Explanation:
hope it's helpful
have a nice day!!!! ;)
Three liquids that do not mix are poured into a cylindrical container with a diameter of 10.0 cm. The densities and volumes of the liquids are as follows.
Liquid 1: ????1 = 2.80 ✕ 103 kg/m3 and V1 = 2.00 ✕ 10−3 m3
Liquid 2: ????2 = 1.00 ✕ 103 kg/m3 and V2 = 1.50 ✕ 10−3 m3
Liquid 3: ????3 = 0.600 ✕ 103 kg/m3 and V3 = 1.00 ✕ 10−3 m3
Determine the pressure on the bottom of the container.
Answer:
P = 9622.9 Pa = 9.62 KPa
Explanation:
First, we will calculate the mass of all three liquids:
m = ρV
where,
m = mass of liquid
ρ = density of liquid
V = Volume of liquid
FOR LIQUID 1:
m₁ = (2.8 x 10³ kg/m³)(2 x 10⁻³ m³) = 5.6 kg
m₂ = (1 x 10³ kg/m³)(1.5 x 10⁻³ m³) = 1.5 kg
m₃ = (0.6 x 10³ kg/m³)(1 x 10⁻³ m³) = 0.6 kg
The total mass will be:
m = m₁ + m₂+ m₃ = 5.6 kg + 1.5 kg + 0.6 kg
m = 7.7 kg
Hence, the weight of the liquids will be:
W = mg = (7.7 kg)(9.81 m/s²) = 75.54 N
Now, we calculate the base area:
A = πr² = π(0.05 m)²
A = 7.85 x 10⁻³ m²
Now the pressure will be given as:
[tex]P = \frac{F}{A}\\\\P = \frac{75.54\ N}{7.85\ x\ 10^{-3}\ m^2}[/tex]
P = 9622.9 Pa = 9.62 KPa
A grindstone in the shape of a solid disk has a shaft attached to allow a force to be exerted on. The grindstone has a diameter of 0.650m and a mass of 55.0 kg. The shaft is 0.300 m from the center of the stone and has a mass of 4.00 kg. The grindstone has a motor attached and it is rotating at 450rev/min at a run when the motor is shut off. The grindstone comes to rest in 9.50 s.
a. What is the moment of inertia of the grindstone including the shaft?
b. What is the angular acceleration of the grindstone?
c. What average torque is done by friction to bring the grindstone to rest?
d. How many revolutions has it taken the grindstone to come to rest?
e. What is the linear acceleration of the shaft in this process?
f. What is the linear distance that the shaft traveled?
Answer:
a) 3.265 kg-m^2
b) - 4.96 rad/s^2
c) 16.1944 N-m
d) 35.625
e) - 1.488 m/s^2
f) 67.1175 m.
Explanation:
Given data:
Diameter of grindstone ( D ) = 0.650 m , Radius ( R ) = 0.325 m
mass of grindstone ( M ) = 55 kg
Radius of shaft ( r ) = 0.300 m
mass of shaft ( m ) = 4 kg
Initial Angular velocity = 450 rev/min = f = 7.5 rev/s = w =15π rad/s
time ( t ) = 9.50 secs
a) Determine the moment of inertia of the grindstone including the shaft
moment of inertia of grindstone = MR^2 / 2 = 55* (0.325)^2 / 2 = 2.905
moment of inertia of shaft = mr^2 = 4 *0.3^2 = 0.36
∴ moment of inertia including shaft = 2.905 + 0.36 = 3.265 kg-m^2
b) Determine the angular acceleration of the grindstone
∝ = - 15π / 9.5 ( i.e. angular velocity / time )
= - 4.96 rad/s^2 ( deceleration value )
c) Determine average torque done by friction to bring the grindstone to rest
Torque ( I * ∝ ) = 3.265 x 4.96 = 16.1944 N-m (magnitude)
d) Determine the number of revolutions before grindstone comes to rest
Total revolutions N before grindstone comes to rest
= ( f1 + f2)* t /2 = 7.5 * 9.5 / 2 = 35.625
Note : f2 = 0 as it comes to rest
f1 = 7.5 rev/s
e) Determine the Linear acceleration of the shaft in this process
This can be calculated using this relation
r * ∝ = 0.3 x (- 4.96 ) = - 1.488 m/s^2
f) Determine the linear distance travelled by the shaft
This can be calculated with the relation below
r * 2 * π * N = 0.3 * 2π * 35.625 = 67.1175 m.
_PCi5 + _H20 + _H3PO4 _HCI
A.) 1,3,1,5
B.) 2,4,1,5
C.) 1,4,1,5
D.) 2,1,2,5
Answer:
C
Explanation:
because of the kids are doing well and staying in the middle of the day early to get a new phone number for home now deep
Which of the following best explains the greater difficulty in stopping a 1000-kg car
moving at 174 km/h compared to an identical 1000-kg car moving at 100 km/h?
Stopping a 1000 kg car moving at a speed of 174 km/h will be more difficult than stopping an identical car moving at 100 km/h.
This can be explained by using Newton's second law of motion which says that the rate of change of momentum of a body is equal to the force applied.
We know that momentum of a body is the product of its mass and velocity.The momentum of the car moving at 174 km/h = [tex]1000\, kg \times 174\, km/h = 174000 \, kg.km/h[/tex].The momentum of the car moving at 100 km/h = [tex]1000\, kg \times 100\, km/h=100000 kg.km/h[/tex].Therefore, the car moving at 174 km/h has higher momentum and from Newton's second law of motion, a higher force would be required to stop this car.
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Please help I’ll mark you brainliest
Answer:
Percentage:
Rr = 50% because it's 2/4 (for both or 25% each since you have them separate)
rr = also 50%, because it's also 2/4.
Phenotype:
Rr = heterozygous
rr = "hozygous" recessive
In addition, RR is "hozygous" dominant
Explanation:
They said the hozygous is a swearword LOL.
For a damped oscillator with a mass of 360 g, a spring constant 130 N/m and a damping coefficient of 86 g/s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 11 cycles
Answer:
A/A₀ = 0.5106
Explanation:
To do this, we need to use several formulas and expressions. First, we need to know the period of time of the oscillator. This can be calculated using the following expression:
ω = 2π/T (1)
But angular frequency (ω) can be also be calculated using:
ω = √(k/m) (2)
Using (1) and (2), we can solve for the period T:
2π/T = √(k/m) (mass in kg)
2π/T = √(130/0.360)
2π/T = √361.11
2π/T = 19.003
T = 2π/19.003 = 0.331 s
Now, the amplitude A at a time t, is:
A = x exp(-bt/2m) (3)
At time 0, A = x. so A₀ = x
The problem states that we have 11 cycles respect to the initial amplitude, so expression (3) can be rewritten as:
A = x exp(-b(17t/2m)) using b as kg/s = 0.086 kg/s
Replacing the data we have:
A = x exp(-0.086(17*0.331)/2*0.36)
A = x exp(-0.086 * 7.815)
A = x exp(-0.6721)
A = 0.5106x (4)
Now, doing the ratio with the innitial we have:
A / A₀ = 0.5106x / x
The ratio is:
A/A₀ = 0.5106Hope this helps
A microscope has an objective lens with diameter 1.04 cm. You wish to resolve an object 9.09 micrometers in size. You are using visible light of wavelength 562 nm. Using the Rayleigh criterion, what is the distance that the objective lens must be from the object to resolve it
Answer:
L = 0.1379 m = 13.79 cm
Explanation:
The Rayleigh criterion establishes that two objects are separated when the maximum of diffraction for slits coincides with the minimum of the other point, therefore the expression for the diffraction
a sin θ = m λ
the first zero occurs when m = 1
let's use trigonometry to find the angle
tan θ = y / L
as in these experiments the angles are very small
tan θ = sin θ /cos θ = sin θ
sin θ = y / L
we substitute
a y /L = λ
In the case of circular aperture the system must be solved in polar coordinates, for which a numerical constant is introduced
a y / L = 1.22 λ
L = a y / 1.22 λ
We search the magnitudes to the SI system
a = 1.04 cm = 1.04 10⁻² m
y = 9.09 10⁻⁶ m
λ = 562 10⁻⁹ m
let's calculate
L = [tex]\frac{1.04 \ 10^{-2} \ 9.09 \ 10^{-6} }{1.22 \ 562 \ 10^{-9} }[/tex]
L = 1.379 10⁻¹ m
L = 0.1379 m = 13.79 cm
The chart lists the masses of four planets.
Planetary Masses
Planet
Mass
Neptune
1.02 x 1026
Uranus
8.68 x 1025
Mars
6.42 x 1023
Venus
4.87 X 1024
According to evidence that supports Einstein's general
theory of relativity, which list shows the planets that
would cause curvature in space-time from the least
amount of curvature to the greatest?
O Mars, Venus, Uranus, Neptune
O Neptune, Uranus, Venus, Mars
O Neptune, Uranus, Mars, Venus
O Venus, Mars, Uranus, Neptune
Answer:
I think it's A
Explanation:
It's definitely not B on edge
The correct option for the given question about Einstein's general
theory of relativity is Option A) Mars, Venus, Uranus, Neptune.
What is the Einstein's general theory of relativity?Albert Einstein established that the rules of physics apply to all non-accelerating observers in his theory of special relativity.He also demonstrated that the speed of light in a vacuum remains constant regardless of the velocity of an observer.The theory may be used to anticipate everything and describes how objects behave in space and time. For instance: that if there are black holes or not, if Gravity can causes light to bend, The way Mercury behaves when it is in orbit and many more interesting things.As a Conclusion, we can state that the planets who would cause curvature in space time from the least amount of curvature to the greatest will be in order Mars, Venus, Uranus, Neptune.
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A balloon with a charge of 4.0 µC is held a distance of 0.70 m from a second balloon having the same charge. Calculate the magnitude of the repulsive force. (Hint: µ is a metric prefix that means micro or 10 -6 so 4.0 µC means 4.0 microcoulombs
Answer:
Im sorry I do not know how to do this, I hope you will be able to figure it out
Explanation:
Your _________ is simply your weight-to-height ratio which can be calculated using certain body measurements.
O health
O size
O weight loss
O BMI
Answer:
SiZe
Explanation:
bad bunny is the best!!!
A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?
Answer:
f = 3.09 Hz
Explanation:
This is a simple harmonic motion exercise where the angular velocity is
w² = [tex]\frac{k}{m}[/tex]
to find the constant (k) of the spring, we use Hooke's law with the initial data
F = - kx
where the force is the weight of the body that is hanging
F = W = m g
we substitute
m g = - k x
k = [tex]- \frac{m g}{x}[/tex]
we calculate
k = [tex]- \frac{9.8 m}{- 2.6 \ 10^{-2}}[/tex]
k = 3.769 10² m
we substitute in the first equation
w² = [tex]\frac{ 3.769 \ 10^2 \ m }{m}[/tex]
w = 19.415 rad / s
angular velocity and frequency are related
w = 2πf
f = [tex]\frac{w}{2\pi }[/tex]
f = 19.415 / 2pi
f = 3.09 Hz
The Chernobyl reactor accident in what is now Ukraine was the worst nuclear disaster of all time. Fission products from the reactor core spread over a wide area. The primary radiation exposure to people in western Europe was due to the short-lived (half-life 8.0 days) isotope 131I
131 I, which fell across the landscape and was ingested by grazing cows that concentrated the isotope in their milk. Farmers couldn't sell the contaminated milk, so many opted to use the milk to make cheese, aging it until the radiactivity decayed to acceptable levels. How much time must elapse for the activity of a block of cheese containing 131I 131 I to drop to 1.0% of its initial value?
Answer:
The correct answer is "53.15 days".
Explanation:
Given that:
Half life of [tex]131_{I}[/tex],
[tex]T_{\frac{1}{2} }= 8 \ days[/tex]
Let the initial activity be "[tex]R_o[/tex]".and, activity to time t be "R".To find t when R will be "1%" of [tex]R_o[/tex], then
⇒ [tex]R=\frac{1}{100}R_o[/tex]
As we know,
⇒ [tex]R=R_o e^{-\lambda t}[/tex]
or,
∴ [tex]e^{\lambda t}=\frac{R_o}{R}[/tex]
By putting the values, we get
[tex]=\frac{R_o}{\frac{R}{100} }[/tex]
[tex]=100[/tex]
We know that,
Decay constant, [tex]\lambda = \frac{ln2}{T_{\frac{1}{2} }}[/tex]
hence,
⇒ [tex]\lambda t=ln100[/tex]
[tex]t=\frac{ln100}{\lambda}[/tex]
[tex]=\frac{ln100}{\frac{ln2}{8} }[/tex]
[tex]=53.15 \ days[/tex]
In addition to producing images, ultrasound can be used to heat tissues of the body for therapeutic purposes. An emitter is placed against the surface of the skin; the amplitude of the ultrasound wave at this point is quite large. When a sound wave hits the boundary between soft tissue and bone, most of the energy is reflected. The boundary acts like the closed end of a tube which can lead to standing waves. Suppose 0.7 MHz ultrasound is directed through a layer of tissue at a bone 0.55 cm below the surface. Remember, sound waves in the body travel at 1540 m/s. Explain.
Answer:
ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]
Explanation:
In this experiment the system can be approximated as a tube with one end open and the other closed.
The open end is where the ultrasound emitter is and the closed end where the limit between the tissue and the bone is, the length of the tube is L = 0.55 cm = 5.5 10⁻³ m
a node is formed at the closed end and a belly at the open end, so the resonance has the form
λ = 4L 1st harmonic
λ = 4/3 L third harmonic
λ = 4/5 L fifth harmonic
λ = 4L / (2n + 1) n = 0, 1, 2, (2n + 1)
This wave is a standing wave therefore energy density remains in place
P = 1/2 ρ v (w S₀)²
angular velocity is related to frequency
w = 2π f
we substitute
E = P = 2π² ρ v f² S₀²
if this energy per unit area is transformed into heat
E = m c_e DT
let's use the concept of density
ρ = m / V
m = ρ V
if there are no losses in the system
½ π² ρ v f² S₀² = ρ V c_e ΔT
ΔT = [tex]\frac{\pi ^2 \ v \f^2 S_o^2}{2V \ c_e}[/tex]
When analyzing this expression the temperature increase is
* quadratic at the frequency and maximum amplitude of the wave
* proportional to the speed of the wave in the tissue
* inversely proportional to tissue volume
we can approximate the volume of the tissue to the volume of a cylinder tube
V = π r² L
ΔT = [tex]\frac{\pi \ v \ f^2 S_o^2 }{r^2 \ L \ c_e}[/tex]
we calculate
ΔT = π 1450 (0.7 10⁶)² S₀² /( r² 5.5 10-3 c_e)
ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]
Please help me ASAP!!
Answer:
the ball lose kentic energy and gains potential energy rolling upward
Answer:
c
Explanation:
when a ball is rolling down it loses potential and gains kinetic, but in this case since the ball is going upwards it is losing kinetic energy and gaining potentail like a roller coaser
the density of brick is 1,600 kg/m3. what is the mass of a brick with a volume of 0.0006 m3? WILL MARK BRAINLIEST
Answer:
.0000004
Explanation:
The mass of a brick with a volume of 0.0006 m³ and a density of 1600 kg/m³ is 0.96kg.
HOW TO CALCULATE MASS?The mass of a substance can be calculated by multiplying the density of the substance by its volume. That is;
Mass = density × volume
According to this question, the density of brick is 1,600 kg/m3 and it has a volume of 0.0006m³. The mass is calculated as follows:
Mass = 0.0006 × 1600
Mass = 0.96kg
Therefore, the mass of a brick with a volume of 0.0006m³ and a density of 1600 kg/m³ is 0.96kg.
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when a light beam emerges from water into air, the average light speed ____.
a. increases
b. decreases
c. remains the same
d. not provided
Answer:
Increases
Explanation:
The light is emerging from a denser medium to rarer medium so the average light speed increases.
When a light beam emerges from water into air, the average light speed increases.
What is refraction?The bending of a ray passes at an angle from one medium to another in which the speed will be different, as when light passes from air into water and also from water into air can be known as refraction.
When light travels from water to air, the angle from water to air the light gets speeds up and then its changes direction. The light turns away from the normal line. Because the light moves from a denser medium to the rarer medium. But from air to water the speed will be decreased because in that case the light travels from the rarer medium to the denser medium.
Hence, the light beam when passes, the average speed gets increased.
So, Option A is the correct answer.
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The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated withsilver so that the inner surface is reflective. When light is incident on the uncoated side it passes throughthe lens, reflects off the silver coating, and passes back through the lens. The overall effect is that of amirror with focal length 5.0 cm. What is the index of refraction of the lens material
Answer:
n = 1.4
Explanation:
Given,
R1 = 18 cm, R2 = -18 cm
From lens makers formula
1/f = (n - 1)(1/18 + 1/18) = (n-1)/9
f = 9/(n-1)
Power, P = 1/f ( in m) = (n-1)/0.09
Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens
Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09
Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05
n = 1.4
Help pls... 100points
1 Answer:
The pitch of a sound depends on
a. frequency
b. intensity
c. amplitude
d. quality
2 Answer:
The loudness of a sound depends on
a. frequency
b. intensity
c. overtones
d. timbre
3 Answer:
The number of sound waves per unit time is called
a. amplitude
b. wavelength
c. speed
d. frequency
4 Answer:
The blending of fundamental tone and overtones produces a sound's
a. intensity
b. pitch
c. timbre
d. loudness
5 Answer:
Sonar is used
a. to locate objects in water
b. to determine underwater distances
c. for both a and b
d. for neither a nor b
6 Answer:
Sounds of frequencies below 20 Hz are called
a. fundamentals
b. overtones
c. ultrasonic sounds
d. infrasonic sounds
7 Answer:
A sound is music if it has
a. a definite pitch
b. a pleasing quality
c. rhythm
d. all of these characteristics
8 Answer:
The stirrup is a part of the
a. outer ear
b. middle ear
c. inner ear
d. cochlea
9 Answer:
An example of a percussion instrument is a (an)
a. trumpet
b. harp
c. drum
d. flute
10 Answer:
The speed of sound depends on
a. temperature
b. density
c. elasticity
d. all of these factors
11 Answer:
Sounds with frequencies above 20,000 Hz are called _____.
12 Answer:
The science of sound is called _____.
13 Answer:
High-frequency waves are used in a sound navigation and ranging
system called _____.
14 Answer:
Sounds produced when sections of a string vibrate separately are
called _____.
15 Answer:
Sound without pleasing quality, identifiable pitch, and repeating
patterns is called _____.
16 Answer:
The speed of sound decreases with increasing temperature.
a. true
b. false
17 Answer:
Sound waves are transverse.
a. true
b. false
18 Answer:
The decibel is a unit of pitch.
a. true
b. false
19 Answer:
A change in the pitch of a sound due to motion of source or
receiver (observer) is called the Doppler effect.
a. true
b. false
20 Answer:
Guitars and cellos are examples of percussion instruments.
a. true
b. false
Answer:
Explanation:
1. a as pitch depends on frequency
2. b as loudness depends on intensity
3. d as that is definition of frequency
4. c as that is definition of timbre
5. c as sonar can do both
6. d as that is definition of infrasound
7. d
8. b as that is one of three bones in the middle ear
9. c as you hit the drum to make sound
10. d
11. ultrasound
12. acoustics
13. ultrasound
14. harmonics? or resonance?
15. noise
16. b
17. b
18. b
19. a
20. b
What is the answer to this problem
Answer:
Material that allow the electrons to move freely in order to produce a current
Please mark as brainliest if answer is right
Have a great day, be safe and healthy
Thank u
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18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de
heno desde una altitud de 60.0 m. Si la paca de heno pesa 175 N, ¿cuál es el momentum
de la paca antes de que golpee el suelo?
Answer:
Definimos momento como el producto entre la masa y la velocidad
P = m*v
(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)
Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.
Peso = m*9.8m/s^2 = 175N
m = (175N)/(9.8m/s^2) = 17.9 kg
Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.
Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:
Vx = 36m/s
Mientras que para la velocidad vertical, usamos la conservación de la energía:
E = U + K
Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)
Entonces al principio solo hay energía potencial:
U = m*g*h
donde:
m = masa
g = aceleración gravitatoria
h = altura
Sabemos que la altura inicial es 60m, entonces la energía potencial es:
U = 175N*60m = 10,500 N
Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:
10,500N = (m/2)*v^2
De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.
√(10,500N*(2/ 17.9 kg)) = 34.25 m/s
La velocidad vertical es 34.25 m/s
Entonces el vector velocidad se podrá escribir como:
V = (36 m/s, -34.25 m/s)
Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.
Reemplazando esto en la ecuación del momento obtenemos:
P = 17.9kg*(36 m/s, -34.25 m/s)
P = (644.4 N, -613.075 N)
THe Wilson cloud chamber is used to study the appearance of individual atoms?
Answer:
The Wilson cloud chamber is used to study the direction, speed, and distance of charged particles. Explanation; ... The Wilson cloud chamber works by producing a super-saturated vapor, as explained by florin.
Explanation:
I hope it's help u
Which of the following statements correctly describes the index of refraction of a material? Select all that apply.
The index of refraction is the ratio of the speed of light in a material to the speed of light in a vacuum.
The index of refraction of a material must be greater than 1.
The index of refraction is the ratio of the speed of light in a vacuum to the speed of light in the material.
The index of refraction of a material must be less than 1.
The index of refraction of water is less than the index of refraction of air.
Answer:
Option C and D only
Explanation:
Option A is incorrect because refractive index of a material is the ratio of speed of light in vacuum to the speed of light in a any given medium
Option B is correct as the speed of light in vacuum is always greater than the speed of light in any given medium.
Option C is correct
Option D is incorrect
Option E is incorrect because the denser the medium the more is the refractive index. Water is denser than air, hence it should have more refractive index as compared to that of air.
A boy is pulling a sled with a net force of 10 N. If the mass of the sled is 20 kg, what is the acceleration of the sled?
Answer:
0.5 m/s
Explanation:
acceleration= force times mass
Given the amount of force applied on the sled as well as its mass, the acceleration of the sled is 0.5m/s².
What is force?A force is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.
From Newton's Second Law, force is expressed as;
F = m × a
Where is mass of object and a is the acceleration
Given the data in the question;
Force applied F = 10N = 10kgm/s²Mass of the sled = 20kgAcceleration a = ?F = m × a
10kgm/s² = 20kg × a
a = 10kgm/s² ÷ 20kg
a = 0.5m/s²
Given the amount of force applied on the sled as well as its mass, the acceleration of the sled is 0.5m/s².
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A flat circular coil of wire having 400 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic field of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?
Answer:
6.8 N.m
Explanation:
The computation of the magnitude of the magnetic torque on the coil is given below:
Given that
n = 400
d = 6.0 cm
Current is I = 7.0 A
Angle is [tex]\theta[/tex] = 30 degree
Now
We know that
the magnitude of the magnetic torque is
= nIABsin[tex]\theta[/tex]
= (400) (7.0) π ÷ 4 (0.06m)^2 sin(90° - 30°)
As
[tex]\theta[/tex] = (90° - Ф)
= (400) (7.0) π ÷ 4 (0.06m)^2 sin 60°
= 6.8 N.m
two 0.5 kg carts, one red and one green, sit about half a meter apart on a low friction track, you push on the red one with the constant force of 4N for 0.17m and then remove your hand. the cart moves 0.33 m on the track and then strikes the green cart. what is the work done by you on the two cart system?
Answer:
The work done by you on the two cart system is 2 N-m
Explanation:
Work done is the product of force and displacement.
W = F * D
Substituting the given values we get -
W =
[tex]4 * (0.17+0.33)\\= 2[/tex]
The work done by you on the two cart system is 2 N-m
b
Jasmine plays a note of wavelength 22 cm on her clarinet. The speed of
sound in air is 340 m/s. What is the frequency of this note? Give your
answer to 3 significant figures.
Answer:
Frequency = 1,550Hz
Explanation:
To solve this we can use the equation: [tex]f=\frac{v}{\lambda}[/tex]
(frequency = velocity/wavelength).
We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m
Now we can substitute these values into the formula and calculate to solve:
[tex]f=\frac{340}{0.22} \\\\f=1545.454...[/tex]
Simplify to 3 significant figures:
f = 1,550Hz
(Which I believe is just below a G6 if you were interested)
Hope this helped!
Osteoporosis is a condition that makes bones more
A. heavy
B. creaky
C. fragile
D. strong
answer is C
the answer for this question is c
8. Consider a capacitor that is made of two large conducting plates that are rectangular in shape (1 cm by 6 cm), aligned parallel to each other, and separated by an air-filled gap of 0.001 cm. This capacitor is included in a circuit where a battery provides 15,000 V of potential difference. When the capacitor is fully charged in this circuit, what is the c
Answer: [tex]7.96\ \mu C[/tex]
Explanation:
Given
The dimension of the plate is [tex]1\ cm\times 6\ cm[/tex]
The gap between the plate is [tex]0.001\ cm[/tex]
Voltage applied [tex]V=15,000\ V[/tex]
The capacitance of the capacitor is
[tex]C=\dfrac{\epsilon_o A}{d}\\\\C=\dfrac{8.85\times 10^{-12}\times 1\times 6\times 10^{-4}}{10^{-5}}\\\\C=53.1\times 10^{-11}\ F[/tex]
Charge acquired by the capacitor
[tex]\Rightarrow Q=CV\\\Rightarrow Q=53.1\times 10^{-11}\times 15,000\\\Rightarrow Q=796.5\times 10^{-8}\\\Rightarrow Q=7.96\times 10^{-6}\ C[/tex]
Light is traveling in glass, and hits a glass/unknown surface. In the glass the light beam is making an angle of 45.0 o with the normal to the surface. The glass has an index of refraction of 1.52. (A) If the refracted light ray leaves the glass at an 55o from the normal, what is the index of refraction for the unknown surface
Answer:
n = 1.31
Explanation:
When a ray of light crosses the separation surface between two transparent media, there exists a fixed relationship between the indexes of refraction of both media, related with the angles of incidence and refraction, which is known as Snell's Law.The Snell's Law can be written as follows:[tex]n_{i} * sin( \theta_{i}) = n_{r} * sin( \theta_{r}) (1)[/tex]
In our case the ray is incident from the glass, so ni = n glass = 1.52The angle of incidence is the angle that the ray makes with the normal to the separation surface, so θi=45º.The angle of refraction is the angle that the refracted ray makes with the normal, so θr= 55ºReplacing by the values in (1), and solving for nr, we have:[tex]n_{r} =\frac{n_{i} * sin \theta_{i} }{sin \theta_{r} } = \frac{sin (45)*1.52}{sin (55)} = 1.31 (2)[/tex]
use a trigonometric equation to determine the leg of this triangle
C=90°
A=30°
c=10m
What is a?
Answer: 5
Explanation: B is for sure 60°, c* cosB = 10*1/2 =5