Answer:
N1 Sin θ1 = N2 Sin θ2 Snell's Law
For blue light
Sin θ2 = N1 / N2 sin 35.8 = 2.444 / 1.531 * .5850 = .9339
θ2 = 69.04 deg
For red light
Sin θ2 = 2.410 /1.520 * .5850 = .9275
θ2 = 68.05 deg
The separation is .01 deg
3. Which vehicle has more momentum? *
1 point
A.
B
VA= 25 m/s
Ug = 20 m/s
ma= 15000 Kg me = 15000 Kg
Ο Α
О.
B
Answer:
the one with v = 25 m/s
Explanation:
Momentum = m * v
if they both have the same mass (15000 kg) , then the one with the higher v has more momentum...I think A= 25 m/s
A ball of mass 0.4kg is dropped from a height of 30m. Determine the potential energy on impact to the ground
Waldo gets stopped by the police.
Answer:
At least they found him.
Explanation:
can someone help with the vocabulary of the relative age of rocks
A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s2. What is the speed of the car after it has traveled 55 m?
Answer:
The speed is =15.91ms−1. The time is =6.92s
Explanation:
Which set of terms is in the proper order, from left to right, for labeling the
reactants and products in this reaction?
H3PO4 + 3KOH + 3H20 + K3P04
A. Base, salt, water, acid
O B. Salt, acid, water, base
C. Acid, base, water, salt
O D. Base, acid, water, salt
In what order do the stages of the birth process occur? delivery of the placenta, dilation and thinning of the cervix, delivery of the fetus uterine contractions, delivery of the placenta, delivery of the fetus delivery of the fetus, delivery of the placenta, delivery of the cervix dilation and thinning of the cervix, delivery of the fetus, delivery of the placenta
The stages of the baby birth are Dilation, strong contractions, and delivery of the baby and the placenta.
What are the stage of the birth process ?Dilation, intense contractions, and delivery of the baby and placenta are the phases of a baby's birth.
The stages of the birth process are as follows;
1. The first stage includes the latent phase, in which cervical dilation begins (3 to 5 cm) along with mild contractions, and the active phase, in which strong contractions occur along with complete or nearly complete dilation.
2. The second stage is fetus expulsion or delivery, in which the cervix is fully dilated, and the baby is born.
3. The third stage is placenta delivery or expulsion, which occurs as the placenta is delivered.
4. Dilation, intense contractions, and delivery of the baby and placenta are the phases of labor and delivery.
Hence,the stages of the baby birth are Dilation, strong contractions, and delivery of the baby and the placenta.
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Answer:
the answer is -dilation and thinning of the cervix, delivery of the fetus, delivery of the placenta
Explanation:
i took a quiz and got it right
any form of stored energy is described as
Given the information in the velocity vs. time graph, what is the displacement of the object after 1 second?
In the velocity vs. time graph, the displacement of the object after 1 second is 3 m.
What is displacement?The displacement of an object is the change in position of the object.
From the graph, the displacement of the object after 1 second is calculated as follows;
x = vt
where;
v is velocity = 3 m/st is time = 1 secondx = 3 x 1 = 3 m
Thus, in the velocity vs. time graph, the displacement of the object after 1 second is 3 m.
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5. For the following pairs of masses, which one has the largest gravitational attraction between masses? Which one has the smallest gravitational attraction between masses? Explain.
Answer:
MOST : E LEAST : D
Explanation:
Grav attraction = G m1 m1 / r^2
G and 'r' are the same for all of the pictures, so the one picture with the biggest mass has the MOST attraction and the one with the least mass has the least attraction.
The pair with the largest gravitational attraction is the Earth and the Moon.
The pair with the smallest gravitational attraction is the Earth and a grain of sand.
The gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In other words, the more massive the objects, the stronger the gravitational force between them. The closer the objects are, the stronger the gravitational force between them.
The Earth and the Moon are the most massive objects in the pair. The Earth has a mass of about 5.972 × 10²⁴ kg, and the Moon has a mass of about 7.34 × 10²² kg. The distance between the Earth and the Moon is about 384,400 km.
The grain of sand is much less massive than the Earth or the Moon. The mass of a grain of sand is about 10⁻⁶ kg. The distance between the Earth and a grain of sand is much greater than the distance between the Earth and the Moon.
Therefore, the gravitational attraction between the Earth and the Moon is much stronger than the gravitational attraction between the Earth and a grain of sand.
Here is a table of the gravitational attractions for each pair of masses:
Pair of masses Gravitational attraction
Earth and Moon 1.98 × 10²⁰ N
Earth and a grain of sand 1.0 × 10⁻¹⁷ N
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Which best ilustrates the way in which radiation transfers thermal energy?
Answer:
B) Warm radiates to Cool
Explanation:
Radiation transfers thermal energy without need for contact, where areas with higher thermal energy (hot) transfer energy to areas with lower thermal energy (cold)
Bacteria vary somewhat in size, but a diameter of 2.9 μm is not unusual.
A) What would be the volume (in cubic centimeters) of such a bacterium, assuming that it is spherical?
B) What would be the surface area (in square millimeters) of such a bacterium, assuming that it is spherical?
Answer:
a) 6.4 x 10^-12 cm^3
b) 17 x 10^-6 mm^2
Explanation
a). The shape is assumed to be spherical The volume = volume of a sphere = \frac{4}{3} \pi r^3
3
4
πr
3
V = \frac{4}{3}*3.142* 1.15^3
3
4
∗3.142∗1.15
3
= 6.3715 μm^3
1 μm^3 = 10^-12 cm^3
6.3715 μm^3 = 6.3715 x 10^-12 cm^3
==> 6.4 x 10^-12 cm^3
Answer:
See below
Explanation:
Volume of a sphere = 4/3 pi r^3
we are given diameter r = 1/2 dia = 1.45 u-meter = 1.45 x 10^-4 cm
4/3 pi (1.45 x 10^-4)^3 = 1.277 x 10^-11 cm^3 = VOLUME
1.45 u-meter = .00145 mm
Surface area = 4 pi r^2 = 4 pi (1.45 x 10^-3 ) ^2 = 2.64 x 10^-5 mm^2
Throughout the reflection, make sure you have a
down menus to complete the statements.
copy of the Student Guide and your data tables. Use the drop-
In Part I of this experiment, the
was intentionally manipulated. This was the independent
variable. The dependent variable measured was the
In Part II of this experiment, the
was intentionally manipulated. This was the
independent variable. The dependent variable measured was the
Answer:
The dependent variable is the variable that is studied while the independent variable is the variable that is being manipulated
Answer: In this experiment, the
✔ fan speed
was intentionally manipulated. This was the independent variable.
The dependent variable measured was the
✔ acceleration
.
Explanation:
A 85.0kg man and a 65, 0kg woman are riding a Ferris wheel with a radius of 20.0m. What is the Ferris wheels tangential velocity if the net centripetal force on the woman is 115N
Answer:
The Ferris wheel's tangential (linear) velocity if the net centripetal force on the woman is 115 N is 3.92 m/s.
Explanation:
Let's use Newton's 2nd Law to help solve this problem.
F = maThe force acting on the Ferris wheel is the centripetal force, given in the problem: [tex]F_c=115 \ \text{N}[/tex].
The mass "m" is the sum of the man and woman's masses: [tex]85+65= 150 \ \text{kg}[/tex].
The acceleration is the centripetal acceleration of the Ferris wheel: [tex]a_c=\displaystyle \frac{v^2}{r}[/tex].
Let's write an equation and solve for "v", the tangential (linear) acceleration.
[tex]\displaystyle 115=m(\frac{v^2}{r} )[/tex][tex]\displaystyle 115 = (85+65)(\frac{v^2}{20})[/tex][tex]\displaystyle 115=150(\frac{v^2}{20} )[/tex][tex].766667=\displaystyle(\frac{v^2}{20} )[/tex][tex]15.\overline{3}=v^2[/tex][tex]v=3.9158[/tex]The Ferris wheel's tangential velocity is 3.92 m/s.
How many times is the motion of the object stopped?
Answer:
4
Explanation:
We can say the motion of the object is stopped when its velocity is 0. That means we just have to check how many times the line of the graph intersects the x-axis. Hence, we can say that the object of the motion is stopped 4 times.
The physical property that determines that how easily heat and electricity pass through a material is?
A. conductivity
B. density
C. weight
D. hardness
Answer: Hello! I'm Jungkook. Here is your answer.....
A. conductivity
Explanation:
Hope this helps! Anneyong/Bye!
xoxoKookie
(See picture) may I have help!!?
How long will it take you to ride your bike 120 miles if you are riding at 5mph?
Legolas shoots 333 arrows at once from his bow. He has 177177177 arrows.
How many times can Legolas shoot his bow before he needs more arrows?
Answer:
532063
Explanation:
If he's shooting 333 arrows per shot and has 177177177 arrows, you need to divide to find the number of arrows he can shoot total, ignoring the decimal since you can't shoot a partial arrow.
177177177/333=532063
A long conducting cylindrical pipe has a length L
The charge distribute on the inner surface with the radius r₁ will be zero,In the region r₁<r<r₂ will also zero. While on the outer surface with radius r₂ will be -q.
What is Gauss law?The total electric flux out of a closed surface is equal to the charge contained divided by the permittivity,
According to Gauss Law. the electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane perpendicular to the field.
From the Gauss law, it is stated that the field inside the conductor is zero. Charge will lie only on the surface.
(a)The charge distribute on the inner surface with the radius r₁ will be zero.
(b) In the region r₁<r<r₂ will also zero.
(c)The outer surface with radius r₂ charge distribution will be -q.
Hence, the charge distribute on the inner surface with the radius r₁ will be zero,
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Determine the net force required to accelerate a 2,160 kg truck from 0 to 27 m/s in 10 s
∑F=m.a = m.(Δv/Δt)
∑F=2.16 x (27-0/10)
∑F=5.832 N
A 20.0 kg trunk is pushed across the floor of a moving van by a horizontal force. If the coefficient of kinetic friction between the trunk and the floor is 0.255, what is the magnitude of the frictional force opposing the applied force?
The magnitude of the frictional force opposing the applied force is 50 N.
What is the frictional force?The friction is caused due to the relative motion between two uneven contacting surfaces.
Given is the mass of trunk m = 20kg, kinetic coefficient of friction μ = 0.255.
The normal force acting on the trunk N = mg
N = 20 kg x 9.81m /s²
N = 196.2 Newton
The friction force is
f =μN
f = 0.255 x 196.2
f =50.031 N
Thus, the magnitude of the frictional force opposing the applied force is approximately 50N.
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A (1.28×10-5 C) charge is moving at (5.63×107 m/s) perpendicular to a magnetic field
of (8.91×10-4 T). What is the magnitude of the magnetic force on the charge?
A. 6.42×10 -1 N
B. 5.02×10 4 N
C. 7.21×10 2 N
D. 1.14×10 -8 N
Hello!
We can use the following equation for magnetic force on a moving particle:
[tex]F_B = qv \times B[/tex]
[tex]F_B[/tex] = Magnetic Force (? N)
q = Charge of particle (1.28 × 10⁻⁵C)
v =velocity of particle (5.63× 10⁷m/s)
B = Strength of magnetic field (8.91 × 10⁻⁴T)
This is a cross-product, so the magnetic force depends on the SINE of the angle between the particle's velocity vector and the magnetic field vector.
Since the charge is moving PERPENDICULAR to the field, the angle between the velocity and magnetic field is 90°. The sine of 90° = 1, so we can simplify the equation to:
[tex]F_B = qvB(1) = qvB\\[/tex]
Plug in the values and solve.
[tex]F_B = (1.28*10^{-5})(5.63*10^7)(8.91*10^{-4}) = \boxed{\text{A. } 0.6421 N}[/tex]
someone help me answer this.
Bug Walking on Pivoted Ring A ring of radius R and mass my lies on its side on a frictionless table. It is pivoted to the table at its rim. A bug of mass my walks on the ring with constant speed v relative to the ring, starting at the pivot, when the ring is initially at rest. Take k to point out of the page. top view frictionless table ring bug Pluot Point (a) What is the angular velocity of the ring when the bug is halfway around? Express you answer in terms of some or all of the following: mi, m2, u, R and k. (b) What is the angular velocity of the ring when the bug is back at the pivot? Express you answer in terms of some or all of the following: mı, m2, v, R and k.
The angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].
What is angular velocity?The velocity of a particle when moving in the circular path.
Let speed of the bug with respect to ground is u.
Speed of bag with respect to ring will be
v = u - (- Rω) =
Then, u = v- Rω...............(1)
Angular momentum of ring and bug will remain conserved.
Initial momentum: L ring + Lbug =0
Final momentum: -2m₁ R²ω + m₂uR =0...............(2)
Using equation (1) and (2), the angular velocity expression will be
ω =[m₂v / {(2m₁ +m₂)R}] in positive z direction
Thus, the angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].
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At left A red ball in a box with arrows pointing away from the ball in all directions. In the middle, a blue ball in a box with arrows pointing toward the ball from all directions. At right, a grey ball in a box. Which object represents a negatively charged particle? Which object represents a positively charged molecule? Which object represents an uncharged molecule? Which object will not move when in an electric field?
Answer: The answers are, 1) B. 2) A. 3) C. 4) C.
here is the proof maybe u will understand it more than the numbers and letters lm.ao
Explanation:
The red ball with arrows pointing away is a negatively charged particle, the blue ball is a positively charged molecule, and the grey ball is an uncharged molecule. Electric fields don't move uncharged molecules.
Analyse each scenario to identify representations:
A negatively charged particle is the red ball in a box with all arrows pointing away from it. The red ball generates an electric field, and negatively charged particles radiate electric field lines. The arrows indicate how a positive test charge would move in the electric field.
A positively charged molecule is the blue ball in a box with all arrows pointing to it. An electric field converges on the blue ball, signifying its destination. The electric field lines point towards positively charged particles, indicating the direction a positive test charge would move.
A boxed grey ball depicts an uncharged molecule. The ball has no arrows. Uncharged molecules do not emit or experience electric fields. The grey ball is electrically neutral because it lacks electric pitch lines.
An electric field will not move the grey ball in a box (uncharged molecule). Electric fields do not interact with uncharged molecules. Thus, an electric field will keep the grey ball still.
Science and engineering require understanding these representations and their electric field behaviour. Scientists and engineers can create and manipulate electrical systems, electronics, and other technologies that use electric charge interactions by recognising charged particles, uncharged molecules, and electric field effects.
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If the trend changed toward traditional (pre-World War II) families, how would that affect women’s rights?
If the trend changed toward traditional (pre-World War II) families, the women’s rights are employment in manufacturing sector.
What is World war?
The war between two countries to take over each other's kingdom using weapons to kill each other.
Before the world war, the army needs armor, weapons, guns and tanks. Their manufacturing is only possible with many workers to work for long hours. If the men are not enough, then women are given opportunities to work with them.
Thus, the women’s rights are employment in manufacturing sector when trend changed traditional families.
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Find the velocity using the d/t graph below
Answer:
-10/3
Explanation:
the slope of the graph is negative, the velocity is slowing. so we know the positive answers are not correct. next we just need to figure out how much of a slope the line has. it's down 10 and over 3... soo
-10/3 looks good .. the format for the fraction is rise / run . in this case rise is negative, but other wise it's the same.
Two 74.0-kg hockey players skating at 6.15 m/s collide and stick together.
If the angle between their initial directions was 90 ∘ , what is their speed after the collision?
If the angle between their initial directions was 90 ∘ , their speed after the collision will be 4.35 m/s.
What is inelastic collision?When two moving bodies after colliding stick to each other and move together is called the inelastic collision.
In inelastic collision , momentum is remains conserved.
Initial momentum =Final momentum
m₁u₁ +m₂u₂ = (m₁ +m₂ )v
where, u₁ and u₂ represents initial velocities of two hockey players and v₁ and v₂ shows the final velocities.
m₁ =m₂ =m and u₁ = u₂ =u
Substitute the values from the question, we get
2mu = 2mv
u =v
u = v x cos (θ/2)
u = 6.15 cos(90/2)
u = v = 4.35 m/s
Thus, the speed after collision is 4.35 m/s.
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How do the stomach and the small intestines work together to digest the food? (2 points)
The stomach breaks down the food and passes it to the small intestine to be digested further.
The stomach passes the digested food from the small intestine to the large intestine.
The small intestine turns the solid food into a liquid mixture and sends it to the stomach.
The stomach digests the food by using digestive juices made by the small intestine.
Answer:
A) The stomach breaks down the food and passes it to the small intestine to be digested further.
Explanation:
I am pretty sure it is the first option because there is no way the stuff you eat can be turned into a liquid, it wouldn't make sense for the stomach to pass the digested food from small intestine to large intestine, so it would have to be the first one. A out of all of the options makes more sense and the stomach has to pass the digested food somewhere.
I hope I helped you
QUESTION 4 A ball is dropped from a height. If it takes 0.2 s to cross the last 6 m before hitting the ground, find the approximate height from which it is dropped.
Answer:
S = V t + 1/2 g t^2 where the ball has a speed of V and falls for .2 sec
6 = V t + 4.9 * .04 = V * .2 + .2
V = (6 - .2) / .2 = 29 m/s speed entering area
T = V / g = 29 / 9.8
T = 2.96 sec time to reach speed V
H = 1/2 g t^2 time to fall a distance H
H = 4.9 * 2.96^2 = 42.9 m height from which ball was dropped
Check using first equation:
S = 29 * .2 + 4.9 + .2^2 = 6 m
