Answer:
Between 18 months and 2 years
THiS is THE CORRECT ANSWER!
Explanation:
Middle toddlers start using words to express meaning and might say “mama” to get their mother’s attention or “baba” to signal for a bottle. At this age, children learn new words weekly or even daily.
The answer "between 12 and 18 months" is INCORRECT!!!
Between 12 and 18 months
Young toddlers start making word-like sounds, such as “mama,” “dada,” and “baba.” They seem to enjoy making these sounds and repeat them over and over again. But they do not yet use the sounds to express meaning.
Help! This is Physics: PLEASE!!!! ASAP
Explanation:
i thnk its about how to make the work and you should use this formula W=Fd and because it doesnt tell us about mass of person so I preper to decide A.
Longitudinal waves and transverse waves are alike in that they only move small amounts, and do not travel with their wave. They differ in that they transfer energy parallel to the direction, and perpendicular to the direction of the wave motion, respectively.
longitudinal and transverse waves exhibit similarities in terms of their lack of significant movement and non-transport of matter. However, their fundamental distinction lies in the orientation of particle oscillations and the direction of energy transfer: parallel for longitudinal waves and perpendicular for transverse waves.
Longitudinal waves and transverse waves share certain similarities but also exhibit distinct characteristics in terms of their movement and energy transfer.
Both types of waves propagate through a medium but differ in the orientation of their oscillations and the direction of energy transfer.
Longitudinal waves move particles of the medium parallel to the direction of wave propagation. As the wave travels, the particles oscillate back and forth along the same axis as the wave's motion.
Examples of longitudinal waves include sound waves, where air particles vibrate parallel to the direction of the sound wave, creating areas of compression and rarefaction.
In contrast, transverse waves exhibit oscillations perpendicular to the direction of wave motion. The particles in the medium move up and down or side to side, perpendicular to the wave's propagation.
Examples of transverse waves include electromagnetic waves, such as light waves or radio waves, where the electric and magnetic fields oscillate perpendicular to the direction of wave travel.
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Which will a positively charged objects attract
module 1 question 11
A cheetah can accelerate from rest to a speed of 26.5 m/s in 6.50 s. What is its acceleration?
m/s2
The acceleration of the cheetah is 4.07 m/s^2.
Acceleration is the rate of change of the velocity of an object with respect to time. The unit of acceleration is meter per seconds squared (m/s^2). The equation of acceleration is given below.
a = (v_f - v_i)/t
Where a = acceleration, v_f = final velocity, v_i = initial velocity and t = time.
Here, the final velocity = 26.5 m/s, initial velocity = 0 m/s (cheetah is at rest) and time = 6.5 s. The acceleration of the cheetah will be,
a = (26.5 - 0)/6.5
a = 4.07 m/s^2
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20N
10N
40N
30N
Balanced or Unbalanced?
Answer:
Unbalanced
Explanation:
_______ is a judgement about how true or false a conditional statement is.
A. Related conditional statement
B. Biconditional statement
C. Truth Value
D. Conditional statement
E. Hypothesis
F. Conclusion
G. Definition
Answer: F. Conclusion
Explanation:
Answer:
I think the answer is C. Truth Value
Explanation:
a Truth Value is the attribute assigned to a proposition in respect of its truth or falsehood, which in classical logic has only two possible values
A 1100 kg safe is 1.8 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 42 cm .
Part A: What is the spring constant of the spring?
Express your answer with the appropriate units.
If the torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 41 N · m, what minimum force must be exerted by the mechanic at the end of a 24 cm-long wrench to loosen the nut?Answer in units of N.
We have that the Force is mathematically given as
F=170.833N
ForceQuestion Parameters:
The torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 41 N · m,the end of a 24 cm-long wrench to loosen the nutGenerally the equation for the Force is mathematically given as
\(F=\frac{T}{d}\)
Therefore
F=41/0.24
F=170.833N
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compare the times of all sunsets during the same period what do you observe
Answer:
Theres no given?
Explanation:
Well, whatever.
I observed the shift of their sunset time.
Examples like:
January to June = their sunset time increased while
July to December = their sunset time decreased
(7%) Problem 14: A robot cheetah can jump over obstacles. Suppose the launch speed is vo = 4.74 m/s, and the launch angle is 0 = 25.5
degrees above horizontal.
What is the maximum height in meters?
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distance of 1.2 cm. What is the electric field strength?
Answer:
\(E = 2.84 * 10^5 N/C\)
Explanation:
The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.
Let us find the acceleration:
\(v^2 = u^2 + 2as\)
\((4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2\)
Electric force is given as the product of charge and electric field strength:
F = qE
where q = electric charge
E = Electric field strength
Force is generally given as:
F = ma
where m = mass
a = acceleration
Equating both:
ma = qE
E = ma / q
For an electron:
m = 9.11 × 10^{-31} kg
q = 1.602 × 10^{-19} C
Therefore, the electric field strength of the electron is:
\(E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C\)
What are fitness assessments designed to do?
diagnose medical conditions
screen for risk of heart disease
treat injuries
identify specific injuries
Answer: Fitness assessments are designed to screen for risk of heart disease.
Explanation:
Fitness assessments are medical examinations that are designed to measure a person's physical fitness and identify any health risks they may have. These assessments may include tests of strength, endurance, flexibility, and cardiovascular fitness. One of the primary objectives of a fitness assessment is to screen for the risk of heart disease, which is a major health concern that can be prevented or treated through exercise and other lifestyle changes. While fitness assessments may identify specific injuries or medical conditions, their primary focus is on evaluating a person's overall health and fitness.
What is the average velocity of the student between the times t = 0 s and t = 12 s?
Choose 1 answer:
12m/s
0m/s
-0.17m/s
0.17m/s
Answer:
12m/s
Explanation:
mark as a brainlist
You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air resistance)?
Explanation:
Given that,
A ball is tossed straight up with an initial speed of 30 m/s
We need to find the height it will go and the time it takes in the air.
At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :
v = u +at
Here, a = -g
v = u -gt
i.e. u = gt
\(t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s\)
So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds
Let d is the maximum distance covered by it.
\(d=ut-\dfrac{1}{2}gt^2\)
Putting all values
\(d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m\)
Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.
Which of these best defines weather? (3 points) a Atmospheric conditions over 30-year period Ob Day-to-day condition of the atmosphere Ос Long-term condition of the atmosphere O d Average atmospheric condition.
Answer:
b Day-to-day condition of the atmosphere
Explanation:
Weather is short term, Climate is long term
Answer:
Average atmospheric condition.
Explanation:
A cement block accidentally falls from rest from the ledge of a 81.5-m-high building. When the block is 14.0 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way
Answer:
0.41s
Explanation:
solve for t and y
,........
25. (a) What does temperature measures ?
Answer:
Temperature is a measure of the average kinetic energy of the particles in an object.
Explanation:
Often heat and temperature are used interchangeably — this is wrong. While the two concepts are related, temperature is distinct from heat. Temperature describes the energy of a system, whereas heat refers to the energy transferred between two objects at different temperatures.
The first P-wave of an earthquake travels 5600 kilometers from the epicenter and arrives at a seismic station at 10:05 a.m. At what time did this earthquake occur?
Ahhhhhh I have a Regent's test in 2 hours and I don't know how to solve this type of question! Any help would be appreciated.
Anyone know what the steps to do this are? I dont even need an answer, just how to get to it. Thank you!
The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.
The p-waves travel with a constant velocity of 7 km/s
The time can be calculated by using the formula
t = d / v
where
T1 = 10:05 a.m
d is the distance they take to travel from the epicenter
v is the speed of the p-waves
On average, the speed of p-waves is
v = 7 km/s
d = 5600 km (given)
Substituting the values in the formula;
t = d / v
t = 5600 ÷ 7
t = 800 seconds
Converting into minutes,
t = 800 ÷ 60
t = 13.3
≈ 13 mins
T1 - 13 mins = T2
10:05 - 13 mins = 9.52 am
It means the earthquake occurred prior 13 minutes, that is at 9.52 am.
Therefore, the earthquake occurred at 9.52 am.
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Two 4.72 kg masses are 3.12 m apart on a frictionless table. Each has 85.5 microCoulombs of charge. What is the initial acceleration of each mass if they are released and allowed to move?
We are given the following information
Mass of objects: m = 4.72 kg
Distance between objects: r = 3.12 m
Charge: q = 85.5 μC
We are asked to find the initial acceleration of each mass.
Recall from Newton's second law of motion,
\(F=m\cdot a\)Where F is the force between two masses, m is the mass, and a is the acceleration.
First, let us find the force between the two masses.
Recall from Coulomb's law,
\(F=\frac{k\cdot q_1\cdot q_2}{r^2}\)Where k is the Coulomb's law constant that is k = 9×10⁹ Nm²/C²
Substitute the given values into the above formula
\(\begin{gathered} F=\frac{9\times10^9\cdot85.5\times10^{-6}\cdot85.5\times10^{-6}}{3.12^2} \\ F=6.7587\;N \end{gathered}\)Finally, the initial acceleration of each mass is
\(\begin{gathered} F=m\cdot a \\ a=\frac{F}{m} \\ a=\frac{6.7587}{4.72} \\ a=1.43\;\frac{m}{s^2} \end{gathered}\)Therefore, the initial acceleration of each mass is 1.43 m/s^2
If you know the answer please tell me ASAP
3. Fulcrum left
Explanation:
Answer and I will give you brainiliest
Please heeeelp
Answer:
T = 4.905[N]
Explanation:
In order to solve this problem we must perform a sum of forces on the vertical axis.
∑Fy = 0
We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.
\(T-m*g=0\\T=0.5*9.81\\T=4.905[N]\)
Cs-124 has a half-life of 30.8 s
A) If we have 7.5 μg initially, how many Cs nuclei are present?
B) How many nuclei are present 2.6 min later?
A) There are 3.70 x 10¹⁶ Cs nuclei present initially and B) There are 2.27 x 10¹⁵ Cs nuclei present 2.6 min later.
A) To solve this problem, we can use the following equation,
\(N = N'2^{\frac{-t}{T_{1/2}}}\)), number of Cs nuclei present is N, initial number of Cs nuclei present is N', elapsed time is t, half-life of Cs-124 is T½. First, we need to convert the initial mass of Cs-124 to the number of nuclei present,
7.5 μg Cs-124(1g/10⁶μg)(6.022x10²³nuclei/1g)
= 4.52 x 10¹⁵ Cs-124 nuclei.
Using the equation above, we can find the number of Cs-124 nuclei present after 0 s,
\(N = 4.52 * 10^{15} * 2^{\frac{-0}{30.8}}}\)
= 4.52 x 10¹⁵ Cs-124 nuclei
Therefore, initially, there are 4.52 x 10¹⁵ Cs-124 nuclei present.
B) After 2.6 min (156 s), we can again use the same radioactivity equation to find the number of Cs-124 nuclei present,
\(N = 4.52 * 10^{15} * 2^{\frac{-156}{30.8}}}\)
N = 1.60 x 10¹⁴ Cs-124 nuclei
Therefore, 2.6 min later, there are 1.60 x 10¹⁴ Cs-124 nuclei present.
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Identify two types of motion where an object's speed remains the same while it continues to change direction
Answer:
motion in which acceleration is orthogonal to travel directionmotion in which speed is constantExplanation:
1) Any motion in which the acceleration is orthogonal to the direction of travel will have this characteristic:
circular motion in a plane
motion of a charged particle in a magnetic field perpendicular to the direction of travel
__
2) Motion in which the speed is constant, regardless:
motion of a photon through a varying gravitational field
5. Elements combine to form
O compounds
O molecules
atoms
mixtures the gen
Elements combine to form compounds. The correct answer is A.
Elements combine with each other in chemical reactions to form compounds. A compound is a substance composed of two or more different elements chemically combined in a fixed ratio by mass. For example, water is a compound composed of hydrogen and oxygen in a fixed ratio of 2:1 by mass.
B. Molecules are formed when two or more atoms combine chemically, but these atoms can be of the same element or different elements. For example, oxygen gas (O2) is a molecule composed of two oxygen atoms.
C. Atoms are the smallest unit of matter that retains the properties of an element. Atoms do not combine to form other atoms or compounds.
D. Mixtures are composed of two or more substances physically mixed together, but the substances retain their individual properties and can be separated by physical means. Mixtures are not formed by the chemical combination of elements. Examples of mixtures include air and saltwater.
Therefore, The correct answer is option A.
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A 3.7-mm-diameter wire carries a 20 A current when the electric field is 8.2×10−2 V/m .
What is the wire's resistivity? (in Ωm)
Answer:
ρ = 4.4 10⁻⁸ Ω m
Explanation:
For this exercise let's start by finding the value of the resistance using Ohm's law
V = I R
R = V / I
The voltage is related to the electric field
V = E L
let's substitute
R = E L / I
R = 8.2 10⁻² / 20
R = 4.1 10⁻³ L
now we can use the resistance relation
R = ρ L / A
the area of a circular wire is
A = π r² = π d² / 4
ρ = R π d² / (4 L)
let's calculate
ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)
ρ = 4.4 10⁻⁸ Ω m
The resistivity of the wire is 4.4 10⁻⁸ Ω m.
We know that;
V = I R
R = V / I
The voltage the electric field can be connected using the formula;
V = E L
Hence;
R = E L /I
R = 8.2 10⁻² / 20
R = 4.1 10⁻³ L
For resistivity;
R = ρ L / A
the area can be obtained from;
A = π r² = π d² / 4
ρ = R π d² / (4 L)
ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)
ρ = 4.4 10⁻⁸ Ω m
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What is the displacement (in miles, with direction) from the
Aquarium to the Cemetery?
Look up "Everything You Need To Know About Math In One Big Fat Notebook pdf." It's the best thing I've ever been given, I have it with me in math class all the time and I've aced every test. I have it with me right now and it has everything I've ever been taught about math in it so it might help you.
a car accelerates from rest to a speed of 36 km/hour in 10 seconds. find
the acceleration
the distance travelled
the speed at the end of the 5th second
Answer: ok, so there should be no speed at the 5th second but there was somehow it was the car fault
∞ω∞
Which general equation represents an exothermic reaction?
O A. reactants - energy → products
OB. reactants products + energy
OC. reactants + products → energy
OD. reactants + energy
products
Answer:
A
Explanation:
The general equation which represents an exothermic reaction is :
reactants - energy → products
This is because on undergoing an exothermic reaction, heat energy is released, which means that the initial energy of the reactants is lost (subtracted).
One disadvantage to experimental research is that experimental conditions do not always reflect reality.
Please select the best answer from the choices provided
T
F
Answer:
It's true I took the test on Edge.
Explanation:
Answer:
True
Explanation:
Got it right on edg
Find the mass of the meter stick (100 cm). For this problem you do not have to convert the grams to kilograms ornewtons.Remember:Only respond with a numerical answer. No units are required.
The center of the stick from the left edge of the stick is at,
\(\begin{gathered} d=\frac{100}{2} \\ d=50\text{ cm} \end{gathered}\)The mass 125 g is at the distance of 25 cm from the hanging.
Thus, the distance of center of the stick from the hanging is,
\(\begin{gathered} d^{\prime}=50-25 \\ d^{\prime}=25\text{ cm} \end{gathered}\)As the stick is balanced,
Thus, the net moment due to the mass of the stick and given 125 g mass is zero.
\(125\times g\times25-m\times g\times25=0\)where g is the acceleration due to gravity,
Substituting the known values,
\(\begin{gathered} 125\times9.8\times25-m\times9.8\times25=0 \\ (125-m)\times9.8\times25=0 \\ (125-m)=0 \\ m=125\text{ g} \end{gathered}\)Thus, the mass of the stick is 125 grams.
In the terms of kilogram, the value of the mass is,
\(\begin{gathered} m=125\times10^{-3}\text{ kg} \\ m=0.125\text{ kg} \end{gathered}\)The value of the mass in newton is,
\(\begin{gathered} W=m\times g \\ W=0.125\times9.8 \\ W=1.225\text{ N} \end{gathered}\)Thus, the value of the weight of the stick is 1.225 N.