Please Help with this
Answer:
Answer will be the 2nd one. I think it will be the Answer
When a reaction that occurs when a person experiences very strong emotion especially those associated with a perceived threat is called.
Answer:
The correct answer is - fight or flight response or reaction.
Explanation:
According to Cannon, the fight or flight reaction is a strong emotion experienced by a person especially people experiences associated with a perceived threat.
This reaction takes place in form of arousal in a perceived threat or stress conditions. It helps in maintaining the body environment in panic or stressed conditions.
Light from the Sun is converted into
.... by plants.
Answer:
Explanation:
The answer is chemical energy
Which structure is represented by letter C?
Choose 1 answers
A. Cell wall
B. Cell membrane
C. Ribosome
D. Cytosol
Answer: C. Ribosome
Explanation: Ribosome is responsible for protein synthesis. I got it right on khan academy :) hope this helps have a blessed day.
The structure which is represented by the letter C is known as free ribosomes. Thus, the correct option for this question is C.
What are Ribosomes?Ribosomes may be defined as a type of cell organelle which is spherical and glandular in shape. They occur freely in the matrix or remain bound with ER.
The major components of ribosomes may include RNA and protein. Ribosomes were first discovered by Palade in 1953. There are two types of ribosomes, ie. the 70s and 80s.
Ribosomes play an important function in the process of protein synthesis. The structure which is represented by the letter A is known as the cell membrane. The structure which is represented by the letter B is known as Cytosol.
Therefore, the structure which is represented by the letter C is known as free ribosomes. Thus, the correct option for this question is C.
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When you are high up in the air you
have
greater potential energy
less potential energy
Answer:
Its Greater potential energy because the air is high up and that makes high energy power
Explanation:
1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is initially at rest. After the collision, block A continues to the right at 1.0 m/s. The mass of block A is 2.0 kg and the mass of block B is 3.0 kg. The positive x-direction is to the right. a. Find the final velocity of block B. Show your work. b. Is the sign of the change in momentum of block A as a result of the collision positive, negative, or is it zero
Answer: 2.67 m/s
Explanation:
Given
Mass of block A is [tex]m_a=2\ kg[/tex]
mass of block B is [tex]m_b=3\ kg[/tex]
The initial velocity of block A [tex]u_a=5\ m/s[/tex]
the initial velocity of block B is [tex]u_b=0[/tex]
After collision velocity of block A is [tex]v_a=1\ m/s[/tex]
Conserving momentum
[tex]m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s[/tex]
The momentum of block A after the collision is [tex]P_a=2\times 1=2\ kg.m/s[/tex]
Therefore, there is no change in sign.
A bat emits a sonar sound wave
(343 m/s) that bounces off a
mosquito 8.42 m away. How
much time elapses between
when the bat emits the sound
and when it hears the echo?
(Unit = s)
Please help I am very confused with this topic
The time between when the bat emits the sound and when it hears the echo is 0.05 s
From the question given above, the following data were obtained:
Velocity of sound (v) = 343 m/s
Distance (x) = 8.42 m
Time (t) =?We can obtain obtained the time as illustrated below:
v = 2x / t
343 = 2 × 8.42 / t
343 = 16.84 / t
Cross multiply
343 × t = 16.84
Divide both side by 343
t = 16.84/343
t = 0.05 sThus, the time between when the bat emits the sound and when it hears the echo is 0.05 s
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Answer:
0.0491
Explanation:
Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move with the wind, but getting them to be windborne requires the moss to shoot the spores upward. Some species do this by using a spore-containing capsule that dries out and shrinks. The pressure of the air trapped inside the capsule increases. At a certain point, the capsule pops, and a stream of spores is ejected upward at 3.6 m/sm/s, reaching an ultimate height of 20 cm.
Required:
What fraction of the initial kinetic energy is converted to the final potential energy?
Answer:
U / K = 0.30
Explanation:
For this exercise we must calculate the energy at the two points.
Initial. Where are the spores
K = ½ m v²2
Final. Higher
U = m g h
the fraction of energy is
U / K = 2gh /v²
let's calculate
U / K = 2 9.8 0.20 / 3.6²
U / K = 0.30
therefore 30% of the energy is lost
what are 3 types of contact forces
There are different types of contact forces like normal Force, spring force, applied force and tension force.
Calculate the standard Gibbs energy of the reaction CO(g) + CH3CH2OH(l) → CH3CH2COOH(l) at 298 K, using the values of standard entropies and enthalpies of formation. The data is given below.
CO(g) : ΔfH0 = −110.53 kJ mol-1, Sm0=197.67 J K-1 mol−1 at 298 K. CH3CH2OH(l) : ΔfH0 = −277.69 kJ mol-1, Sm0=160.7 J K-1 mol−1 at 298 K. CH3CH2COOH(l) : ΔfH0 = −510 kJ mol-1, Sm0 = 191 J K-1 mol−1 at 298 K.
Answer:
-7.2 * 10^4 kJ mol-1
Explanation:
First we obtain the change in enthalpy for the reaction;
ΔHrxn= ΔHproducts - ΔHreactants
ΔHrxn=[( −510 ) - (−110.53) + (−277.69)]
ΔHrxn= -121.78 * 3 J mol-1
The we obtain the entropy change of the reaction
ΔSrxn= ΔSproducts - ΔSreactants
ΔSrxn= [(191) - (197.67) + (160.7)]
ΔSrxn= -167.37 J K-1 mol−1
Then we calculate ΔG at 298 K
ΔG = ΔH - TΔS
ΔG = ( -121.78 * 3) - (298) (-167.37)
ΔG = -7.2 * 10^4 kJ mol-1
Aliens from the planet Mars can have long or short antennas. The allele for a long antenna is DOMINANT, and the allele for a short antenna is RECESSIVE. Which of the following is the pair of alleles an alien would have to inherit to have a short antenna?
Recessive, dominant
Dominant, dominant
Recessive, recessive
Dominant, recessive
Answer:
Recessive, dominant
Explanation:
umm thats the answer
what is the definition of oligopoly?
Answer:
Its a state of limited competition, in which a market is shared by a small number of producers or sellers.
The Sun's chemical composition was about 70%% hydrogen when it formed, and about 13%% of this hydrogen was available for eventual fusion in the core. (The rest remains in layers of the Sun where the temperature is currently too low for fusion.)Use the given data to calculate the total mass of hydrogen available for fusion over the lifetime of the Sun.
Answer:
1.8 × 10^29 kg
Explanation:
The computation of the total mass is shown below;
As we know that
The total mass of the sun is
= 2 × 10^30 kg
Since 70% of the mass involves hydrogen gas, so total mass of the hydrogen gas is
= 70% × 2 × 10^30 kg
= 1.4 × 10^30 kg
And, 13% available for fusion, so the hydrogen gas mass is
= 13% × 1.4 × 10^30 kg
= 1.8 × 10^29 kg
Which of the following is not true of a topic sentence?
A It should support or reinforce the thesis statement in some way.
B. Every body paragraph should have one.
C. It should summarize the main idea of the paragraph.
O D. It should summarize the main idea of the essay.
Answer: its d
Explanation:
What are the two forces acting on the upper magnet?
A. The magnet's weight and the magnetic force attracting
it to the lower magnet.
B. Two magnetic forces: one attracting it to the lower
magnet and another pushing it away from the lower
magnet.
C. The magnet's weight and the magnetic force pushing it
away from the lower magnet.
D. The upper magnet's weight and mass.
Answer:
b
Explanation:
I'm not so sure but it makes sense
The brightness of a star depends on what
two things?
A. age and color
B. temperature and age
C. size and age
D. size and temperature
Please help The half-life of argon-44 is 12 minutes. Suppose you start with 20 atoms of
argon-44 and wait 12 minutes. How many atoms of argon-44 will be left?
A. 20 atoms
B. 40 atoms
C. 5 atoms
D. 10 atoms
Answer:
D. 10 atoms
Explanation:
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ANSWER IS = B. 60 KG
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Which of the following technologies uses electromagnetic radiation with the
shortest wavelength?
O A. Wireless communications transmitter
O B. Infrared sensor on a DVD player
C. Microwave industrial drying equipment
O D. Compact fluorescent lightbulb
Answer:
A) wireless communications transmitter
Explanation:
Answer:
Compact fluorescent lightbulb
Explanation:
This is because it has the shortest wavelength
Not only did Skid get shot out of a cannon when he was a clown in the circus, but he was also launched into the air by a vertically compressed spring. Skid of 55 kg is standing on a massless platform attached to a spring of constant 8700 N/m that is compressed 125 cm. Skid is launched from rest. He eventually leaves the platform and reaches a peak high above the platform. (a) What is the velocity of Skid when he leaves the platform
Answer:
v = 16.11 m / s
Explanation:
For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation
starting point. When the spring is compressed
Em₀ = K_e + U = ½ k x² + m g x ’
final point. The point where it leaves the platform
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
½ k x² + m g x ’= ½ m v²
v² = [tex]\frac{k}{m}[/tex] x² + g x
let's calculate
v² = [tex]\frac{8700}{55}[/tex] 1.25² + 9.8 1.25
v² = 247.159 + 12.25 = 259.409
v = 16.11 m / s
A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the Horizontal. Find the distance the ball travels before it hits the ground.
Answer:
The horizontal distance traveled by the ball before it hits the ground is 64.42 m.
Explanation:
Given;
initial velocity, u = 27.0 m/s
angle of projection, θ = 30⁰
The horizontal distance traveled by the ball before it hits the ground is known as Range;
[tex]Range = \frac{u^2 sin(2\theta)}{g} \\\\Range = \frac{(27^2)sin(2 \times 30)}{9.8} \\\\Range = 64.42 \ m[/tex]
Therefore, the horizontal distance traveled by the ball before it hits the ground is 64.42 m.
: Suppose somebody, using the same apparatus which you used, measured I = 45.5 ma, and V = 8.2 volts on some resistor. Using your recorded uncertainties for the 50 ma and 10-volt scales, what would be the maximum % uncertainty in R if it were calculated from the Ohm’s Law Equation (1)? Use calculus methods to answer this question if you can.
Answer:
R = (18 ± 2) 10¹ Ω
ΔR = 2 10¹ Ω
Explanation:
Ohm's law relates voltage to current and resistance
V = i R
R = [tex]\frac{V}{i}[/tex]V / i
the absolute error of the resistance is
ΔR = | [tex]| \frac{dR}{DV} | \ \Delta V + | \frac{dR}{di} | \ \Delta i[/tex]
the absolute value guarantees the worst case, maximum error
ΔR = [tex]\frac{1}{i} \Delta V+ \frac{V}{i^2} \Delta i[/tex]
The error in the voltage let be approximate, if we use a scale of 10 V, in general the scales are divided into 20 divisions, the error is the reading of 1 division, let's use a rule of direct proportion
ΔV = 1 division = 10 V / 20 divisions
ΔV = 0.5 V
The current error must also be approximate, if we have the same number of divisions
Δi = 50 mA / 20 divisions
Δi = 2.5 mA
let's calculate
ΔR = [tex]\frac{1}{45.5 \ 10^{-3}} \ 0.5 + \frac{8.2}{(45.5 \ 10^{-3})^2 } \ 2.5 \ 10^{-3}[/tex]
ΔR = 10.99 + 9.9
ΔR = 20.9 Ω
The absolute error must be given with a significant figure
ΔR = 2 10¹ Ω
the resistance value is
R = 8.2 / 45.5 10-3
R = 180 Ω
the result should be
R = (18 ± 2) 10¹ Ω
A hammer strikes one end of a thick iron rail of length 8.80 m. A microphone located at the opposite end of the rail detects two pulses of sound, one that travels through the air and a longitudinal wave that travels through the rail. (The speeds of sound in air and in iron are 343 m/s and 5950 m/s, respectively.)
Required:
Find the separation in time between the arrivals of the two pulses.
Answer:
ΔT = 0.02412 s
Explanation:
We will simply calculate the time for both the waves to travel through rail distance.
FOR THE TRAVELING THROUGH RAIL:
[tex]T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s[/tex]
FOR THE WAVE TRAVELING THROUGH AIR:
[tex]T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s[/tex]
The separation in time between two pulses can now be given as follows:
[tex]\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\[/tex]
ΔT = 0.02412 s
Question below in photo!! Please answer! Will mark BRAINLIEST! ⬇⬇⬇⬇⬇⬇⬇
Answer:
it shows the Crest of the wave
Recall your experimental setup from Lab 05A: a constant force was applied to a disc by attaching a mass to a light string wrapped around a mass-less pulley and hanging the mass over the edge of the apparatus. In the lab, you used energy conservation arguments to derive an expression for the angular velocity of the disc after the mass had fallen a distance x . Your goal now is to use kinematics and dynamics to confirm your expression. Use the following symbols throughout this question: m is the mass of the hanging mass, M is the mass of the disc, r is the radius of the pulley, R is the radius of the disc, x is the distance the mass has fallen, and g is the acceleration due to gravity. What is the linear acceleration of the mass after it has fallen a distance x
Answer:
w = [tex]\sqrt{\frac{2gy}{r^2 + \frac{1}{2} R^2 } }[/tex]
Explanation:
For this exercise let's start by applying Newton's second law to the mass with the string
W - T = m a
In this case, as the system is going down, we will assume the vertical directional down as positive.
T = W - m a
Now we apply Newton's second law for rotational motion to the pulley of radius r. We will assume the positive counterclockwise rotations
∑ τ = I α
T r = I α
the moment of inertia of the disk is
I = ½ M R²
angular and linear acceleration are related
a = α r
we substitute
T r = (½ m R²) (a / r)
T = ½ m ([tex]\frac{R}{r}[/tex] )² a
we write our two equations
T = W - m a
T = ½ m ([tex]\frac{R}{r}[/tex] )² a
we solve the system of equations
W - m a = ½ m (\frac{R}{r} )² a
m g = m a [ 1 + ½ (\frac{R}{r} )² ]
a = [tex]\frac{g}{ 1 + \frac{1}{2} (\frac{R}{r})^2 }[/tex]
this acceleration is constant throughout the trajectory, so with the angular and lineal kinematics relations
w² = w₀² + 2 α θ
v² = v₀² + 2 a y
as the system is released its initial angular velocity is zero
w² = 0 + 2 α θ
v² = 0 + 2 a y
we look for the angular acceleration
a =α r
α = a / r
α = [tex]\frac{g}{r (1 + \frac{1}{2} (\frac{R}{r})^2 }[/tex]
we look for the angle, remember that they must be measured in radians
θ = s / r
in this case we approximate the arc to the distance
s = y
θ = y / r
we substitute
w = [tex]\sqrt{2 \frac{g}{ r( \frac{1}{2} (\frac{R}{r})^2 } \frac{y}{r} }[/tex]
w = [tex]\sqrt{\frac{2gy}{r^2 + \frac{1}{2} R^2 } }[/tex]
for the simple case where r = R
w = [tex]\sqrt{ \frac{2gy}{ \frac{3}{2} R^2 } }[/tex]
w = [tex]\sqrt{ \frac{4}{3} \frac{gy}{R^2} }[/tex]
What is the wavelength of the wave in the image to the left?
Graph 1
Answer:
There is no graph i can aswer it soww
Explanation:
What major region does this profile
most likely represent?
Given the information on the graph, we can confirm that this profile represents the Ocean floor profile.
What causes the spike in the graph?The spike in the graph demonstrates a sharp change in depth. This is caused by the irregularities we expect to find on the ocean floor. Many parts of the ocean floor are deeper than others, caused by the movement of tectonic plates over time.Therefore, we can confirm that the profile belongs to the ocean floor profile, firstly because of the information provided, as well as it being consistent with the irregularities in the depth of the ocean floor.
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2.) Assume you decided to build a cabin off the electrical grid in a remote region of the United States, and your cabin would require 720 kWh per month to easily meet your needs. You decide to produce your electricity using photovoltaic cells and store it using a battery system so that it is available when you wish to use it. If every photovoltaic cell has the potential to produce 0.25 kilowatts (250 watts), and they produced electricity for 24 hours a day, how many photovoltaic cells would you need to meet your electricity needs for a month
Answer:
4 photovoltaic cells are required to meet monthly energy consumption of the cabin.
Explanation:
Let suppose that PV generation system produces energy at constant energy, the number of photovoltaic system to meet energy consumption of the cabin is:
[tex]n = \frac{E}{\dot E \cdot t}[/tex] (1)
Where:
[tex]E[/tex] - Energy consumption of the PV system, in kilowatt-hours.
[tex]\dot E[/tex] - Power generation of the PV cell, in kilowatts.
[tex]t[/tex] - Working times, in hours.
If we know that [tex]E = 720\,kWh[/tex], [tex]\dot E = 0.25\,kW[/tex] and [tex]t = 720\,h[/tex], then the number of photovoltai cells is:
[tex]n = \frac{E}{\dot E \cdot t}[/tex]
[tex]n = 4[/tex]
4 photovoltaic cells are required to meet monthly energy consumption of the cabin.
explain melting and freezing using the kinetic theory of matter
As a liquid is cooled its molecules lose kinetic energy and their motion slows. When they've slowed to where intermolecular attractive forces exceed the collisional forces from random motion, then a phase transition from liquid to solid state takes place and the material freezes
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A student flexes her neck to 40 degrees from vertical while studying for her final exam. She pivots her head around the upper vertebra, approximated as point P (in the middle of the picture). Her head has a mass of 4.5 kg, and its center of mass is 11 cm away from point P. Her neck muscles are 1.5 cm from point P, measured perpendicular and pulling straight down. How much force do the neck muscles have to exert to keep the head still
Answer:
[tex]T=208N[/tex]
Explanation:
From the question we are told that:
Neck flex angle [tex]\angle_n=40 \textdegree[/tex]
Head mass [tex]M_h=4.5kg[/tex]
Center of mass [tex]M_c= 11cm[/tex]
Distance of neck from P [tex]d_p=1.5cm[/tex]
Let
[tex]T_g=Gravitational\ force\ torque\\T_{nm}=Tension\ on\ neck\ muscle\ torque[/tex]
Generally the net Torque T_n is mathematically given by
[tex]T_n=0[/tex]
Therefore
[tex]T_g-T_{nm}=0[/tex]
Generally the equation for Torque T is mathematically given by
[tex]T=\frac{M_h*g*l_g}{d_p}[/tex]
where
[tex]l_g=M_c*sin\angle_n[/tex]
[tex]l_g=11*10^{-2}*sin40[/tex]
[tex]l_g=0.07m[/tex]
Therefore
[tex]T=\frac{4.5*9.8*0.07}{1.5*10^{-2}}[/tex]
[tex]T=208N[/tex]