The correct answer is D. 1.55 hm
We need to convert dm to hectometers,
We know that,
1 decimeter = 0.001 hectometers
1 dm = 0.001 hm
Using this conversion we get
(1550 × 1) dm = (1550 × 0.001) hm
1550 dm = 1.55 hm
It means 1550 dm can expressed as 1.55 hectometers.
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There are a variety of waves from light waves to mechanical waves. Waves can exhibit different effects like the Doppler Effect.
All light waves behave in a similar manner. They either get transmitted, reflected, absorbed, refracted, polarized, diffracted, or scattered based off of the composition of the object and the wavelength of the light.
According to Wikipedia, “One important property of mechanical waves is that their amplitudes are measured in an unusual way, displacement divided by (reduced) wavelength. When this gets comparable to unity, significant nonlinear effects such as harmonic generation may occur, and, if large enough, may result in chaotic effects.” Mechanical waves are chaotic and its “amplitudes” are measured unusually.
Diffraction is when light bends around objects and spread after passing out through small openings. “Diffraction occurs with all waves, including sound waves, water waves, and electromagnetic waves such as light that the eye can see.”-Wikipedia. Here is the formula to Diffraction: <em>d </em>sin <em>θ </em>= <em>nλ</em>
Doppler effect can occur for any type of wave like sound or water waves. An example of this is when we hear a police car with its sirens on, coming towards us. The closer you are to the police car, the higher the wavelength, but the farther away you are, the lower the wavelength.
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Answer:
P=3.42×10^-6 J/s
Explanation:
From the kinematics of motion with constant acceleration we know that :
vf^2=vi^2+2*a(xf-xi)
Where :
• vf , vi, are the the final and the initial velocity of the electron
• a is the acceleration of the electron
• xf , xi are the final and the initial position of the electron .
Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.
Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m
vf^2 =vi^2+2*a(xf-xi)
vf^2-vi^2=2*a(xf-xi)
2*a(xf-xi)= vf^2-vi^2
a = (vf^2-vi^2)/2(xf-xi)
Pluging known information to get :
a = (vf^2-vi^2)/2(xf-xi)
= 1.411 × 10^17
From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m
so,
vf^2 =vi^2+2*a(xf-xi)
vf^2 =5.312× 10^7
From the following Eq. we can calculate the time elapsed in this motion .
xf =xi+vi*t+1/2*a*t
xf =xi+vi*t+1/2*a*t
t=√2(xf-xi)/a
t=3.765×10^-10 s
now we can use the power P Eq.
P=W/Δt => ΔK/Δt
Where: the work done W change the kinetic energy K of the electron ,
ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2
P=1/2*m*vf^2-1/2*m*vi^2/Δt
P=3.42×10^-6 J/s
Answer:
0.444atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (
P2 = final pressure (
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question,
P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm
P2 = ?
V1 = 80L
V2 = 160L (double of V1)
T1 = 34°C = 34 + 273 = 307K
T2 = 0°C = 0 + 273 = 273K
Using P1V1/T1 = P2V2/T2
0.999 × 80/307 = P2 × 160/273
79.92/307 = 160P2/273
Cross multiply
307 × 160P2 = 79.92 × 273
49120P2 = 21818.16
P2 = 21818.16 ÷ 49120
P2 = 0.444
P2 = 0.444atm
Metallic elements can exist on their own as individual atoms.
Nonmetals usually exist as molecules, combining with atoms of themselves.
Nonmetals can exist on their own as individual atoms.
Explanation:
Metallic elements can exist on their own as individual atoms. They have a large radius and are stable enough to freely exist as single atoms. The metallic bonding in them gives them stability. Example Gold, Silver, Platinum e.t.c
Non-metals are usually found in combined as state as molecules. They are usually joined together by covalent attraction. The bonds are formed with the atoms shares their valence electrons and their octet is completed. For example nitrogen gas, oxygen gas.
Most noble gases are non-metals that exist as individual mono-atomic gases on their own. The gases are stable and have an octet/duet configuration. For example helium gas, neon gas e.t.c
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