Answer:
Option 2
Explanation:
Given
Two 2-Ohm resistors are in series and these two resistors are in parallel with a 4-Ohm resistor
Equivalent resistance of two resistors in series = R1 + R2 = 2+2 = 4 Ohm
Equivalent resistance of two 4 Ohm resistors is parallel =1/ (1/4 +1/4) = 2 Ohm
Voltage = 12 Volts.
Hence, current = V/R = 12/2 = 6 Amp
Option 2 is correct
These steps are followed when using the half-life of carbon-14 to determine
the age of an object that contains carbon. What is the correct order of these
steps?
A. Use the half-life of carbon-14 to determine the number of half-lives
that have passed.
B. Measure the ratio of parent nuclei to daughter nuclei.
C. Use the number of half-lives that have passed to determine the age
of the object.
A. A,B,C
B. A,C,B
0 0
C. B, A,C
D. C, A, B
Answer: a different one is a.b.c
Explanation: still for ape.x
The correct order to determine the age of the an object using carbon-14 is C, A, B. Thus, option D is correct.
What is half life?
The half-life time is defined as the time taken by the radioactive element to reduce one half of its initial value. It is denoted by t(1/2).
To measure the age of an object, a radioactive isotope called carbon-14 is used. The half-life of carbon-14 is 5,730 years. All the objects in the universe consumes carbon in their lifetime and hence, carbon-14 is used to measure the age of the objects.
The process of determining the age of objects using carbon-14 is called Radiocarbon dating. All living organisms consume carbon in means of food and from atmosphere and when the plant and animals dies, the radioactive carbon atoms start decaying.
When it starts decaying, by using Carbon-14 the age of an object is calculated. The age is estimated by measuring the amount of carbon-14 present in the sample and comparing this carbon with the reference Carbon-14 isotope.
The amount of carbon in preserved plants is identified by:
f(t) = 10e {₋ct}
t = time in years when the plant dies( t= 0)
c = the amount of carbon-14 remaining in preserved plants.
The steps include to find the age of an object is :
1. Use the number of half-lives that have passed to determine the age of the object.
2. Use the half-life of carbon-14 to determine the number of half-lives that have passed.
3.Measure the ratio of parent nuclei to daughter nuclei.
Hence, from these steps the age of an object is determined. Therefore the correct solution is D) C, A, B.
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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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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.
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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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.
What is the light speed formula?
If ' c ' is the speed of light, then the formula for it is . . .
c = 299,792,458 meters per second
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].
Learning Goal: To understand the concept of moment of inertia and how it depends on mass, radius, and mass distribution.
In rigid-body rotational dynamics, the role analogous to the mass of a body (when one is considering translational motion) is played by the body's moment of inertia. For this reason, conceptual understanding of the motion of a rigid body requires some understanding of moments of inertia. This problem should help you develop such an understanding.
The moment of inertia of a body about some specified axis is I = cmr^2, where c is a dimensionless constant, m is the mass of the body, and r is the perpendicular distance from the axis of rotation. Therefore, if you have two similarly shaped objects of the same size but with one twice as massive as the other, the more massive object should have a moment of inertia twice that of the less massive one. Furthermore, if you have two similarly shaped objects of the same mass, but one has twice the size of the other, the larger object should have a moment of inertia that is four times that of the smaller one.
Two spherical shells have their mass uniformly distrubuted over the spherical surface. One of the shells has a diameter of 2 meters and a mass of 1 kilogram. The other shell has a diameter of 1 meter. What must the mass m of the 1-meter shell be for both shells to have the same moment of inertia about their centers of mass?
Answer:
m₂ = 4 kg
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
for bodies with high symmetry it is tabulated, for a spherical shell
I = 2/3 m r²
in this case the first sphere has a radius of r₁ = 2m and a mass of m₁ = 1 kg, the second sphere has a radius r₂ = 1m.
They ask what is the masses of the second spherical shell so that the moment of inertia of the two is the same.
I₁ = ⅔ m₁ r₁²
I₂ = ⅔ m₂ r₂²
They ask that the two moments have been equal
I₁ = I₂
⅔ m₁ r₁² = ⅔ m₂ r₂²
m₂ = (r₁ / r₂) ² m₁
let's calculate
m₂ = (2/1) ² 1
m₂ = 4 kg
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]
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
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
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 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].
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!!
While diving in cancun Mexico where the seawater has a density of 1,015 kg/m3 Nana observed that her pressure meter device reading was 3.75 atm. The reading at sea level is standard 1.0 atm. At what depth is she diving when the meter read 3.75 atm g
Answer:
The depth of the diver is 28.01 m
Explanation:
Given;
density of the seawater, ρ = 1,015 kg/m³
standard sea level pressure, P₀ = 1.0 atm = 101,325 Pa
the final reading of her pressure, P₁ = 3.75 atm = 379968.75 Pa
acceleration due to gravity, g = 9.8 m/s²
Let the depth she was diving at the final pressure = h
This depth is calculated as;
P₁ = P₀ + ρgh
P₁ - P₀ = ρgh
[tex]h = \frac{ P_1 \ - \ P_o}{\rho g} = \frac{379968.75 \ - \ 101325}{1015 \ \times \ 9.8} = 28.01 \ m[/tex]
Therefore, the depth of the diver is 28.01 m
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
A record is spinning on a turntable. A record is a uniform disk of mass 1.00 kg and a radius of 0.13 m that spins around an axis through its center. The record is initially spinning at 10 rad/s. Then the motor is turned off and the record slows to a stop with constant angular acceleration. As the record is slowing down it spins through 6.37 revolutions. What is the magnitude of the net torque acting on the record as it slows down
Answer:
T = 0.01 Nm
Explanation:
First, we will calculate the angular acceleration of the disk:
[tex]2\theta\alpha = \omega_f^2-\omega_i^2[/tex]
where,
θ = angular displacement = (6.37 rev)(2π rad/1 rev) = 40.02 rad/s
α = angular acceleration = ?
ωi = initial angular speed = 10 rad/s
ωf = final angular speed = 0 rad/s
Therefore,
[tex](2)(40.02\ rad/s)\alpha = (0\ rad/s)^2-(10\ rad/s)^2[/tex]
α = -1.25 rad/s²
negative sign shows deceleration
α = 1.25 rad/s²
Now, we will calculate the moment of inertia of disk:
[tex]I = \frac{1}{2}mr^2[/tex]
where,
I = Moment of Inertia = ?
m = mass of disk = 1 kg
r = radius of disk = 0.13 m
Therefore,
[tex]I = \frac{1}{2} (1\ kg)(0.13\ m)^2[/tex]
I = 0.00845 kg.m²
Now, the torque can be given as:
T = Iα
T = (0.00845 kg.m²)(1.25 rad/s²)
T = 0.01 Nm
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 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]
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:
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 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.
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]
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)
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 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.
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.
Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fully compressed compare to the energy of the system at the moment immediately before the box hits the ground? Justify your answer.
Answer:
the energy when it reaches the ground is equal to the energy when the spring is compressed.
Explanation:
For this comparison let's use the conservation of energy theorem.
Starting point. Compressed spring
Em₀ = K_e = ½ k x²
Final point. When the box hits the ground
Em_f = K = ½ m v²
since friction is zero, energy is conserved
Em₀ = Em_f
1 / 2k x² = ½ m v²
v = [tex]\sqrt{ \frac{k}{m} }[/tex] x
Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.
Based on the law of conservation of energy, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.
What is the energy in a compressed spring?The energy in a compressed spring is elastic potential energy given by the formula:
Ek = 1/2 Kx^2where
K is spring constant x is displacement of the springWhat is the kinetic energy of a body?The kinetic energy of a body is the energy the body the has due to it's motion.
Kinetic energy, KE, is givenby the formula below:
KE = 1/2mv^2How does the energy of the system when the spring is fully compressed compare to the energy of the system at the moment immediately before the box hits the ground?From the law of conservation of energy, the total energy in a closed system is conserved.
Based on this law, all the energy in the compressed spring is converted to the kinetic energy of the box just before it reaches the ground.
Therefore, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.
Learn more about conservation of energy at: https://brainly.com/question/381281
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
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.
To learn more about humidity, click:
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