Answer: 0.0146m
Explanation: The formula that defines the velocity of a simple harmonic motion is given as
v = ω√A² - x²
Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.
The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA
One half of maximum speed = speed of motion
3ωA/2 = ω√A² - x²
ω cancels out on both sides of the equation, hence we have that
A/2 = √A² - x²
(0.0169)/2 = √(0.0169² - x²)
0.00845 = √(0.0169² - x²)
By squaring both sides, we have that
0.00845² = 0.0169² - x²
x² = 0.0169² - 0.00845²
x² = 0.0002142
x = √0.0002142
x = 0.0146m
Kinetic energy = (1/2) (mass) (speed)²
BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second. So we'll have to
make that conversion.
KE = (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²
= (725.5) · (48 · 1000 · 1 / 3,600)² (kg) · (km·m·hr / hr·km·sec)²
= (725.5) · ( 40/3 )² · ( kg·m² / sec²)
= 128,978 joules (rounded)
<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
Answer:
D. Top is emission; bottom absorption.
Explanation:
Emission and spectrum of elements are due to the element absorbing or emitting wavelength of e-m energy. Elementary particles of elements can absorb energy from a ground state to enter an excited state, creating an absorption spectrum, or they can lose energy and fall back to a lower energy state, creating an emission spectrum. A simple rule to differentiate between an emission and an absorption spectrum is that: "all absorbed wavelength is emitted, but not all emitted wavelength is absorbed."
From the image, the lines indicates wavelengths. We can see that all of the wavelengths of the bottom absorption spectrum coincides with some of the wavelength of the upper emission wavelengths.
Answer:
average speed
Explanation:
The directions were different, so the velocities could not be the same.
However, the magnitude of the velocity (speed) was 56/2 = 28 m/s for the first car, and 84/3 = 28 m/s for the second car. These<em> average speeds are the same</em>.