answer all please!!!!!!!!!!✌​

Answer:

Clear-cutting eliminates all trees leaving none behind to maintain soil fertility or house native animals.

Explanation:

edge 2022

A forestry/logging practice known as clear-cutting, clear felling, or clear-cut logging involves consistently cutting down the majority or all of the trees in a given region. Foresters utilize it to cultivate particular types of forest ecosystems and to advance particular species, in addition to harvesting shelter wood and seed trees.

The removal of a forest or tree stands from land before it is used for a purpose other than as a forest is known as deforestation or clearing of forests. The use of forest land for urban, agricultural, or ranch purposes is referred to as deforestation. In tropical rainforests, there is the greatest concentration of deforestation. Forests currently cover about 31% of the Earth's land area. With half of that loss occurring in the previous century, this is a third less forest cover than there had before agriculture's growth.

As a result of clear-cutting, no trees are left to protect the topsoil or serve as a haven for wildlife. In turn, this causes soil erosion and soil fertility to be destroyed, as well as the removal of the trees required to produce seeds for natural forest development to continue. Animal populations are also pushed out of their natural habitats by clear-cutting, forcing them to look for new homes.

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Answer:

The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s

Explanation:

Free Fall Motion

A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is [tex]g = 9.8 m/s^2.[/tex]

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The distance traveled by a dropped object is:

[tex]\displaystyle y=\frac{gt^2}{2}[/tex]

Given a grapefruit free falls from a tree and hits the ground t=0.72 s later, we can calculate the height it fell from:

[tex]\displaystyle y=\frac{9.8\cdot 0.72^2}{2}[/tex]

y = 2.54 m

The final speed is computed below:

[tex]vf=9.8\cdot 0.72[/tex]

vf = 7.06 m/s

The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s

Answer:

Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.

Explanation:

Hope this helps! :]

Answer:

C. Surface 1 is blacktop, Surface 2 is gravel, and

Surface 3 is ice.

Explanation:

yesssss I am agree ......

Answer:

Yes, I agree. The pressure exerted upon the floor is larger for high heel shoes than flat shoes, hence causing more damage to the floor.

Explanation:

Pressure is force/area. High heel shoes have a sharp point at which they touch the floor, while flat shoes have a flatter surface at which they touch the floor. Due to you exerting the same force anyway, the area is smaller for high heel shoes.

As you might know, the smaller a denominator is, the larger the number is. As an example, 3/1 is larger than 3/2. 3/1 is 3, and 3/2 is 1.5. Hence, the force is larger for high heel shoes, and it will cause more damage.

Hope this helped!

Answer:

The resistance of the lamp is 4Ω.

Explanation:

You have to apply voltage formula :

V = I × R

R = V ÷ I

R = 240 ÷ 60

R = 4 Ω

Answer:

5.3 cm

Explanation:

This question is an illustration of real and apparent distance.

From the question, we have the following given parameters

Real Distance, R = 8.0cm

Refractive Index, μ = 1.5

Required

Determine the apparent distance (A)

The relationship between R, A and μ is:

μ = R/A

i.e.

Refractive Index = Real Distance ÷ Apparent Distance

Substitute values in the above formula

1.5 = 8/A

Multiply both sides by A

1.5 * A = A * 8/A

1.5A = 8

Divide both side by 1.5

1.5A/1.5 = 8/1.5

A = 8/1.5

A = 5.3cm

Hence, the letters would appear at a distance of 5.3cm

Answer:

B because thats the answer

Answer:

Vf = 210 [m/s]

Av = 105 [m/s]

y = 2205 [m]

Explanation:

To solve this problem we must use the following formula of kinematics.

[tex]v_{f} =v_{o} +g*t[/tex]

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (released from the rest)

g = gravity acceleration = 10 [m/s²]

t = time = 21 [s]

Vf = 0 + (10*21)

Vf = 210 [m/s]

Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)

The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)

Av = (210 + 0)/2

Av = 105 [m/s]

To calculate the distance we must use the following equation of kinematics

[tex]v_{f} ^{2} =v_{o} ^{2} +2*g*y\\\\(210)^{2} = 0 + (2*10*y)[/tex]

44100 = 20*y

y = 2205 [m]

Answer:

electromagnetic force

Explanation:

Answer:

The range of the projectile is 60 meters

Explanation:

To determine the range/distance of the projectile, the formula for velocity is used;

velocity = distance/time

where velocity is 15 m/s

time is 4 seconds

distance is unknown

From the formula above, distance is made the subject and thus

distance = velocity × time

distance = 15 × 4

distance = 60 m

The range of the projectile is 60 meters

Answer:

99.48

Explanation:

99.48

Answer:

a. 23.9 m/s b. 69.58°

Explanation:

a. Since the space vehicle is moving at a constant velocity of 22.4 m/s in the y direction relative to the space station, its initial vertical velocity v = 22.4 m/s.

Also, since the space vehicle moves in the y direction, its initial horizontal velocity u = 0 m/s.

Since its acceleration a = 0.206 m/s² for time, t = 40.5 s, we find the final horizontal velocity u' from u' = u + at.

Substituting the values of the variables into the equation, we have

u' = u + at

u' = 0 m/s  + 0.206 m/s² × 40.5 s

= 8.343 m/s

≅ 8.34 m/s

The resultant velocity relative to the space station V = √(v² + u'²)

= √[(22.4m/s)² + (8.34 m/s)²]

= √[501.76 m²/s² + (69.56 m²/s²]

= √[571.32 m²/s²]

= 23.9 m/s

b. The direction of the vehicle's velocity relative to the space station is thus θ = tan⁻¹(v/u')

= tan⁻¹(22.4 m/s/8.34 m/s)

= tan⁻¹(2.6959)

= 69.58°

Explanation:

Please mark me as the brainliest answer

Answer:

The electrostatic force is larger by a factor of 8.988E9 / 6.674E-11 = 1.35E20

Explanation:

William gave you all the ingredients, but not the answer.  Being both 1/r^2 forces (resulting from massless mediators in QFT), and with unit charges in the problem (1 coulomb, 1 Kg), separated by unit distance (1m), the only nontrivial numerical values in the problem are the constants of proportionality: Coulomb's constant (k) and Newton's gravitational constant (G).  So to "compare and constrast": the ratio of the forces is simply the ratio of these constants.  The electromagnatic force is 1.35 X 10^20 times stronger than the gravitational force.  Assuming positive charge on both objects (the problem is ambiguous on this), they are repelled, whereas the much weaker gravitational force is attractive.  (Gravity only has one kind of "charge" - it's unsigned - and the force is always attractive).  

The problem doesn't say the objects are pointlike: if they have some extent and are either conductive or made of some dielectric, then things get messy because the charge distribution on the object won't be uniform then, but save that for grad school. :-)

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 200 × 9.8

We have the final answer as

Hope this helps you

Answer:

D

Explanation:

their octet is complete they don't react with any thing

Argon and krypton have similar properties, and they have a complete octet, which means that they do not react with anyone. Hence, option D is correct.

The phrase "inert gas" is a bit misleading because, in some circumstances, these gases can really be reactive. As a result, these gases are typically referred to as noble gases in the context of chemistry and materials science.

The term "noble" has historically been used in chemistry (and earlier in alchemy) to characterize the resistance of some metals to chemical reaction, and the term "noble gas" denotes the same resistance.

These are the inert gases which are mentioned in the periodic table.

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Answer:

Workdone is 5734.06Nm.

Explanation:

Given the following data;

Force applied = 125N

Angle = 35°

Distance = 56m

To find the workdone by the lawnmower, we would first of all find the horizontal component of the force applied.

[tex] Horizontal force, Fx = mgCosd[/tex]

Where;

mg = weight = 125N

Substituting into the equation, we have;

[tex] Fx = 125 * Cos35[/tex]

[tex] Fx = 125 * 0.8192[/tex]

Fx = 102.39N

Workdone is given by the formula;

[tex] Workdone = force * distance[/tex]

[tex] Workdone = 102.39 * 56[/tex]

Workdone = 5734.06Nm

Therefore, the work done by the lawnmower is 5734.06Nm.

The correct answer is A physical change

Explanation:

Jorge's experiment shows water at different temperatures; in this experiment, it is expected at low temperatures such as -20°C water is in solid-state (ice), at medium temperatures such as 40°C water is in a liquid state (liquid water), and at high temperatures such as 120°C water is in gaseous state (water vapor). This implies during this experiment the changing factor is the physical state (solid, gas, or liquid), and this is a physical change because only the physical properties of water change but not its composition or identity. According to this, the correct answer is physical change.

Answer:

Interplanting is the practice of planting a fast-growing crop between a slower-growing one to make the most of your garden space. ... Intercropping enables you to boost the health of all plants because it can enhance soil fertility and cooperation among different plants.

Intercropping refers to large, often commercial, farming operations and is the procedure of planting alternating crop rows. Growing two or more crops in close proximity is known as intercropping.

Starting to grow two or more crops in close vicinity is known as intercropping.

The most common goal of intercropping is to increase yield on a given plot of land by utilizing resources that would otherwise go unused by a single crop.

Intercropping is the cultivation of more than one crop in the same space at the same time. Intercropping is the simultaneous planting of two or more species in a mixture, or the interplanting of one species during the growth of another.

The terms are frequently and regionally used interchangeably. Intercropping is the practice of planting alternating crop rows in large, often advertising, farming operations.

Thus, this is the difference between intercropping and interplanting.

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Answer:

a

Yes it clears

b

 [tex]b= 0.19 \ m[/tex]

c

 No it does not clear

d

[tex]z= 0.86 \ m[/tex]

Explanation:

From the question we are told that

  The speed at which the player serves the ball is [tex]v = 23.6 \ m/s[/tex]

  The height of the ball above the ground is  [tex]h = 2.3 7 \ m[/tex]

  The distance of the net is  [tex]d = 12 \ m[/tex]

   The height of the net is [tex]H = 0.9 \ m[/tex]

Generally the time taken for the ball to reach the net is mathematically represented as

     [tex]t = \frac{d}{v}[/tex]

=>  [tex]t = \frac{12}{23.6}[/tex]

=>  [tex]t = 0.508 \ s[/tex]

Generally the change in height of the ball after t is mathematically represented as

     [tex]\Delta h = ut + \frac{1}{2} gt^2[/tex]

Here u is the initial velocity which is zero given that the ball was at rest initially

So

     [tex]\Delta h = 0* t + \frac{1}{2} * 9.8 * 0.50 8^2[/tex]

=>  [tex]\Delta h =1.28 \ m[/tex]

Generally the new height of the ball is mathematically evaluated as

      [tex]s= h-\Delta h[/tex]

=>   [tex]s = 2.37 - 1.28[/tex]

=>   [tex]s = 1.09 \ m[/tex]

From the value obtained we see that [tex]s > H[/tex] hence the ball clears the net

Generally the distance between the center of the ball and the top of the net is mathematically represented as

      [tex]b = s - H[/tex]

=>   [tex]b = 1.09 - 0.90[/tex]

=>   [tex]b= 0.19 \ m[/tex]

Given that the ball makes an angle of [tex]5^o[/tex] with the horizontal  , the velocity along the x-axis is  

      [tex]v_x = v cos(5)[/tex]

=>   [tex]v_x = 23.6 cos(5)[/tex]

=>   [tex]v_x = 23.5 \ m/s[/tex]

The velocity along the y-axis is  

      [tex]v_y = v sin(5)[/tex]

=>   [tex]v_y = 23.6 sin(5)[/tex]

=>   [tex]v_y = 2.06 \ m/s[/tex]      

Generally the time taken for the ball to reach the net is

      [tex]t = \frac{d}{v_x}[/tex]

=>   [tex]t = \frac{12}{23.5}[/tex]

=>   [tex]t =0.508 \ s[/tex]

Generally the change in height of the ball after t seconds is  

     [tex]c = v_yt + \frac{1}{2}gt^2[/tex]

=>  [tex]c = 2.06 * 0.508 + \frac{1}{2}* 9.8 * 0.508 ^2[/tex]

=>  [tex]c = 2.33[/tex]

Generally the new height of the ball after time t seconds is  

     [tex]e = h - c[/tex]

=>   [tex]e = 2.37 - 2.33[/tex]

=>   [tex]e = 0.04 \ m[/tex]

From the value obtained we see that [tex]e < H[/tex] hence the ball does not clear the net

Generally the distance between the center of the ball and the top of the net is mathematically represented as

      [tex]z = H-e[/tex]

=>   [tex]z = 0.90 - 0.04[/tex]

=>   [tex]z= 0.86 \ m[/tex]

(a) Yes, the ball clears the net.

(b) The distance between the center of the ball and the top of the net is 0.203 m.

(c) No, the ball does not clear the net.

(d) Now, the distance between the center of the ball and the top of the net is -0.85 m.

When any object or body is launched with some initial velocity and making some angle with the horizontal, the body travels in a parabolic path. It is known as the projectile motion.

Given,

The horizontal distance traveled by the ball is 12 m.

The height of the top of the net is 0.90 m.

The height of the horizontal launch of the ball is 2.37 m.

The time for the horizontal motion of the projectile that is the ball is,

[tex]\begin{aligned} {{v}_{0x}}&={{S}_{x}}t \\ t&=\dfrac{{{S}_{x}}}{{{v}_{0x}}} \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}\cos 0{}^\circ } \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}} \end{aligned}[/tex]

The equation for the vertical motion of the projectile can be solved by substituting the above result.

[tex]\begin{aligned} y&={{y}_{0}}+{{v}_{0y}}t-\frac{1}{2}g{{t}^{2}} \\ y&={{y}_{0}}+\left( {{v}_{0}}\sin 0{}^\circ \right)\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)-\frac{1}{2}g{{\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)}^{2}} \\ \left( 0.90\text{ m}+h \right)&=\left( 2.37\text{ m} \right)+0-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( \frac{12\text{ m}}{23.6\text{ m/s}} \right)}^{2}} \\ h&=2.37\text{ m}-\text{1}\text{.267 m}-0.90\text{ m} \\ &=0.203\text{ m} \end{aligned}[/tex]

Now, consider the case when the ball is thrown with an angle [tex]\bold{5^{\circ}}[/tex] with the horizontal.

The horizontal component of the initial velocity is [tex]\bold{v_0 \cos 5{}^\circ.}[/tex]

The vertical component of the initial velocity is [tex]\bold{v_0 \sin 5{}^\circ.}[/tex]

For the horizontal distance traveled by the ball in this case, the time taken can be calculated as below,

[tex]\begin{aligned} {t}'&=\frac{{{{{S}'}}_{x}}}{{{{{v}'}}_{0x}}} \\ &=\frac{12\text{ m}}{\left( 23.6\text{ m/s} \right)\cos 5{}^\circ } \\ &=0.51\text{ s} \end{aligned}[/tex]

Now, the vertical distance above the ground, y’, traveled by the projectile till reaching the net can be determined as,

[tex]\begin{aligned} {y}'&={{{{y}'}}_{0}}+{{{{v}'}}_{0y}}{t}'-\frac{1}{2}g{{{{t}'}}^{2}} \\ &=0\text{ m}-\left( \left( 23.6\text{ m/s} \right)\sin 5{}^\circ \right)\left( 0.51\text{ s} \right)-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( 0.51\text{ s} \right)}^{2}} \\ &=-1.05\text{ m}-1.28\text{ m} \\ &=-2.32\text{ m} \end{aligned}[/tex]

The height above the top of the net can be determined by adding the above result with (2.37 m - 0.90 m) which is the height of the net’s top relative to the launch position.

[tex]\begin{aligned} {h}'&=\left( 2.37\text{ m}-0.90\text{ m} \right)+{y}' \\ &=\left( 2.37\text{ m}-0.90\text{ m} \right)-2.32\text{ m} \\ &=-0.85\text{ m} \end{aligned}[/tex]

Thus, the distance between the center of the ball and the top of the net is -2,32 m.

When the ball leaves the racquet at 5.00° below the horizontal, the distance between the center of the ball and the top of the net is -0.85 m.

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Answer:100 g of ca

Explanation:

Answer:

In a chemical reaction, only the atoms present in the reactants can end up in the products. No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange amd form new bonds to make the products.

Answer:

b

Explanation:

Answer:

The first law, an object will not change its motion unless a force acts on it.

The second law, the force on an object is equal to its mass times its acceleration.

The third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Answer:

Explanation:

hope it helps:)

Answer: a. 126.21secs

b. 12.621 meters.

Explanation:

Given data:

Speed of tortoise = 0.1m/s.

Speed of hare = 2m/s.

Solution:

a. Distance traveled = Speed* Time

Speed of tortoise = 0.1 m/s

Speed of hare = 20*0.1 m/s = 2 m/s

2 minutes = 2* 60 s = 120 s

Let the time taken for the race be t seconds.

• Distance moved by tortoise

= (0.1 /s)* (t s)

= 0.1*t meter

•Hare has run for a time of (t - 120)s.

distance moved by hare

= Speed * Time

= (2 m/s)*(t- 120)s

= (2t - 240) meter.

Since hare is 20 cm (0.2 m) behind the tortoise, therefore

(0.1*t - 0.2) meter

= (2t - 240) meter

0.1*t - 0.2 = 2t - 240

Collect like terms

239.8 = 1.9t

Divide both sides by 1.9

t = 126.21secs

The race lasted for 126.21secs

b. Length of race

= Distance moved by tortoise

= 0.1*126.21 meter

= 12.621 meter

The length of the race is 12.621 meters.

The question is incomplete, the complete question is;

A student is measuring the volumes of nectar produced by a flowering plant for an experiment. He measures nectar from 50 flowers using a graduated cylinder that measures to the nearest millilitre (mL). Which statement describes a change that can help improve the results of his experiment?

A.) His measurements will be more precise if he takes measurements from an additional 100 flowers. B.) His measurements will be more accurate if he uses a graduated cylinder that measures to the nearest tenth of a mL. C.) His measurements will be more precise if he uses a graduated cylinder that measures to the nearest tenth of a mL. D.) His measurements will be more accurate if he takes measurements from an additional 100 flowers.

Answer:

His measurements will be more accurate if he uses a graduated cylinder that measures to the nearest tenth of a mL.

Explanation:

In the measurements of volume using most graduated cylinders, the cylinders are calibrated to the nearest tenth owing to the uncertainty in the measurement of volume.

Hence if a cylinder has measures to the nearest milliliter(mL), then he can improve his experiment by using a graduated cylinder that measures to the nearest tenth of a mL

Answer:

1.7

2.4

Explanation:

Answer:

1.7

2.4

Explanation:

Answer:

Ro = 133 [kg/m³]

Explanation:

In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

[tex]Ro = m/V[/tex]

where:

m = mass [kg]

V = volume [m³]

We will convert the units of length to meters and the mass to kilograms.

L = 15 [cm] = 0.15 [m]

t = 2 [mm] = 0.002 [m]

w = 10 [cm] = 0.1 [m]

Now we can find the volume.

[tex]V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ][/tex]

And the mass m = 4 [gramm] = 0.004 [kg]

[tex]Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}][/tex]