Answer:
aim at prisma and will have all colors
Explanation:
Answer:
If the vision position is above the actual image location then the light travel from the object in such a way that the angle of incidence is less than the angle of reflected ray which means that the reflected beam is above the incident beam.
Explanation:
please help, no links please! I dont understand stand this question and im going to cry
Which part of this system is in the gas phase?
Answer:
I'm pretty sure its helium
A star with the same mass and diameter as the sun rotates about a central axis with a period of about 24.0 days. Suppose that the sun runs out of nuclear fuel and collapses to form a white dwarf star with a diameter equal to that of the earth. Assume the star acts like a solid sphere and that there is no loss of mass in the process. You will need some data from the inside front cover of you text. (a) What would be the new rotation period (s) of the star? (b) What is the ratio of final to initial kinetic energies (Kf /Ki)?
Answer:
a) w = 2.52 10⁷ rad / s, b) K / K₀ = 1.19 10⁴
Explanation:
a) We can solve this exercise using the conservation of angular momentum.
Initial instant. Before collapse
L₀ = I₀ w₀
Final moment. After the collapse
L_f = I w
angular momentum is conserved
L₀ = L_f
I₀ w₀ = I w (1)
The moment of inertia of a sphere is
I = 2/5 m r²
we take from the table the mass and diameter of the star
m = 1,991 10³⁰ kg
r₀ = 6.96 10⁸ m
r = 6.37 10⁶ m
to find the angular velocity let's use
w = L / T
where the length of a circle is
L = 2π r
T = 24 days (24 h / 1 day) (3600 s / 1h) = 2.0710⁶ s
we substitute
w = 2π r / T
wo = 2π 6.96 10⁸ / 2.07 10⁶
wo = 2.1126 10³ rad / s
we substitute in equation 1
w = [tex]\frac{I_o}{I}[/tex]
w = 2/5 mr₀² / 2/5 m r² w₀
w = ([tex]\frac{r_o}{r}[/tex]) ² wo
w = (6.96 10⁸ / 6.37 10⁶) ² 2.1126 10³
w = 2.52 10⁷ rad / s
b) the kinetic energy ratio
K = ½ m w²
K₀ = ½ m w₀²
K = ½ m w²
K / K₀ = (w / wo) ²
K / K₀ = 2.52 10⁷ / 2.1126 10³
K / K₀ = 1.19 10⁴
In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 32.0 nm long) with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring.
Required:
a. Find the ratio of the frequency with the virus attached ( fS+V) to the frequency without the virus (fS) in terms of mV and mS, where mV is the mass of the virus and mS is the mass of the silicon sliver.
b. In some data, the silicon sliver has a mass of 2.13×10^-16 g and a frequency of 2.04×10^15 Hz without the virus and 2.85×1014 with the virus. What is the mass of the virus in grams?
Answer:
a) m_v = m_s (([tex]\frac{w_o}{w}[/tex])² - 1) , b) m_v = 1.07 10⁻¹⁴ g
Explanation:
a) The angular velocity of a simple harmonic motion is
w² = k / m
where k is the spring constant and m is the mass of the oscillator
let's apply this expression to our case,
silicon only
w₉² = [tex]\frac{K}{m_s}[/tex]
k = w₀² m_s
silicon with virus
w² = [tex]\frac{k}{m_s + m_v}[/tex]
k = w² (m_v + m_s)
in the two expressions the constant k is the same and q as the one property of the silicon bar, let us equal
w₀² m_s = w² (m_v + m_s)
m_v = ([tex]\frac{w_o}{w}[/tex])² m_s - m_s
m_v = m_s (([tex]\frac{w_o}{w}[/tex])² - 1)
b) let's calculate
m_v = 2.13 10⁻¹⁶ [([tex]\frac{20.4}{2.85}[/tex])² - 1)]
m_v = 1.07 10⁻¹⁴ g
In 2-3 complete sentences, analyze how scientists know dark matter and dark energy exist.
Answer:
It doesn't interact with baryonic matter and it's completely invisible to light and other forms of electromagnetic radiation, making dark matter impossible to detect with current instruments. But scientists are confident it exists because of the gravitational effects it appears to have on galaxies and galaxy clusters.
Explanation:
red light from a He-Ne laser is at 590.5 nm in the air. it is fired at an angle of 31.0 to horizontal at a flat transparent crystal of calcite (n= 1.34 ar this frequency) .find the wavelength and frequency of the light inside the crystal and the angle from horizontal that it travels inside the calcite crystal.
Answer:
7374.4
Explanation:
I took the test
(filler so I can post)
Swordfish are capable of stunning output power for short bursts. A 650 kg swordfish has a cross-sectional area of 0.92 m2 and a drag coefficient of 0.0091- very low due to some evolutionary adaptations. Such a fish can sustain a speed of 30 m/s for a few seconds. Assume seawater has a density of 1026 kg/m3. a) How much power does the fish need to put out for motion at this high speed
Answer:
the required or need power is 115960.57 Watts
Explanation:
First of all, we take down the data we can find from the question, to make it easier when substituting values into formulas.
mass of swordfish m = 650 kg
Cross - sectional Area A = 0.92 m²
drag coefficient C[tex]_D[/tex] = 0.0091
speed v = 30 m/s
density p = 1026 kg/m³
Now, we determine our Drag force F[tex]_D[/tex]
Drag force F[tex]_D[/tex] = [tex]\frac{1}{2}[/tex] × C[tex]_D[/tex] × A × p × v²
Next, we substitute the values we have taken down, into the formula.
Drag force F[tex]_D[/tex] = [tex]\frac{1}{2}[/tex] × 0.0091 × 0.92 × 1026 × (30)²
Drag force F[tex]_D[/tex] = 4.294836 × 900
Drag force F[tex]_D[/tex] = 3865.3524
Now, we determine the power needed P[tex]_w[/tex]
P[tex]_w[/tex] = F[tex]_D[/tex] × v
we substitute
P[tex]_w[/tex] = 3865.3524 × 30
P[tex]_w[/tex] = 115960.57 Watts
Therefore, the required or need power is 115960.57 Watts
17. 53 A small grinding wheel is attached to the shaft of an electric motor that has a rated speed of 3600 rpm. When the power is turned off, the unit coasts to rest in 70 s. The grinding wheel and rotor have a combined weight of 6 lb and a combined radius of gyration of 2 in. Determine the average magnitude of the couple due to kinetic friction in the bearings of the motor.
Answer:
0.337 lb-in
Explanation:
From the law of conservation of angular momentum,
L' = L" where L = initial angular momentum of system and L" = final angular momentum of system
Now L = Iω + Mt where Iω = angular momentum of shaft + wheel and Mt = impulse on system due to couple M.
L' = Iω' + (-Mt) (since the moment about the shaft is negative-anticlockwise)
L' = Iω' - Mt where Iω' = angular momentum of shaft at t' = 0 + wheel and Mt = impulse on system due to couple M in time interval t = 70 s.
L" = Iω" where Iω" = angular momentum of shaft at t" = 70 s.
Now I = moment of inertia of system = mk² where m = mass of system = W/g where W = weight of system = 6 lb and g = acceleration due to gravity = 32 ft/s². So, m = W/g = 6lb/32 ft/s² = 0.1875 lb-s²/ft and k = radius of gyration = 2 in = 2/12 ft = 1/6 ft.
So, I = mk² = (0.1875 lb-s²/ft) × (1/6 ft)² = 0.00521 lb-ft-s², ω' = initial angular speed of system = 3600 rpm = 3600 × 2π/60 = 120π rad/s = 377 rad/s, ω" = final angular speed of system = 0 rad/s (since it stops), t' = 0 s, f" = 70 s and M = couple on system
So,
Iω' - Mt" = Iω"
Substituting the values of the variables into the equation, we have
Iω' - Mt" = Iω"
0.00521 lb-ft-s² × 377 rad/s - M × 70 s = 0.00521 lb-ft-s² × 0 rad/s"
0.00521 lb-ft-s² × 377 rad/s - 70M = 0
1.964 lb-ft-s = 70M
M = 1.964 lb-ft-s/70 s
M = 0.0281 lb-ft
M = 0.0281 lb × 12 in
M = 0.337 lb-in
How much energy is supplied to a 9 V bulb if it is switched on for 3 minutes and takes a current of
0.2 A ?
Answer:
0.01j
Explanation:
the energy equals the work done by the bulb.
Workdone=
[tex]workdone = \frac{power}{time} [/tex]
power=voltage×current
=9×0.2
=1.8 W.
THEREFORE,
time=3×60
= 180s
workdone=1.8/180
=0.01 j