Answer:
ultraviolet light
plz mark me as brainliest.
Answer:
Ultra violet
Explanation:
Can someone tell me anything useful about energy management in the human body?
Answer:
The human body carries out its main functions by consuming food and turning it into usable energy. Immediate energy is supplied to the body in the form of adenosine triphosphate (ATP). Since ATP is the primary source of energy for every body function, other stored
Explanation:
this what teacher explain to us
A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surface, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33.
Part A: The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? t= 313nm
Part B: Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? t=125nm
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t[tex]_{min[/tex] = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t[tex]_{min[/tex] = 750 / 2(1.20)
t[tex]_{min[/tex] = 750 / 2.4
t[tex]_{min[/tex] = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t[tex]_{min[/tex] = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t[tex]_{min[/tex] = 750 / 4(1.50)
t[tex]_{min[/tex] = 750 / 6
t[tex]_{min[/tex] = 125 nm
Therefore, the minimum thickness be now will be 125 nm
Which of the following happens to
density as air pressure decreases?
С C
A. Density increases.
B. Density stays the same.
C. Density decreases.
D. There is no correlation between air pressure and
density.
Explanation:
As pressure increases, with temperature constant, density increases. Conversely when temperature increases, with pressure constant, density decreases. Air density will decrease by about 1% for a decrease of 10 hPa in pressure or 3 °C increase in temperature.
How many gallons of water does it take to produce the following:
a. Cheeseburger
b. Pound of butter
c. A pair of jeans
Answer:
a. 660 gallons
b.665 gallons
c. 1,800
A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 20
kg. The carriage has
energy. Calculate it
Answer:
Energy in carriage (Potential energy) = 4,116 J
Explanation:
Given:
Mass of baby = 20 kg
Height = 21 m
Find:
Energy in carriage (Potential energy)
Computation:
The energy accumulated in an object as a result of its location relative to a neutral level is known as potential energy.
In carriage accumulated energy is potential energy.
Energy in carriage (Potential energy) = mgh
Energy in carriage (Potential energy) = (20)(9.8)(21)
Energy in carriage (Potential energy) = 4,116 J
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel during 30 seconds of rotation?
Each minute, the disk completes 100 revolutions, so a point on the rim traverses a distance of 100 times the circumference of the disk and would have a linear speed of
100 rev/min
= (100 rev/min) × (2π × 50 cm/rev) × (1/100 m/cm) × (1/60 min/s)
= 5π/3 m/s ≈ 5.236 m/s
Then after 30 s of rotation, the point would have traveled a distance of
(5π/3 m/s) × (30 s) = 50π m ≈ 157.08 m
What is the unit of measurement of mass and weight?
Answer:
kilogram
In the International System of Units (SI), the kilogram is the basic unit of mass, and the newton is the basic unit of force. The non-SI kilogram-force is also a unit of force typically used in the measure of weight.
why is potassium and sodium considered as reactive metals?
Answer:
because they are found freely in nature uncombined so they are highly reactive with other elements
What is the order of the events for the water cycle on a typical warm day?
А
rain, snow, sleet
B
precipitation, evaporation, rain
с
evaporation, condensation, precipitation
D
condensation, evaporation, precipitation
How would increasing the pressure of this reaction affect the equilibrium
Explanation:
c because there is element
Answer:
C. H2 and N2 would react to produce more NH3
Explanation:
A.P.E.X
A 2.0 kg breadbox on a fric-
tionless incline of angle u 40 is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k 120 N/m, as shown in
Fig. 8-43. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the in- cline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s acceleration at the instant the box momentarily stops?
2. Plastic is a great conductor of charge so it moves quicker.
True
False
Answer:
the answer is false
Explanation:
plastic doesnt conduct anything
Which of the following actions will increase the current induced in a wire by a
magnetic field?
Answer:
The induced current can be increased in the coil in the following ways: By increasing the strength of the magnet. By increasing the speed of the magnet through the coil.
Explanation:
Which device converts electric energy into mechanical energy?
O A. An electromagnet
O B. A motor
O C. A transformer
O D. A generator
Answer:
B motor
Explanation:
A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 2ft/sec. Find the velocity of the top of the ladder at time t=1.
Answer: 0.516 ft/s
Explanation:
Given
Length of ladder L=20 ft
The speed at which the ladder moving away is v=2 ft/s
after 1 sec, the ladder is 5 ft away from the wall
So, the other end of the ladder is at
[tex]\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft[/tex]
Also, at any instant t
[tex]\Rightarrow l^2=x^2+y^2[/tex]
differentiate w.r.t.
[tex]\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s[/tex]
Your boss asks you to design a drone that begins its flight near the surface and rises to 9600 m. At the surface it will fly through air having a density of 1.23 kg per cubic meter and at its highest altitude the air density will become 0.62 kg per cubic meter. If the flight velocity near sea level is 45 mph, then how fast will in need to go at its highest altitude to maintain the same lift. Assume the coefficient of lift remains constant.
Answer:
[tex]63.38\ \text{mph}[/tex]
Explanation:
L = Lift force
[tex]\rho[/tex] = Density of air
A = Surface area
v = Velocity
[tex]v_1[/tex] = 45 mph
[tex]\rho_1=1.23\ \text{kg/m}^3[/tex]
[tex]\rho_2=0.62\ \text{kg/m}^3[/tex]
Coefficient of lift is given by
[tex]CL=\dfrac{2L}{\rho v^2A}\\\Rightarrow \rho=\dfrac{2L}{CL v^2A}[/tex]
So
[tex]\rho\propto \dfrac{1}{v^2}[/tex]
[tex]\dfrac{\rho_1}{\rho_2}=\dfrac{v_2^2}{v_1^2}\\\Rightarrow v_2=\sqrt{\dfrac{\rho_1}{\rho_2}}\times v_1\\\Rightarrow v_2=\sqrt{\dfrac{1.23}{0.62}}\times 45\\\Rightarrow v_2=63.38\ \text{mph}[/tex]
The velocity at the required altitude should be [tex]63.38\ \text{mph}[/tex] to maintain the same lift.
One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 42.9o angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.
Answer:
x = 0.455 L
Explanation:
For this exercise we must use the rotational equilibrium condition
Σ τ = 0
it has two forces, the first is perpendicular to the rod, so its stub is
τ₁ = F₁ L
the second force is applied with an angle, so we can use trigonometry to find its components
sin θ = F_parallel / F₂
cos θ = F_perpendicular / F₂
F_parallel = F₂ sin θ
F _perpendicular = F₂ cos θ
torque is
τ₂ = F_perpendicular x + F_parallel 0
the parallel force is on the rod therefore its distance is zero
we apply the equilibrium equation
τ₁ - τ₂ = 0
F₁ L = F₂ cos θ x
x = [tex]\frac{L}{cos \theta} \ \frac{F_1}{F_2}[/tex]
let's calculate
x = [tex]\frac{L}{cos \ 42.9} \ \frac{2.00}{6.00}[/tex]
x = 0.455 L
What voltage would be measured across the 15 ohm resistor?
A)
2.5 volts
B)
5.0 volts
C)
7.5 volts
D)
10 volts
Answer:
7.5 volts
Explanation:
I did it on USA Testprep
When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s
what is the speed and direction of the other 15 kilogram piece of the explosive!
Answer:
speed: 35m/s
direction: left
Explanation:
Assuming the right side is the positive direction:
before explosion:
P = mv = 0
after explosion:
P' = 15P + 5P
(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')
P' = 0.75mv1' + 0.25mv2'
P' = (15kg)v' + (5kg)(105m/s)
P' = 525kg/m/s + (15kg)v1'
P = P'
525kg/m/s + (15kg)v1' = 0
(15kg)v1' = -525kg/m/s
v1' = -35m/s
speed = |-35| = 35m/s
direction is to the left since the right side is the positive direction.
true or false
The Total electric potential due to two or more charges is equal to the algebraic sum of the potentials due to the individual charges.
Answer:
i guess the answer is false
Why is it harder to breathe on a
mountain?
A. The air pressure is so high the lungs can't expand.
B. The air is denser and oxygen can't flow easily into the
lungs.
C. The denser oxygen molecules sink below the
surrounding air.
D. The air is less dense so there are fewer oxygen
molecules.
The liquid emerges into a vertical jet as it drains from the container, with the velocity profile in the jet remaining uniform. The outlet of the container is located 2.0 m above ground, and the radius of the emerging liquid jet changes with vertical distance from the bottom of the container as it accelerates under the action of gravity. Neglecting viscous losses and surface tension effects in the liquid jet, what is the velocity of the water jet as it strikes the ground when the container begins to drain
Answer:
6.26 m/s
Explanation:
Since we are neglecting viscous losses and surface tension effects in the liquid jet, by conservation of energy, the potential energy loss of the jet = kinetic energy gain of the jet
So, mgh = 1/2mv² where m = mass of water in jet, g = acceleration due to gravity = 9.8 m/s², h = height of outlet = 2.0 mand v = velocity of liquid jet
So, mgh = 1/2mv²
gh = 1/2v²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 2.0 m)
v = √(39.2 m²/s²)
v = 6.26 m/s
would it be m/s or kg?
Answer:
m.s
Explanation:
Calculate the terminal velocity of
the following nain drops faning
through air (a) one with a diameter
of 0.3cm 6 one with a a diameter
of o. Olm. Take the density of
water to be looo Kym3 and the
eis cosity of air to be ixlos pas.
The buoyancy effect of the air
may be ignored)
A +0.0129 C charge feels a 4110 N
force from a -0.00707 C charge. How
far apart are they?
[?] m
Answer:
r = 14.13 m
Explanation:
Given that,
Charge 1, q₁ = +0.0129 C
Charge 2, q₂ = -0.00707 C
The force between charges, F = 4110 N
We need to find the distance between charges. The formula for the force between charges is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Where
r is the distance between charges
So,
[tex]r=\sqrt{\dfrac{kq_1q_2}{F}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 0.0129 \times 0.00707 }{4110 }} \\\\r=14.13\ m[/tex]
So, the distance between charges is equal to 14.13 m.
Answer:
14.13 m
Explanation:
acellus
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 41 mm , while nonathletes' stretch only 33 mm .
Hello. Your question is incomplete. However, I managed to find it completely on the internet and I realized that you forgot to mention that the question asks you for the maximum energy difference between velovistas and non-athletes, considering that the spring constant for the tendon of the two groups is equal to 33n/mm.
To make this calculation you will need to use Hooke's law, using the formula: ¹/2*K*x², where "K" will be the value of the spring constant for the tendon and "X" will be the value of the sprinter and non-athlete terms.
So for the sprinter we will have the calculation:
¹/2*33*41² -------> 0,5*33*1681 = 27736. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
For the non-athlete we will have the calculation:
¹/2*33*33² -------> 0,5*33*1089 = 17968. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
Now, to reach the final result, you only need to subtract the two values presented by the sprinter and the non-athlete.
27736.5 - 17968.5 = 9768 Nmm
Will give brainliest!
Describe how heat is moving in the image and label each as Radiation, Conduction, or Convection.
Radiation / Conduction / Convection
Answer:
well in the pot there is conventional heat, the pot itself is giving off conductable heat, and the radiational heat is coming from the stove.
You are at a train yard observing trains (because why not). You see a train car (let's call it car 1) moving to the right ( x direction) towards a stationary train car (let's call this one car 2). Car 1 has an initial velocity of 15.0 m/s. A helpful train employee tells you that Car 1 also has a mass of 1,825 kg and Car 2 has a mass of 2,645 kg. Car 1 gently collides with Car 2, allowing them to connect. After the collision the two train cars stay connected. You can assume that there is no friction in the system. If you have never see train cars connect, you can watch the first 25ish seconds of this video to see two train cars couple. However, these cars have friction, so they stop - unlike our problem. What is the Final Velocity of the system consisting of Car 1 and Car 2
Answer:
6.12 m/s
Explanation:
Using the law of conservation of momentum
momentum before collision = momentum after collision
m₁v₁ + m₂v₂ = (m₁ + m₂)V (since the train cars become attached to each other) where m₁ = mass of car 1 = 1,825 kg, m₂ = mass of car 2 = 2,645 kg, v₁ = initial velocity of car 1 = + 15.0 m/s (positive since it is moving in the positive x direction), v₂ = initial velocity of car 2 = 0 m/s (since it is initially stationary) and V = velocity of both cars after collision,
So, m₁v₁ + m₂v₂ = (m₁ + m₂)V
m₁v₁ + m₂(0 m/s) = (m₁ + m₂)V
m₁v₁ + 0 = (m₁ + m₂)V
V = m₁v₁/(m₁ + m₂)
substituting the values of the other variables into the equation, we have
V = 1,825 kg × 15.0 m/s/(1,825 kg + 2,645 kg)
V = 27375 kgm/s/ 4470kg
V = 6.124 m/s
V ≅ 6.12 m/s
Convert (a) 50 oF, (b) 80 oF, (c) 95 oF to Celsius
. Indiana Jones needs to ascend a 10-m-high building. There is a large hose filled with pressurized water hanging down from the building top. He builds a square platform and mounts four 4-cm-diameter nozzles pointing down at each corner. By connecting hose branches, a water jet with a velocity, u, can be produced from each nozzle. Jones, the platform, and the nozzles have a combined mass of 150 kg. Determine the minimum water jet velocity, u (m/s), needed to raise the system.
Answer:
u = 14 m / s
Explanation:
For this exercise let's use conservation of energy
starting point. On the floor just when u speeding out
Em₀ = K = ½ m v²
final point. When on top of the building, no speed
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ m v² = m g h
v = u
u = [tex]\sqrt{2 g h}[/tex]
u= [tex]\sqrt{2 \ 9.8 \ 10}[/tex]
u = 14 m / s