the improvement of fine motor skills during middle and late childhood is due to which reason?

Answer:

A. It could be depleted quickly.

Explanation:

It's Non-renewable, meaning it's hard to get back quickly, and using it at a high rate makes it run out well.. quicker.

Answer:

a

Explanation:

At the time of the summer solstice, the noontime altitude of the sun is at its highest point, around 90°.

Altitude is the height above sea level. It is typically measured in either metres or feet. In aviation, altitude can also refer to the vertical distance between an aircraft and a certain reference point on the ground. Altitude can be used to determine the air pressure, temperature, and density of the air. Altitude can also be used to calculate the distance a plane can travel without refueling. Altitude can play an important role in the weather of an area, as air pressure and temperature tend to decrease with altitude. Altitude can also affect the type of vegetation found in an area. In mountain regions, the altitude can have a dramatic effect on the climate, creating distinct areas of vegetation and wildlife. Altitude can also affect the types of crops that can be grown in an area, depending on the air pressure, temperature, and precipitation.

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The centre of mass of the region is bounded by y=9-x^2 y=5/2x, and the z-axis is (3.5, 33/8). Formulae used to find the centre of mass are as follows:x bar = (1/M)*∫∫∫x*dV, where M is the total mass of the system y bar = (1/M)*∫∫∫y*dVwhere M is the total mass of the system z bar = (1/M)*∫∫∫z*dV, where M is the total mass of the systemThe region bounded by y=9-x^2 and y=5/2x, and the z-axis is shown in the attached figure.

The two curves intersect at (-3, 15/2) and (3, 15/2). Thus, the total mass of the region is given by M = ∫∫ρ*dA, where ρ = density. We can assume ρ = 1 since no density is given.M = ∫[5/2x, 9-x^2]∫[0, x^2+5/2x]dAy bar = (1/M)*∫∫∫y*dVTherefore,y bar = (1/M)*∫[5/2x, 9-x^2]∫[0, x^2+5/2x]y*dA= (1/M)*∫[5/2x, 9-x^2]∫[0, x^2+5/2x]ydA...[1].

The limits of integration in the above equation are from 5/2x to 9-x^2 for x and from 0 to x^2+5/2x for y.To evaluate the above integral, we need to swap the order of integration. Therefore,y bar = (1/M)*∫[0, 3]∫[5/2, (9-y)^0.5]y*dxdy...[2].

The limits of integration in the above equation are from 0 to 3 for y and from 5/2 to (9-y)^0.5 for x.Substituting the values and evaluating the integral, we get y bar = (1/M)*[(9-5/2)^2/2 - (9-(15/2))^2/2]= (1/M)*(25/2)...[3].

Also, the x coordinate of the center of mass is given by,x bar = (1/M)*∫∫∫x*dVTherefore,x bar = (1/M)*∫[5/2x, 9-x^2]∫[0, x^2+5/2x]x*dA= (1/M)*∫[5/2x, 9-x^2]∫[0, x^2+5/2x]xdA...[4].

The limits of integration in the above equation are from 5/2x to 9-x^2 for x and from 0 to x^2+5/2x for y.To evaluate the above integral, we need to swap the order of integration. Therefore, x bar = (1/M)*∫[0, 3]∫[5/2, (9-y)^0.5]xy*dxdy...[5].

The limits of integration in the above equation are from 0 to 3 for y and from 5/2 to (9-y)^0.5 for x.

Substituting the values and evaluating the integral, we get x bar = (1/M)*[63/8]= (1/M)*(63/8)...[6]Thus, the centre of mass of the region is bounded by y=9-x^2 y=5/2x, and the z-axis is (3.5, 33/8).

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Answer:

Explanation:

roles of nutrient in our body​

The force of friction acting on him is 122 N. The correct answer is option C

Friction is a force that opposes the motion of a static or a moving object.

Given that Barry slides across an icy pond. The coefficient of kinetic friction between his shoes and the ice is 0.15. If his mass is 83 kg

The given parameters are;

The normal reaction N on the body = mg

N = 83 x 9.8

N = 813.4 N

The Frictional force formula is  \(F_{r}\) = μN

\(F_{r}\) = 0.15 x 813.4

\(F_{r}\) = 122.01 N

Therefore, the force of friction acting on him is 122 N approximately.

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The required force to push the box straight up at a constant speed is zero.

The net force on the box is calculated by applying Newton's second law of motion as follows;

∑F = 0

F - Ff = ma

(at constant speed, acceleration is zero)

\(F- F_f = ma\\\\F - 0 = 0\\\\F = 0\)

Thus, the required force to push the box straight up at a constant speed is zero.

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Answer:

F_B = 0.12N

Explanation:

In order to calculate the magnetic force on the charge, you use the following formula:

\(\vec{F_B}=q\vec{v}\ X\ \vec{B}\)          (1)

q: charge of the particle = 4μC = 4*10^-6 C

v: speed of the charge = 7.5*10^4 m/s

B: magnitude of the magnetic field = 0.4T

The direction of the motion of the charge is perpendicular to the direction of the magnetic field. Then, the magnitude of the magnetic force is:

\(F_B=qvBsin90\°\\\\F_B=(4*10^{-6}C)(7.5*10^4 m/s)(0.4T)=0.12N\)

The magnetic force on the charge is 0.12N

A. Assume incompressible flow

Since volumtric flow rate will be same in both section

Q1 = Q2

A1*V1 = A2*V2

V2 = A1*V1/A2

V2 = 6.9*10^-4*288*10^-2/(3.3*10^-4)

V2 = 6.02 m/sec

B. using bernoulli's equation:

P1/rho + v1^2/2 + g*h1 = P2/rho + v2^2/2 + g*h2

Since pipes are horizontal

h1 = h2

P1/rho + v1^2/2 = P2/rho + v2^2/2

P2 = P1 + (v1^2 - v2^2)*rho/2

P2 = 1.5*10^5 + (2.88^2 - 6.02^2)*1100/2

P2 = 1.35*10^5 Pa

The rate of change in position of an object in any course. Speed is measured because the ratio of distance to the time wherein the distance became included. Speed is a scalar amount because it has handiest route and no significance.

Velocity can be notion of as the rate at which an object covers distance. A quick-moving object has a high speed and covers a notably big distance in a given amount of time, whilst a sluggish-shifting item covers a relatively small amount of distance in the same amount of time.

The primary unit or the S.I. unit of velocity is m/s or ms?¹.

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This UHF radio wave has a frequency of roughly 1.6 GHz.

The magnetic and electric fields oscillate in time and are perpendicular to one another in a radio wave. The wavelength,, is the separation between locations where both fields are zero. The wavelength in this instance is listed as 0.188 m.

The formula: relates a wave's wavelength and speed to its frequency.

Frequency equals speed/wavelength

Around 3 x 108 m/s is the speed of light in a hoover, which is extremely close to the speed of radio waves in the air. Hence, in this instance, we can consider the radio wave's speed to be equal to the speed of light.

When we enter the specified values into the formula, we obtain:

frequency = (3 x 10^8 m/s) / 0.188 m

frequency = 1.6 x 10^9 Hz or 1.6 GHz

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W= F x D indicates the relationship between force, distance and work.

Work is defined as an activity in which one exerts force to do something. Work is also refers to an activity in which a person do something regularly to earn a livelihood. People looking for work, a particular task, duty, or assignment often being a part of some larger activity.

The definition of work in physics shows its relationship to energy. When work is done, energy is transferred from one place to another. Work is also known as the product of the force in the direction in which  displacement is covered and displacement.

So we can conclude that Work is the product of Force and Distance.

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Answer:

1.4×10⁵ m/s

Explanation:

Applying,

K = mv²/2............. Equation 1

Where K = Energy of the satellite, m = mass of the satellite, v = speed of the satellite.

make v the subject of the equation

v = √(2K/m)............ Equation 2

From the question,

Given: K = 9.8×10⁹ J, m = 1000 g = 1 kg

Substitute these values into equation 2

v = √(2×9.8×10⁹/1)

v = √(1.96×10¹⁰)

v = 1.4×10⁵ m/s

This means that the two players move together at a final velocity of 4.70 m/s after the collision.

m1v1 + m2v2 = (m1 + m2)vf

Plugging in the values given in the problem, we get:

(100.0 kg)(4.70 m/s) + (m2)(0 m/s) = (100.0 kg + m2)vf

Simplifying, we get:

470 kg m/s = (100.0 kg + m2)vf

Dividing both sides by (100.0 kg + m2), we get:

vf = 4.70 m/s

A collision occurs when two or more objects come into contact with each other, often resulting in a sudden and forceful impact. Collisions can occur between any type of object, including cars, trains, airplanes, particles, or even celestial bodies like planets and asteroids. The effects of a collision can range from minor damage to catastrophic destruction, depending on the speed, mass, and angle of impact.

A collision refers to a situation where two different inputs produce the same output in a hash function or encryption algorithm. This can lead to security vulnerabilities, as it may be possible for an attacker to create malicious inputs that produce the same hash as legitimate inputs, thereby bypassing security checks and gaining unauthorized access to sensitive information. Preventing collisions is a critical concern in the design and implementation of secure cryptographic systems.

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Answer:

here you have to find the range

Explanation:

\(range=\frac{u^{2}sin2\theta }{g}\)

\(\frac{4.22^{2} * sin2(63) }{10}\)

1.4431m

In the longitudinal wave ,B represents the phenomenon of rarefaction. Rarefaction refers to the region of a sound wave where the pressure of the medium is lower than its normal value.

Rarefaction is a term used to describe a decrease in the density or pressure of a substance, such as a gas or liquid. In the context of sound waves, rarefaction refers to the region of a sound wave where the pressure of the medium is lower than its normal value, causing the particles of the medium to be spread further apart than usual.

Sound waves are composed of regions of compression and rarefaction that alternate in a regular pattern as the wave travels through a medium. In a compressional (longitudinal) sound wave, the particles of the medium are pushed together in regions of compression, while they are spread apart in regions of rarefaction. These changes in pressure and density cause the wave to propagate through the medium.

In general, rarefaction can occur in any medium, not just in sound waves. For example, in a gas, rarefaction can be caused by a decrease in pressure, temperature or density. In a liquid, rarefaction can be caused by a decrease in pressure or density. Rarefaction waves can be observed in many natural phenomena, such as atmospheric pressure waves, seismic waves, and waves on the surface of water.

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Answer: C. 39 s

Explanation:

We know the constant speed is 0.22 m/s. We have to get to 8.5 m. We divide 8.5 m by 0.22 = 38.6. After we estimate, 6 is greater than 5, so 39 s.

Clouds generally form when air is lifted, causing cooling and condensation of water vapor into visible droplets or ice crystals.

Cloud formation primarily occurs when air is lifted. When air rises, it expands and cools, leading to a decrease in its capacity to hold moisture. As the air cools, the water vapor it contains begins to condense into visible water droplets or ice crystals, forming clouds.

While warming of air can indirectly contribute to cloud formation by promoting atmospheric instability and convection, it is the lifting of air that plays a more direct role in cloud formation. Similarly, the addition of water vapor through evaporation provides the necessary moisture content for clouds to form but is not the sole factor.

The greenhouse effect, on the other hand, is not directly linked to cloud formation. The greenhouse effect refers to the trapping of heat in the Earth's atmosphere by certain gases, such as carbon dioxide and methane. While the greenhouse effect influences Earth's climate and weather patterns, it does not directly determine cloud formation. Clouds can form under a variety of greenhouse effect conditions, depending on factors such as temperature, humidity, and atmospheric dynamics.

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The shuttle is under a force of 1.625 105N from the fuel. The force on a body expressed as a change in momentum is,

                     F= Δ\(\frac{p}{t}\)  (1)

Where,

F is the force that the shuttle is being subjected to from the fuel.

P represents the change in the shuttle's momentum.

t stands for the overall amount of time needed.

Considering our query,

ΔP= \(=325000=3.25X 10^{5} kg-m/s\)

t = 2 seconds

Equation (1) yields the following results when the necessary values are substituted:

\(F= \frac{3.25 10^{5} }{2}\)

\(F= 1.625\) x \(10^{5}\)N

Hence, 1.625105N is the force the fuel is exerting on the shuttle. The amount of motion a body has is referred to as momentum. Given that momentum relies on both velocity and the vector of the body's motion, it really is quantified as "mass speed". As acceleration and mass are both scalar quantities, vector quantities include momentum.

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Given:

The mass of the oil drop is

The number of electrons is n = 11

The distance between the plates is

To find

(a) The charge absorbed by the oil drop.

(b) The potential difference between the plates.

Explanation:

The charge absorbed by the oil drop can be calculated by the formula

Here, e is the charge of the electron whose magnitude will be

On substituting the values, the charge absorbed by the oil drop will be

(b) The potential difference can be calculated as

Answer:

hello! Jaesuk~Sakai here!

Explanation:

The four group 3 elements are scandium, yttrium, lanthanum and actinium. “lanthanide” refers to the elements between lanthanum and lutetium. That term does not encompass scandium and yttrium. Scandium and yttrium act pretty much identically to lanthanides.

happy to help!!

Remember to always stay happy no matter what happens! ^^

In the spectral sequence of star types, each category has been divided into ten intervals. The sun is classified as a G2 star.

The spectral sequence of star types is a sequence of star classes that are characterized by their spectra (the range of electromagnetic radiation they produce). The sequence is arranged in alphabetical order and is divided into seven main categories (O, B, A, F, G, K, M), with each category divided into ten subcategories, ranging from 0 to 9, with 0 being the hottest and 9 being the coolest. The Sun is classified as a G2 star, which is a yellow dwarf star, G2 is a subcategory of the G class, which is, in turn, a subcategory of the main G-K-M spectral class. The Sun's surface temperature is about 5,500 degrees Celsius (9,932 degrees Fahrenheit), and it has a yellow color, hence its classification as a yellow dwarf star.

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The original speed of the truck is 19.7 m/s.

To find the original speed of the truck, we can use the equation of motion for constant acceleration:

v = v0 + at,

where v0 is the initial velocity, v is the final velocity, a is the acceleration, and t is the time.

Since the truck is slowing down, its acceleration is negative, and we can find it using the final velocity and time:

a = (v - v0) / t = (2.90 m/s - v0) / 8.60 s.

Next, we can use the distance-velocity relationship for constant acceleration:

d = v0t + 1/2 at²,

where d is the distance traveled.

Substituting the expression for a into the above equation and solving for v0, we get:

v0 = √(v² + 2ad) = √(2.90² + 2 × a × 44.0) m/s.

Substituting the values for v, a, and d into the expression for v0, we get:

v0 = √(2.90² + 2 × (-(2.90 m/s - v0) / 8.60 s) × 44.0) m/s.

Solving this equation for v0, we get:

v0 = 19.7 m/s.

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the resultant force is F=F1-F2.

F1 = 20 N, and F2 = 50 N

In this instance, two different resultant forces are feasible.

when the two are at an angle of A with one another.

See the outcome in case 1 of the image if both are facing the same way. if they are both facing the opposite directions. when both are at a zero-degree angle from one another If they are parallel to one another and moving in the same direction, the resultant force is F=F1+F2. If they are parallel but moving in the opposite direction, the resultant force is F=F1-F2.

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Perfect inelasticity governs the collision. Both athletes go north at a speed of 2 m/s just after the tackle. Prior to the tackle, the 90 kg athlete was moving at a speed of 6 m/s south.

In an inelastic collision, momentum is preserved, hence we may apply the equation: (m1 * v1) plus (m2 * v2) equals (m1 + m2) * vf.

The following is the result of substituting the above values: (90 kg * v1) + (120 kg * (-4 m/s)) = (90 kg + 120 kg) * 2 m/s

When we simplify the equation, we obtain: 90v1 - 480 = 210 90v1 = 690\sv1 = 7.67 m/s They move at a speed of -4 m/s. As a result, the player weighing 90 kg was moving at the following speed right before the tackle: south

Nevertheless, the question specifically asks for the northward velocity shortly before the tackle, therefore we must adjust the sign:

North: v1 = -11.67 m/s plus 2 m/s equals -9.67 m/s

Prior to the tackle, the 90 kg athlete was moving at a speed of 6 m/s south.

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Answer:

a) We can use the following formula to calculate the acceleration of the object:

a = (v_f - v_i) / t

where a is the acceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time interval. Substituting the given values, we get:

a = (90 m/s - 40 m/s) / (2 min * 60 s/min) = 0.83 m/s^2

The force applied causing it to speed up can be found using Newton's second law of motion:

F = m * a

where F is the net force, m is the mass of the object, and a is the acceleration. Substituting the given values, we get:

F = 0.05 kg * 0.83 m/s^2 = 0.042 N

Therefore, the force applied causing the object to speed up is 0.042 N.

b) The work done by this force can be calculated using the following formula:

W = F * d

where W is the work done, F is the net force, and d is the displacement of the object. The displacement of the object is given by:

d = 670 m

Substituting the given values, we get:

W = 0.042 N * 670 m = 28.14 J

Therefore, the work done by the force is 28.14 J.

c) The work done by friction can be calculated using the formula:

W_friction = F_friction * d

where W_friction is the work done by friction, F_friction is the force of friction, and d is the displacement of the object. The force of friction can be calculated using:

F_friction = μ * F_norm

where μ is the coefficient of friction and F_norm is the normal force. The normal force is equal to the weight of the object, which is given by:

F_weight = m * g

where m is the mass of the object and g is the acceleration due to gravity. Substituting the given values, we get:

F_weight = 0.05 kg * 9.81 m/s^2 = 0.49 N

The normal force is equal in magnitude to the weight of the object, so we have:

F_norm = F_weight = 0.49 N

Substituting the given coefficient of friction, we get:

F_friction = 0.34 * 0.49 N = 0.17 N

The work done by friction can now be calculated by substituting the values we have found:

W_friction = 0.17 N * 670 m = 113.9 J

Therefore, the work done by friction is 113.9 J.

d) The net work done can be calculated as the sum of the work done by the applied force and the work done by friction:

W_net = W_applied + W_friction

Substituting the values we have found, we get:

W_net = 28.14 J + 113.9 J = 142.0 J

Therefore, the net work done is 142.0 J.

The distance of the surface illuminated by the light of the given wavelength is 5,269.32 m.

The distance of the surface illuminated by the light of the given wavelength is calculated as follows;

D = (rd)/(0.61λ)

where;

D = (0.45 x 0.005) / (0.61 x 700 x 10⁻⁹)

D = 5,269.32 m

Thus, the distance of the surface illuminated by the light of the given wavelength is 5,269.32 m.

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Sir Isaac Newton organized the visible colors into a chart known as the color spectrum or Newton's color wheel.

Newton conducted experiments with light and prisms in the 17th century, leading to his groundbreaking work on the nature of light and color. He observed that when white light passes through a prism, it separates into a continuous band of colors. This band of colors, arranged in a specific order, became known as the color spectrum. Newton's color spectrum consists of the colors red, orange, yellow, green, blue, indigo, and violet. He arranged these colors in a circular shape, resembling a wheel, where each color transitions smoothly into the next. This representation of colors was a significant contribution to our understanding of the visible light spectrum and paved the way for further investigations into the properties of light and color by subsequent scientists.

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sir isaac newton organized the visible colors into chart known as the_____.

The ball will travel more distance when projected or kicked at an angle of 35 degrees.

The range of the projectile or the horizontal distance traveled by the ball is calculated by applying the following kinematic equation as shown below.

R = ( u² sin (2θ ) ) / g

where;

The horizontal distance traveled by the ball when projected at 35 degrees is calculated as;

R = ( 13.9² x sin ( 2 x 35 ) ) / ( 9.8 )

R = 18.53 m

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