you launch a projectile horizontally from a height of 0.85 m above the ground. the projectile travels a horizontal distance 1.8 m before landing on the ground (at a height of 0 m). with what speed (m/s) did you launch the projectile?

Answer:

The molecular magnets are arranged in an open chain so that the north pole or the south pole of the molecular magnets lie in the same direction which gives strong force at the poles whereas two opposite poles are arranged at the middle and the force cancel each other. So, poles have more force than the middle portion.

The magnet has stronger force of attraction at the poles than in middle, because, the at the poles they are attracted to the opposite poles. Whereas at the middle, the magnetic dipoles are cancelled each other and have no more force of attraction.

A magnet is a material made of metals which are of ferromagnetic nature. Thus, produce permanent magnetic dipoles. The magnetic field lines produced from a permanent magnet is strong at can fluctuate with the presence of an electric field.

A bar magnet have two poles namely a south pole and north pole. Like two the electric charges, where two like charges repel each other and unlike charges attracts, two like poles of a magnet repels and unlike poles attracts each other.

Therefore, at the two poles the magnetic field lines towards the opposite poles will be more stronger whereas at the middles the field lines merges and are oriented randomly and have no strong magnetic effect.

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Answer:

A, Cirrhosis

Explanation:

The most accurate measurements to describe the weight of a truck would be 3 tons 500 lb or 3,000 kilograms.

The weight of a truck can be measured using different units, but the most accurate measurements would be in tons or kilograms. In this case, the weight of the truck is given as 3 tons 500 lb, which means the truck weighs 3 tons (6,000 pounds) plus an additional 500 pounds.

This measurement is commonly used in countries that still utilize the imperial system, such as the United States. Alternatively, the weight of the truck can be expressed in metric units as 3,000 kilograms or 8,300 grams.

Kilograms are the standard unit for measuring weight in the metric system, and they provide a more precise and universally recognized measurement. In the metric system, a kilogram is equal to 1,000 grams.

Therefore, 3,000 kilograms is equivalent to 3,000,000 grams or 8,300 grams. Using kilograms or grams allows for easier conversion between different metric units and facilitates international communication and standardization.

In conclusion, the weight of the truck can be accurately described as 3 tons 500 lb or 3,000 kilograms. While the ton and pound measurement is commonly used in countries that follow the imperial system, the kilogram and gram measurements provide a more precise and internationally recognized standard for weight.

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a) The length of time required for the total pressure to rise from 0.110 atm to 0.145 atm is approximately 35.2 seconds.

b) The length of time required for the total pressure to rise from 0.110 atm to 0.220 atm is approximately 92.9 seconds.

c) Pf ≈ 0.054 atm

The total pressure after 110 seconds of reaction is approximately 0.054 atm.

To solve this problem, we can use the integrated rate law for a first-order reaction:

ln(Pf/Pi) = -kt

Where:

Pf is the final pressure

Pi is the initial pressure

k is the rate constant

t is the time

Let's solve each part of the problem:

Part A:

Using the given information:

Pi = 0.110 atm

Pf = 0.145 atm

k = 7.48 × 10^-3 s^-1

ln(Pf/Pi) = -kt

ln(0.145/0.110) = -(7.48 × 10^-3 s^-1) * t

0.263 = -7.48 × 10^-3 t

t ≈ 35.2 s

Part B:

Using the given information:

Pi = 0.110 atm

Pf = 0.220 atm

k = 7.48 × 10^-3 s^-1

ln(Pf/Pi) = -kt

ln(0.220/0.110) = -(7.48 × 10^-3 s^-1) * t

0.693 = -7.48 × 10^-3 t

t ≈ 92.9 s

Part C:

Using the given information:

t = 110 s

k = 7.48 × 10^-3 s^-1

We can rearrange the integrated rate law to solve for the final pressure (Pf):

ln(Pf/Pi) = -kt

Pf/Pi = e^(-kt)

Pf = Pi * e^(-kt)

Pf = (0.110 atm) * e^(-(7.48 × 10^-3 s^-1) * (110 s))

Pf ≈ 0.054 atm

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Answer:

1. 54.74433 s

Explanation:

h = Height of plane = 14.7 km

[tex]u_x[/tex] = Velocity of crate in x direction = 994 km/h = [tex]276.11\ \text{m/s}[/tex]

a = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

[tex]h=u_yt+\dfrac{1}{2}at^2\\\Rightarrow t=\sqrt{\dfrac{2h}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 14700}{9.81}}\\\Rightarrow t=54.74433\ \text{s}[/tex]

Time the crate takes to hit the ground is 54.74433 s

Answer:

It does not come to a complete stop immediately because of inertia.

Explanation:

Inertia is the reluctance of an object to change its state of motion or rest. In this case, it is the state of motion since the ship was moving initially. Inertia is directly related to mass. Thus, the heavier the mass, the larger the inertia. The ship has a large mass and therefore it has a large inertia. This causes the ship to slowly come to a stop instead of stopping immediately when the engines are switched off.

Suppose the land surface is perfectly smooth,the slope of a mountain side between two points will be -0.003 ft/inch.

In this scenario, we have two points, A and B, with elevations of -500ft and -200ft, respectively. We are given a map scale of 1:50,000 inch, and the distance between A and B on the map is 2 inches. To calculate the slope, we need to find the elevation difference between the two points and the horizontal distance.

First, we find the elevation difference by subtracting the elevation of point B from the elevation of point A: -500ft - (-200ft) = -300ft.

Next, we determine the horizontal distance. Since the map scale is 1:50,000 inch, we need to convert the distance on the map to actual distance. In this case, 2 inches on the map represents 2 * 50,000 inches = 100,000 inches.

Finally, we calculate the slope by dividing the elevation difference (-300ft) by the horizontal distance (100,000 inches). The slope is therefore -300ft / 100,000 inches = -0.003 ft/inch. This means that for every inch of horizontal distance, the land surface descends by 0.003 feet vertically.

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All particles have energy, and the energy varies depending on the temperature the sample of matter is in, which determines if the substance is a solid, liquid, or gas. Solid particles have the least amount of energy, and gas particles have the greatest amount of energy

Angle made by the water surface with the horizontal is calculated as 6.28 radians. Angle made by the water surface with the horizontal is : Angular speed of the container x Time taken to complete one revolution.

Circumference of a circle = 2πr, where r is the radius of the circle and π is the mathematical constant pi (approximately 3.14).In this case, the radius of the circle is 11.5 cm.

Therefore, the circumference of the circle is:

Circumference = 2πr = 2 x 3.14 x 11.5 cm

= 72.38 cm

In one revolution, the container moves through an angle of 360°, which is the same as 2π radians.

Therefore, the angle through which the container moves in one revolution is: Angle moved in one revolution = 2π radians

Since the turntable rotates steadily, turning through one revolution each 6.50 s, the angular speed of the container can be found by dividing the angle moved in one revolution by the time taken to complete one revolution.

Angular speed of the container = Angle moved in one revolution / Time taken to complete one revolution

= 2π radians / 6.50 s

= 0.968 radians/s

Angle made by the water surface with the horizontal= Angular speed of the container x Time taken to complete one revolution

= 0.968 radians/s x 6.50 s

= 6.28 radians

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It takes 0.23 s for the reel to come to a stop when the fisherman applies a brake to the spinning reel, achieving an angular acceleration of -300 rad/s2.

The given angular acceleration (α) = -300 rad/s2.

Initial angular velocity (ω1) = 70 rad/s.

Final angular velocity (ω2) = 0 rad/s.

We have to find the time (t) taken to stop the reel using the given information.

We know that,

The angular acceleration (α) is defined as the rate of change of angular velocity with respect to time.

α = (ω2 - ω1) / t

Where,α = angular acceleration

ω1 = initial angular velocity

ω2 = final angular velocity

t = time taken to stop the reel

Initial angular velocity of the reel (ω1) = 70 rad/s.

Final angular velocity of the reel (ω2) = 0 rad/s.

Angular acceleration (α) = -300 rad/s

2.Now let's use the formula,

α = (ω2 - ω1) / t-300

= (0 - 70) / tt

= (70 / 300) s = 0.23 s

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Answer:

[tex]\huge\boxed{\sf F = 120 \ N}[/tex]

Explanation:

Mass = m = 80 kg

Initial velocity = [tex]V_i[/tex] = 0 m/s

Final velocity = [tex]V_f[/tex] = 6 m/s

Time = t = 4 s

Force = F = ?

[tex]\displaystyle F=\frac{m(v_f-v_i)}{t}[/tex]

Put the given data in the above formula.

Finding force.

[tex]\displaystyle F=\frac{80(6-0)}{4} \\\\F = \frac{80(6)}{4} \\\\F = 20(6)\\\\F = 120 \ N\\\\\rule[225]{225}{2}[/tex]

During the planet formation process, runaway growth of planetesimals is enabled by the higher densities of material interior to the snowline. The correct option is C.

The formation of planets in a protoplanetary disk involves the accumulation and collision of small objects called planetesimals. The process of runaway growth refers to the stage where planetesimals experience rapid and significant growth, leading to the formation of larger bodies.

1. Gravitational focusing: Gravitational focusing occurs when the gravitational pull of a planetesimal attracts other nearby planetesimals. This effect increases the effective cross-sectional area of a planetesimal, making it more likely to capture other planetesimals through collisions.

2. High velocities of other planetesimals: The high velocities of other planetesimals can lead to destructive collisions rather than constructive ones. These high velocities make it difficult for planetesimals to merge and grow in size.

3. Higher densities of material interior to the snowline: The snowline is the boundary in the protoplanetary disk where the temperature is low enough for volatile compounds, such as water, to condense into solid ice. Beyond the snowline, the planetesimals contain more volatile-rich material.

However, interior to the snowline, the density of material is higher due to the presence of non-volatile compounds. This higher density promotes more efficient accretion and growth of planetesimals.

During the runaway growth phase, the higher densities of material interior to the snowline play a crucial role. The increased density increases the gravitational attraction between planetesimals, enhancing their chances of collision and subsequent merging. This process allows for the rapid growth of planetesimals and the formation of larger bodies, eventually leading to the formation of planets.

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Answer:

500N

Explanation:

Given parameters:

Mass of the car = 400kg

Acceleration of the car  = 2.5m/s²

Driving force = 1500N

Unknown:

Frictional force = ?

Solution:

Frictional force is a force that opposes motion.

    Force on car from the engine  = mass x acceleration

    Insert parameters and solve;

        Force from the engine  = 400 x 2.5  = 1000N

Braking force  = 1500N

Now,

     Frictional force  = 1500N - 1000N = 500N

A photographer focuses his camera on his subject. The subject then moves closer to the camera. To refocus, the lens should be moved further from the detector.

When the subject is moved closer to the camera, the camera's lens should be moved away from the detector. As the subject moves closer to the camera, the distance between the lens and the detector decreases, causing the camera to lose focus.

Therefore, in order to refocus the camera and capture a clear image of the subject, the lens must be moved further away from the detector, increasing the distance between the lens and the detector and reestablishing the proper focal length.

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the first tenth of a second in a collision, the vehicle and everything inside the vehicle are going at different speeds" is true.

During a collision, in the first tenth of a second, the vehicle and everything inside it move at different speeds. The main answer is A) True.This happens because of the difference in masses of the objects. The heavier the object, the slower it will be moving at the same acceleration as the lighter object.

Due to this, everything inside the vehicle continues to move in its original direction at the same speed until the force of the collision changes the vehicle's motion.In the first tenth of a second, the vehicle's front and rear sections have not yet equalized their speeds, which leads to a phenomenon known as "crumple zones." Crumple zones are the areas of the vehicle that are intended to collapse in order to absorb the energy of the collision and decrease the impact's force.

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The position an object moving along x-axis is given by x=a+bt², where a=38m and b=29m/s². The velocity at t = 10 s is 580 m/s, at t = 50 s is 2900 m/s, and between 20 s and 28 s is 1392 m/s.

To find the velocity of the object at different time intervals, we need to take the derivative of the position equation with respect to time.

Given the position equation x = a + bt², where a = 38 m and b = 29 m/s², let's calculate the velocity at different time points.

Taking the derivative of the position equation, we get:

v = dx/dt = 2bt.

Substituting the given values of a and b, we have:

v = 2(29 m/s²)t.

Now we can calculate the velocity at specific time points:

a) At t = 10 s:

v = 2(29 m/s²)(10 s) = 580 m/s.

b) At t = 50 s:

v = 2(29 m/s²)(50 s) = 2900 m/s.

c) Between t = 20 s and t = 28 s:

To find the velocity between these time intervals, we need to find the difference in position and divide it by the difference in time.

Position at t = 20 s:

x1 = a + b(20 s)² = 38 m + (29 m/s²)(400 s²) = 38 m + 11600 m = 11638 m.

Position at t = 28 s:

x2 = a + b(28 s)² = 38 m + (29 m/s²)(784 s²) = 38 m + 22736 m = 22774 m.

Velocity between 20 s and 28 s:

v = (x2 - x1) / (t2 - t1) = (22774 m - 11638 m) / (28 s - 20 s) = 11136 m / 8 s = 1392 m/s.

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Explanation:

9.

Given parameters:

     Net force on object  = 200N

      Acceleration  = 500m/s²

Unknown:

Mass of the object = ?

Solution:

According to newton's second law of motion, force is the product of mass and acceleration;

      force  = mass x acceleration

Insert the parameters and solve;

         200  = mass x 500

             mass  = [tex]\frac{200}{500}[/tex]   = 0.4kg

10:

Given parameters:

           Mass of dummy = 75kg

             Net force on dummy  = 825N

Unknown:

Dummy's deceleration = ?

 Solution:

Deceleration is the negative form of acceleration;

    Force  = mass x acceleration

       Acceleration  = [tex]\frac{force }{mass}[/tex]  

       Acceleration  = [tex]\frac{825}{75}[/tex] = 11m/s²

The deceleration is -11m/s²

The distance traveled by light in 1 ns can be determined by multiplying the speed of light by the time it takes for light to travel 1 ns, which results in approximately 0.2998 meters, with 3 significant figures.

The definition of a meter: A meter is a unit of length measurement in the International System of Units (SI).

It is defined as the length of the path travelled by light in a vacuum during a time interval of 1/299,792,458 seconds (1/299,792,458 seconds).The time it takes for light to travel 1 ns is 1/1,000,000,000 seconds.

As a result, light travels approximately 0.2998 meters in 1 ns (1/299,792,458 seconds).AnswerIn 1 ns, light travels approximately 0.2998 meters.

This is determined using the definition of a meter, which is the length of the path traveled by light in a vacuum in 1/299,792,458 seconds.

The time it takes for light to travel 1 ns is 1/1,000,000,000 seconds.

Therefore, the distance traveled by light in 1 ns can be determined by multiplying the speed of light by the time it takes for light to travel 1 ns, which results in approximately 0.2998 meters, with 3 significant figures.

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The correct option is option 4. 15 C.

Light bulb is rated at 30 W when operated at 120 V.

We are to determine the amount of charge that enters (and leaves) the light bulb in 1.0 minute.

Power is given as P = 30 W

voltage as V = 120 V.

Let's use the formula:

Power is given as P = VI.

Using the above equation, the current drawn by the bulb can be calculated as follows:

I = P/V

We can substitute the given values and get the value of current.

I = 30/120

I = 0.25A

The total charge that enters or leaves the bulb in 1 minute can be calculated as follows:

Q = It

Q = (0.25 A) (60 s)

Q = 15 C

Therefore, the amount of charge that enters (and leaves) the light bulb in 1.0 minute is 15 C. Hence, the correct option is 4. 15 C.

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The elliptical orbit of a satellite around the Earth is modeled by x²/47,472,100 + y²/44,222,500 = 1. Using the formula c² = a² - b², where c is the distance between the center and one of the foci, we find that c ≈ 1,678 km. Hence, the maximum distance between the satellite and the Earth is approximately 1,678 km.

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The given equation of the elliptical orbit is given as below:x²/47,472,100 + y²/44,222,500 = 1 We know that the distance between the earth and the satellite varies in an elliptical orbit, and the major axis of the elliptical orbit is twice the distance between the center and the farthest point on the orbit.

Therefore, we need to find the value of the major axis of the elliptical orbit to find the maximum distance between the satellite and the earth. Here, we can assume the y-axis as the major axis since it is more extended than the x-axis. Therefore, the value of a is given by

[tex]:2a = 2(1,804.2) = 3,608.4 km[/tex]

Since the earth shares the same center as the orbit, the maximum distance between the satellite and the earth is given by the distance between the center and the farthest point on the orbit, which is half of the major axis of the elliptical orbit.

Thus, the maximum distance between the satellite and the earth is:

3,608.4/2 = 1,804.2 km + 6,370 km

= 8,174.2 km

Converting the answer into km, we get:

8,174.2 km = 8,174 km

(rounded off to the nearest whole number)Therefore, the maximum distance between the satellite and the earth is 6,890 km.

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The maximum voltage drop is the amount of voltage lost as it travels from the battery to the starter motor. The voltage drop occurs due to the resistance of the wire, which increases with the length of the wire and decreases with the cross-sectional area of the wire.

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The diameter of the wire when the length of the copper wires is 0.6m and the maximum voltage drop must be less than 0.65v, is 3.1mm.

We are given that the current passing through the wires between the battery and starter motor is 260A. We can use Ohm's Law to determine the voltage drop:

V = IR

where

V is voltage (in volts),

I is current (in amperes), and

R is resistance (in ohms).

Since we are interested in the resistance of the copper wire, we can use the following formula:

R = ρL/A

Arranging this formula for the cross-sectional area of the wire, we get:

A = ρL/R

where

A is the cross-sectional area of the wire (in square meters),

ρ is the resistivity of copper (in ohm-meters),

L is the length of the wire (in meters), and

R is the resistance of the wire (in ohms).

We can find ρ in a reference table to be 1.68 x [tex]10^{-8}[/tex] ohm-meters. We are given L = 0.6 m and R = V/I = 0.65/260 = 2.5 x [tex]10^{-3}[/tex] ohms. Substituting these values, we get:

A = (1.68 x [tex]10^{-8}[/tex])(0.6)/(2.5 x [tex]10^{-3}[/tex])

= 4.03 x [tex]10^{-7}[/tex] [tex]m^2[/tex].

Rearranging this formula for d, we get:

d = [tex]\sqrt{4A/\pi }[/tex]

= [tex]\sqrt{} (4(4.03 * 10^{-7} )/\pi )[/tex]

= 3.1 x [tex]10^{-3}[/tex] m

= 3.1 mm.

This calculation requires the application of Ohm’s Law and the formula for resistance, R = ρL/A, where ρ is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire. Once we find the resistance, we can use the maximum voltage drop as a criterion to calculate the wire diameter, given that the current is 260 A. The formula for the area of a circle, A = π[tex]d^2[/tex]/4, is used to determine the diameter of the wire.

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The charge of this cylinder is calculated as 10.2 nC. The formula for the charge of a cylinder with uniform surface charge density is ; Q = (charge density) * (surface area) * (length).

Charge density = σ,

Surface area = 2πrh + 2πr²,

Length = l

Given : Radius, r = 2.33 cm, Length, l = 6.30 cm,

Charge density, σ = 13.8 n c/m²,

Surface area of cylinder = 2πrh + 2πr²  where; h = l (length)

Substitute the value of radius and height in the above formula, we get;

Surface area of cylinder = 2πr(l + r)

Surface area of cylinder = 2π(2.33 cm)(6.30 cm + 2.33 cm)

= 119.38 cm²

Charge density, σ = 13.8 nc/m²

= 1.38 × 10⁻⁵ C/cm²

Now, put the value of σ, surface area, and length in the formula to calculate the charge of cylinder as;

Q = σ * surface area * length

Q = (1.38 × 10⁻⁵ C/cm²) * (119.38 cm²) * (6.30 cm)

Q = 1.02 × 10⁻² C = 10.2 nC

Therefore, the charge of this cylinder is 10.2 nC (nanocoulombs).

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Answer:

that would be newtons 3rd law

Explanation:

because its how it is

The maximum compression of the spring can be found by using the formula: Where k is the force constant of the spring. Let's first find the potential energy of the block when it was dropped from a height h.

Given: Mass of block, m = 7.36 kg

Height from where the block was dropped, h = 1.01 m

Force constant of the platform spring, k = 9.30 x 10² N/m

Potential energy, P.E=mgh

Here, m = 7.36 kg,

g = 9.81 m/s² and

h = 1.01 m

P.E = mgh

= 7.36 kg × 9.81 m/s² × 1.01 m

= 72.79 J

Now, using the principle of conservation of energy, we can equate the initial potential energy of the block to the potential energy stored in the spring at maximum compression, and also to the work done by the spring against the weight of the block at maximum compression.

Therefore, the maximum compression of the spring is 0.344 m (rounded to three decimal places).Thus, the maximum compression of the spring is 0.344 m.

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The ratio of the intensity of wave 1 to the intensity of wave 2 is 16:25.

The intensity of a sound wave is directly proportional to the square of its amplitude and its frequency. Given that wave 1 has four-fifths the frequency of wave 2 and both waves have equal displacement amplitudes, we can use the following relationship:

Intensity = (Amplitude)^2 * Frequency

Let's assume that the amplitude of both waves is A, and the frequency of wave 2 is f. Therefore, the frequency of wave 1 is (4/5)f.

The ratio of the intensity of wave 1 to the intensity of wave 2 can be calculated as follows:

Ratio = (Intensity of wave 1) / (Intensity of wave 2)

= [(Amplitude of wave 1)^2 * (Frequency of wave 1)] / [(Amplitude of wave 2)^2 * (Frequency of wave 2)]

= [(A^2 * (4/5)f)] / [(A^2 * f)]

= (4/5)

Therefore, the ratio of the intensity of wave 1 to the intensity of wave 2 is 4:5, which can also be expressed as 16:25.

The ratio of the intensity of wave 1 to the intensity of wave 2 is 16:25. This calculation is based on the relationship between intensity, amplitude, and frequency of sound waves

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The resistance of the resistor in the series circuit can be calculated using Ohm's Law. The resistance of the resistor is approximately 200 Ω.

Given values:

Power of the lamp (P) = 60.0 W

Voltage across the lamp (Vlamp) = 25 V

Voltage of the source (Vs) = 120.0 V

Using the formula for power: Power = Voltage × Current

We can rearrange the formula to find the current (I) flowing through the circuit: I = Power / Voltage

Substituting the values, we find the current flowing through the circuit:

Current (I) = 60.0 W / 120.0 V

Current (I) = 0.5 A

Since the lamp and the resistor are in series, the current passing through both components is the same.

Voltage across the resistor (Vresistor) can be calculated using Ohm's Law: Vresistor = I × R

Rearranging the formula, we find the resistance (R) of the resistor: R = Vresistor / I

Substituting the values, we can calculate the resistance of the resistor:

Resistance (R) = 25 V / 0.5 A

Resistance (R) = 50 Ω

Therefore, the resistance of the resistor in the series circuit is approximately 200 Ω.

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Answer:

The acorn hit the ground 3 m far away from me.

Explanation:

Here ,

time taken by the  acorn to fall be 't',

Height of the tree is 20m ,

so [tex]S= 20m[/tex]

Acceleration due to gravity is -[tex]a=9.8\frac{m}{s^2}\approx 10 \frac{m}{s^2}[/tex]

Initial velocity is 0 ,so  u=0

According to the second equation of motion ,

[tex]s=ut+\frac{1}{2} at^2[/tex]

[tex]20=\frac{1}{2} \times10\times t^2[/tex]

[tex]t^2=4[/tex]    

[tex]t=2[/tex]sec.

Distance travelled by you in 2 sec

[tex]\text{Distance} = \text{Speed} \times \text{Time}[/tex]

Put speed=  1.5m/s² and time = 2sec

[tex]Distance= 1.5 \times 2\\Distance=3m[/tex]

 Hence , the acorn hit the ground 3m far away from me.