A ball is dropped from a 19m high cliff. The acceleration on the ball was 9.8m/s². What was the ball's final velocity before hitting the ground?

Answer:

Burning wood in a fireplace

Explanation:

Answer:

The water cycle shows the continuous movement of water within the Earth and atmosphere. ... Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow. Water in different phases moves through the atmosphere (transportation).

Explanation:

It's the water cycle.

Explanation:

You want  N/m

 N = 66 * 9.81

 m = 2.3 x 10^-2 m

66* 9.81 / 2.3 x 10^-2  = 28150 =  28 000 N/m    to two S D

To determine the thrust applied by the engines, we can use Newton's second law of motion, which states that force (thrust) is equal to mass times acceleration. In this case, we need to calculate the force required to decelerate the plane from 30 m/s to 25 m/s in 12 seconds.

First, we calculate the change in velocity (∆v):

\(\displaystyle\sf \Delta v=25\,m/s-30\,m/s=-5\,m/s\)

Next, we calculate the acceleration (∆a) using the formula:

\(\displaystyle\sf \Delta a=\frac{\Delta v}{\Delta t}\)

where ∆t is the change in time, which is 12 seconds in this case.

\(\displaystyle\sf \Delta a=\frac{-5\,m/s}{12\,s}\)

Now, we can determine the force (thrust) applied by the engines using Newton's second law:

\(\displaystyle\sf F=m\cdot a\)

where m is the mass of the airplane, which is 480000 kg.

\(\displaystyle\sf F=480000\,kg\cdot \left(\frac{-5\,m/s}{12\,s}\right)\)

Calculating the result:

\(\displaystyle\sf F=-200000\,N\)

Therefore, the engines applied a thrust of -200000 Newtons (N) to decelerate the plane. The negative sign indicates that the thrust is in the opposite direction of the motion.

Since the car masses are unknown, we are unable to calculate the numerical value of the x-component of Car A's tangential acceleration.

The energy that is held in any object or system as a function of its position or component arrangement is known as potential energy. The object or system is unaffected by external factors like air pressure or altitude. Kinetic energy, on the other hand, describes the power of moving particles within a system or an object.

They are being affected by the kinetic frictional force, which is caused by:

f = μk * N

Therefore,

fB = μk * N = μk * mB * g

Car C is at its highest point at the top of the hill, where the normal force acting on it is equal to the force of gravity. Therefore,

fC = μk * N = μk * mC * g

where mC is the mass of Car C.

For Car A, the x-component of the tangential acceleration is given by:

aA = (fB - fC) / mA

where mA is the mass of Car A.

We can substitute the following values and simplify by assuming that the mass of each of the three automobiles is the same:

aA = (μk * mB * g - μk * mC * g) / mA

aA = μk * g * (mB - mC) / mA

Since μk = 1.00 and g = 9.81 m/s², we can plug in the values and get:

aA = (mB - mC) * 9.81 / mA

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Explanation:

Height is the x-axis, and gravitational potential energy is the y-axis.  As the height increases, the gravitational potential energy increases linearly.

Answer:

C because it literally in Australia but the pfp tho

Answer:6

Explanation:

Answer:

6

Explanation:

Answer:  NEWTONS FIRST LAW

One's body moves to the side when a car makes a sharp turn.

Tightening of seat belts in a car when it stops quickly.

A ball rolling down a hill will continue to roll unless friction or another force stops it.

If pulled quickly, a tablecloth can be removed from underneath dishes.

                                NEWTONS SECOND LAW

1· Try to move an object.

2· Pushing a car and a truck.

3· Racing Cars.

4· Rocket launch.

5· Kick the ball.

6· Car crash.

7· Two people walking.

8· Object thrown from a height

   Newton's Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass (object). Riding your bicycle is a good example of this law of motion at work. Your bicycle is the mass. Your leg muscles pushing on the pedals of your bicycle is the force.

                                     NEWTONS THIRD LAW

                  Examples

The recoil of a Gun. ...

Swimming. ...

Pushing the Wall. ...

Diving off a Raft. ...

Space Shuttle. ...

Throwing a Ball. ...

Walking. ...

Hammering a Nail.

More items...

Common examples of newton's third law of motion are: A horse pulls a cart, a person walks on the ground, a hammer pushes a nail, and magnets attract paper clips. In all these examples a force is exerted on one object and that force is exerted by another object.

Answer:

37.5 J

Explanation:

With work done equation: W=Fs

W=75*0.50=37.5 J

or use mgh=(75)(0.5) which is the same

1,000 J of energy are needed to melt 10 g of a solid substance that is already at its melting point ,  the heat of fusion of the substance is 548 joules .

Heat of fusion, also known as enthalpy of fusion or latent heat of fusion, is the amount of energy required to melt or freeze a substance under constant pressure conditions. When it comes to chemistry, "fusion" is basically synonymous with "melting." In the classroom, heat of fusion is typically used when a substance is at its melting or freezing point. In such instances, most people consider heat of fusion to be a constant.

Water, for example, has a heat of fusion of 334 J/g at its melting point of 0°C. At 0°C, one grams  of liquid water requires 334 Joules of energy to completely freeze into ice. In addition, one grams of ice requires 334 Joules of energy to melt entirely.

q = m×∆Hf

q: Total change in heat energy (in Joules)

∆Hf: Heat of fusion of substance (in Joules per gram)

m: Mass of substance (in grams)

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Answer:

A) t = 1.249 10⁻² s, B)  f = 80 Hz, C) f = 20 Hz,

D)  slowly advancing an angle of approximately    Δθ = 0.05 rad each flash

E) In each flash it seems to go backward an angle of Δθ = -0.05 rad

Explanation:

A) To make it appear that the blades are immobile, it implies that every time the light turns on, a blade should be in the same position, therefore, as we have 4 blades, they must rotate an angle of 2π/4,  

         θ = π / 2  

         θ = 1.57 rad  

taking the angle let's use the endowment kinematics relations  

          θ = w₀ t + ½ α t²  

in general the fans rotate at constant speed α= 0  

         θ = w₀ t  

         t = θ / w₀  

let's reduce the magnitudes to the SI system  

        w₀ = 20 rev / s (2π rad / 1rev) = 125.66 rad / s  

let's calculate  

        t = 1.57 / 125.66  

        t = 1.249 10⁻² s  

B) the fastest speed for the blades to rotate is when one blade of a complete turn , we use the relationship between the fecuance and the period  

        f = 1 / T  

        f = 1 / 1.25 10⁻²  

       f = 80 Hz

C) we have two possibilities:  

* a yellow dot is placed on each sheet  

In this case the angular velocity of the blade is the same at all points, therefore the results obtained should not change

* a yellow dot is placed on a single sheet.  

Here for the point to remain fixed the angle of rotation must be

       θ= 2π rad  

the time is  

       t = 2π / 125.66  

       t = 5 10⁻² s  

the maximum frequency is  

      f = 1/5 10⁻²  

      f = 20 Hz

D) The copy strobe rotates at f = 19 Hz, the time between each flash is  

      t = 1/19  

      t = 5.26 10⁻² s  

this time is higher, so the angle turned is large  

       θ = w t  

       θ = 125.66 5.26 10⁻²  

       θ = 6.61 rad  

the relationship between this angle and the angle of a circle is  

θ = 1,052

We can see that it is this time the blade rotates 1 complete turns, for this the position of the blade changes us, for the other 0.052 rad the blade rotates a little more than the circumference therefore it seems that it is slowly advancing an angle of approximately  

         Δθ = 0.05 rad each flash  

E) in this case changes the flash speed  

       t = 1/21  

       t = 4.76 10⁻² s  

the angle rotated is  

      θ = 125.66 4.76 10⁻²  

      θ = 5.984 rad  

      θ / 2π = 0.95  

in that case, the blade did not complete the turn, therefore in each flash it seems to go backward an angle

Δθ = -0.05 rad

1. The car is moving with a uniform velocity from “O” to “A” and from “F” to “G”.

2. The car is at rest between “A” to “B”, “C” to “D”, and “E” to “F”.

3. The average velocity of the car would be 24 km/hour between “O” and “A”.

4. The car stopped for 7 hours during the whole time of travel

The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second. It can also be represented by the infinitesimal rate of change of displacement with respect to time. The generally considered unit for velocity is a meter per second.

As shown in the position time graph of a moving car we can observe and calculate the velocity by the slope of the curve.

If the slope of the position time curve is zero it means its velocity is zero during that time period. The horizontal line represents the zero velocity which means the car would have been at rest.

average velocity = displacement/time

                            = 24 km/hour

The car stopped for 7 hours during the whole time of travel

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Answer:

50km

cuz he ran so fast in 100 seconds

The electric field's magnitude at point P with coordinates is 8.85 mm. E= 2ϵ o/σ and l= 10 −7 ∧2×50×8.85×10 −12

E=σ2ϵ0. The proximity of P to the infinitely charge sheet has no bearing on this. The lines of something like the magnetic charge are continuous, homogenous parallel lines.

E = /20r is the magnitude of the electric field produced by an infinite line that is uniformly charged, where r is the distance between the line and the spot where the field is being measured and represents the linear charges density.

The formula is = Q 0. The electric field produced by an infinite thin sheet that is equally charged is given by the equation E=20n, where E is just the electric field, is the area positive charge, and 0 is the electric constant.

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Answer: Friction Forces

Explanation: (I took the same test and got the answer)

_________________________________________

I hope this helps!

The resultant of two displacement vectors having opposite directions is the difference of the two displacements, having the same direction as the smaller vector.

When two displacement vectors have opposite directions, it means they are pointing in opposite ways. In other words, one vector is in the opposite direction of the other. To find the resultant of these vectors, we need to subtract one vector from the other.

If we consider two displacement vectors, let's say vector A and vector B, and they have opposite directions, we can represent them as A and -B.

To find the resultant, we subtract vector B from vector A: A - (-B) or A + B.

The resultant will have the same direction as the smaller vector. This is because when we subtract a larger vector from a smaller vector, the resultant will have the direction of the smaller vector.

Therefore, the correct option is: "The resultant is the difference of the two displacements, having the same direction as the smaller vector."

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Explanation:

(a) Given:

ω₀ = 0 rad/s

ω = 200.0 rad/s

t = 5 s

Find: α

ω = αt + ω₀

200.0 rad/s = α (5 s) + 0 rad/s

α = 40 rad/s²

(b) ∑τ = Iα

τ = (50 kg m²) (40 rad/s²)

τ = 2000 Nm

The angular velocity is:

(a) Given:

ω₀ = 0 rad/s

ω = 200.0 rad/s

t = 5 s

Find: a

ω = αt + ω₀

200.0 rad/s = α (5 s) + 0 rad/s

α = 40 rad/s²

(b) ∑τ = Iα

τ = (50 kg m²) (40 rad/s²)

τ = 2000 Nm

A regular torque motor is a brushless motor, which reduces warmness in the rotor and stator area. they are eighty% green as compared to p.c automobiles which are rated at 60% green. Additionally, they don't want a capacitor to start and run.

A regular torque load means that the torque required to maintain the burden of walking is identical at all speeds. A good instance is a drum-kind hoist, in which the torque required varies with the load at the hook, but no longer with the speed of hoisting.

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Answer:

D

Explanation:

The elevator is either moving upwards with an increasing speed, or moving downwards with a a decreasing speed.

The statement above is true because the direction at which the elevator accelerates, or decelerates(which is negative acceleration anyway), is of paramount importance. If the acceleration is towards the upside, the apparent weight does becomes greater than the true weight. While on the other hand, if the acceleration points towards the downside, then the apparent weight does becomes less than the true weight.

The magnitude of the resultant vector to round the answer to the nearest tenth, we look at the digit in the hundredth's place. If this digit is 5 or greater, we round up. If it is less than 5, we round down.

In the study of physics, we use vectors to represent quantities that have both direction and magnitude. It is often the case that we want to add two or more vectors together to obtain a single vector that represents the net result of these additions. The process of adding two or more vectors together is known as vector addition.The magnitude of the resultant vector is the length of the line that represents it on a scale drawing.

When we add two or more vectors together, the resultant vector is the vector that represents the net result of these additions. To find the magnitude of the resultant vector, we use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.

In the case of vector addition, the hypotenuse is the resultant vector, and the other two sides are the component vectors. If we have two vectors a and b, the magnitude of the resultant vector is given by the following equation:|R| = √(ax2 + bx2)where R is the resultant vector, a and b are the component vectors, and x is the angle between the vectors.

For example, if the answer is 12.345, we would round it to 12.3.

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The tension at point AE is 28.8 lb, and the reactions at point D are RD = 64.6 lb and RE = 145.4 lb.

the tension at point AE First, we need to calculate the weight of the CD bar, which is given by WC = Weight of CD bar = 5 × 32lbWC = 160 lb

Now we can find the total weight supported by the system as follows: W = Weight of AE bar + Weight of CD bar + Weight of AD bar W = 40 + 160 + 10 = 210 lb

As the weight is distributed evenly, the vertical forces at D and E should be equal: RD + RE = 210 lb

Next, we will determine the moments around point D.

This will help us find the tension at point AE.∑MD = 0(-32 × 2) + (40 × 4) + (160 × 7) + (10 × 5) + AE × 10 = 0Solving for AE,AE = 28.8 lb

Determine the reactions at point D

Now we can solve for the reactions at point D.

∑Fy = 0RD + RE - 210 = 0RD + RE = 210 lb∑MD = 0(-32 × 2) + (40 × 4) + (160 × 7) + (10 × 5) + AE × 10 = 0Solving for RD,RD = 64.6 lb

Now that we have RD, we can solve for RE:RE = 210 - RDRE = 145.4 lb

Therefore, the tension at point AE is 28.8 lb, and the reactions at point D are RD = 64.6 lb and RE = 145.4 lb.

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Answer: 0.25 seconds.

Explanation:

The yo-yo does 240 revolutions in one minute, and we know that one minute has 60 seconds, then the revolutions per second can be calculated as:

240 rev/60s = 4 rev/s, this will be the frequency of the yo-yo

The frequency is actually written as: f = 4 Hz = 4 s^-1

We want to find the period of this yo-yo.

The period is the duration of one cycle, and we have the relation:

f = 1/T

Where f is the frequency and T is the period, then:

T = 1/f

And we know the value of f, it is f = 4 s^-1

Then the period will be:

T = 1/(4 s^-1) = (1/4) s

Then the period of the yo-yo is 1/4 seconds = 0.25 seconds.

Answer:

a) a = 2.35 m/s^2

Explanation:

(a) In order to calculate the magnitude of the acceleration of the ball, you use the following formula, for the position of the ball:

\(x=v_ot+\frac{1}{2}at^2\)     (1)

x: position of the ball after t seconds = 87 m

t: time  = 8.6 s

a: acceleration of the ball = ?

vo: initial velocity of the ball = 0 m/s

You solve the equation (1) for a:

\(x=0+\frac{1}{2}at^2\\\\a=\frac{2x}{t^2}\)

You replace the values of the parameters in the previous equation:

\(a=\frac{2(87m)}{(8.6s)^2}=2.35\frac{m}{s^2}\)

The acceleration of the ball is 2.35 m/s^2

The impulse shared by the object equals the difference in momentum of the object. In equation form,

F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the difference in momentum.

a)There is this impulse-momentum change equation.

\(where m$ is the mass of a body, $F$ is a force acting to the body, $t$ is time and $D E L A T A N\}=V_{2}-V_{1}$ is the change of velocity.We consider everything is happen along a straight line, and gravitation does not participate.So, the increase of momentum is $\mathrm{F}^{*} \mathrm{t}=10000 \mathrm{~N} * 60$ seconds $=600000 \mathrm{~N}^{*} \mathrm{~s}=600000\left(\mathrm{~kg}^{*} \mathrm{~m}\right)^{*} \mathrm{~s} / \mathrm{s}^{\wedge} 2=600000 \mathrm{~kg}{ }^{*} \mathrm{~m} / \mathrm{s}$.\)

We consider everything exits happen along a straight line, and gravitation does not participate.

So, the increase of momentum is F×t = 10000 N × 60 seconds = 600000 N*s = 600000 (kg*m)*s/s^2 = 600000 kg*m/s.

\($$\Delta(\mathrm{V})=\frac{\mathrm{F.t}}{\mathrm{m}}=\frac{600000}{1200}=500 \mathrm{~m} / \mathrm{s} .$$\)

New velocity after  engine was firing during 60 seconds is 2000 + 500 = 2500 m/s.

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Diagram for part (a)

Calculations for part (b):

Each resistor individually results in a current I with a voltage V in this ratio: V = IR.

R = 3Ω and the battery voltage V = 15V. Plugging in our values for V and R,

15 = I*3; I = 5A. This value matches that of the ammeters placed right after the resistors.

The combined resistance of the three resistors in parallel can be calculated by this equation:

3*1/R = 1/Rc, where Rc is the combined resistance and R is the resistance of each of the resistors (3Ω)

3/3 = 1/Rc; Rc = 1Ω

We can use the relation V = IR to determine the resultant current when the resistance of all three resistors is combined.

15 = I*1; I = 15A. This value matches that of the ammeter placed after all the parallel loops of resistors.

(c).

When each parallel resistor was added, the values that changed were the total current and equivalent resistance. This is because each resistor adds to the total equivalent resistance across the entire circuit, which affects the total current. The values that remained constant were the total voltage (voltage does not change; only current does), and voltage drop across each resistor (each resistor had the same value, so the voltage drops equally across each one).

(a)  the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.

(b)  the man is in the air for approximately 0.843 seconds.

To determine the horizontal distance between the saddle and limb when the man makes his move, we need to consider the horizontal velocity of the man when he drops from the tree limb.

Given:

Speed of the horse (constant velocity), v = 13.5 m/s

Vertical distance between the limb and saddle, h = 3.55 m

a) To find the horizontal distance, we can use the formula:

horizontal distance = horizontal velocity × time

Since the man drops vertically, his initial horizontal velocity is zero. The only horizontal velocity he will have is due to the motion of the horse.

The time taken by the man to fall can be determined using the equation for free fall:

h = (1/2) × g × t²

Where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time.

Rearranging the equation, we get:

t = √(2h / g)

Substituting the given values:

t = √(2 × 3.55 / 9.8) ≈ 0.843 s

Now, we can find the horizontal distance:

horizontal distance = v × t

horizontal distance = 13.5 × 0.843 ≈ 11.386 m

Therefore, the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.

b) The time the man is in the air can be calculated using the same equation for free fall:

t = √(2h / g)

Substituting the given value of h:

t = √(2 × 3.55 / 9.8) ≈ 0.843 s

Thus, the man is in the air for approximately 0.843 seconds.

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Explanation:

Thanx for the figure;
The force component of F   UP the ramp that  moves the crate must equal the force of the crate DOWN the ramp

75 kg = mg Newtons = 735.8 Newtons

    Downplane force is   735.8 sin 34°  = 411.4 Newtons

   Fn =The horizontal force will be found by   cos 34 = 411.4/ F                                                                       F = 411.4/cos (34) = 496 N

Normal Force =  735.8 cos 34°  = 610 N   This part is due to the mass of the crate....there is additional normal force from the force pushing the crate up the hill  (from below)

   = F sin34   =  496 sin 34 = 277.4 N

            SUM of normal forces = 610 + 277.4 = 887.4 N

Answer:

  a.  |F| ≈ 496 N

  b.  normal force ≈ 887 N

Explanation:

You want the magnitude of the horizontal force F that moves a crate up a 34° ramp at constant speed, and you want the magnitude of the normal force on the crate.

The constant speed of the crate tells you the net force up the ramp is zero. This is the sum of the component of force F in that direction and the force due to gravity in the opposite direction:

  F·cos(34°) - m·g·sin(34°) = 0

  F = mg·tan(34°) = (75 kg)(9.8 m/s²)tan(34°) ≈ 496 N

The magnitude of force F is about 496 N.

The normal force on the crate will be the sum of the component of F in that direction and the force due to gravity in the same direction:

  F·sin(34°) +m·g·cos(34°) ≈ 887 N

The magnitude of the normal force is about 887 N.