Answer:
b I'm pretty sure sorry if I'm wrong
Answer:
I think the answer is B
Explanation:
The warm air turns cold and then it goes back to clouds
The low pressure area near Earth's equator is filled by cool air moving in from
А
Europe and South America
B
the North and South Pole
с
the Prime Meridian
D
the Atlantic and Pacific Ocean
the number of perpandicular components of a force is
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
What is a transfer of energy called?
A. Displacement
B. Acceleration
C. Work
D. Torque
Find the wavelength of light which is capable of ionizing a hydrogen atom?
Answer:
The correct answer is - 91.4 nm
Explanation:
According to Bohr's model, the minimum wavelength to ionize Hydrogen atom from n= 1 state is expressed as:
(h×c)/λ=13.6eV
here,
h - Planck constant
c - the speed of light
λ - wavelength
Placing the value in the formula for the wavelength
(6.626×10^−34J.s × 3×10^8 m/s)/λ = 13.6 ×1.6 × 10^−19 J
λ≈91.4nm
Thus, the correct answer would be = 91.4 nm
A 1.65-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 16.0 N. (a) What are the frequencies of the first three allowed modes of vibration
Answer:
Explanation:
mass per unit length ρ = .100 / 1.65 = .0606 . kg /m
length of wire L = 1.65 m
For fundamental frequency , the expression is as follows
n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]
L = 1.65 , T = 16 n and m = .0606
n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]
= 4.9 /s .
This is fundamental frequency .
other mode of vibration ( first three ) will be as follows
4.9 x 2 = 9.8 /s ,
4.9 x 3 = 14.7 /s .
A hand dryer blows heated air downwards out of the exit duct at a velocity of 4 m/s. The temperature and density of the ambient air at the inlet are 15 C and 1.23 kg/m3, while at the outlet it has temperature 35 C and density 1.15 kg/m3 The blower power is 10.0 W and the heater power is 715 W. Consider the inlet to be at the large mass of ambient air which has negligible velocity.
a) What is the pressure at the outlet? 4 m/s, 35 C
b) You will be applying the energy equation. Why can you ignore any height differences in this situation?
c) If the specific heat of air C-1000 J/(kg K), where Δυ-C Δ T, find the change in internal energy per unit mass from the inlet to outlet.
d) Find the mass flow rate through the dryer.
e) What is the power loss in the system?
f) What is the loss in the system?
g) What is the head loss in the system?
h) What is the total loss coefficient of the system, referred to the outlet velocity?
i) If there were no heater, would the temperature of gas at the outlet be higher, the same, or lower than the inlet? Explain why.
Answer:
nzkdjdksishdjsdjjdjnzkskejeoueeuieeoyrie ryrhdhcksodopdncndnszdkhfoeosheiwuef wokxkzdkjdoeehxjbxbdkeiehdhdhddjjddjdhhdhdhhhjdhfjdjjfjddhdheudiwiehdbejwowud
Explanation:
isos
A 64.0 cm long cord is vibrating in such a manner that it forms a standing wave with two antinodes. (The cord is fixed at both ends.) Which harmonic does this wave represent
Answer:
the wave represents the second harmonic.
Explanation:
Given;
length of the cord, L = 64 cm
The first harmonic of a cord fixed at both ends is given as;
[tex]f_o = \frac{V}{2L}[/tex]
The wavelength of a standing wave with two antinodes is calculated as follows;
L = N---> A -----> N + N ----> A -----> N
Where;
N is node
A is antinode
L = N---> A -----> N + N ----> A -----> N = λ/2 + λ/2
L = λ
The harmonic is calculated as;
[tex]f = \frac{V}{\lambda} \\\\f = \frac{V}{L} = 2(\frac{V}{2L} ) = 2(f_o) = 2^{nd} \ harmonic[/tex]
Therefore, the wave represents the second harmonic.
L = λ
2. Plastic is a great conductor of charge so it moves quicker.
True
False
Answer:
the answer is false
Explanation:
plastic doesnt conduct anything
Light rays enter a transparent material. Which description best describes what happens to the light rays
When light encounters transparent materials, almost all of it passes directly through them. Glass, for example, is transparent to all visible light. ... Most of the light is either reflected by the object or absorbed and converted to thermal energy. Materials such as wood, stone, and metals are opaque to visible light.
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.
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
. Calculate
the length of this wire required to make a 10 Ω coil.
Answer:
the length of the wire is 134.62 m.
Explanation:
Given;
resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm
cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²
resistance of the wire, R = 10Ω
The length of the wire is calculated as follows;
[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]
Therefore, the length of the wire is 134.62 m.
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 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
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?
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]
Which one the answer to this question
What is the importance of using locally available resources in creating art?
Answer:
please give me brainlist and follow
Explanation:
Using locally available resources for art help in the preservation of environment. A significant and practical aspects of art is material significance. The items used by artists while making an art piece affects both the form and the material. Every material delivers something special in the creative process.
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.
Scientists have concluded that the uppermost part of the mantle is partially-molten. Which observation helped them reach this conclusion?
Answer:
P and S waves slow down when they reach this layer. The asthenosphere, also known as the magma chamber, is the uppermost component of the mantle. This layer is partially molten and is a ductile zone in a tectonically poor state.
It's almost hard and seismic waves move through the asthenosphere at a slow rate. The fragile lithosphere and the uppermost portion of the asthenosphere are assumed to be rigid.
seismic waves travel more quickly through denser materials and therefore generally travel more quickly with the depth it moves more slowly through a liquid than a solid. Molten areas within the Earth slow down P waves and stop S waves because their shearing motion cannot be transmitted through a liquid. Partially molten areas may slow down the P waves and attenuate or weaken S waves.
hope this helps...
S and P wave slow down and stop in the uppermost part of the mantle. - For this, scientists have concluded that the uppermost part of the mantle is partially-molten.
What is mantle?A planetary body's mantle is a layer that is surrounded by the crust on top and the core underneath. The largest and most substantial layer of a planetary body, mantles are often comprised of rock or ice. Planetary bodies that have undergone density differentiation typically have mantles. Mantles are found on all terrestrial planets (including Earth), many asteroids, and a few planetary moons.
Between the crust and the outer core, there is a silicate rock layer known as the Earth's mantle. Despite being mostly solid, it behaves like a viscous fluid over geological time. Oceanic crust is created by the partial melting of the mantle at mid-ocean ridges, and continental crust is created by the partial melting of the mantle at subduction zones.
Learn more about mantle here:
https://brainly.com/question/28827790
#SPJ2
Please answer this for 15 points please don’t put in a link.
Answer:
c. Double Replacement
Explanation:
As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.
Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2
and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.
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)
need help ASAP!!!!!!!!!!!
Answer:
The equation says that due to variation in temperature is
delt T = .59 m/s / C = 16 C * .59 m/s = 9.44 m/s
So v = 332 m/s + 9.44 m/s = 341 m/s (to three significant figures)
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
Which of the
following
DECREASES
as you go UP a
mountain?
A. climate
B. altitude
C. amount of oxygen
D. buoyancy
Answer:
C. Amount of oxygen
Explanation:
Options A and D are invalid as they aren't affecting factors.
Option B is false as the altitude increases as you go up a mountain.
Option C is true as the air pressure (atmospheric pressure) is inversely proportional to the height/altitude of the mountain.
A solenoid that is 93.9 cm long has a cross-sectional area of 17.3 cm2. There are 1270 turns of wire carrying a current of 7.80 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).
Answer:
[tex]65.6\ \text{J/m}^3[/tex]
[tex]0.11\ \text{J}[/tex]
Explanation:
B = Magnetic field = [tex]\mu_0 \dfrac{N}{l}I[/tex]
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi10^{-7}\ \text{H/m}[/tex]
N = Number of turns = 1270
[tex]l[/tex] = Length of solenoid = 93.9 cm = 0.939 m
[tex]I[/tex] = Current = 7.8 A
A = Area of solenoid = [tex]17.3\ \text{cm}^2[/tex]
Energy density of a solenoid is given by
[tex]u_m=\dfrac{B^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{(\mu_0 \dfrac{N}{l}I)^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{\mu_0N^2I^2}{2l^2}\\\Rightarrow u_m=\dfrac{4\pi\times 10^{-7}\times 1230^2\times 7.8^2}{2\times 0.939^2}\\\Rightarrow u_m=65.6\ \text{J/m}^3[/tex]
The energy density of the magnetic field inside the solenoid is [tex]65.6\ \text{J/m}^3[/tex]
Energy is given by
[tex]U_m=u_mAl\\\Rightarrow U_m=65.6\times 17.3\times 10^{-4}\times 0.939\\\Rightarrow U_m=0.11\ \text{J}[/tex]
The total energy in joules stored in the magnetic field is [tex]0.11\ \text{J}[/tex].
An enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm to take some readings while you stay dry at home. While Clark is at the center of the storm, he sees and hears lightning strike a tree that is 150 m from where he is standing. You are 127 km from the tree. How long does it take for the sound to reach Clark
Answer:
t = 0.437 s
Explanation:
The speed of sound is a constant that is worth v = 343 m / s
v = d / t
t = d / v
the time it takes for the sound to reach Clark at d = 150 m is
t = 150/343
t = 0.437 s
This same sound takes much longer to reach you
t₂ = 127 10³/343
t₂ = 370 s
If our atmosphere had a uniform density of 1.25 kg/m3 all the way up to a border with empty space above, that border would be Answer km above sea level. The pressure at sea level is 1 atm = 105 N/m2 and g = 10 m/s2. Enter your answer as an integer.
Answer:
The border is 8km above sea level.
Explanation:
We know that:
Density = 1.25 kg/m^3
Pressure = 10^5 N/m^2
g = 10m/s^2
Now, suppose that we have a virtual rectangle, such that its bases have an area of 1m^2 and the rectangle has a height equal to H.
This virtual figure has a volume V = 1m^2*H, and it is filled with air (which we know that has a density 1.25 kg/m^3)
Then the total mass inside that volume is:
M = (1.25 kg/m^3)*V = (1.25 kg/m^3)*(1m^2*H)
The weight of this mass is:
W = g*M = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)
And if we divide the weight in a given surface, let's say 1 m^2, we get the pressure per square meter, which we know is equal to 10^5 N/m^2
then:
P = 10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)
Whit this equation we can find the value of H.
10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)
10^5 N = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)
(10^5 N)/(10 m/s^2) = (1.25 kg/m^3)*(1m^2*H)
(10^4 kg) = (1.25 kg/m^3)*(1m^2*H)
(10^4 kg)/( 1.25 kg/m^3) = 1m^2*H
8,000 m^3 = 1m^2*H
(8,000 m^3)/(1m^2) =H
8,000 m = H
And we want this answer in km, knowing that 1,000m = 1km
8,000m = 8km = H
The border is 8km above sea level.
Height of boundaries is 8.2 km
Given that:Normal density = 1.25 kg/m³
1 atm = 101325 N/m²
Find:Height of boundaries
Computation:Pressure = Height × Density × Gravitational acceleration
101325 = Height × 1.25 × 9.8
101325 = Height × 12.25
Height of boundaries = 101325 / 12.25
Height of boundaries = 8271.42 m
Height of boundaries = 8.2 km
Learn more:https://brainly.com/question/23358029
Light containing two different wavelengths passes through a diffraction grating with 1,250 slits/cm. On a screen 17.5 cm from the grating, the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum for the longer wavelength. If the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen, what are the wavelengths in the light
Answer:
[tex]\lambda_s =6.43*10^-4m[/tex]
Explanation:
From the question we are told that:
Diffraction grating [tex]N=1250slits/cm[/tex]
Distance b/w Screen and grating length [tex]d_{sg}=17.5 cm[/tex]
Distance b/w neighboring maxima and Screen [tex]d_{ms}=8.44[/tex]
Generally the equation for grating space is mathematically given by
[tex]d(g)=\frac{1}{N}[/tex]
[tex]d(g)=\frac{100}{1250}[/tex]
[tex]d(g)=0.08[/tex]
Generally the equation for small angle approximation is mathematically given by
[tex]\triangle y=\frac{\lambda d}{L}[/tex]
Therefore for longest wavelength
[tex]\lambda _l=\frac{8.44*10^{-3}*(0.08)}{0.175m}[/tex]
[tex]\lambda _l=3.858*10^{-3}[/tex]
Therefore the third order maximum equation for the shorter wavelength as
[tex]\lambda_s =\frac{1}{6} \lambda_l[/tex]
[tex]\lambda_s =\frac{1}{6} (3.858*10^-^3)[/tex]
[tex]\lambda_s =6.43*10^-4m[/tex]
The wavelengths in the light is given as
[tex]\lambda_s =6.43*10^-4m[/tex]
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.