Answer:
Im so so sorry but I do now know how to do this
Explanation:
Biodiversity decline poses a problem in an ecosystem because
Answer:
Biodiversity decline continues due to a rapidly expanding human population. Habitat is damaged in order to meet growing needs for agriculture, urban development, water and materials. Fish, wildlife and plants are overharvested, despite mounting evidence that many harvesting practices are unsustainable.
A spring is hung from the ceiling. When a coffee mug is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation
Answer:
Explanation:
In equilibrium , weight of mug is equal to restoring force .
mg = kx where m is mass of mug , k is spring constant and x is extension .
k / m = g / x = 9.8 ms⁻² / .025 m
= 392
frequency of oscillation n = [tex]\frac{1}{2\pi}\sqrt{\frac{k}{m} }[/tex]
[tex]n=\frac{1}{2\pi}\sqrt{392 }[/tex]
= 4.46 per second.
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
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.400 rev/s. What is its angular velocity (in rev/s) after a 22.0 kg child gets onto it by grabbing its outer edge
Answer:
The final angular velocity is rev/s is 0.293 rev/s.
Explanation:
Given;
mass of the merry-go-round, m₁ = 120 kg
radius of the merry-go-round, r = 1.8 m
initial angular velocity, ω = 0.4 rev/s
mass of the child, m₂ = 22 kg
Apply the principle of conservation angular momentum to determine the final angular velocity;
[tex]I_i= I_f\\\\\frac{1}{2} m_1r^2 \omega _i = \frac{1}{2} m_1r^2 \omega _f + m_2r^2 \omega _f\\\\ \frac{1}{2} m_1r^2 \omega _i =( \frac{1}{2} m_1r^2 + m_2r^2 )\omega _f\\\\\omega _f = \frac{ \frac{1}{2} m_1r^2 \omega _i}{\frac{1}{2} m_1r^2 + m_2r^2} \\\\\omega _f = \frac{ \frac{1}{2} m_1 \omega _i}{\frac{1}{2} m_1 + m_2}\\\\\omega _f = \frac{0.5 \ \times \ 120\ kg \ \times \ 0.4\ rev/s}{0.5 \ \times 120\ kg \ \ + \ \ 22 \ kg} \\\\\omega _f = 0.293 \ rev/s\\[/tex]
Therefore, the final angular velocity is rev/s is 0.293 rev/s.
Pedro is planning to model how changes in weather affect evaporation from lakes for his first experiment he wants to test how humidity affects the evaporation rate. he places one beaker with 300 mL of water in a dry area. and places another beaker with 300 mL of water near a humidifier which of the following variables does Pedro need to control during his experiment
A. humidity only
B. humidity and evaporation rate
C. volume of water and tempature
D. volume of water only
Pedro needs to control the variables such as volume of water and temperature during his experiment. So, option C.
What is meant by humidity ?The amount of water vapor in the air is known as humidity. The humidity will be high if there is a lot of water vapour in the atmosphere.
Water can evaporate even at very low temperatures, but as the temperature rises, the rate of evaporation increases.
More surface molecules per unit of volume may be able to escape from a substance with a larger surface area, so it will evaporate more quickly.
The control variables in an experiment are the variables that the experimenter intends to keep constant always so as to limit their effect on the measurements of the relationship between the dependent and the independent variable.
Therefore, in order to have a proper measurement of the effect of humidity on evaporation rate, other variables such as temperature, and the volume of the water in the experiment investigations which affect evaporation rate by the provision of heat, (temperature) and their heat capacity, the volume, etc. should be controlled.
Hence,
Pedro needs to control the variables such as volume of water and temperature during his experiment. So, option C.
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Plz help w answer 1:/ confused ash
Answer:
I would say d I had the same question yesterday and I got it correct so hope that helps
Science questions!! Please help!!
Post Assessment on Investigating the Immune System
please help!! Please choose the right answers!! Dont guess if you dont know the answers!!
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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You are asked to design a spring that will give a 1070 kg satellite a speed of 3.75 m/s relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00g. The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will all be negligible.
(a) What must the force constant of the spring be?
(b) What distance must the spring be compressed?
Answer:
[tex]380697.33\ \text{N/m}[/tex]
[tex]0.138\ \text{m}[/tex]
Explanation:
m = Mass rocket = 1070 kg
v = Velocity of rocket = 3.75 m/s
a = Acceleration of rocket = 5g
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
The energy balance of the system is given by
[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\Rightarrow kx=\dfrac{mv^2}{x}\\\Rightarrow kx=\dfrac{1070\times 3.75^2}{x}\\\Rightarrow kx=\dfrac{7250}{x}[/tex]
The force balance of the system is given by
[tex]ma=kx\\\Rightarrow m5g=\dfrac{7250}{x}\\\Rightarrow x=\dfrac{7250}{1070\times 5\times 9.81}\\\Rightarrow x=0.138\ \text{m}[/tex]
The distance the spring must be compressed is [tex]0.138\ \text{m}[/tex]
[tex]k=\dfrac{7250}{x^2}\\\Rightarrow k=\dfrac{7250}{0.138^2}\\\Rightarrow k=380697.33\ \text{N/m}[/tex]
The force constant of the spring is [tex]380697.33\ \text{N/m}[/tex].
Which diagram shows magnets that will attract each other? 2 bar magnets side by side with their long axes vertical, both red S on top and blue N on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with blue N on top and red S on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with blue N on top and red S on bottom and the bottom magnet with red S on top and blue N on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with red S on top and blue N on bottom.
2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with red S on top and blue N on bottom. this diagram shows magnets that will attract each other. Hence option D is correct.
What is Magnet ?A permanent magnet is an item constructed of magnetised material that generates its own persistent magnetic field. A refrigerator magnet, for example, is commonly used to hold notes on a refrigerator door. Ferromagnetic (or ferrimagnetic) materials are those that can be magnetised and are strongly attracted to a magnet. These include the elements iron, nickel, and cobalt, as well as their alloys, some rare-earth metal alloys, and naturally occurring minerals such as lodestone. Although ferromagnetic (and ferrimagnetic) materials are the only ones that are strongly attracted to a magnet and are widely thought to be magnetic, all other substances respond weakly to a magnetic field via one of many different forms of magnetism.
Hence option D is correct.
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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]
Two creatures sit on a horizontal frictional rotating platform. The platform rotates at a constant speed. The creatures do not slip off as it rotates.
ASSUME:
Red has a mass of 5 kg
Red is 1.5 m from the center
Red has a speed of 9 m/s
Blue has a mass of 25 kg
Blue has a speed of 1.8 m/s
The force of friction on Red is EQUAL to the force of friction on Blue
DETERMINE:
How far from the center is Blue
Answer:
M v^2 / R = centripetal force
For Red: M v^2 / R = 5 * 9^2 / 1.5 = 270
For Blue M v^2 / R = 270 = 25 * 1.8^2 / Rb
So Rb = 25 * 1.8^2 / 270 = .3 m
A 10 kg medicine ball is thrown at a velocity of 15 km/hr ( m/s) to a 50 kg skater who is
at rest on the ice. The skater catches the ball and subsequently slides with the ball across the
ice.
Complete Question
A 10 kg medicine ball is thrown at a velocity of 15 km/hr ( m/s) to a 50 kg skater who is at rest on the ice. The skater catches the ball and subsequently slides with the ball across the ice.
Calculate the kinetic energy after collision(in joules).
Answer:
[tex]K.E=70.23J[/tex]
Explanation:
From the question we are told that:
Mass of ball [tex]m_b=10kg[/tex]
Speed [tex]V_{b1}=15 km/hr ( m/s)[/tex]
[tex]V_{b1} = 4.1667 m/s[/tex]
[tex]V_{b1} = 4.1667 m/s[/tex]
Mass of Skater [tex]m_s=50kg[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_sV_{s1}+m_bV_{b1}=(m_s+m_b)V[/tex]
[tex]V=\frac{m_sV_{s1}+m_bV_{b1}}{(m_s+m_b)}[/tex]
[tex]V=\frac {50+10*4.1667}{(50+10)}[/tex]
[tex]V=1.53m/s[/tex]
Generally the equation for Kinetic energy is mathematically given by
[tex]K.E=\frac{1}{2}(m_s+m_b)V^2[/tex]
[tex]K.E=\frac{1}{2}(50+10)(1.53)^2[/tex]
[tex]K.E=70.23J[/tex]
Therefore kinetic energy K.E after collision is given as
[tex]K.E=70.23J[/tex]
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.
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
XD
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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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)
g 2. In a laboratory experiment on standing waves a string 3.0 ft long is attached to the prong of an electrically driven tuning fork which vibrates perpendicular to the length of the string at a frequency of 60 Hz. The weight (not mass) of the string is 0.096 lb. a) [5 pts] What tension must the string be under (weights are attached to the other end) if it is to vibrate in four loops
Answer:
The tension in string will be "3.62 N".
Explanation:
The given values are:
Length of string:
l = 3 ft
or,
= 0.9144 m
frequency,
f = 60 Hz
Weight,
= 0.096 lb
or,
= 0.0435 kgm/s²
Now,
The mass will be:
= [tex]\frac{0.0435}{9.8}[/tex]
= [tex]0.0044 \ kg[/tex]
As we know,
⇒ [tex]\lambda=\frac{2L}{n}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2\times 0.9144}{4}[/tex]
⇒ [tex]=0.4572 \ m[/tex]
or,
⇒ [tex]v=f \lambda[/tex]
⇒ [tex]=0.4572\times 60[/tex]
⇒ [tex]=27.432 \ m/s[/tex]
Now,
⇒ [tex]v=\sqrt{\frac{T}{\mu} }[/tex]
or,
⇒ [tex]T=\frac{m}{l}\times v^2[/tex]
On putting the above given values, we get
⇒ [tex]=\frac{0.0044}{0.9144}\times (27.432)^2[/tex]
⇒ [tex]=\frac{752.51\times 0.0044}{0.9144}[/tex]
⇒ [tex]=3.62 \ N[/tex]
A bat emits a 40 kHz chirp to locate flying insects. If the speed of sound is 340 m/s and a bat hears the echo from the moth after 0.6 seconds, then how far away is the moth?
102 m
Explanation:
The time 0.6 sec is the time it took for the sound to travel from the bat to the moth and back. So it took 0.3 sec for the sound to reach the moth. From the definition of speed, the distance of the moth d to the bat is given by
v = d/t ---> d = vt = (340 m/s)(0.3 sec) = 102 m
Convert 125 mL to L
Answer:
1 L = 1000 mL
125 mL = 125/1000 = 0.125 L
Answer:
0.125
Explanation:
divide by 1 000 to convert mL to liters
A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 25.0 m/s. Then the vehicle travels for 39.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.
(a) How long is the self-driving car in motion (in s)?
(b) What is the average velocity of the self-driving car for the motion described? (Enter the magnitude in m/s.) m/s
Answer:
[tex]56.5\ \text{s}[/tex]
[tex]21.13\ \text{m/s}[/tex]
Explanation:
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time
s = Displacement
Here the kinematic equations of motion are used
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}[/tex]
Time the car is at constant velocity is 39 s
Time the car is decelerating is 5 s
Total time the car is in motion is [tex]12.5+39+5=56.5\ \text{s}[/tex]
Distance traveled
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}[/tex]
[tex]s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}[/tex]
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}[/tex]
The total displacement of the car is [tex]156.25+975+62.5=1193.75\ \text{m}[/tex]
Average velocity is given by
[tex]\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}[/tex]
The average velocity of the car is [tex]21.13\ \text{m/s}[/tex].
A student wants to determine the speed of sound at an elevation of one mile. To do this the student performs an experiment to determine the resonance frequencies of a tube that is closed at one end. The student takes measurements every day for a week and gets different results on different days. Which of the following experiments would help the student determine the reason for the different results?
a. Repeating the experiment on several 10 degree C days and several 20 degree C days
b. Repeating the experiment using a wider range of frequencies of sound
c. Repeating the original experiment for an additional week
d. Repeating the experiment using a longer tube
Answer:
The correct answer is a
Explanation:
The speed of a sound wave depends on the square root of the modulus of compressibility and the density of the medium.
For the same medium, the speed of sound depends on the temperature of the fora
v = [tex]v_o \ \sqrt{1 + \frac{T}{273} }[/tex]
Therefore, the different results that are obtained are due to changes in temperature. The correct answer is a
since this way it has the values of the speed of sound for each temperature, for which it can compare with the results obtained from the trip.
Which of the following creates the night-and-day cycle experienced on Earth?
A) Orbit of Earth around the Sun
B) Rotation of the Sun
C) Rotation of the Earth
D) Tilt of the axis of Earth
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]
Place the balloon in a bell jar. If available also add some shaving cream and fresh marshmallows. Ask the instructor for help if you are unfamiliar with this apparatus. The motor of the vacuum pump will remove air from the bell jar when it is turned on. Make a prediction about what you think will happen to the balloon as air is removed from the bell jar.
Answer:
The balloon will collapse
Explanation:
When air is removed from the bell jar, the balloon will collapse if the internal pressure from the balloon does not balance the atmospheric pressure from the surroundings.
An electron is accelerated through 1.90 103 V from rest and then enters a uniform 1.80-T magnetic field.
(a) What is the maximum magnitude of the magnetic force this particle can experience?
Answer:
https://www.slader.com/discussion/question/an-electron-is-accelerated-through-240-times-103-v-from-rest-and-then-enters-a-uniform-170-t-magnetic-field-what-are-a-the-maximum-and-b-the-9e425fbd/
( Here is solution)
A ray of light traveling in water hits a glass surface. The index of refraction of the water is 1.33, and that of the glass is 1.50. At what angle with the plane of the surface must the incident ray strike the glass in order that the polarization of the reflected ray is the greatest
Answer:
[tex]\mu=41.5\textdegree[/tex]
Explanation:
From the question we are told that:
Water index of refraction [tex]i_w=1.33[/tex]
Glass index of refraction [tex]i_g=1.50[/tex]
Generally the equation for Brewster's law is mathematically given by
[tex]\theta=tan^{-1}(\frac{i_g}{i_w})[/tex]
[tex]\theta=tan^{-1}(\frac{1.50}{1.33})[/tex]
[tex]\theta=48.44 \textdegree[/tex]
Therefore Angle of incident to plane \mu (normal at 90 degree to the surface)
[tex]\mu=90\textdegree-\theta[/tex]
[tex]\mu=90\textdegree-48.44\textdegree[/tex]
[tex]\mu=41.5\textdegree[/tex]
Determine how would the frequency of the pendulum change if it was taken to the moon by finding the ratio of its frequency on the moon fM to its frequency on the earth fE. Suppose that gE is the free-fall acceleration on the earth and gM is the free-fall acceleration on the moon.
Express your answer in terms of some or all of the variables l, m, gE, gM.
fM/fE = ?
For the pendulum taken to the moon, The frequency change that would occur is mathematically given as
[tex]\frac{Fmoon}{Fearth}=0.408[/tex]
What frequency change would occur to the pendulum if it was taken to the moon?Generally, the equation for the Time period is mathematically given as
[tex]T=2\pi\sqrt{L/g}[/tex]
Therefore
[tex]\frac{Fmoon}{Fearth}=\frac{\sqrt{g/6L}}{\sqrt{g/6L}}\\\\\frac{Fmoon}{Fearth}=\sqrt{1/6}[/tex]
[tex]\frac{Fmoon}{Fearth}=0.408[/tex]
In conclusion, The frequency change
[tex]\frac{Fmoon}{Fearth}=0.408[/tex]
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Answer:
.408
Explanation:
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
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!