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
Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric eld E, which is orientedperpendicular to a uniform magnetic eld B. If the plates are separated byd= 2:0 mm and the value of the magnetic eld isB= 0:60T. Calculatethe potential di erence, between the capacitor plates, required to allow aparticle
This question is incomplete, the complete question is;
A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.
Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.
Hint : ΔV = Ed
Answer:
the required potential difference, between the capacitor plates is 600 V
Explanation:
Given the data in the question;
B = 0.60 T
d = 2.0 mm = 0.002 m
v = 5.0 × 10⁵ m/s.
since particle pass straight through without deflection.
F[tex]_{net[/tex] = 0
so, F[tex]_E[/tex] = F[tex]_B[/tex]
qE = qvB
divide both sides by q
E = vB
we substitute
E = (5.0 × 10⁵) × 0.6
E = 300000 N/C
given that; potential difference ΔV = Ed
we substitute
ΔV = 300000 × 0.002
ΔV = 600 V
Therefore, the required potential difference, between the capacitor plates is 600 V
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)
When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV.
What is the maximum kinetic energy K0 of the photoelectrons when light of wavelength 310 nm falls on the same surface?
Use h = 6.63×10−34 J⋅s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.
Answer:
Explanation:
energy of photon having wavelength of 400 nm = 1237.5/400 eV
= 3.1 eV.
Maximum kinetic energy of photoelectrons = 1.1 eV .
Threshold energy Ф = 3.1 - 1.1 = 2 eV .
energy of photons having wavelength of 310 nm = 1237.5 / 310 eV = 4 eV .
Maximum kinetic energy of photoelectrons = energy of photons - Threshold energy
= 4 - 2 = 2 eV .
Required kinetic energy K₀= 2 eV.
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
A cart of mass m is moving with negligible friction along a track with known speed v1 to the right. It collides with and sticks to a cart of mass 4m moving with known speed v2 to the right. Which of the two principles, conservation of momentum and conservation of mechanical energy, must be applied to determine the final speed of the carts, and why
Answer:
conservation of linear momentum
We were told that two objects became stuck together hence we have to use the principle of conservation of momentum to obtain the final velocities of the carts.
What is conservation of momentum ?The principle of conservation of momentum lets us know that the momentum before collision is equal to the momentum after collision. As such we can write; m1u1 + m2u2 = m1v1 + m2v2.
We can use this thus principle to obtain the final speeds of the carts since the two objects that collided became stuck together.
Learn more about conservation of momentum: https://brainly.com/question/11256472
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:
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
What does it mean if the reflected beam is above the incident beam? What does it mean if reflected beam is below the incident beam?
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:
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The block is suspended by wires from the ceiling and is initially at rest. After the bullet is embedded in the block, the block swings up to a maximum height of 0.295 cm above its initial position. What is the velocity of the bullet on leaving the gun's barrel
Answer:
The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.
Explanation:
Given;
mass of the bullet, m₁ = 2.47 g = 0.00247 kg
mass of the wooden block, m₂ = 2.43 kg
initial velocity of the wooden block, u₂ = 0
height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m
let the initial velocity of the bullet on leaving the gun's barrel = v₁
let final velocity of the bullet-wooden block system after collision = v₂
Apply the principle of conservation of linear momentum;
Total initial momentum = Total final momentum
m₁v₁ + m₂u₂ = v₂(m₁ + m₂)
0.00247v₁ + 2.43 x 0 = v₂(2.43 + 0.00247)
0.00247v₁ = 2.4325v₂ -------(1)
The kinetic energy of the bullet-block system after collision;
K.E = ¹/₂(m₁ + m₂)v₂²
K.E = ¹/₂ (2.4325)v₂²
The potential energy of the bullet-block system after collision;
P.E = mgh
P.E = (2.4325)(9.8)(0.00295)
P.E = 0.07032
Apply the principle of conservation of mechanical energy;
K.E = P.E
¹/₂ (2.4325)v₂² = 0.07032
1.21625 v₂² = 0.07032
v₂² = 0.07032 / 1.21625
v₂² = 0.0578
v₂ = √0.0578
v₂ = 0.24 m/s
Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;
0.00247v₁ = 2.4325v₂
0.00247v₁ = 2.4325 (0.24)
0.00247v₁ = 0.5838
v₁ = 0.5838 / 0.00247
v₁ = 236.36 m/s
Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.
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
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
Which form of energy increases when a spring is compressed?
Answer:
When the spring compresses, elastic potential energy increases.
Answer:
the answer is b
Explanation:
elastic potential energy
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