This is a __________ reaction and the product(s) is/are
A) single replacement; zinc chloride and hydrogen gas.
B) double replacement; zinc oxide and hydrogen gas.
C) synthesis; zinc chloride and hydrogen gas.
D) decomposition; zinc chloride and water.

The new acceleration of gravity is D) 1/16 g

Explanation:

The magnitude of the acceleration of gravity in the gravitational field of a planet is given by

[tex]g=\frac{GM}{r^2}[/tex]

where

G is the gravitational constant

M is the mass of the planet

r is the distance of the object from the centre of the planet

In this problem, the acceleration of gravity g on the surface of the planet (when r=R) is

[tex]g=\frac{GM}{R^2}[/tex]

Then the object is moved to a distance of

r' = 4R

Substituting into the original equation, we can find what is the new acceleration of gravity:

[tex]g'=\frac{GM}{(4R)^2}=\frac{1}{16}(\frac{GM}{R^2})=\frac{1}{16}g[/tex]

So, the acceleration of gravity has decreased by a factor 16.

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Final answer:

When an object is moved from a distance of R to 4R from the center of a planet, the gravitational acceleration it experiences decreases to 1/16th of the original value, making the correct answer D) 1/16g.

Explanation:

The question deals with the change in gravitational acceleration experienced by an object when its distance from the center of a planet is increased. According to Newton's law of universal gravitation, the gravitational force, and hence the acceleration, is inversely proportional to the square of the distance between the objects. Therefore, if the distance is increased from R to 4R, the acceleration will decrease by a factor of the square of the increase in distance. This is calculated as (1/4)^2 or 1/16th of the original acceleration.

So, the correct answer is D) 1/16g. As a comparison, g on the surface of Earth is approximately 9.8 m/s², and when discussing acceleration in multiples of g, it helps to envision the effect of this force under various conditions, such as creating an acceleration of 1 g using centripetal forces.

Answer:

I just did this question and it's A and D

The correct options are :

(A) The germinating corn seed produced more energy than the non-germinating corn seed.

(C) The germinating corn seed consumed more oxygen than the non-germinating corn seed.

information from the graph:

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Answer:

The horizontal velocities of the projectile for both cases are going to remain the same.

Explanation:

Taking the horizontal direction as x-axis and the vertical component as y-axis.

Let V be the object's velocity and θ be the angle formed with the horizontal.

Therefore,  

The horizontal component of velocity is,

                      [tex]V_{x}=V Cos\theta[/tex]

                      [tex]V_{y}=V Sin\theta[/tex]

Due to the force of gravity the vertical component of velocity changes. But there is no influence of force in the horizontal component other than the air resistance.

Hence, the horizontal component of the velocities of the projectile remains the same.

Answer:

D. Newton's Third Law of Motion

Explanation:

Newton's law of gravity is definitely not applicable to your hands. So we can cross this bad boy out

Newton's First Law is F=MA (force equals mass times acceleration). This is basically the root of most physics but it isn't the reason for your hand being red after hitting a wall.

Newton's Second law deals with velocities and forces, so even though you are apply a force your are not changing the velocity of the wall much.

Newton's Third Law basically says that for whatever force you apply to an object, that object will apply an equal and opposite force back to you. This is why your hand gets red. When you slap the wall with all your strength, the wall hits your hand back with the same amount of force. The 2nd law can also be seen when you're trying to push a desk and it won't budge. You are pushing on it, but the desk is pushing back. (there are multiple other factors applicable like friction but we physicists like to ignore them :) )

I hope this helps!

In the described chemical reaction, the reactants are ammonia (NH3) and carbon dioxide (CO2). These reactants are combined in the body to produce urea (CH4N2O), which is the product.

In the chemical reaction described in the passage, the reactants are the substances that combine to produce another substance. From the passage, we can identify that ammonia (NH3) and carbon dioxide (CO2) are combined by the liver to produce urea (CH4N2O). Hence, in this chemical reaction, ammonia and carbon dioxide are the reactants, while urea is the end product or the product of the reaction.

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The velocity of the ping-pong ball must be 8109 m/s

Explanation:

The momentum of an object is given by

[tex]p=mv[/tex]

where

m is the mass of the object

v is its velocity

For the bowling ball, we have

m = 9 kg

v = 3.28 m/s

so its momentum is

[tex]p=(9)(3.28)=29.5 kg m/s[/tex]

We want the ping-pong ball to have the same momentum, so:

p' = 29.5 kg m/s

And the mass of the ping-pong ball is

m = 3.638 g = [tex]3.638\cdot 10^{-3} kg[/tex]

Therefore, its velocity must be:

[tex]v=\frac{p}{m}=\frac{29.5}{3.638\cdot 10^{-3}}=8109 m/s[/tex]

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The velocity of the Ping-Pong ball can be found by setting the momentum of the bowling ball equal to the momentum of the Ping-Pong ball and solving for its velocity. Using the given masses and velocities, we can calculate the velocity of the Ping-Pong ball in m/s.

In order for two balls to have the same momentum, the momentum of the bowling ball must be equal to the momentum of the Ping-Pong ball. The momentum of an object is calculated by multiplying its mass by its velocity. Therefore, to find the velocity of the Ping-Pong ball, we can use the equation: momentum of bowling ball = momentum of Ping-Pong ball.

Given: mass of bowling ball = 9 kg, velocity of bowling ball = 3.28 m/s, mass of Ping-Pong ball = 3.638 g = 0.003638 kg (convert g to kg)

Momentum of bowling ball = mass of bowling ball × velocity of bowling ball

Momentum of Ping-Pong ball = mass of Ping-Pong ball × velocity of Ping-Pong ball

Setting the two momentum equations equal to each other, we can solve for the velocity of the Ping-Pong ball:

(9 kg) × (3.28 m/s) = (0.003638 kg) × (velocity of Ping-Pong ball)

Dividing both sides of the equation by 0.003638 kg will give us the velocity of the Ping-Pong ball:

velocity of Ping-Pong ball = (9 kg × 3.28 m/s) / 0.003638 kg

Calculating this expression will give us the velocity of the Ping-Pong ball in m/s.

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Answer:

Velocity

Explanation:

                                   E = P.E + K.E

                                  P.E = mgh

                                  K.E = ½ mv²

Answer:

Elements in a period share the highest unexcited electron energy level. There are more elements in some periods than others because the number of elements is determined by the number of electrons allowed in each energy sub-level.

Explanation:

The tension in the string that accelerates a 5 kg cart to 2 m/s² is equal to 10 Newton.

Explanation:

Tension:

Tension is a force exerted by the rope, Tension refers to the pulling force transmitted axially by the means of a string, cable or a chain.

In the given case, Tension in the string is simply the pulling force it applies on the cart. There is no special formula for tension in the string.

Newton's second law is often stated as F = ma, which means the force (F) acting on an object is equal to the mass (m) of an object times the acceleration (a) produced in it due to that force.

Mass of the cart = 5 kg

Acceleration produced = 2 m/s²

Tension in the string = F = ma

= 5*2

= 10 kg m/s²

= 10 N

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Answer:

An unbalanced force.

Explanation:

Teacher: "An object in motion will stay in motion unless..."

Auggie : "Acted on by an unbalanced force."

Teacher: "Very good."

Answer:

3.2 m

Explanation:

we know

Hmax = V²/2g

= 8² / 2*10 = 3.2

Answer:

The ball will get upto the height of 3.26 meters.

Explanation:

Initial velocity of the ball = u = 8 m/s

final velocity of the ball = v = 0 m/s ( upward)

Acceleration due to gravity ,a= -g =[tex]-9.8 m/s^2[/tex]

Height of the ball when kicked = s

Using third equation of motion :

[tex]v^2-u^2=2as[/tex]

[tex](0 m/s)^2-(8 m/s)^2=2\times (-g)\times s[/tex]

[tex](0 m/s)^2-(8 m/s)^2=2\times (-9.8)\times s[/tex]

[tex]s=\frac{(-8 m/s)^2}{2\times -9.8 m/s^2}=3.26 m[/tex]

The ball will get upto the height of 3.26 meters.

a) The net force on the system is 883.5 N

b) The tension in the rope is 505.5 N

c) The applied force on the snowmobile is 1875.7 N

Explanation:

a)

To solve this first part, we just analyze all the forces acting in the horizontal direction on the snowmobile+sled system, and we apply Newton's second law, which states that:

[tex]\sum F = (m+M)a[/tex]

where

[tex]\sum F[/tex] is the net force on the system

m = 150 kg is the mass of the sled

M = 315 kg is the mass of the snowmobile+rider

[tex]a=1.9 m/s^2[/tex] is the acceleration

Therefore, solving for [tex]\sum F[/tex], we find the net force on the system:

[tex]\sum F = (150+315)(1.9)=883.5 N[/tex]

b)

We can write now the equation of the forces acting on the sled only. We have:

[tex]T-F_f = ma[/tex]

where:

T is the tension in the rope between the sled and the snowmobile, which is pulling the sled forward

[tex]F_f[/tex] is the force of friction acting on the sled

m = 150 kg is the mass of the sled

[tex]a=1.9 m/s^2[/tex] is the acceleration

The force of friction can be written as

[tex]F_f = \mu_k mg[/tex]

where

[tex]\mu_k = 0.15[/tex] is the coefficient of kinetic friction of the sled on ice

Substituting into the previous equation and solving for T, we find the tension:

[tex]T-\mu_k mg = ma\\T=ma+\mu_k mg=(150)(1.9)+(0.15)(150)(9.8)=505.5 N[/tex]

c)

We can now write the equation of the forces acting on the snowmobile, and we have:

[tex]F_a - T - F_F = Ma[/tex]

where:

[tex]F_a[/tex] is the applied force

T = 505.5 N is the tension in the rope, which pulls the snowmobile backward

[tex]F_F[/tex] is the force of friction on the snowmobile

M = 315 kg is the mass of the snowmobile+rider

[tex]a=1.9 m/s^2[/tex] is the acceleration

The force of friction can be written as

[tex]F_F = \mu_k Mg[/tex]

where

[tex]\mu_k = 0.25[/tex] is the coefficient of kinetic friction of the snowmobile on ice

Substituting into the previous equation and solving for [tex]F_a[/tex], we find:

[tex]F_a = Ma+T+\mu_k Mg=(315)(1.9)+505.5+(0.25)(315)(9.8)=1875.7 N[/tex]

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The plank moves the same distance as the point on the circumference of the wheels, which is 1.00 meter, because there is no slipping and the plank doesn't tip.

When the wheels roll 1.00 meter from their initial position without slipping, the distance moved by the plank will be equivalent to the circumference of a circle traced by the outer edge of the wheels. The outer radius of each wheel is 0.25 meters, thus the circumference C can be calculated using the formula C = 2πr, which gives us C = 2π(0.25) = 1.57 meters. Since the wheels have rolled 1.00 meter, which is less than a full revolution, we need to find what fraction of a revolution 1.00 meter corresponds to and then calculate the corresponding linear distance that the plank would move.

To find the fraction of a revolution, we divide 1.00 meter by the circumference to get approximately 0.637 revolutions. Since the plank's movement corresponds to the rotation of the wheels and because there's no slippage, the plank will move the same linear distance as the edge of the wheels over this fraction of a revolution. Therefore, the plank moves approximately 0.637 times the wheel's circumference, which is 0.637 × 1.57 meters ≈ 1.00 meter.

The plank moves the same distance as the point on the circumference of the wheels, which in this case is 1.00 meter.

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Answer:

The bomb will remain in air for 17.5 s before hitting the ground.

Explanation:

Given:

Initial vertical height is, [tex]y_0=1500\ m[/tex]

Initial horizontal velocity is, [tex]u_x=688\ m/s[/tex]

Initial vertical velocity is, [tex]u_y=0(\textrm{Horizontal velocity only initially)}[/tex]

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is [tex]S=y-y_0=0-1500=-1500\ m[/tex]

Acceleration in the vertical direction is due to gravity, [tex]g=-9.8\ m/s^2[/tex]

Therefore, the displacement of the bomb is given as:

[tex]S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s[/tex]

So, the bomb will remain in air for 17.5 s before hitting the ground.

A particular isotope of plutonium has 94 protons, 147 neutrons, and a mass number of 241.

Explanation:

An atom is identified by two numbers:

Mathematically, this can be rewritten as

[tex]Z=p[/tex]

[tex]A=p+n[/tex]

where

p is the number of protons

n is the number of neutrons

For the isotope of plutonium in this problem, we have

[tex]Z=p=94[/tex], since it has 94 protons

[tex]A=p+n=241[/tex] (mass number)

From the second equation we get

[tex]n=241-p[/tex]

And substituting [tex]p=94[/tex], we find the number of neutrons:

[tex]n=241-94=147[/tex]

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Answer:

92, 147, 227

Explanation:

got it right on edge2022

Answer:

[tex]V_{c} = V_{b}  = V_{a}[/tex]

Explanation:

Let [tex]V_{a}[/tex],  [tex]V_{b}[/tex],  [tex]V_{c}[/tex] be the speed of the balls which was fired at an angle of 45°, 60°, 90° respectively from the horizontal surface, when the balls crosses the dashed horizontal line a few meters away from the place where it is fired initially

After firing the only force that is acting on the ball will be the force of gravity assuming that there is no air resistance, so the acceleration will be g in the downward direction

Using the below formula we can calculate the final velocity of the ball after travelling some distance

v² - u² = 2×a×s

where v is the final velocity of the ball

u is the initial velocity of the ball

s is the distance between the points where initial velocity is u and final velocity is v

a is the acceleration of the ball

From this we get

v² = u² + 2×a×s

Here in this case we are considering in vertical direction and in this direction for all balls a = -g as g is acted on the balls opposite to the direction of motion

s is same for all balls

Let u be the initial speed of all balls

speed is same for all balls but not initial velocity in the vertical direction

For the ball that is fired from an angle of 90°, initial velocity is u

For the ball that is fired from an angle of 60°, initial velocity is u×sin60° = u×(√3 ÷ 2)

For the ball that is fired from an angle of 45°, initial velocity is u ×sin45° = u×(1 ÷ √2)

As a and s in the formula are constant for all particles, the final velocity will depend on their initial velocities

We can observe that as angle with horizontal is increasing the initial velocity is also increasing

∴ Final velocity in the vertical direction of the ball fired from 90° > Final velocity in the vertical direction of the ball fired from 60° > Final velocity in the vertical direction of the ball fired from 45°

As there is no acceleration in horizontal direction initial velocity in horizontal direction =  final velocity in horizontal direction

In horizontal direction

velocity of the ball fired from 90° is 0

velocity of the ball fired from 60° is u×cos60° = u÷2

velocity of the ball fired from 45° is u ×sin45° = u×(1 ÷ √2)

Speed = √((velocity in horizontal direction)² +(velocity in vertical direction)²)

For all balls we get speed² = u² - 2×g×s

∴ Final speed of all balls is same

Final answer:

The ball fired straight up at an angle of 90 degrees will have the highest speed when it crosses the horizontal line, followed by the ball fired at an angle of 45 degrees, and then the ball fired at an angle of 60 degrees.

Explanation:

The speeds of the three balls can be determined by analyzing their motion in the vertical and horizontal directions. Since the vertical motion is independent of the horizontal motion, the ball fired straight up at an angle of 90 degrees will have the highest speed when it crosses the horizontal line. This is because it is only influenced by gravity in the vertical direction and does not have any horizontal motion to slow it down. The ball fired at an angle of 45 degrees will have a slightly lower speed, while the ball fired at an angle of 60 degrees will have the lowest speed. This is because their horizontal velocities are not as high as the ball fired straight up. Therefore, the ranking from largest to smallest speed is: Vc (fired straight up), Va (fired at 45 degrees), and Vb (fired at 60 degrees).

Answer:

26

Explanation:

Answer:

false

Explanation:

Answer:

TRUE

In a wire, negatively charged electrons move, and positively charged atoms don't. Electrical engineers say that, in an electrical circuit, electricity flows one direction: out of the positive terminal of a battery and back into the negative terminal.

A lot of people think of electron flow as electrons moving along a wire freely like cars go down a highway.

Explanation:

Explanation:

The given problem should be solved using basic sedimentary laws proposed by Nicolas Steno and other group of geologists.

Some of the laws that would find application in this problem are:

Based on this premise, we can derive the geologic column for the area:

                       Chart A                                               Chart B

                        Erosion                                              Erosion

1                             C                                                          a

2                             E                                                         e

3                             H                                                         h

3                         Erosion                                                   m

4                           A                                                         Erosion

5                            B                                                              c

6                            F                                                               j

7                            D                                                              l

8                            L                                                              g

9                            G                                                              k

10                           J                                                              f

11                            i                                                                i

12                                                                                          d

13                                                                                           b

The above was derived using the sedimentary laws for relative dating.

Observations 1

In Record A unconformity will occur between layers H and A

    Record B unconformity will occur between layers c and m

An unconformity is a huge break in the stratigraphic record. It is usually a time gap as a result of non-deposition of sediments, erosion of a rock layers amidst other factors.

The type of unconformity here is called an angular unconformity.

In this kind of unconformity, parallel beds are deposited on the tilted or folded layer below resulting in an angular discordance of some sort.

The surface between H and A is an erosional surface of the tilted beds in record A. This was followed by the deposition of new beds that are parallel to the tilted ones below.

In record be, a folded and eroded surface separate c and m.

Analysis 2

In record A, the rock layer K is older than F but younger than D. This rock layer has been tilted. This interpretation is based on the law of superposition of strata. If the sequences are reconstructed without the tilt, layer K will be on top of D and F will be on top of K.

Analysis 3

Layer c is younger than j but older than the parallel lying m on top of it. This geologic structure is called a synform. In synform, the rock strata has a concave shape. Usually, we can assume that this is a synformal syncline in which the youngest bed is at the core of the fold. This pits layer c as the youngest in the folded beds. The age increases outward in a synform.

Analysis 4

Layers C and E in record A are both eroded beds. Layer C has been eroded to the level of layer E.

We can establish that when layer C was exposed and eroded, the overburden on the earth was relieved and part of bed E re-adjusted up to the level of E.

Analysis 5

Reconstructing the geologic history of A

To form a sedimentary rock sequence, a basin must be available. Sedimentary basins are natural depressions on the earth crust in which sediments collects and are deposited.

Sediments i - A were deposited layer by layer into the basin. They became compacted and lithified to form sedimentary rocks.

With the passage of time, these layers became tilted by natural forces. This could be an earthquake, stress e.t.c

After the tilting, a period of non-deposition occurred in which the forces of denudation prevailed to cause the erosion of the bed when it was uplifted and exposed.

After this period, layers H-C was then deposited, compacted and lithified. The whole rock sequence was then uplifted exposing some rock layers. Erosion washed parts of layers C and E to what we can see in the present day.

Analysis 6

In record B, sequences b-c were laid in the basin of deposition where they were transformed into sedimentary rocks.

The rocks were then subjected to a compressive stress regime. The layers buckled and a synformal sycline fold formed.

After this period, the rock was uplifted and exposed. Erosion caused the washing away of parts of the top layer.

A period of deposition shortly followed and parallel layers m-a were deposited. This layer was then transformed into sedimentary rocks.

After this the layer became exposed and part of a was eroded to its present day level.

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Due to missing visuals, it's impossible to provide factual answers. Remember, rock layers are in order of their age, youngest at top, oldest at bottom, taking into account any geological disruptions. This concept should help in interpreting the charts and layers.

Apologies, but it's impossible to assist effectively as we cannot view your Page 1 or the Rock Records you're referring to. However, I can guide you on how to interpret rock layers. Always remember that the youngest layer of rock is located at the top, and the oldest at the bottom, unless disturbed by a geological event like a tilting, erosion or folding. Look for signs of these disturbances - tilting might have shifted the layers from their horizontal plane, erosion can create gaps in the sequence, and folding could have distorted the layers. If a layer cuts across another (like an intruding magma chamber or a fault), it is younger. This should assist you in interpreting both Chart A and Chart B. Consider these tips while studying the age of a certain layer (e.g., layer K or C) and while making conclusions on how the group of rock layers formed.

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To determine the volume displaced by the 54 g sample of brass, use the water displacement method and calculate the density by dividing the mass by the volume.

To determine the volume displaced by the 54 g sample of brass, you need to use the water displacement method. Submerge the brass sample in water and measure the volume of water displaced. The volume of water displaced will be equal to the volume of the brass sample. Then, to calculate the density of the brass sample, divide the mass of the sample (54 g) by the volume of the sample.

For example, if the volume of water displaced is 30 mL, then the volume of the brass sample is also 30 mL. The density of the brass sample would be calculated as 54 g/30 mL = 1.8 g/mL.

Answer:

Sulfate

The sulfate or sulphate (see spelling differences) ion is a polyatomic anion with the empirical formula SO2−4. Sulfate is the spelling recommended by IUPAC, but sulphate is used in British English. Salts, acid derivatives, and peroxides of sulfate are widely used in industry. Sulfates occur widely in everyday life. Sulfates are salts of sulfuric acid and many are prepared from that acid.

Sulfide

Sulfide (systematically named sulfanediide, and sulfide(2−)) (British English sulphide) is an inorganic anion of sulfur with the chemical formula S2− or a compound containing one or more S2− ions. It contributes no color to sulfide salts. As it is classified as a strong base, even dilute solutions of salts such as sodium sulfide (Na2S) are corrosive and can attack the skin. Sulfide is the simplest sulfur anion.

Answer:

Answer:

The 'b x h' side of the brick exerts more pressure on the ground, P =  1428.57 N/m²    

Explanation:

Given data,

The dimension of the brick, l = 10 cm = 0.1 m

                                              b = 7 cm = 0.07 m

                                              h = 5 cm = 0.05 m

The force exerted by the brick on ground, F = 5 N

The brick has three 2 x 3 sides, b x h, l x h, and l x b

The area of the side, A₁ = b x h

                                        = 0.07 x 0.05

                                         = 0.0035 m²

The area of the side, A₂ = l x h

                                         = 0.1 x 0.05

                                          = 0.005 m²

The area of the side, A₃   = 0.1 x 0.07

                                          = 0.007 m²

The pressure exerted by the surface A₁ on the ground, P₁ = F/A₁

                                                                                                 = 5 / 0.0035

                                                                                                  = 1428.57 N/m²

The pressure exerted by the surface A₁ on the ground, P₂ = F/A₂

                                                                                                  = 5 / 0.005

                                                                                                  = 1000 N/m²  

The pressure exerted by the surface A₁ on the ground, P₃ = F/A₃

                                                                                                 = 5 / 0.007

                                                                                                  = 714.29 N/m²  

Hence, the 'b x h' side of the brick exerts more pressure on the ground, P =  1428.57 N/m²                                                

A race track has 47520 inches.

Explanation:

Given that,

Length if the race track is 0.750 miles.

We know that, one mile is 5280 feet.

One feet is 12 inches.

To convert length of the track from miles to inches follow these steps:

Step 1:

Convert into feet:

Multiply 0.750 miles to 5280 feet.

Then the race track’s length is 3960 feet.

Step 2:

Convert into inch:

Multiply 3960 feet to 12 inches.

Then the race track’s length is 47520 inches.

Therefore, length of the race track in inches is 47520 inches.

Answer:

for physics the answer is 47500

The potential energy of the skater is 2352 J

Explanation:

The gravitational potential energy of a body is the energy possessed by the body due to its position in a gravitational field. Near the Earth's surface, it can be calculated as

[tex]PE=mgh[/tex]

where

m is the mass of the body

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity near the Earth's surface

h is the height of the body relative to the ground

For the skater in this problem,

m = 60 kg

h = 4 m

Substituting, we find her potential energy:

[tex]PE=(60)(9.8)(4)=2352 J[/tex]

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Answer:

A. case studies, because it's one of the 5 groups that are part of the qualitative methods

Answer:

Explanation: 0.02

Recall: μ mg = ma

μ = ?

m = 10N

g = 10m\s2

a = 0.2m\s

Substituting the values , we have

μ x 10 x 10 = 10 x 0.2

100  μ = 2

μ = 2/100

μ = 1/50

μ = 0.02

μ = 0.02

Final answer:

The coefficient of friction between the mass and the table is calculated using Newton's second law and the formula for the coefficient of friction, which results in a value of 0.08.

Explanation:

Calculation of Coefficient of Friction

To calculate the coefficient of friction, we can use Newton's second law which states that force equals mass times acceleration (F = ma). Given that a force of 10 N is applied to a 10 kg mass and the resulting acceleration is 0.2 m/s², we can deduce that there must be a frictional force opposing the applied force. The equation simplifies to F = ma + frictional force (f). Plugging in the known values, we get 10 N = (10 kg)(0.2 m/s²) + f, which results in f = 10 N - (10 kg)(0.2 m/s²) = 10 N - 2 N = 8 N.

The weight of the mass can be calculated using weight = mass × gravity, which gives us a weight of (10 kg)(10 m/s²) = 100 N. The normal force on a horizontal surface is equal to the weight of the object, so it's also 100 N. The coefficient of friction (mu) is calculated using the formula mu = frictional force / normal force. Therefore, mu = 8 N / 100 N = 0.08.

The coefficient of friction between the mass and the table is 0.08.

Answer:

Total moment of inertia when arms are extended: 1.613 [tex]kg\,m^2[/tex]

Explanation:

This second part of the problem could be a pretty complex one, but if they expect you to do a simple calculation, which is what I imagine, the idea is just adding another moment of inertia to the first one due to the arms extended laterally and use the moment of inertia for such as depicted in the image I am attaching.

In that image:

L is the length from one end to the other of the extended arms (each 0.75m from the center of the body) which gives 1.5 meters.

m is the mass of both arms. That is: twice 5% of the mass of the person: which mathematically can be written as: 2 * 0.05 * 56.5 = 5.65 kg

Therefore this moment of inertia to be added can be obtained using the formula shown in the image:

[tex]I_z=\frac{1}{12} \,m\,L^2\\\\I_z=\frac{5.65\,*\,1.5^2}{12} \\I_z=1.05937\,kg\,m^2[/tex]

Now, one needs to add this to the previous moment that you calculated, resulting in:

0.554 + 1.059 = 1.613 [tex]kg\,m^2[/tex]

Answer:

Thunder and lightning may become connected in the mind because they often occur together. This is an example of (A) law of similarity

Explanation:

"Gestalt law of similarity" states that objects or things that shares visual characteristics like shape, size, color, texture, value and orientation can be perceived as a unified group.

Thus thunder and lightning can be placed in the same group as lightning causes thunder. When passing from the clouds to that of the surface, a lightning strike pops open up a little void in the atmosphere, called a channel. After the lightning has disappeared, the atmosphere falls back in and produces a sound wave we perceive as a thunder.

Answer:

d. continuity

Explanation:

The law of continuity says that two events can become connected to each other if they happen close together. For example, thunder and lightning may become connected in the mind because they often occur together.

Answer:work done can be calculated by the following formula;

Explanation:

To solve this we must be knowing each and every concept related to work. Therefore, 1,425J of work Russell do to overcome the force of friction.

Work is defined as "Energy transfer that happens when an item is moved across a distance by such an external force, at least a portion that is exerted inside the direction of displacement" in terms of physics.

If there is a constant force operating on the block, but a difference between the force's direction as well as the displacement it causes. In this situation, the force F applies at an angle of to the displacement d.

The formula for work can be given as work

Work = force ×displacement

force=95.0 N

displacement = 15.0 m

substituting all the given values in the above equation, we get

Work =95.0 × 15.0 m

Work =1,425J

Therefore, 1,425J of work Russell do to overcome the force of friction.

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The amount of energy transferred is 4320 J

Explanation:

Power is defined as the amount of energy transferred per unit of time:

[tex]P=\frac{E}{t}[/tex]

where

P is the power

E is the amount of energy transferred

t is the time

In this problem, we have:

P = 720 W is the power

t = 6 s is the time elapsed

Solving for E, we find the energy transferred needed to push the box:

[tex]E=Pt=(720)(6)=4320 J[/tex]

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