamma rays
What are electromagnetic waves?These are waves that can propagate (i.e travel) through space while transferring energy. They travel through space with the speed of light (i.e 3×10⁸ m/s)
Examples of electromagnetic waves includes
Gamma rayX-rayUltraviolet Light Infrared Radio wave MicrowaveFrom the examples given above, it should be noted that gamma ray has the shortest wavelength and the longest frequency.
With the above information in mind, we can conclude that the order that shows increasing frequency for gamma rays, microwaves, visible light, and X-rays is:
microwaves, visible light, X-rays, gamma rays
Learn more about electromagnetic waves:
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what is force,momentum,and velocity.
Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object.
Momentum is force or speed of movement.
Velocity defines the path of the motion of the frame or the object
Why would the kinetic energy at the bottom of the track be less than the potential energy at the top of the track?
Answer:
This is because of friction and heat lost.
Explanation:
What is not the ideal location for a radioactive waste storage facility?
A.
in an area that already has a lot of background radiation
B.
in an area that has few earthquakes
C.
far away from ground water
D.
in an area that is unpopulated with people
Answer:
D
Explanation:
9. Mr. Smith went skiing in Maine last weekend. He traveled 523 kilometers to Sugarloaf from
Leominster. His average speed was 109 km/hr. How long did it take Mr. Smith to hit the slopes?
Answer:
Time taken by Mr. smith = 4.80 hour (Approx.)
Explanation:
Given:
Distance travel by Mr. smith = 523 kilometer
Average speed of Mr. smith = 109 km/hr
Find;
Time taken by Mr. smith
Computation:
Time taken = Distance cover / Speed
Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.
smith
Time taken by Mr. smith = 523 / 109
Time taken by Mr. smith = 4.798 hr
Time taken by Mr. smith = 4.80 hour (Approx.)
6. If an object accelerates at 3m/s/s, how long does it take for the object to travel at a speed of 12 m/s.
Answer:
4 seconds
Explanation:
Assuming that the object started from rest,
v = at
--> t = v/a = (12 m/s) / (3 m/s^2)
= 4 seconds
If acceleration is zero what statement about velocity is
true *
A)Velocity is zero
B)Velocity is constant
C)Velocity cannot be determined
D) Velocity is changing
Answer: A
Velocity is zero because the acceleration isn't affected, and velocity is the rate of change, so it can't be any other options.
Answer:
B)Velocity is constantExplanation:
If an object moves with a velocity and there is no acceleration, then the velocity remains constant. His velocity after five second will be equal to his initial velocity.#keeplearning dude:)Determine the poles of the magnet. Look at the three compass readings that are on top of the magnet. Label the
end the compass points away from as "S" (south), and the other end that the compass points toward as "N" (north).
Record these poles in Figure 1.
Continue
Intro
Answer:
the red pointer on the magnet ( grey region) : points towards north
red pointer outside the magnet ( white region) is pointing towards south
Explanation:
please see the attached image
The first extra-solar planet around a main sequence star (i.e., not a neutron star or white dwarf) was found around the star 51 Pegasi in 1995. The large planet causes a measureable motion of the star around the center of mass of the system. Pegasi 51's orbital motion had a period of 4 days indicating a very large planet very close to the star. A large planet of mass M around a small star of mass 4M. The distance between planet and star is L. Both the planet and star will orbit around the center of mass of the system (marked by the red X).
Required:
Where is the center of mass of the star-planet system?
Answer:
[tex]r_{cm}[/tex]= 1/5 L
Explanation:
To find the center of mass of the system let's use
[tex]r_{cm} = \frac{1}{M}[/tex] ∑ r_i x_i
where m is the total mass of the system
let's apply this expression to our case
Let's set the reference frame on the star
[tex]r_{cm} = \frac{1}{M +4M} ( 4M 0 + M L)[/tex]
r_{cm} = [tex]\frac{1}{5}[/tex] L
[tex]r_{cm}[/tex]= 1/5 L
Hey, the breast center is 1/5 of the distance between the star and the planet.
1. Objects become electrically charged as a result of the transfer of
Answer:
Electron
Explanation:
An object can become electrically charged when it gains or loses an electron. Because an electron is negatively charged, when an object gains an electron it becomes negatively charged. Also, when it gives up an electron, it becomes positively charged. This positive charge is because the atom has one proton more than electron. In a neutral atom, the number of the proton is equal to the number of the electron. An electron is negatively charged, and a proton is positively charged.
A 0.5kg ball of clay originally moving at 6 m/s strikes a wall and comes to rest in 0.25s, what is the magnitude of the impulse given to the ball of clay?
A) 0.75 kg m/s
B) 1.5 kg m/s
C) 3.0 kg m/s
D) 12 kg m/s
Answer:
C I did USA testprep
Explanation:
While flying at an altitude of 5.75 km, you look out the window at various objects on the ground. If your ability to distinguish two objects is limited only by diffraction, find the smallest separation between two objects on the ground that are distinguishable. Assume your pupil has a diameter of 4.0 mm and take ???? = 460 nm.
Answer:
the smallest separation between two objects is 0.8067 m
Explanation:
Given the data in the question;
Altitude h = 5.75 km = 5750 m
Diameter D = 4.0 mm = 0.004 m
λ = 460 nm = 4.6 × 10⁻⁷ m
Now, Using Rayleigh criterion for Airy disks resolution.
we know that, Minimum angular separation for resolving two points is;
θ = 1.22λ / D
so we substitute
θ = (1.22 × 4.6 × 10⁻⁷) / 0.004
θ = 5.612 × 10⁻⁷ / 0.004
θ = 1.403 × 10⁻⁴ rad
so minimum separation [tex]d_{min[/tex] = θh
so we substitute
[tex]d_{min[/tex] = (1.403 × 10⁻⁴) × 5750 m
[tex]d_{min[/tex] = 0.8067 m
Therefore, the smallest separation between two objects is 0.8067 m
g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller pipe that is 10.0 cm in diameter. Determine: a) The speed of the water flowing through the 10.0 cm diameter pipe. b) The mass of water that flows through the larger pipe in 1.00 minute. c) The mass of water that flows through the smaller pipe in 1.00 minute.
Answer:
a) v₂ = 30 m/s
b) m₁ = 12600 kg
c) m₂ = 12600 kg
Explanation:
a)
Using the continuity equation:
[tex]A_1v_1 = A_2v_2[/tex]
where,
A₁ = Area of inlet = π(0.15 m)² = 0.07 m²
A₂ = Area of outlet = π(0.05 m)² = 0.007 m²
v₁ = speed at inlet = 3 m/s
v₂ = speed at outlet = ?
Therefore,
[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]
v₂ = 30 m/s
b)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₁ = mass of water flowing in = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]
m₁ = 12600 kg
c)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₂ = mass of water flowing out = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]
m₂ = 12600 kg
1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20
Answer:
It will go up to 93.75 m before it is moving at 20 m/s
Explanation:
As we know that
[tex]v^2 - u^2 = 2aS[/tex]
here v is the final speed i.e 20 m/s
u is the initial speed i.e 5 m/s
a is the acceleration due to gravity i.e 2 m/s^2
Substituting the given values in above equation, we get -
[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters
Pause
He
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be:
OA. 40 m/s
OB. 9.8 m/s2
OC 2.5 m/s2
OD. 0.40 m/s2
Answer:
C
Explanation:
a=f/m
10Kgm/s2/4kg
2.5m/s2
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity, or g. It is a vector quantity whose strength or magnitude is determined by the norm and whose direction correlates with a plumb bob.
Given in the question, force 10 N and mass 4.0 Kg the acceleration is,
a = 10/4 = 2.5 m/sec²
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².
To learn more about acceleration refer to the link:
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The density of 1 kilogram of gold is
Answer:
0.02 kg/cm³
Explanation:
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
pls help
What is the frequency range of UV light? Of infrared light?
Answer:
violet region (UV) is wavelength less than 400 nm, the most common is between 200 and 400 nm,
The infrared (IR) range wavelengths greater than 700 nm
Explanation:
The spectral region near the visible is what humans can see with our eyes.
The range of the ultra violet region (UV) is wavelength less than 400 nm, the most common is between 200 and 400 nm, this radiation is responsible for tanning and skin burns.
The infrared (IR) range is a very wide range that begins at wavelengths greater than 700 nm and continues up to approximately 25,000 nm (1012 Hz).
A sound wave travels with a velocity of 1.5 m/s and has a frequency of 500 Hz. What is its wavelength?
If a runner is running at 100 meters per minute at the exact moment they cross the finish line of a race, it is the
instantaneous speed.
O True
O False
Answer:
Hello! Your answer would be, O False
Explanation:
Hope I helped! Brainiest plz!♥ Have a nice morning! Hope you make a 100%! -Abby
Answer:
False is the correct answer.
Explanation:
plz mark me as brainliest.
How much heat is needed to warm 365 mL of water in a baby bottle from 240C to 38C
Answer:98.6 degrees Fahrenheit
Explanation:
how can we know that atmosphere exert pressure explain with figure
Can you please help me ?
Answer:
Distance is 300 and displacement is 100
Explanation:
Distance= 100+100+100=300
Displacement=100
plz mark me as brainliest.
What is the period of a wave with a speed of 20.0 m/s and a frequency of 10.0 Hz?
im confused hold on imma send you a link to the answerExplanation:
Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 132 W/m2 . You may want to review (Pages 897 - 898) . Part A What is the angle from vertical of the axis of the second polarizing filter
Answer:
θ = 32.4º
Explanation:
For this exercise let's use Malus's law
I = Io cos² θ
in this case it indicates that the incident intensity is 370 W/m², when the first polarization passes, only the radiation with the same polarization of the polarizer emerges, that is, vertical
I₀ = 370/2 = 185 W / m²
this is the radiation that affects the second polarizer, let's apply the expression of Maluz
θ = cos⁻¹ ([tex]\sqrt{\frac{I}{I_o} }[/tex])
θ = cos⁻¹ ([tex]\sqrt{132/185}[/tex])
θ = cos⁻¹ (0.844697)
θ = 32.4º
Two risks of exposure to High levels of UV radiation
Answer:
uv radiation cause cancer
uv radiation effect our eyes
Answer:
you can get
1:skin cancer
2:eye damage
3:skin damage
4:immune system suppression
choose which two u want
hope this helped
:)
Explanation:
Which electromagnetic wave has the shortest wavelength?
Which electromagnetic wave would have a wavelength approximately the size of the tip of a pencil?
Which electromagnetic wave has the longest wavelength?
Answer:
shortest - gamma Ray's
size of a pencil tip : most likely microwaves
longest - radio waves
Gamma rays have the shortest wavelength whereas Radio waves have the longest wavelength. The wave whose wavelength is about the size of the tip of a pencil is infrared ray.
What is an Electromagnetic spectrum?The whole electromagnetic waves are divided into several classes based on their frequencies or energies as follows, in the order of increasing frequency:
Radio waves
Microwaves
Infrared
Visible
Ultraviolet
X-rays
Gamma rays
The Gamma rays are the shortest wavelength group with frequencies greater than 3 × 10²⁰ Hz.
The Radio waves are the longest wavelength group. Their frequency is lower than 3 × 10⁷ Hz.
The tip of a pencil is about 0.5mm in width.
Calculate the frequency of the wave whose wavelength is about the size of the tip of a pencil i.e. λ = 0.5 mm,
ν = 3×10⁸/(0.5×10⁻³) = 6×10¹¹ Hz.
The infrared rays have frequencies in the range of 3×10¹¹ Hz to 3×10¹³ Hz. So, the wave is infrared.
Infrared waves have a wavelength about the tip of a pencil.
Learn more about the electromagnetic spectrum here
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A whole set of birdfeeders are designed using conservation of Angular Momentum to spin when a squirrel jumps on them. This can throw the squirrel off (though not all squirrels give up that easily - see this video for an example). A bird, landing, doesn't cause the same problem. A squirrel, with a mass of 3.00 kg launches itself at the bird feeder with a velocity of 3.40 m/s. The bird feeder has a radius of 6.30 cm and a Moment of Inertia of 2.00 kg m2. Initially the bird feeder is not rotating at all, but starts rotating when the squirrel lands on the outer edge (at the same radius as described above). You can assume that the squirrel is small compared to the size of the bird feeder radius (not true in the video, but it does make this a bit easier for out calculations). What is the angular velocity of the bird feeder - squirrel system after the squirrel lands on it
Answer:
w = 0.319 rad / s
Explanation:
This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.
initial instant. Before the squirrel jumps
L₀ = m v r
final instant. After the trough and the squirrel are together
L_f = (I_fetter + I_ardilla) w
angular momentum is conserved
L₀ = L_f
m v r = (I_fetter + I_ardilla) w
w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]
the moment inercial ofbody is
I_thed = 2.00 kg m²
We approach the squirrel to a specific mass
I_ardilla = m r²
we substitute
w = m v r / ( I_[feefer + m r²)
let's calculate
w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )
w = 0.6426 / 2.0119
w = 0.319 rad / s
The speed of surface waves in water decreases as the water becomes shallower. Suppose waves travel across the surface of a lake withn a speed of 2m/s and a wavelength of 1.5m. When these waves move into a shallower part of the lake, their speed decreases to 1.6m/s, though their frequency remains the same. Find the wavelength of the wave in the shallower water.
Answer:
The correct solution is "1.2 m".
Explanation:
The given values are:
Wavelength of waves,
λ = 1.5 m
Speed of waves on surface,
V = 2 m/sec
Speed of waves in water,
V₁ = 1.6 m/sec
As we know,
⇒ [tex]V=f\times \lambda[/tex]
or,
⇒ [tex]f=\frac{V}{\lambda}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2}{1.5}[/tex]
⇒ [tex]=1.33 \ Hz[/tex]
hence,
⇒ [tex]\lambda_1=\frac{V_1}{f}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{1.6}{1.33}[/tex]
⇒ [tex]=1.2 \ m[/tex]
Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human is taken to be 90.0 years, a spaceship would need to achieve some minimum speed vmin to deliver a living human being to this galaxy. How close to the speed of light would this minimum speed be? Express your answer as the difference between vmin and the speed of light c.
Answer:
[tex]0.0018833\ \text{m/s}[/tex]
Explanation:
[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]
[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]
[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
We have the relation
[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]
[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]
The required answer is [tex]0.0018833\ \text{m/s}[/tex].
The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves past a space dock at 0.5c. As rocket ship Adventure passes by the space dock, the ship's captain flashes a flashlight at 1.20-s intervals as measured by space-dock personnel.
Required:
How often does the flashlight flash relative to the captain?
Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
