Answer:
The answer is "1.26".
Explanation:
[tex]D=18^{\circ}[/tex]
The refractive index is:
[tex]\to \mu=2\sin(30^{\circ}+\frac{D}{2})\\\\[/tex]
[tex]=2\sin(30^{\circ}+\frac{18^{\circ}}{2})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(39^{\circ})\\\\=2 \times 0.63\\\\=1.26[/tex]
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
Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that:________
a. the star is a white dwarf.
b. the star is periodically ejecting gas into space, every 127 hours.
c. the star is a Cepheid variable.
d. the star is a member of an eclipsing binary star system.
Answer:
d. the star is a member and also a part of an eclipsing binary star system.
Explanation:
If any star happens to be brighter for an extended period of time, however, at some times, it becomes dimmer, is due to the fact that the star is being overshadowed (hiding behind another star that is known as eclipse).
The above-mentioned eclipsing binary star system is essentially what has been defined. It occurs when two stars' orbit planes are so similar that one star will obscure (the light) of the other.
Thus, option D is correct.
Planet X has a moon similar to Earth’s moon.
Which path would this moon’s orbit take?
An ideal horizontal spring-mass system has a mass of 1.0 kg and a spring with constant 78 N/m. It oscillates with a period of 0.71 seconds. When this same spring-mass system oscillates vertically instead, the period is _______ seconds. Enter 2 significant figures (a total of three digits) and use g = 10.0 m/s2 if necessary.
Answer:
T = 0.71 seconds
Explanation:
Given data:
mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.
We have to calculate time period when this same spring-mass system oscillates vertically.
As we know
[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]
This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.
Therefore, T = 0.71 seconds
Application question: In Lancaster county, Pennsylvania, it is common for members of the Amish community to use windmills to pump water from underground to fill a tank for drinking water. The wind causes the turbine blades to spin, rotating a shaft, which is transferred through some gears to operate a pump, which pumps water up from deep below the ground to fill an above ground tank. Identify the energy conversions happening at each step below.
Wind blows______causing the turbine to turn, rotating shaft works pump_____energy.
Motion of water moving up from well______energy.
Water in tank which is positioned 5 feet above the ground level______potential energy.
Answer:
Wind blows__wind energy ____causing the turbine to turn, rotating shaft works pump__mechanical___energy.
Motion of water moving up from well___kinetic energy___energy.
Water in tank which is positioned 5 feet above the ground level______potential energy
Explanation:
Wind has in it wind energy which is then used to rotate the turbine shaft which is a form of mechanical work and hence possess mechanical energy
Moving water posses kinetic energy and when this water is kept at some height it possess potential energy
03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft
and a frequency is 0.59 HZ. Find the perind and spring constant, the maximum speed and
acceleration of the mass, the speed and acceleration when the displacement is 6 cm, compute the
kinetic and the potential energy when the position is 6 cm
Answer:
a) T = 1.69 s, b) k = 0.825 N / m, c) v = 1.46 feet/s, d) a = 5.41 ft / s²,
e) v = - 1,319 ft / s, a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J
Explanation:
In a mass-spring system with simple harmonic motion, the angular velocity is
w = [tex]\sqrt{\frac{k}{m} }[/tex]
a) find the period
angular velocity, frequency, and period are related
w = 2π f = 2π / T
f = 1 / T
T = 1 / f
T = 1 / 0.59
T = 1.69 s
b) the spring constant
w = 2π f
w = 2π 0.59
w = 3.70 rad / s
w² = k / m
k = w² m
k = 3.70² 0.060
k = 0.825 N / m
c) the maximum speed
simple harmonic movement is described by the expression
x = A cos (wt + Ф)
speed is defined by
v =[tex]\frac{dx}{dt}[/tex]
v = -A w sin (wt + fi)
the speed is maximum when the cosine is ± 1
v = A w
v = 0.394 3.70
v = 1.46 feet/s
d) maximum acceleration
a = [tex]\frac{dv}{dt}[/tex]
a = - A w² cos wt + fi
the acceleration is maximum when the cosine is ±1
a = A w²
a = 0.394 3.70²
a = 5.41 ft / s²
e) velocity and acceleration for x = 6 cm
let's reduce the cm to feet
x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot
Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s
let's use the expression for the velocity
v = -A w sin (0 + Фi)
0 = - A w sin Ф
so sin Ф = 0 which implies that Фi = 0
the equation of motion is
x = A cos wt
x = 0.394 cos 3.70t
we substitute
0.1969 = 0.394 cos 370t
3.70 t = cos⁻¹ (0.1969 / 0.394)
let's not forget that the angle is in radians
3.70, t = 1.047
t = 1.047 / 3.70
t = 0.2826 s
we substitute this time in the equation for velocity and acceleration
v = - Aw sin wt
v = - 0.394 3.70 sin 3.70 0.2826
v = - 1,319 ft / s
a = - A w² cos wt
a = - 0.394 3.70² cos 3.70 0.2826
a = - 2.70 ft / s²
f) the kinetic and potential energy at this point
K = ½ m v²
let's slow down to the SI system
v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s
K = ½ 0.060 0.402²
K = 4.8 10⁻³ J
U = ½ k x²
U = ½ 0.825 0.06²
U = 1.49 10⁻³ J
What is the magnitude of the gravitational force acting on a
1.0 kg object which is 1.0 m from another 1.0 kg object?
Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!
bit.[tex]^{}[/tex]ly/3a8Nt8n
mdjxjxjcjfkfjjdksklqlakzjxjxkkskakMmznxkxkdkd?
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
I hope this helped!+*
A radioactive material produces 1160 decays per minute at one time, and 4.0 h later produces 170 decays per minute. whats the half life
Answer:
Half life is 3.23 hours
Explanation:
Given
Decay rate at starting = 1160 decays per minute
Decay rate after 4 hours = 170 decays per minute
As we know know
[tex]N = N_0 *e ^{\Lambda *T}[/tex]
Substituting the given values, we get -
[tex]170 = 1160 *e ^{-4*\Lambda}\\0.1465 = e ^{-4*\Lambda}\\-0.834 = -4 * \Lambda\\\Lambda = 0.834/4\\\Lambda = 0.2085[/tex]
Also
[tex]t_{1/2} = \frac{ln2}{\Lambda}[/tex]
Substituting the given values we get -
[tex]t_{1/2} = =0.693/0.2085\\= 3.23[/tex]hours
ILL GIVE BRAINLIST PLS In which circuit does charge reverse direction many times per second?
A. A DC circuit
B. A combined circuit
C. A parallel circuit D. An AC circuit
Answer: D. An AC circuit
Explanation:
I took it on a test and it was correct ; )
two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg
and the incline is µ1 = 0.20 and that between the block of mass 4.0 kg and the
incline is µ2 = 0.30. Find the acceleration of 2.0 kg block. ( g = 10m/s^2).
Answer:
The acceleration of 2.0 kg block is 2.7 m/s²
Explanation:
Since, µ₁ < µ₂ acceleration of 2 kg block down the plane will be more than the acceleration of 4 kg block, if allowed to move separately. But, as the 2.0 kg block is behind the 4.0 kg block both of them will move with same acceleration say a. Taking both the blocks as a single system:
Force down the plane on the system
= (4 + 2) g sin30°
= (6)(10)(½)
= 30N
Force up the plane on the system
= µ₁ (2)(g)cos30° + µ₂ (4)(g)cos30°
= (2µ₁ + 4µ₂) g cos30°
= (2 × 0.2 + 4 × 0.3)(10)(√3/2)
≈ 13.76 N
∴ Net force down the plane is F
F = 30 - 13.76
F = 16.24 N
∴Acceleration of both the blocks down the
plane will b a
a = F ÷ (4 + 2)
a = 16.24 ÷ 6
a = 2.7 m/s²
Thus, The acceleration of 2.0 kg block is
2.7 m/s²
-TheUnknownScientist
Explanation:
2.7m/s2
I hope its helpful
Which type of wires are ferromagnetic metals?
cooper
aluminum
string
A running Marites launched the egg she
stole as she was about to be caught with
a velocity of 25 m/s in a direction making
an angle of 20° upward with the
horizontal
a) What is the maximum height reached by
the egg?
b) What is the total flight time (between
launch and touching the ground) of the
egg?
c) What is the horizontal range (maximum
* above ground) of the egg?
d) What is the magnitude of the velocity
of the egg just before it hits the ground?
Answer:
a) y = 3.73 m, b) t = 1.74 s, c) R = 40.99 m,
d) vₓ = 23.49 m/s, v_y = -8.5 m / s
Explanation:
This is a projectile launching exercise, we start by breaking down the initial velocity
sin θ = v_{oy} / v₀
cos θ = v₀ₓ / v₀
v_{oy} = v₀ sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 25 sin 20 = 8.55 m / s
v₀ₓ = 25 cos 20 = 23.49 m / s
a) when the egg reaches the maximum height its vertical speed is zero
v_y² = v_{oy}² - 2 g y
0 = v_[oy}² - 2g y
y = v_{oy}² / 2g
y = [tex]\frac{8.55^2}{2 \ 9.8 }[/tex]
y = 3.73 m
b) flight time
y = v_{oy} t - ½ g t²
the time of flight occurs when the body reaches the ground y = 0
0 = (v_{oy} - ½ g t) t
The results are
t₁ = 0s this time is for using the body star
v_{oy} - ½ g t = 0
t = [tex]\frac{2v_{oy}^2}{g}[/tex]
t = 2 8.55 / 9.8
t = 1.74 s
c) the range
R = v₀² sin 2θ / g
R = 25² sin (2 20) / 9.8
R = 40.99 m
d) speed at the point of arrival
horizontal speed is constant
vₓ = v₀ₓ = 23.49 m/s
vertical speed is
v_y = Iv_{oy} - g t
v_y = 8.55 - 9.8 1.74
v_y = -8.5 m / s
If acceleration is zero what statement about velocity is true *
A)Velocity is zero
B)Velocity is constant
C)Velocity cannot be determined
D) Velocity is changing
Answer:
Option"B" is correct.
Explanation:
when a body move with constant velocity then acceleration is zero.
If the acceleration of an object is 0, it means the velocity of an object is constant and not changing with respect to time,
So, the correct option will be :
=》B)Velocity is constant
whem completing an emergency Roaside stop,it is necessary to put on your parking brake
A. True
B. False
Answer:
trueeeeeeee..........mmmm...........
a. Why don't we hear the sound of oscillation of second
pendulum?
Answer:
because the frequency is too low for humans to hear.
Explanation:
Answer:
we cannot hear the sound produced due to vibrations of a seconds' pendulum.
Explanation:
This is because the frequency of sound produced as a result of vibrations of seconds' pendulum is 0.5Hz0.5Hz which is infrasonic sound.
How much work will a 500 watt motor do in 10 seconds?
Answer:
50j
Explanation:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Answer:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Explanation:
The universe cooled after the Big Bang.At some point hydrogen atoms combined to form helium.What is this process called?
Answer:
Nuclear fusion
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)
Describing a Wave
What does a wave carry?
Answer:
Waves carry energy from one place to another.
Explanation:
Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals.
Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel to the plane. The coefficient of kinetic friction between each mass and the plane has the same value μ.k At the highest point is:______
a. KEA > KEB
b. KEA = KEB
c. KEA < KEB
d. The work done by F on A is greater than the work done by F on B.
e. The work done by F on A is less than the work done by F on B.
Answer:
The correct answer is option (A) that is KEA > KEB .
Explanation:
Let us calculate -
If the object is straighten up and inclined plane , the work done is
[tex]W=F_d- F_f_r_id-F_gh[/tex]
[tex]W=F_d-\mu_kmgdcos\theta-mgdsin\theta[/tex]
The change in kinetic energy is ,
[tex]\Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2[/tex]
At the top of the inclined plane , the velocity is zero
So,
[tex]\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2[/tex]
[tex]\Delta KE=-\frac{1}{2}m\nu_0^2[/tex]
From the work energy theorem , we have [tex]W=-\Delta K[/tex] in case of friction , so
[tex]\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
[tex]KE=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
For object A-
[tex]KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
For object B
[tex]KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta[/tex]
[tex]KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)[/tex]
Thus , larger mass is going to mean less total work and a lower kinetic energy .
From the above results , we get
[tex]KE_A >KE_B[/tex]
Therefore , option A is correct .
Which statement best explains why objects are pulled toward Earth’s center?
Answer:
Earth has a much greater mass than objects on its surface
An object is placed 12.0 cm from a thin diverging lens with a focal length of 4 cm. Which one of the
following statements is true concerning the image?
A. The image is virtual and 3.0 cm from the lens.
B. The image is real and 6.0 cm from the lens.
C. The image is virtual and 12 cm from the lens.
D. The image is real and 12 cm from the lens.
Answer:
soluble soluble soluble soluble
Explanation:
solublesolublesolublesolublesolublesolublesoluble dguhjjewugbcsbdc csyuhjci
which of the following best defines spring constant ?
a. the amount of force needed to extend or compression og the spring.
b. the amount of force needed every 1 meter of stretch or compression of the spring.
c. the amount of energy needed to extend or compress a spring for every 1 kilogram of mass of the spring.
d. the amount of energy needed for every 1 meter of stretch or compression of the spring.
Answer:
A
Explanation:
an object is moving at 60m/s and has a mass of 5 kg what is its momentum
Answer:
300
Explanation:
the momentum is 300
p=mv
p=5×60
5×60 =300
What does the Curl-up test assess?
O A.
Body composition
ОВ.
Muscular strength and endurance
O C. Flexibility
D.
Cardiovascular fitness
HURRRY
(it’s pe not physics)
Answer:
assesses C.) muscular endurance
Explanation:
Scientists are constantly exploring the universe, looking for new planets that support life similar to the life on
Earth. A new planet that supports life would have all of the following characteristics except -
A. a gaseous atmosphere.
B. an orbiting moon.
C. liquid water.
D. protection from radiation.
A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.
What is a Planet?An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.
By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.
The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.
To know more about Planet:
https://brainly.com/question/14581221
#SPJ2
Classify each change (which can be manipulated within the green box) according to its effect on the wavelength.
a. Decrease frequency
b. Decrease damping
c. Decrease amplitude
d. Increase frequency
e. Increase amplitude
f. Increase damping
g. Shortens wavelength
Answer:
Explanation:
The classification will be made into 3 categories, which are
Ones that shortens wavelengths
Ones that lengthens wavelengths
Ones that has no effect on wavelengths
Shortens wavelengths -> Increase frequency
Lengthens wavelengths -> Decrease frequency
No effect -> Increase amplitude, decrease amplitude, increase damping, decrease damping.
What happens when Earth rotates on its axis and how long does it take
Answer:
You get Day and Night
It takes 24 hour
Answer:
Explanation:
The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.
A doorknob is a type of wheel and axle. In a doorknob, the door handle acts as the wheel. The handle is connected to a cylinder, called a spindle, which acts as the axle. When the handle turns, the spindle turns in the same direction. The spindle is located inside the handle and pulls the latch back and forth, allowing the door to open.
In the doorknob shown above, when the handle is rotated a distance of 84 millimeters, the spindle is rotated a distance of 14 millimeters. What is the mechanical advantage of this doorknob?
A. 504
B. 6
C. 84
D. 14
