A ball was kicked horizontally off a cliff at 12.0 m/s, how high was the cliff if the
ball landed 66.0 m from the base of the cliff?
cliff height = _m

Answer:

Theres no given?

Explanation:

Well, whatever.

I observed the shift of their sunset time.

Examples like:

January to June = their sunset time increased while

July to December = their sunset time decreased

The effort distance of a lever should be increased to lift a heavier load because it provides a mechanical advantage, allowing for easier lifting of the load.

The effort distance of a lever should be increased to lift a heavier load because it allows for a mechanical advantage that compensates for the increased weight.

In a lever system, the effort distance is the distance between the point of application of the input force (effort) and the fulcrum, while the load distance is the distance between the point of application of the output force (load) and the fulcrum. The mechanical advantage of a lever is determined by the ratio of the load distance to the effort distance.

By increasing the effort distance, the mechanical advantage of the lever system is increased. This means that for the same input force (effort), a greater output force (load) can be achieved. When dealing with a heavier load, a higher mechanical advantage is required to overcome the increased resistance.

By increasing the effort distance, the lever system can effectively multiply the applied force, making it easier to lift the heavier load. This allows for the redistribution of force and facilitates the efficient use of human effort in various applications, such as in construction, engineering, and even everyday tools like scissors and pliers.

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Answer:

Friction is when 2 solids move against each other. The cause of friction is adhesion, and surface roughness. Surface roughness is when a surface is rough enough that is causes friction against another surface. Adhesion is when 2 surfaces collide because of thier molecular force.

Answer:

a) 4652 cal

b) 8000 cal

Explanation:

Amount of heat transferred

Q1 = mL(v)

Q1 = 8.63 * 539

Q1 = 4652 cal

Amount of heat transferred to the water

Q2 = mcΔT

Q2 = 100 * 1 * (100 - 20)

Q2 = 8000 cal

Q = Q1 + Q2

Q = 4652 + 8000

Q = 12652 cal

b)

Heat transferred to the copper bowl

Q(b) = m(b) * c(b) * ΔT

Q(b) = 0.13 * 0.0923 * (100 - 20)

Q(b) = 0.96 cal

c)

Original heat of the cylinder

Q(c) = Q + Q(b)

m(c) * c(c) * ΔT = Q + Q(b), making ΔT subject of the formula

ΔT = (Q + Q(b))/ (m(c) * c(c))

ΔT = (12652 + 0.96) / (0.42 * 1)

ΔT = 12652.96/0.42

ΔT = 30126.1

Answer:

2.11 m/s

Explanation:

V=disp/time

V=(0.52m+3.7m)/(2)

V=4.22/2

V=2.11 m/s

Explanation:

There are two answers

m/(s/s)=m

or

(m/s)/s=m/s²

12. The time taken for the journey is 400 s

13. The time taken for the train is 200 s

14. The time taken is 7 h

15. The time taken for the beetle is 12 s

16. The time taken for the journey is 0.0068 h

The time taken in each case as given by the question can be obtain as follow:

12. The time taken for the journey

Time taken = Distance / Speed

Time taken = 26000 / 65

Time taken = 400 s

13. The time taken for the train

Time taken = Distance / Speed

Time taken = 3200 / 16

Time taken = 200 s

14. The time taken to travel

Time taken = Distance / Speed

Time taken = 672 / 96

Time taken = 7 h

15. The time taken for the beetle

Time taken = Distance / Speed

Time taken = 1.08 / 0.09

Time taken = 12 s

16. The time taken for the journey

Time taken = Distance / Speed

Time taken = 35 / 5147

Time taken = 0.0068 h

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For a longitudinal wave, the compression region, is the one represented by the densely packed particles with high pressure and the region with loosely packed particles is called rarefaction. Hence, the first part is C he dense region and the second one with some space between the dots is labeled as R.

Longitudinal waves are a type of mechanical waves passing through a medium. Unlike electromagnetic waves, they cannot be passed through vacuum.

In a longitudinal wave, the oscillation of particles is along the direction of wave propagation. The wave is composed of high pressure regions and low pressure regions called compressions and rarefactions respectively.

The regions where, particles are densely packed and shows the thick dote are labelled as compressions and the regions where, some space between particles are labeled as rarefactions.

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a)  the average power of the elevator motor during this period is 0.1696 hp

b) The power during an upward trip with constant speed is 16.13 horsepower.

To calculate the average power of the elevator motor during the period of acceleration, we need to find the work done by the motor and divide it by the time taken.

Given:

Mass of the elevator (m) = 682 kg

Acceleration (a) = (1.80 m/s - 0) / 3.10 s = 0.5806 m/s²

Time taken for acceleration (t) = 3.10 s

(a) First, let's calculate the displacement (d) using the formula for uniformly accelerated motion:

d = 0.5 * a * t^2

= 0.5 * 0.5806 m/s² * (3.10 s)^2

= 1.0153 m

Next, we can calculate the work done (W) by the elevator motor:

W = m * a * d

= 682 kg * 0.5806 m/s² * 1.0153 m

= 391.55 J

Now, to find the average power (P), we divide the work done by the time taken:

P = W / t

= 391.55 J / 3.10 s

= 126.36 W

To convert the power to horsepower, we can use the conversion factor: 1 horsepower (hp) = 745.7 watts.

Therefore, the average power of the elevator motor during this period is:

P = 126.36 W / 745.7

= 0.1696 hp

(b) During an upward trip with constant speed, the elevator does not accelerate, so the power required is only to counteract the force of gravity and friction. The power during an upward trip with constant speed is equal to the power required to overcome the force of gravity and friction.

The force of gravity (Fg) can be calculated using:

Fg = m * g

= 682 kg * 9.8 m/s²

= 6683.6 N

The power (P) required is given by the formula:

P = Fg * v

= 6683.6 N * 1.80 m/s

= 12030.5 W

To convert the power to horsepower:

P = 12030.5 W / 745.7

= 16.13 hp

Therefore, the power during an upward trip with constant speed is 16.13 horsepower.

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The pressure after the temperature has risen to 40.0 °C, assuming that there are no appreciable leaks or changes in volume is 4.35×10⁵ Pa

From the question given above, the following data were obtained:

The pressure the temperature has risen to 40 °C can be obtained as follow:

P₁V₁ / T₁ = P₂V₂ / T₂

Volume = contant

P₁ / T₁ = P₂ / T₂

4.00×10⁵ / 288 = P₂ / 313

Cross multiply

P₂ × 288 = 4.00×10⁵ × 313

Divide both sides by 288

P₂ = (4.00×10⁵ × 313) / 288

P₂ = 4.35×10⁵ Pa

Thus, the pressure at 40 °C is 4.35×10⁵ Pa

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Answer:

true

Explanation:

these are not the only parts of the atmosphere, i dont know the full list but i know these arent the only parts

Answer:

D) Slide an object of known mass across a rough surface, using a constant applied force that can be measured by a force sensor. Place a motion detector behind the object so that its speed can be measured as it slides across the surface. Repeat the experiment for different applied forces.

Explanation:

"The motion detector will provide information about the object’s speed as a function of time as it slides as a result of the applied force. The information about the object’s speed as a function of time can be used to determine the acceleration of the object. The force sensor measures the applied force exerted on the object, and the mass of the object is known. Therefore, this experiment can be used to determine how an object’s mass is related to the net force exerted on the object and the acceleration of the object."

It cannot be A because we need an acceleration will be determined by gravity.

It cannot be B because the term constant speed means that there is no net force, which is required by the initial question.

It cannot be C because the experiment is good for determining the coefficient of friction but not for determining how the mass relates to the acceleration.

It must be D because the object is moving and we have a motion detector, we can graph the acceleration vs time graph. So D allows you to have a lot of the different acceleration values which helps with determining the relationship between acceleration and the mass.

We are given the weight of two people, one on the moon and the other on earth. To determine the mass we will use the following formula:

Where:

Now, we solve for the mass "m" by dividing both sides by "g":

Now, for the case of the moon we have that the acceleration of gravity is:

Plugging in the values:

Solving the operations:

Now, for the case of the earth we have:

Plugging in the values:

Solving the operations:

Therefore, the greater mass is the mass of Bob.

Answer:

proportional tax is.the answer.

Explanation:

because both girls tax is equal.

The speed of the car is 6 m/s. The acceleration vector at point B has a direction of (-1, -1) and a magnitude of approximately 16.97 m/s².

We can start by finding the speed of the toy car. Since it is moving with constant speed around the circle, its speed is the same at points A and B. To find the speed, we can use the fact that the velocity vector has a magnitude equal to the speed:

|v| = √((6 m/s)²) = 6 m/s

So the speed of the car is 6 m/s.

Next, we can find the direction of the velocity vector at point B. We know that the car is moving around a circle centered at the origin, and that point B is on the circle. Therefore, the velocity vector at point B is tangent to the circle and perpendicular to the line connecting the origin to point B.

The line connecting the origin to point B is given by:

y = (0 - 3)/(0 - (-3)) * (x - (-3)) + 0

y = -x + 3

The velocity vector at point B is therefore perpendicular to this line, which means it has a direction given by the vector (1, -1).

Finally, we can find the acceleration vector at point B. Since the car is moving with constant speed around a circle, it is undergoing uniform circular motion, which means it is accelerating towards the center of the circle. The magnitude of the acceleration is given by:

a = v² / r

where v is the speed and r is the radius of the circle. We don't know the radius of the circle, but we can find it using the fact that point B lies on the circle. The distance from the origin to point B is:

d = √((-3 - 0)² + (0 - 3)²) = 3√(2) m

Therefore, the radius of the circle is:

r = d / 2 = (3√(2)) / 2 m

Substituting in the values for v and r, we get:

a = (6 m/s)² / ((3√(2)) / 2 m) ≈ 16.97 m/s²

To find the direction of the acceleration vector, we can use the fact that it is pointing towards the center of the circle. The center of the circle is at the origin, so the acceleration vector at point B is given by the vector (-3, 0) minus the vector (0, 3), which is:

(-3, 0) - (0, 3) = (-3, -3)

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Answer:

velocity

Explanation:

the displacement of an object during a specific unit of time.

Explanation:

The second law states that the greater the mass of an object, the more force it will take to accelerate the object. There is even an equation that says Force = mass x acceleration or F=ma. This also means that the harder you kick a ball the farther it will go.

Answer:

The truck and car have the same momentum.

Explanation:

\(m_1\) = Mass of car = \(2000\ \text{lb}=2000\times0.45359237\ \text{kg}\)

\(v_1\) = Velocity of car = \(\dfrac{100}{3.6}\ \text{m/s}\)

\(m_2\) = Mass of truck = \(4000\times0.45359237\ \text{kg}\)

\(v_2\) = Velocity of truck = \(\dfrac{50}{3.6}\ \text{m/s}\)

Momentum of car

\(p_1=m_1v_1\\\Rightarrow p_1=2000\times0.45359237\times \dfrac{100}{3.6}\\\Rightarrow p_1=25199.58\ \text{kg m/s}\)

Momentum of the truck

\(p_2=m_2v_2\\\Rightarrow p_2=4000\times0.45359237\times \dfrac{50}{3.6}\\\Rightarrow p_2=25199.58\ \text{kg m/s}\)

Both the truck and car have the same momentum of \(25199.58\ \text{kg m/s}\).

We're missing one essential piece of information that we need in order to answer this question.  You have not specified what planet the object is falling on.  The answer depends on the gravitational acceleration on that planet, and they're all different.

Without that information, we'll just go ahead and assume that the object is falling to the surface of the Earth.  Wherever on Earth this tense drama is unfolding, the acceleration of gravity is going to be around 9.8 m/s² everywhere.

So THAT's the object's acceleration if there is no air resistance.  The object's MASS makes no difference.  It doesn't matter whether the object is a sparrow feather or a school bus. Heavier objects DO NOT fall faster than light objects.

If there is no air resistance, then ALL objects fall with the same acceleration.    It's called the "acceleration of gravity" on that planet or moon, and you can easily look it up.  It's 9.8 m/s² on Earth, 1.62 m/s² on the Moon, 3.71 m/s² on Mars, 8.87 m/s² on Venus, and 24.8 m/s² on Jupiter.

Answer:

workers must have a legitimate respect for the common good whose advancement they serve.

Explanation:

Answer:

so they don't get fired and they should be over all nice people

Explanation:

Answer:

14 secs and velocity will be 48

Explanation:

The object accelerates for the first 4 s (0 to 4), then moves at a constant speed for the next 4 s (4 to 8), then speeds up again for the next 2 s (8 to 10), moves at another constant speed for 6 s (10 to 16), then slows to a stop for the last 4 s (16 to 20).

In the times where the speed is linearly increasing, the acceleration is constant and equal to the slope of those line segments. Where the speed is constant, acceleration is 0.

Recall that average acceleration is given by

a = ∆v / ∆t

So we get

[i] a = (20 m/s - 0) / (4 s - 0) = 5 m/s²

[ii] a = 0 (because the speed is constant at 12 s)

[iii] Not really sure what is meant by "total acceleration" here... Is it the sum of all accelerations observed here? If so, we know the acceleration 5 m/s² on [0, 4]; it's 0 for both [4, 8] and [10, 16]; and we can compute the acceleration for the remaining two intervals to be

[8, 10]: a = (30 m/s - 20 m/s) / (10 s - 8 s) = 5 m/s²

[16, 20]: a = (0 m/s - 30 m/s) / (20 s - 16 s) = -7.5 m/s²

Then the total might be 2.5 m/s². Maybe.

[iv] We can compute distance traveled by finding the area under the velocity curve. The area under the curve for the first 10 s can be split up into a triangle on [0, 2] with height 20 m/s and base 4 s; a rectangle on [4, 8] with height 20 m/s and base 4 s; and a trapezoid on [8, 10] with "base" lengths 20 m/s and 30 m/s, and "height" 2s. The total area is then

1/2 (20 m/s) (4 s) + (20 m/s) (4 s) + 1/2 (20 m/s + 30 m/s) (2 s) = 170 m

[v] We already have the distance traveled over the first 10 s, so just subtract the area of the trapezoid:

1/2 (20 m/s) (4 s) + (20 m/s) (4 s) = 120 m

[vi] This would be the acceleration we found in [iii] over the interval [16, 20]:

a = (0 m/s - 30 m/s) / (20 s - 16 s) = -7.5 m/s²

[vii] According to the plot, the velocity at 16 s is 30 m/s.

[viii] According to the plot, at 2s the velocity is 10 m/s.

The power wasted in the transmission line connecting the power house to the consumer is 2kW.

The equation Power = Potential difference*Current, or P = VI, can be used to compute the amount of energy that is wasted.

We can see that it is given,

V1 = 600 V

V2 = 580 V

Current= I = 100 A

We need to figure out how much energy is lost in the transmission line that runs from the generator to the consumer.

In physics, power is the amount of energy that is transferred or transformed in a given amount of time.

The International System of Units uses the watt, or one joule per second, as the unit of power.

The formula for the voltage across the transmission line is V = V1 - V2.

Thus, V = 600 - 580

V = 20V

We've been told that I equals 100 A.

The power loss equation is given by P = VI, where V is the potential difference and I is the current.

Power loss = VI, therefore becomes

P = 20 × 100

P = 2000 W

P = 2kW

Therefore, 2 kW of power is lost in the transmission line that connects the power source to the customer.

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Answer: A sonic boom is a sound associated with shock waves created when an object travels through the air faster than the speed of sound. Sonic booms generate enormous amounts of sound energy, sounding similar to an explosion or a thunderclap to the human ear. The crack of a supersonic bullet passing overhead or the crack of a bullwhip are examples of a sonic boom in miniature. Sonic booms due to large supersonic aircraft can be particularly loud and startling, tend to awaken people, and may cause minor damage to some structures. They led to prohibition of routine supersonic flight over land. Although they cannot be completely prevented, research suggests that with careful shaping of the vehicle, the nuisance due to the sonic booms may be reduced to the point that overland supersonic flight may become a feasible option. A sonic boom does not occur only at the moment an object crosses the speed of sound; and neither is it heard in all directions emanating from the supersonic object. Rather the boom is a continuous effect that occurs while the object is travelling at supersonic speeds. But it affects only observers that are positioned at a point that intersects a region in the shape of a geometrical cone behind the object. As the object moves, this conical region also moves behind it and when the cone passes over the observer, they will briefly experience the boom.

Explanation:

As incredible as the Concorde was, the sonic booms created by its supersonic flights were so disruptive that most countries restricted or completely prohibited the aircraft from flying over land.  The sonic boom, at its worst, would be heard as a very loud thunder clap that was right overhead. The force of the boom rattled windows and loosened roof tiles. But even when the sonic boom sounded like a “softer” distant thunder clap, it was distracting to people and caused disruption of sleep and interruptions in activity. Imagine that you are driving on your way to work, and with clear skies overhead, you suddenly hear the sound of thunder. Your immediate responses are most likely surprise, shock, and an instinctive search for the source. Being caught by surprise in certain situations is rather annoying, and in others, potentially dangerous. In 1964, the FAA and NASA conducted a six-month sonic boom research project in Oklahoma City – without warning residents beforehand. The experiment consisted of eight sonic booms, every day, for six months. 15,000 complaints and a class action lawsuit were filed. The government lost on appeals. Great idea, guys, just brilliant. When the Concorde was originally designed, in the early 1960s, governments and airlines around the world lined up to place orders. The plane did an around-the-world publicity trip, and was well-received. But as awareness of the sonic boom effect grew, almost every country banned the aircraft. Only the US, Great Britain, and France allowed the Concorde to enter their airspace, and then only to cities in close proximity to the ocean – NYC, London, Paris, and Washington, DC. The Concorde was specifically designed for supersonic flight (specifically, Mach 2) and was very fuel-inefficient at subsonic speeds (less than Mach 1). Unfortunately, it was thus not feasible to fly at supersonic speed over water and then at subsonic speed over land.

What causes a sonic boom?  

When any object moves, it creates waves in front of and behind it. Think of the waves that a boat creates at its bow and stern. In front, the waves are compressed together as the boat sails forward. Behind, the waves spread out away from the boat. In this case, you only see the waves on the surface of the water, and it appears two-dimensional. Similar principles are at play with aircraft. In front of the nose of a plane, air is pushed together and compressed as the aircraft flies forward. Behind the plane, the air creates waves that radiate out and away in the shape of a cone – three-dimensionally.

Answer:

A sonic boom is caused by the shock waves created when an object travels through the air faster than the speed of sound.

Explanation:

When any object moves, it creates waves in front of and behind it. Think of the waves that a boat creates at its bow and stern. In front, the waves are compressed together as the boat sails forward. Behind, the waves spread out away from the boat. Similar principles are at play with aircraft. In front of the nose of a plane, air is pushed together and compressed as the aircraft flies forward. Behind the plane, the air creates waves that radiate out and away in the shape of a cone – three-dimensionally. Things get interesting, and complicated, when you fly faster than the speed of sound – supersonic flight. The nose of a supersonic aircraft pushes ahead of its forward waves. These waves get in the way of the airplane, causing compression which results in a shock wave. Actually, this creates two shocks, one forming as the aircraft passes the front of the wave and then another as it leaves the wave. The shock wave generated stays mostly behind the aircraft, and radiates out in a cone

The work done on the boxes and the energy spent is 700J while the power required to lift 4 boxes per minute is 2.8kW

This work done on an object is the force required to move that object from point A to point B.

Data Given;

This is the work done to move the boxes from point A to point B.

\(w = F.s\\F = ma = mg = 200N\\w = 200 * 3.5\\w = 700J\)

The work done to move the object is 700J

This is used to calculated the energy used to carry the work done and the formula is given as

\(E = force * displacement\\E = F.s\\E = 200 * 3.5\\E = 700J\)

The energy transferred when he lifts one box is 700J

Power is the rate at which energy is used with respect to time

\(P = E/t\\\)

The energy required to lift 4 boxes in one minutes is

\(700 * 4 = 2800\)

Now we can calculate the power used

\(P = Energy / time\\P = 2800 / 1 \\P = 2800W = 2.8kW\)

The power required to move four boxes in 1 minutes is 2.8kW

From the calculations above, the following was gotten

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Answer: 24km

Explanation:

hope this helps

Answer:

a. 5.901x10⁻³H

b. 2.36x10⁻²V

Explanation:

a. inductance of coil

we calculate the inductance of coil using this formula

L = N²μπA/λ

= [N²μπd²/4]/λ

we substitute values

L = 580² x 4πx10⁻⁷ x π(8x10⁻²)²/ 4x0.36

L = 336400 x 0.000001257 x 0,020096/1.44

L = 0.005901

= 5.901x10⁻³H

b. EMF

|e| = Ldi/dt

= 0.005901 * 4.00

= 0.023604

= 2.36x10⁻²V

If both runner A and runner B start the race together, they will 1.125 m apart at the finish of the race

v = d / t

v = Velocity

d = Distance

t = Time

d = 4.5 m

For runner A,

t = 4.5 min

v = 4.5 / 4.5

v = 1 m / min

For runner B,

t = 6 min

v = 4.5 / 6

v = 0.75 m / min

Runner A finished the race first at 4.5 min. At that time the runner B will have covered,

0.75 = d / 4.5

d = 0.75 * 4.5

d = 3.375 m

Difference in distances = 4.5 - 3.375

Difference in distances = 1.125 m

Therefore, they will 1.125 m apart at the finish of the race

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Answer:

Explanation:

1 km =1000 m

24 km=24/1000

24 km=0.024 m

1 hour= 3600 s

0.6 hours=2160 s

now average velocity=distance/time

v=0.024 m/2160 s

v=1.11*10^-5

another way

v=24 km/0.6 hour

v=40 km/ hour