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How do wetlands reduce flooding and erosion? (Site 1)

The box accelerates to the right due to the applied force of 35 N in the +x direction.

Newton's second law states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. In this case, the net force acting on the box is 35 N in the +x direction, and its mass is 15 kg. Therefore, we can calculate the acceleration using the formula:

acceleration = net force / mass

acceleration = 35 N / 15 kg = 2.33 m/s² (rounded to two decimal places)

Since the box is not accelerating up or down, we can conclude that the force applied is only causing the box to accelerate in the horizontal direction.

Other forces such as gravity and friction are not considered in this scenario. Thus, the 15 kg box will experience an acceleration of approximately 2.33 m/s² in the +x direction due to the applied force of 35 N.

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The correct answer is C. The inner planets are also called terrestrial planets.

Explanation:

Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".

Answer:

The answer is c.) The inner planets are also called terrestrial planets.

Explanation:

The use of the wheel barrow in this case makes it a simple machine

A wheelbarrow is an illustration of a straightforward device that can simplify labour by lowering the force needed to move a large load. The wheel and axle and the lever are two basic machines that are used.

The wheelbarrow's wheel is made of an axle and wheel. The friction between the wheel and the ground is decreased by the wheel's rotation around the axle. It can also be seen as one of the classes of the lever.

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Speed of the rocket under uniform circular motion is 4.97 × \(10^7\) m/s.

A fictitious force that moves in a circle and is directed away from the center of the circle is called centrifugal force. When measurements are taken in an inertial frame of reference, the force does not exist. It only becomes relevant when we switch from a ground/inertial reference frame to a rotating reference frame.

As the rocket is in uniform circular motion, the gravitational force due to earth and centrifugal force must be equal in magnitude.

The magnitude of gravitational force due to earth, \(F_G = \frac{G M m}{R^2}\)

And, the magnitude of centrifugal force, \(F_C = \frac{mv^2}{R}\)

Where,

G = universal gravitational constant =   6.67 * 10-11 N (m/kg)².

M = the mass of the earth = 5.97 x \(10^{24\) Kg.

m =  the mass of the rocket = 0.5 x \(10^6\)kg.

R = the orbit of the rocket = \({8*10^6}\)m.

For uniform circular motion,

\(\frac{G M m}{R^2}\) = \(\frac{mv^2}{R}\)

⇒ v² = \(\frac{GM}{R}\)

⇒v² =\(\frac{6.67 * 10^{-11 } *5.97 x 10^{24 } }{8*10^6}\)

⇒ v = 4.97 × \(10^7\) m/s.

Hence, the speed of the rocket is 4.97 × \(10^7\) m/s.

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Answer:

F = 2250 [N]

Explanation:

In order to solve this problem, we must first use the following equation of kinematics.

\(v_{f}^{2} =v_{o}^{2}-2*a*x\)

where:

Vf = final velocity = 0 (come to rest)

Vo =  initial velocity = 30 [m/s]

a = acceleration or desaceleration [m/s²]

x = distance = 100 [m]

\((0)=30^{2} -2*a*100\\900 = 200*a\\a = 4.5 [m/s^{2}]\)

Now we must use the following equation of kinetics, which is based on Newton's second law that explains that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force [N]

m = mass = 500 [kg]

a = acceleration = 4.5 [m/s²]

\(F = 500*4.5\\F = 2250 [N]\)

Answer:

\(T_0=80695.17162...\)

Explanation:

Given equation:

\(\ln \left(\dfrac{T_0-100}{T_0}\right)=-0.00124\)

To solve the given equation:

\(\textsf{Apply log rules}: \quad e^{\ln (x)}=x\)

\(\implies \dfrac{T_0-100}{T_0}=e^{-0.00124}\)

Multiply both sides by T₀:

\(\implies T_0-100=T_0e^{-0.00124}\)

Add 100 to both sides:

\(\implies T_0=T_0e^{-0.00124}+100\)

Subtract \(T_0e^{-0.00124}\) from both sides:

\(\implies T_0-T_0e^{-0.00124}=100\)

Factor out the common term T₀:

\(\implies T_0(1-e^{-0.00124})=100\)

Divide both sides by \((1-e^{-0.00124})\)

\(\implies T_0=\dfrac{100}{1-e^{-0.00124}}\)

Carry out the calculation:

\(\implies T_0=\dfrac{100}{1-0.99876...}\)

\(\implies T_0=\dfrac{100}{0.001239231...}\)

\(\implies T_0=80695.17162...\)

If A box of mass 210 kg is pulled from rest with a string of tension 1300n inclined at 35° to the horizontal. if the box moved with a speed of 10m/s and the frictional force between the box and surface is 100 n, Then the distance covered by the box is 10.89 meters.

To calculate the distance covered by the box, we need to analyze the forces acting on it and apply the work-energy principle.

Given:

Mass of the box, m = 210 kg

Tension in the string, T = 1300 N

The angle of inclination, θ = 35°

Frictional force, f = 100 N

Initial speed, u = 0 m/s

Final speed, v = 10 m/s

First, let's resolve the tension force into components parallel and perpendicular to the incline. The parallel component of the tension force can be calculated as:

T_parallel = T * cos(θ)

Next, let's calculate the net force acting on the box along the incline. The net force is given by:

Net force = T_parallel - f

Now, using Newton's second law, we can calculate the acceleration (a) of the box:

Net force = m * a

From the given information, we have the final velocity (v), initial velocity (u), and acceleration (a). We can use the following kinematic equation to calculate the distance covered (s):

v^2 = u^2 + 2as

Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Now, let's plug in the given values and calculate the distance covered:

T_parallel = 1300 N * cos(35°) ≈ 1067.35 N

Net force = 1067.35 N - 100 N = 967.35 N

a = (967.35 N) / (210 kg) ≈ 4.61 m/s^2

s = (10 m/s)^2 - (0 m/s)^2 / (2 * 4.61 m/s^2) ≈ 10.89 m

Therefore, the distance covered by the box is approximately 10.89 meters.

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Answer:

B

Explanation:

Velocity=disp/time

V=5m/15s

V=1/3 m/s

Answer:

a) interior,  b) inside,   c) minor

Explanation:

In this exercise you are asked to select the correct words so that the statements have been correct

Electric charges always repel each other when they are of the same sign, in conductors this has the consequence that charges accumulate on the surface and the interior remains without electric charges.  with this we analyze the statements

a) interior

b) inside

c) minor

therefore the phrase would be:

(a) The net charge is always zero ---INTERIOR--- the surface of an isolated conductor.

(b) The electric field is always zero ---INSIDE--- a perfect conductor.

(c) The charge density on the surface of an isolated, charged conductor is highest where the surface is ---MINOR---to)

Answer: When a disrupted part of the ecosystem is left alone so that nature can help restore itself what people are counting on happening is secondary succession

Explanation:

Answer:

39.40 MeV

Explanation:

Determine the minimum possible Kinetic energy

width of region = 5 fm

From Heisenberg's uncertainty relation below

ΔxΔp ≥ h/2 , where : 2Δx = 5fm ,  Δpc = hc/2Δx = 39.4 MeV

when we apply this values using the relativistic energy-momentum relation

E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,

Also in a nuclear confinement ( E, P >> mc )

while The large value will portray a Non-relativistic limit  as calculated below

K = h^2 / 2ma^2 = 1.52 GeV

The water particles moving perpendicular to the source of the sound wave best describes the direction of the water particles.

These are the types of waves whose oscillations are perpendicular of the wave motion.

A sound wave traveling through water will cause ripples to form which are perpendicular to the source of the sound wave due to it being a transverse wave.

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When a car comes to a stop, the car’s brakes create a force that is in the opposite direction as the velocity vector.

A friction force is a type of force that opposes two surfaces that are in contact with each other.

According to Newton's first law of motion, an object at rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force. And the object will tend to move in the direction of the applied force.

So if a car's brake is applied, it will create a friction force that will oppose the motion of the car and by so doing, it will gradually bring the car to a complete stop.

The force created by the brake of the car is friction force and it is in opposite direction to the force of the car.

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An Particle a has Weston practical.

The resistance to motion of one object moving in relation to another is known as friction. It is not regarded as a fundamental force like gravity or electromagnetic, according to the International Journal of Parallel, Emergent and Distributed Systems(opens in new tab).

According to the book Soil Mechanics(opens in new tab), scientists started putting together the laws governing friction in the 1400s.

However, because the interactions are so complex, characterizing the force of friction in various circumstances typically requires experiments and can't be derived from equations or laws alone. There are numerous exceptions to every frictional general rule.

Therefore, An Particle a has Weston practical.

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Answer:

it shows the Crest of the wave

Answer:

No of proton is 13 and nucleus is 13

Answer:

In picture one because string telephones are best heard when there is more tension on the string

Explanation:

Answer:

BS33

Explanation:

its a cable length meter used for measuring length of all kinds of wires.

Answer:

see table below

Explanation: This table should be in your textbook within the first 4 chapters as all chemistry books follow the same "path".

Feel free to ask any other questions!

when a dielectric of constant K is inserted between the plates of one of two identical capacitors connected in parallel and charged with voltage V0, the charges on the capacitors become Q0 and Kε0A/d(1+K), respectively, while the voltages across the capacitors become V0 and KV0, respectively.

How to solve the problem?

When the identical capacitors are connected in parallel to a voltage source, each capacitor acquires a charge Q0. After the voltage source is disconnected and a dielectric of constant K=3.2 is inserted between the plates of one of the capacitors, the system's capacitance changes. The capacitance of a capacitor is given by C = εA/d, where ε is the permittivity of the material between the plates, A is the area of each plate, and d is the distance between the plates.

Let C0 be the initial capacitance of each capacitor, which is equal to ε0A/d, where ε0 is the permittivity of free space. When the dielectric is inserted, the capacitance of the capacitor with the dielectric increases to C' = Kε0A/d, while the capacitance of the other capacitor remains unchanged at C0. Therefore, the total capacitance of the system is Ctot = C0 + C' = ε0A/d + Kε0A/d = ε0A/d(1+K).

The charge on each capacitor after the dielectric is inserted can be found by applying the conservation of charge. Since the voltage source is disconnected, the total charge on the system must remain constant. Therefore, the charge on the capacitor without the dielectric remains Q0, while the charge on the capacitor with the dielectric increases to Q' = CV = (C'/ε0)V0 = Kε0A/d(1+K)V0.

Therefore, the charges on the capacitors are Q0 and Kε0A/d(1+K), respectively.

The voltage across each capacitor can be found using the equation V = Q/C. Therefore, the voltage across the capacitor without the dielectric is V0 = Q0/C0, while the voltage across the capacitor with the dielectric is V' = Q'/C' = Kε0A/d(1+K)V0/(ε0A/d(1+K)) = KV0.

Therefore, the voltages across the capacitors are V0 and KV0, respectively.

In summary, when a dielectric of constant K is inserted between the plates of one of two identical capacitors connected in parallel and charged with voltage V0, the charges on the capacitors become Q0 and Kε0A/d(1+K), respectively, while the voltages across the capacitors become V0 and KV0, respectively.

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Answer:

pressure is an affect which occurs when a force is applied on a surface area

Explanation:

its unit are barr,millbarr,torr,mmhg

Answer:

ΔE = 59.75 A,

Explanation:

Titanium has 3 electrons in its last shell, as it is doubly ionized, it is left with a single electron in this shell, which is why it behaves like a hydrogen-type atom, consequently we can use Bohr's atomic theory

                  rₙ = a₀ /Z     n²

                 Eₙ = 1k e² / 2a₀ (Z² / n²)

Where a₀ is Bohrd's atomic radius so  = 0.529 núm

Let's find out what quantum number n has each orbit

rn = 13.25 A = 1.325 nm

for Titanium with atomic number 22

            n² = Z rₙ / a₀

           n = √ (22 (1.325 / 0.529))

           n = 7.4

since N is an entry we take

           n = 7

rn = 2.12 A = 0.212 nm

           n = √ (22 / 0.529) 0.212

           n = 3

With these values ​​we can calculate the energy of the transition from level ne = 7 to level no = 3

         ΔE = ka e2 Z2 / 2ao (1n02 - 1 / nf2)

          ΔE = 9 10⁹ 1.6 10⁻¹⁹ 22² (2 0.529 10⁻⁹) (1/3² - 1/7²)

          ΔED = 6.5875 10² (0.111 - 0.0204)

          ΔE = 59.75 A

let us be the Planck relation between energy and frequency

          E = h f

the frequency is related to the speed of light

           c = λ f

            f = c / λ

we substitute

           E = h c /y

           E = ΔE

           h c /λ = E

           λ  = 6.63 10-34 3 108 / 59.75

           λ= 3.01939 10⁻²⁴ m

          λ = 3.01939 10⁻²² cm

Tension in the left cable is 4395.9 N.

Tension is defined as a pulling force that acts along the length of a flexible medium like rope or cables.

Here,

The total mass of the riders is given, m = 330 kg

Let the tension in the left cable be T₁ and that in the right cable be T₂.

From the figure,

T₁ cos 15 = T₂ cos 26

T₁ = T₂ cos 26/cos 15

Also,

T₁ sin15 + T₂ sin 26 = mg

Substituting values,

(T₂ cos 26/cos 15) sin 15 + T₂ sin 26 = 330x 9.8

0.241 T₂ + 0.438 T₂ = 3234

0.679 T₂ = 3234

T₂ = 4762.8 N

Therefore, Tension in the left cable, T₁ = 0.930x 4762.8

         T₁ = 4395.9 N

Hence, The tension in the left cable is 4395.9 N

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Your question was incomplete. Attaching the image here.

The mass of the object is approximately 0.457 kg.

The mass of the object attached to the trolley can be calculated using the principle of conservation of momentum. Since the two trolleys couple together and move as a single system after the collision, the total momentum before and after the collision should be the same. Given the mass of one trolley is 0.80 kg and the initial speed is 1.1 m/s, the momentum before the collision is 0.80 kg * 1.1 m/s = 0.88 kg·m/s. After the collision, the total mass is the sum of the two trolleys, and the final speed is 0.70 m/s.

Using the momentum equation, the mass of the object can be calculated as follows:

Total momentum before collision = Total momentum after collision

0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s

Solving for the mass of the object, we get:

0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s

0.88 kg·m/s = 0.56 kg + 0.70 kg * mass of the object

0.88 kg·m/s - 0.56 kg = 0.70 kg * mass of the object

0.32 kg = 0.70 kg * mass of the object

Dividing both sides by 0.70 kg, we find:

mass of the object = 0.32 kg / 0.70 kg = 0.457 kg

The two trolleys collide and couple together, the total momentum before the collision is equal to the total momentum after the collision according to the principle of conservation of momentum.

The momentum of an object is defined as the product of its mass and velocity. In this case, the mass of one trolley is known (0.80 kg) and the initial speed is given (1.1 m/s), allowing us to calculate the momentum before the collision.

After the collision, the two trolleys move together at a new speed (0.70 m/s). By setting the initial momentum equal to the final momentum and solving for the unknown mass of the object, we can find its value.

In the calculation, we subtract the masses of the two trolleys from the total mass in order to isolate the mass of the object.

Dividing the difference in momentum by the product of the known mass and the new speed, we obtain the mass of the object. In this case, the mass of the object is approximately 0.457 kg.

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Answer:

a. 50 Kg hope it helps :)

The  unit of ETA in terms of the basic units of the SI system is:

ETA = [kg] * [s]⁻¹* [m]⁻¹= kg * s / m

The dimension of the coefficient of viscosity (ETA) can be determined by using the dimensions of the other variables in the equation:

F = ETA * A * (dv/dx)

The dimension of force (F) is [M] * [L] * [T]⁻², the dimension of surface area (A) is [L]⁻², and the dimension of velocity gradient (dv/dx) is [L] * [T]⁻¹.

Since the equation must be dimensionally consistent, we can set up the following equation:

[M] * [L] * [T]⁻²= ETA * [L]² * [L] * [T]⁻¹

Solving for the dimension of ETA, we get:

ETA = [M] * [T]⁻¹ * [L]⁻¹

Therefore, the basic unit of mass in the International System of Units (SI) is the kilogram (kg), the basic unit of time is the second (s), and the basic unit of length is the meter (m). Therefore, the unit of ETA in terms of the basic units of the SI system is:

ETA = [kg] * [s]⁻¹* [m]⁻¹= kg * s / m

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