Answer:
a) # lap = 301.59 rad , b) L = 90.48 m
Explanation:
a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm
# turns = 1 turn (24 cm / 0.5 cm)
# laps = 48 laps
Let's reduce to radians
# laps = 48 laps (2 round / 1 round)
# lap = 301.59 rad
b) Each lap gives a length equal to the length of the circle
L₀ = 2π R
L = # turns L₀
L = # turns 2π R
L = 48 2π 30
L = 9047.79 cm
L = 90.48 m
The spool needs to be turned through 2π radians to wrap one even layer of wire. The length of the wound wire can be calculated using the circumference formula.
Part a: To wrap one even layer of wire around the spool, the spool must be turned through an angle of 2π radians because one complete revolution is equal to 2π radians.
Part b: To calculate the length of the wire wound on the spool, we find the circumference of the spool using the formula: circumference = 2πr, where r is the radius of the spool (30 cm). The length of the wound wire is the product of this circumference and the height of the spool (24 cm).
Let v⃗ A be the velocity of the car at point A. What can you say about the acceleration of the car at that point?
Answer and Explanation:
Let the velocity of the car at a point is A is v
We have to tell about the acceleration at that point
Acceleration at that point will be perpendicular to the velocity and directed inside of the track
We can tell anything about the magnitude of the acceleration as magnitude of acceleration is the rate of change of velocity and here we have no information about time
While a car's velocity at a given point only describes its speed and direction at that moment, its acceleration at that point accounts for changes in speed and direction over time. To ascertain acceleration at a point, the velocities before and after that point need to be compared.
Explanation:The velocity of a car at a given point, such as point A, tells us it's direction and speed at that instant. However, the acceleration of the car at that point is a different matter. Acceleration is defined as the change in velocity per unit time, and it doesn't only involve speed, but also direction. A car could be moving at a constant speed, but if it's changing direction, it's considered to be accelerating. Therefore, to ascertain acceleration at point A, we would need to know not just the velocity at point A, but also at points immediately before and after.
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A particular star cluster contains stars all with the same apparent magnitude of +4. Near the cluster (and at the same distance from the Earth) is a single star with an apparent magnitude of +1. The star's brightness appears to be the same as the collective brightness of the entire cluster.
How many stars are in this cluster?
Answer:
3,000 stars.
Explanation:
Subtract the magnitude of the single star from the magnitude of the star cluster.
(+4) - (+1) = +3
Then multiply your answer by 1000
= 3,000 is a stars.
The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.
A. on the half of the road with traffic going in the opposite direction
B. on the half of the road that the vehicle is traveling
C. anywhere on a two-way street
Answer: the correct option is B ( on the half of the road that the vehicle is traveling).
Explanation: according to Georgia Code About Pedestrians; The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is
on the half of the road that the vehicle is traveling.
Answer:
B
Explanation:
The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.
A. on the half of the road with traffic going in the opposite direction
B. on the half of the road that the vehicle is traveling
C. anywhere on a two-way street
According to the law regrading pedestrian, the pedestrian has the right of way immediately she is crossing on the crosswalk and is half of the road that the vehicle is travelling.
These rules are there for the safety of the driver, pedestrians and other road users
Suppose you have an accounting background and you are considered an expert in the field of discovering errors in electronic spreadsheets. You are considered a knowledge engineer if you are part of a team that is developing an expert system to discover errors in spreadsheets. True or false?
Answer: True
Explanation: A knowledge engineer is a professional who works to improve and builds with the aid of science and also implements advanced logic and AI( artificial intelligence) in computer systems in order to investigate particular issues.
A knowledge engineer also helps tries to emulate the judgements of man.
Since you are part of the team involved in discovering errors in the electronic spread sheet you are a knowledge engineer.
Technician A says that the aspect ratio of a tire represents the relationship between the tire's cross-sectional height to its cross-sectional width. Technician B says that low aspect ratios provide a softer ride because they will deflect more over irregular surfaces and under heavy loads. Who is correct?
a. Technician A
b. Technician B
c. Both A and B
d. Neither A nor B
Answer:
Technician A
Explanation:
Often referred to as the profile or series, the aspect ratio of a tire is determined by dividing a tire’s section height by its section width when the tire is: inflated to maximum air pressure, mounted on the approved measuring rim, and under no load. This rules out Technician B.
A tire with a lower aspect ratio responds to lateral force more effectively than a tire with a higher aspect ratio. The aspect ratio affects steering stability. Generally, the shorter the sidewall, or the lower the aspect ratio, the less time it takes to transmit the steering input from the wheel to the tread. The result is quicker steering response. Aspect ratio also affects the tread contact patch. As a rule, a low-profile tire produces a wider tread contact patch. This wider tread contact patch creates a stiffer footprint that reduces distortion and provides improved cornering traction. Aspect ratio also impacts ride. A low-profile tire usually has a stiffer ride than the standard aspect ratio of 75 or more.
Final answer:
A. Technician A is correct about the aspect ratio of a tyre, while Technician B is incorrect about the impact of low aspect ratios on ride softness.
Explanation:
Technician A is correct. The aspect ratio of a tyre indeed represents the relationship between the tyre's cross-sectional height to its cross-sectional width. The aspect ratio is a critical factor in determining the tyre's performance and characteristics. Technician B is incorrect. Low aspect ratios do not provide a softer ride because they will deflect more over irregular surfaces.
In the given case, the aspect ratio primarily affects the tyre's handling, stability, and sidewall height. In conclusion, Technician A provides an accurate explanation regarding the aspect ratio of a tyre in relation to its dimensions and characteristics.
A solar eclipse will occur Group of answer choices
1. at every full Moon. during every new Moon.
2. at full Moon only when the Moon is on the ecliptic.
3. at new Moon only when the Moon is on the ecliptic.
Answer:
3. at new Moon only when the Moon is on the ecliptic.
Explanation:
Solar eclipse is the condition when the moon comes in between the sun and the earth. In this condition the moon casts its shadow on the earth.Whether the eclipse is a total solar eclipse, a partial solar eclipse or an annular solar eclipse depends on various factors, but the position of the moon must be on the same orbital plane as that of the earth's orbit around the sun.The sun is about 400 times larger than the moon in size and the sun is almost 400 times farther from the earth than the moon is, this makes it possible for the moon to cover the sun completely leading to a complete solar eclipse.As we know that the orbit of the earth around the sun and the orbit of the moon around the earth is elliptical which leads to a variation in the distance from their rotating centers, so not of every eclipse the moon covers the sun completely developing an annular eclipse.When the moon is close enough to the earth on the ecliptic but not completely aligned in between the sun and the earth leads to a partial solar eclipse.A solar eclipse occurs during a new Moon, exclusively when the Moon is on the ecliptic path. This is when the Moon moves between the Earth and the Sun and blocks the sunlight. A solar eclipse can't occur during a full Moon.
Explanation:A solar eclipse happens during a new Moon when the Moon is on the ecliptic path. This event occurs when the Moon moves between the Earth and the Sun and blocks the Sun's light. The Moon must be on the ecliptic, which is the apparent path of the Sun's motion on the celestial sphere, for the alignment to be precise. If the Moon is not on this path, then it will not block the Sun, and a solar eclipse will not occur. It's also important to note that during a full Moon, the Moon is opposite the Sun in our sky, so a solar eclipse cannot occur.
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Which two security measures must an engineer follow when implementing layer 2 and layer 3 network design?
Answer:
Utilize the Private VLAN feature to segregate network traffic at Layer 2, and
Utilize the ARP inspection feature to help prevent the misuse of gARP.
Explanation:
An Engineer designing layer 2 and layer 3 network, must Utilize the Private VLAN feature to segregate network traffic at Layer 2 and Utilize the ARP inspection feature to help prevent the misuse of gARP.
Why are summer days longer than winter days on earth?
Answer:
Tilt of the Earth
Explanation:
The Earth's rotational axis is titled at an angle 23.4°. When then northern hemisphere is pointed towards the Sun it is summer in the northern hemisphere and winter in the southern hemisphere.
In the northern hemisphere a day would be longer and gets more sunlight than the southern hemisphere.
The same happens when it is summer in the southern hemisphere.
Answer:
The sun is father away during the winter.
Explanation:
In the northern hemisphere, during the winter it is summer in the southern hemisphere. The sun is farther away for us so the days get shorter in the winter since the sun inches away faster.
A vertical spring has a spring constant of 100 newtons per meter. When an object is attached to the bottom of the spring, the spring changes from its unstretched length of 0.50 meters to a length of 0.65 meters. The magnitude of the weight of the attached object is
a. 1.1N
b. 15N
c. 50N
d. 65N
Answer:
option B
Explanation:
given,
spring constant = k = 100 N/m
distance of unstretched length = 0.5 m
distance after stretched length = 0.65 m
magnitude of weight = ?
We know,
W = K x
x is the displacement
x = 0.65 - 0.5 = 0.15 m
W = K x
W = 100 x 0.15
W = 15 N
Hence, the correct answer is option B
Final answer:
The weight of the attached object is found by multiplying the spring constant (100 N/m) by the stretch in the spring (0.15 m), which equals 15 N. Therefore, the weight of the object is 15 N.
Explanation:
The weight of the attached object can be calculated using Hooke's Law, which states that the force exerted by the spring (F) is equal to the spring constant (k) multiplied by the displacement from the rest position ( extDelta x): F = k extDelta x. In this case, the spring has a spring constant of 100 newtons per meter and the displacement from the original length of 0.50 meters to the new length of 0.65 meters is 0.15 meters.
Thus, F = 100 N/m exttimes 0.15 m, which gives us a force of 15 N. Since the weight of the object is the force exerted by the spring and these two forces are equal and opposite when the spring is at rest with the object hanging, the weight of the attached object is 15 N. Therefore, the correct answer is (b) 15N.
A student attaches a rope to a 20.0 kg box of books. He pulls with a force of 90.0 N at an angle of 30 degrees with the horizontal. To make his job easier, he oiled the bottom of the box, reducing the coefficient of friction to 0.250. What is the magnitude of the acceleration of the box?
Answer:
The magnitude of the acceleration of the box is 2.01 m/s².
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following horizontal forces:
Fr = friction force.
Fx = Horizontal component of the applied force, F.
And we have the following vertical forces:
Fy = vertical component of the applied force.
N = normal force exerted on the box.
W = weight of the box.
According to Newton´s second law:
∑F = m · a
Then, in the horizontal direction:
Fx - Fr = m · a
Where "m" is the mass of the box and "a" its acceleration.
Fx can be obtained by trigonometry (see figure):
Fx = F · cos 30°
Fx = 90.0 N · cos 30°
Fr is calculated as follows:
Fr = μ · N
Where μ is the coefficient of friction and N the normal force.
So, we have to find the magnitude of the normal force.
Using Newton´s second law in the vertical direction:
∑F = N + Fy - W = m · a
Notice that the box has no vertical acceleration, then:
N + Fy - W = 0
Solving for N:
N = W - Fy
The weight is calculated as follows:
W = m · g
Where g is the acceleration due to gravity:
W = 20.0 kg · 9.8 m/s² = 196 N
And the vertical component of the applied force can be obtained by trigonometry:
Fy = F · sin 30°
Fy = 90.0 N · sin 30°
The normal force will be:
N = W - Fy = 196 N - 90.0 N · sin 30°
N = 151 N
Now, we can calculate the friction force:
Fr = μ · N
Fr = 0.250 · 151 N
Fr = 37.8 N
And now, we can obtain the acceleration of the box:
Fx - Fr = m · a
(Fx - Fr) / m = a
(90.0 N · cos 30° - 37.8 N ) / 20.0 kg = a
a = 2.01 m/s²
The magnitude of the acceleration of the box is 2.01 m/s².
Which layer of the atmosphere absorbs the most ultraviolet rays
Answer:
The stratosphere
Explanation:
In the lower portion of the stratosphere there is a high concentration of ozone in relation to other parts of the atmosphere. This region is known as the ozone layer. This layer absorbs most of the Sun's ultraviolet radiation, especially ultraviolet B rays, allowing ultraviolet A rays to pass.
Answer:
TROPOSPHERE
Explanation:
An artery is constricted at one location to 1/2 its normal cross-sectional area.
How does the speed of blood past the constriction compare to the speed of blood flow in the rest of the artery? (Note: Assume ideal fluid flow.)
a) it is 1/4 as fast
b) it is 1/2 as fast
c) it is 2 times as fast
d) it is 4 times as fast
Answer:
v₂ = 2 v₁ blood velocity in the built zone is twice the speed of the normal zone, so the correct response is c
Explanation:
For that problem we must use the fluid continuity equation
A₁ v₁ = A₂ v₂
Where point 1 is in the normal artery and point 2 in the contraction
v₂ = v₁ A₁ / A₂
In the exercise indicates
A₂ = ½ A₁
v₂ = 2 v₁
The blood velocity in the built zone is twice the speed of the normal zone, so the correct response is c
When following a motorcycle, drivers must remember that motorcycles can ____________ than other vehicles.
Answer:
Stooping more Quickly
Explanation: Stooping more quickly means that Motor cycles can easily bend and stop,this is due to it's design, Motor cycle have only two wheels,they contain just two tyres one in front and one at the back,they are not as firm on the ground when compared to other vehicles such Motor car. They have a lower weight which makes them easily bent or stopped once there's any barrier.
Motorcycles can stop quicker than other vehicles.
Explanation:When following a motorcycle, drivers must remember that motorcycles can stop quicker than other vehicles. Due to their smaller size and weight, motorcycles have a shorter stopping distance. This means that drivers should maintain a safe following distance to avoid rear-ending the motorcycle.
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An electron is accelerqated in the uniform field betwen two parallel charged oplates. The separation of the plates is 1.20 cm
Answer:
The speed of electron is [tex]8.7\times10^{6}\ m/s[/tex]
Explanation:
Given that,
Separation of the plate = 1.20 cm
Suppose the field is [tex]E=1.80\times10^{4}\ N/C[/tex].
If the electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.
What the speed does it leave the hole?
We need to calculate the acceleration
Using formula of electric force
[tex]F = qE[/tex]
[tex]ma=qE[/tex]
[tex]a=\dfrac{qE}{m}[/tex]
We need to calculate the speed of electron
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]v^2=2as[/tex]
Put the value of acceleration in the formula
[tex]v^2=2\times\dfrac{qE}{m}\times s[/tex]
Put the value into the formula
[tex]v^2=2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}[/tex]
[tex]v=\sqrt{2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}}[/tex]
[tex]v=8.7\times10^{6}\ m/s[/tex]
Hence, The speed of electron is [tex]8.7\times10^{6}\ m/s[/tex]
What is the difference between balanced and unbalanced forces
Balanced forces are equal and opposite, causing no change in an object's movement, leading to static equilibrium. On the contrary, unbalanced forces change an object's velocity or direction, causing it to move or accelerate.
Explanation:The primary difference between balanced and unbalanced forces lies in their effects on the motion of an object. Balanced forces are equal and opposite forces that act on an object, resulting in no change in its velocity or direction. An example would be a static car or an ice hockey stick lying flat with two equal and opposite forces applied to it. Friction is negligible and the gravitational force is balanced by the support, leading to static equilibrium.
On the other hand, unbalanced forces are not equaled.
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a skater increases her speed uniformly from 2.0 m/s to 10 m/s in 3.0s. what is the acceleratioon of the biker?
Answer:
a= 2.667 m/s²
Explanation:
Given that
Initial velocity ,u= 2 m/s
Final velocity ,v= 10 m/s
Time taken ,t= 3 s
As we know that
v = u + at
a=Acceleration ,u=Initial velocity ,v= Final velocity ,t=time
Now by putting the values in the above equation
v = u + at
10 = 2 + a x 3
8 = 3 a
[tex]a=\dfrac{8}{3}\ m/s^2[/tex]
a= 2.667 m/s²
Therefore the acceleration of the biker will be 2.667 m/s²
Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object of mass M > m, which is initially at rest on the surface?
The collision is elastic.
A. All of the initial kinetic energy of the less-massive object is lost.
B. The momentum of the objects that are stuck together has a smaller magnitude than the initial momentum of the less massive object.
C. The speed of the objects that are stuck together will be less than the initial speed of the less-massive object.
D. The direction of motion of the objects that are stuck together depends on whether the hit is a head on collision.
Answer:
C. The initial momentum should be equal to the final momentum due to the conservation of momentum.
[tex]P_{initial} = mv_0\\P_{final} = (M+m)v_1\\v_1 = \frac{m}{M+m}v_0[/tex]
Since m/(M+m) < 1, v_1 > v_0.
Explanation:
Wrong -> A. Since the smaller particle still moves after the collision, it has a kinetic energy.
Wrong -> B. The total initial momentum is equal to the momentum of the smaller particle. Therefore, the momentum of the objects that stuck together is equal to that of the smaller object.
Wrong -> D. Since the bigger object is initially at rest and the surface is frictionless, the direction of motion will be the same as the direction of the smaller particle.
A body of mass 400 kg is suspended at a lower end of a light vertical chain and is being pulled up vertically. Initially the body is at rest and the pull on the chain is 6000 g (g is 9.8 m/s^2) The pull gets smaller uniformly at the rate of 360g N per each meter through which the body is raised. What is the velocity of the body when it has been raised 10m?
Answer:31.62 m/s
Explanation:
Given
mass of body [tex]m=400 kg[/tex]
Pull on chain is [tex]F_1=6000g N=60 kN[/tex]
Pull get smaller at the rate of [tex]F_2=360g N/m[/tex]
Net Upward Force [tex]F=6000 g-360 g\times 10=24 kN[/tex]
net acceleration [tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{24\times 1000}{m}[/tex]
[tex]a=\frac{24000}{400}[/tex]
[tex]a=60 m/s^2[/tex]
but g is acting downward
[tex]a_{net}=a-g=60-10=50 m/s^2[/tex]
using [tex]v^2-u^2=2 as[/tex]
here initial velocity is zero
[tex]v^2=2\times 50\times 10[/tex]
[tex]v=31.62 m/s[/tex]
The velocity of the body will be "31.62 m/s".
Given:
Mass of body,
m = 400 kgPull on chain,
[tex]F_1= 6000 \ g.N[/tex][tex]F_2 = 360 \ g.N/m[/tex]The net upward force will be:
→ [tex]F = 6000-360\times 10[/tex]
[tex]= 6000-3600[/tex]
[tex]= 2400[/tex]
or,
[tex]= 24 \ kN[/tex]
Now,
The net acceleration will be:
→ [tex]a = \frac{F}{m}[/tex]
[tex]= \frac{24\times 1000}{400}[/tex]
[tex]= 60 \ m/s^2[/tex]
But,
"g" is acting downwards then,
→ [tex]a_{net} = a -g[/tex]
[tex]= 60-10[/tex]
[tex]= 50 \ m/s^2[/tex]
By using,
→ [tex]v^2-u^2=2as[/tex]
[tex]v^2= 2\times 50\times 10[/tex]
[tex]= 1000[/tex]
[tex]v = \sqrt{1000}[/tex]
[tex]= 31.62 \ m/s[/tex]
Thus the above answer is right.
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Unrestrained occupants of a car keep moving during the time the car takes to stop due to ________.
A. centrifugal force
B. gravity
C.inertia
D.precipitation
Answer:
C.inertia
Explanation:
Inertia is a property of matter to maintain its state of motion.
In this case, if the car was moving along with the people in it when the car stopped, the inertia of the people will cause them to continue moving.
This is why in a crash a person can be shot from the car, due to the inertia he has from his previous state of motion.
The complete sentence is: Unrestrained occupants of a car keep moving during the time the car takes to stop due to inertia. (option C)
Unrestrained occupants of a car continue moving forward when the car stops due to inertia, as an object in motion remains in motion unless acted upon by an external force.
Explanation:Unrestrained occupants of a car keep moving during the time the car takes to stop due to inertia. This phenomenon occurs because, according to Newton's first law, an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. In the context of a car, when it suddenly stops, the passengers inside will continue moving forward because their bodies were initially in motion and there is no forward force directly applied to them to cause a deceleration similar to that of the car.
Centripetal force is the force that causes an object moving in a circular path to move towards the center of the path. However, when the force disappears, inertia is what carries the object in a straight line, tangent to the circular path. While centrifugal force is often used in common language to describe the sensation of being pushed outward in a rotating system, it is not an actual force but rather a perceived effect due to inertia.
Earl is using his hands to hold a metal pan 10 centimeters above a hot burner. How can this scenario be changed to demonstrate conduction between the pan and the burner?
Answer:
Earl needs to touch the pan to the burner.
Explanation:
Earl is demonstrating heat transfer by conduction.
Conduction is the transition of heat energy by direct contact. Hotter molecules have more momentum and they move faster.These hot molecules collide with the cooler ones, transfer their energy to cooler particles , this process is continued to pass it on to other molecule near to them.
Earl needs to touch the pan to the burner.
Answer:
D the answer is D
Explanation:
A biker to move from speed of 0 m/s to final speed of 25 m/s in 10 s. What is the acceleration of the bike?
Answer:
a = 2.5 m/s²
Explanation:
given,
initial speed = 0 m/s
final speed = 25 m/s
time = 10 s
acceleration of bike is equal to change in velocity per unit time
[tex]a = \dfrac{\Delta v}{t}[/tex]
[tex]a = \dfrac{v_f-v_i}{t}[/tex]
[tex]a = \dfrac{25-0}{10}[/tex]
a = 2.5 m/s²
hence, acceleration of bike is equal to a = 2.5 m/s²
A 2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per second east. What will be the cart's velocity after a 6.0-newton westward force acts on it for 2.0 seconds and in what direction?
Answer:
v = 2 m/s (West)
Explanation:
Given
m = 2 Kg
v initial = 4 m/s (East)
F = 6 N (West)
t = 2 s
We can use the formula
v = v initial + a*t
if F = m*a ⇒ a = F/m = 6 N / 2 Kg
a = 3 m/s² (West)
then
v = v initial + a*t = (4 m/s) + (-3 m/s²)(2 s)
v = - 2 m/s = 2 m/s (West)
The velocity is a vector quantity. The velocity of the given cart is 2 m/s to the west.
To solve the given problem, use the kinematic equation,
[tex]v = s + at[/tex]........................1
Where,
[tex]s[/tex] - initial speed = 4 m/s
[tex]a[/tex] - acceleration
[tex]t[/tex] - time - 2 s
Acceleration can be calculated by the formula,
[tex]F = ma[/tex] or
[tex]a = \dfrac Fm[/tex]
Where,
m = mass = 2 Kg
F = force = 6 N
So,
[tex]a = \dfrac {6}2\\\\a = 3\rm \ m/s^2[/tex]
Put the values in equation 1,
[tex]v = 4 +3\times 2\\\\v = \bold {2 m/s}[/tex]
Therefore, the velocity of the given cart is 2 m/s.
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A ray of white light moves through the air and strikes the surface of water in a beaker. The index of refraction of the water is 1.33 and the angle of incidence is 30 degrees. All of the following are true EXCEPT:_________
I. the angle of reflection is 30 degrees
II. the angle of refraction is 30 degrees
III. total internal reflection will result, depending on the critical angle
Answer:
All of the following are true EXCEPT, the angle of refraction is 30 degrees
Explanation:
It is given that,
The index of refraction of the water is 1.33 and the angle of incidence is 30 degrees. We know that when a ray of white light moves through the air and strikes the surface of water in a beaker, refraction occurs. Let r is the angle of refraction. It can be calculated using Snell's law as :
[tex]n=\dfrac{sin\ i}{sin\ r}[/tex]
[tex]sin\ r=\dfrac{sin\ i}{n}[/tex]
[tex]sin\ r=\dfrac{sin(30)}{1.33}[/tex]
[tex]r=22.08^{\circ}[/tex]
So, the angle of refraction is 22.08 degrees.
According to the law of reflection, the angle of incidence is equal to the angle of reflection. Since, it is incident at an angle of 30 degrees, so the angle of reflection will be 30 degrees. Also, total internal reflection will result, depending on the critical angle.
So, the wrong statement is the angle of refraction is 30 degrees. It is equal to 22.08 degrees.
What types of compounds are involved in solutions that are used to grow crystals? In your own words please
Answer:
This is a table of common chemicals that produce nice crystals. The color and shape of the crystals are included. Many of these chemicals are available in your home. Other chemicals in this list are readily available online and are safe enough for growing crystals at home or in a school. Recipes and specific instructions are available for hyperlinked chemicals.
Ionic compounds, held together by electrostatic forces, form crystalline structures that exhibit rigid and brittle properties. They tend to have high melting and boiling points, reflecting the strength of the ionic bonds. Though ionic solids are poor conductors in their solid state, they readily dissolve in water, becoming good conductors when melted or dissolved.
Explanation:Compounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds: electrostatic forces of attraction between oppositely charged cations and anions. The properties of ionic compounds shed some light on the nature of ionic bonds. Ionic solids exhibit a crystalline structure and tend to be rigid and brittle; they also tend to have high melting and boiling points, which suggests that ionic bonds are very strong. Ionic solids are also poor conductors of electricity for the same reason the strength of ionic bonds prevents ions from moving freely in the solid state. Most ionic solids, however, dissolve readily in water. Once dissolved or melted, ionic compounds are excellent conductors of electricity and heat because the ions can move about freely.
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The total lifetime of the Sun is about 10 billion years, from when it was born to when it dies, and the half-life of 238U is 4.5 billion years. Suppose a particular meteoroid contained 24 micrograms of uranium-238 (238U) when the Sun was born. If the meteoroid is still in orbit when the Sun dies, it will contain _____
Answer:
a little less than 6 micrograms
Explanation:
The half-life of an element is defined as the time in which half of the isotope emits its radiation and becomes a different element. Therefore after 4.5 billion years the meteoroid will contain 12 micrograms of uranium, after 9 billion years will contain 6 micrograms. After 10 billion years will contain a little less than 6 micrograms.
Consider a telescope with a small circular aperture of diameter 2.0 centimeters.A) If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resolved if their angular separation is 3.0×10−5 radians?
Answer:
[tex]4.91803\times 10^{-7}\ m[/tex]
Explanation:
[tex]\theta[/tex] = Angular seperation = [tex]3\times 10^{-5}\ rad[/tex]
[tex]\lambda[/tex] = Wavelength
We have relation
[tex]sin\theta=1.22\dfrac{\lambda}{d}\\\Rightarrow \lambda=\dfrac{dsin\theta}{1.22}\\\Rightarrow \lambda=\dfrac{2\times 10^{-2}sin(3\times 10^{-5})}{1.22}\\\Rightarrow \lambda=4.91803\times 10^{-7}\ m[/tex]
The maximum wavelength of the telescope is [tex]4.91803\times 10^{-7}\ m[/tex]
The maximum wavelength λ at which the two sources can be resolved is of [tex]1.49 \times 10^{-10} \;\rm m[/tex].
Given data:
The diameter of the telescope is, d = 2 cm = 0.02 m.
The angular separation of the two point source of light is, [tex]\theta = 3.0 \times 10^{-5} \;\rm radians = 3.0 \times 10^{-5} \;\rm radians \times \dfrac{\pi}{180}\\\theta = 5.23 \times10^{-7}^{\circ}[/tex].
The angular separation of two point source is also known as apparent distance between the sources, and it is defined as the angle between the two objects as viewed directly by the observer.
The standard relation for the angular separation of telescope is given as,
[tex]sin \theta = 1.22 \times \dfrac{\lambda}{d}[/tex]
here, [tex]\lambda[/tex] is the maximum wavelength of light.
Solving as,
[tex]sin (5.23 \times 10^{-7}) = 1.22 \times \dfrac{\lambda}{0.02}\\\\\lambda = \dfrac{sin (5.23 \times 10^{-7}) \times 0.02}{1.22} \\\\\lambda = 1.49 \times 10^{-10} \;\rm m[/tex]
Thus, we can conclude that the maximum wavelength λ at which the two sources can be resolved is of [tex]1.49 \times 10^{-10} \;\rm m[/tex].
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Tarzan steps out of his house and grabs a hanging vine to swing to the ground. If his house is 11.3 m above the ground, what is Tarzan's speed when he reaches the ground?
Answer:
v = 14.88 m / s
Explanation:
For this exercise let's use energy conservation. We place the reference system on the ground
Start. Tarzan before jumping
Em₀ = U = m g h
Final. Upon reaching the ground
[tex]Em_{f}[/tex]= K = ½ m v²
Energy is conserved
Em₀ = [tex]Em_{f}[/tex]
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 11.3)
v = 14.88 m / s
Which element has properties of electrical conductivity and luster and exists as a liquid at STP?
Answer:
Mercury - hydrargyrum (Hg)
Explanation:Mercury is a chemical element with the symbol Hg and atomic number 80. It is commonly known as quicksilver and was formerly named hydrargyrum.
Symbol: Hg
Atomic number: 80
Atomic mass: 200.59 u
Melting point: -38.83 °C
Boiling point: 356.7 °C
Electron configuration: [Xe] 4d^14 5d^10 6s^2
Group :12
Ionization energy 1007.1J
Electronegativity :2
Crust Abundance : 0.085ppm
Electron Affinity : Unknown eV
Name: Mercury , the first planet in the solar system (Hg from former name hydrargyrum , from greek hydr - water and argyros silver .
I hope it helps. :).
Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0×1011 pascals. How far ( ΔL) would such a string stretch under a tension of 1500 newtons? Use two significant figures in your answer. Express your answer in millimeters.
Answer:
[tex]\Delta l=0.015m[/tex]
Explanation:
We have given initial length of the steel guitar l = 1 m
Cross sectional area [tex]A=0.5mm^2=0.5\times 10^{-6}m^2[/tex]
Young's modulus [tex]\gamma=2\times 10^{11}Pa[/tex]
Force F = 1500 N
So stress [tex]=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa[/tex]
We know that young's modulus [tex]=\frac{stress}{strain}[/tex]
So [tex]2\times 10^{11}=\frac{3\times 10^{9}}{strain}[/tex]
[tex]strain=1.5\times 10^{-2}=0.015m[/tex]
Now strain [tex]=\frac{\Delta l}{l}[/tex]
[tex]0.015=\frac{\Delta l}{1}[/tex]
[tex]\Delta l=0.015m[/tex]
Answer:
Explanation:
L = 1 m
A = 0.5 mm² = 0.5 x 10^-6 m²
Y = 2 x 10^11 Pa
F = 1500 N
ΔL = ?
Use the formula for the young's modulus
[tex]Y = \frac{FL}{A\Delta L}[/tex]
[tex]\Delta L = \frac{FL}{AY}[/tex]
[tex]\Delta L = \frac{1500\times 1}{0.5\times10^{-6}\times 2\times 10^{11}}[/tex]
ΔL = 0.015 m
ΔL = 0.02 m
When a sound wave travels directly toward a hard wall, the incoming and reflected waves can combine to produce a standing wave. There is an antinode right at the wall, just as at the end of a closed tube, so the sound near the wall is loud. You are standing beside a brick wall listening to a 80Hz tone from a distant loudspeaker.
How far from the wall must you move to find the first quiet spot? Assume a sound speed of 340 m/s.
Answer:
L= 1.06 m
Explanation:
frequency of a wave in a pipe closed at one end is given by
[tex]f= \frac{v}{4L}[/tex]
L= length of the pipe
v= velocity of wave
therefore, the length of the pipe
[tex]L= \frac{v}{4f}[/tex]
[tex]L= \frac{340}{4\times80}[/tex]
L= 1.06 m
L= 1.06 m far from the wall must you move to find the first quiet spot.
The distance from the wall(i.e the pipe length) you need to move to find the first quiet spot is 1.06 m.
What is the frequency of a sound wave in a closed pipe?The smallest standing wave pattern, i.e the first fundamental frequency or harmonic for a closed pipe usually consists of one node and an antinode. When the wavelength of a closed pipe of length L is 4 L, the basic standing wave is formed.
The frequency of a sound wave for a closed pipe at one end may thus be represented using the following formula:
[tex]\mathbf{f = \dfrac{v}{4L}}[/tex]
where;
L = pipe length = ???v = velocity of the wave = 340 m/sf = frequency = 80 HzThus, the length of the pipe can be computed as:
[tex]\mathbf{L = \dfrac{340 \ m/s}{4\times 80 \ Hz}}[/tex]
L = 1.06 m
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