A stone is thrown horizontally with an initial speed of 10m/s from the edge of the cliff. A stop watch measures the stone’s trajectory with time from the top of the hill to the bottom to be 6.7s. What is the height of the cliff?

Answers

Answer 1

Answer:

Answer and steps in the pic

A Stone Is Thrown Horizontally With An Initial Speed Of 10m/s From The Edge Of The Cliff. A Stop Watch

Related Questions

A single living thing.​

Answers

Answer:

What do you mean ma´am/sir?

Explanation:

***ECONOMICS***

A government that wants to increase its GDP would most likely take which
action?
A. Increase the money supply to make it easier to borrow money
Ο Ο
B. Decrease the money supply to slow the growth of inflation
C. Increase taxes on businesses that operate outside the country
O O
D. Decrease taxes on citizens who are poor or unemployed

Answers

Answer:

The correct answer is A. A government that wants to increase its GDP would most likely increase the money supply to make it easier to borrow money.

Explanation:

If the government wanted to increase its GDP, the most appropriate way to do so would be to increase the money supply both through issuance and through a reduction in bank reserve requirements, thereby increasing the circulating money in the hands of society.

This, in turn, would make citizens reinvest that money, increasing economic production and, therefore, the national GDP.

Answer: A. Increase the money supply to make it easier to borrow money

Explanation: I just took the test on Ap ex

How have the owners of the game reserve invested in the local community?

Answers

Answer:They have made community members shareholders so they get a share of the profits, which they use for schools and healthcare clinics

Explanation: Edmentum

Answer:

shareholders of the community get profits and that is used for schools and healthcare clinics. cs. 

Explanation:

A yo-yo is made of two solid cylindrical disks, each of mass 0.055 kg and diameter 0.070 m , joined by a (concentric) thin solid cylindrical hub of mass 0.0055 kg and diameter 0.011 m . Part A Use conservation of energy to calculate the linear speed of the yo-yo when it reaches the end of its 1.1 m long string, if it is released from rest. Express your answer using three significant figures and include the appropriate units.

Answers

Answer: IM 95%sure that the answer is B jus took the test got the answer right

Explanation:

Answer:

sorry I forgot I wish I could help

Plz help me fast WITH EXTRA POINTS AFTER SUBMITTING

Answers

Answer:

4 bobux

Explanation:

one bobux

two bobux

three bobux

four bobux

A 66-N ⋅ m torque acts on a wheel with a moment of inertia 175 kg ⋅ m2. If the wheel starts from rest, how long will it take the wheel to make one revolution?

Answers

Answer:

t = 5.77 s

Explanation:

This exercise will use Newton's second law for rotational motion

            τ = I α

             α = τ / I

             α = 66/175

             α = 0.3771   rad/s²

now we can use the rotational kinematics relations, remember that all angles must be in radians

 

            θ = 1 rev = 2π radians

               

            θ = w₀ t + ½ α t²

as the wheel starts from rest w₀ = 0

             t = √ (2θ/α)

let's calculate

             t = √ (2 2π / 0.3771)

             t = 5.77 s

What is the resultant velocity of a plane that is traveling at 245 m/s North and encounters a tailwind of 55 m/s North?

Answers

Answer:

b

Explanation:

The word acid comes from the Latin word

Answers

Hi :)

The word acid comes from the Latin word acere, which means sour

Hope this helps!
It is acere but for future reference you can search of definition press more and google tells you its origin

How much work is required to move it at constant speed 5.0 m along the floor against a friction force of 290 N?

Answers

Answer:

The answer is 1450 J

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

force = 290 N

distance = 5 m

We have

workdone = 290 × 5

We have the final answer as

1450 J

Hope this helps you

A student studies the effect of an object's speed on its amount of kinetic energy. This graph summarizes the data from the study Kinetic energy Speed Which statement best describes what the graph shows?
A. As speed increases, kinetic energy increases exponentially
B. As speed increases, kinetic energy stays the same
C. As speed decreases, kinetic energy doubles each time.
D. As kinetic energy increases, speed decreases exponentially​

Answers

The answer is A I Hope this answer helps you (i got the question right)

Answer:

A. As speed increases, kinetic energy increases exponentially.

Explanation:

The amount of kinetic energy an object has depends on the speed. Kinetic energy is also known as "motion energy." This being said, if speed is increasing, decreasing, or staying constant, the kinetic energy of the object will too.

Metals that have shine and luster?

Answers

Answer:

luster

Explanation:

Velocity which stone gains when falling from height of 80 m is approximately equal to *

A. 0
B. 1 m/s
C. 8 m/s
D. 40 m/s
E. 300 m/s

Answers

Answer:

40

Explanation:

Time it takes stone to fall from the height of 80 m is approximately equal to *

A. 1 s
B. 2 s
C. 4 s
D. 8 s

Answers

Answer:

D

Explanation:

Answer:

c.4s

Explanation:

6) The magnitude of the force the Sun exerts on Uranus is 1.41 x 1021 newtons. Explain how it is possible for the Sun to exert agreater force on Uranus than Neptune exerts on Uranus.

Answers

Answer and Explanation:

TL: DR The Sun is much more massive than Neptune — more than enough to make up for the somewhat smaller distance between the two planets at the closest approach.

[The surprise in this answer (to me, a non-astronomer), is that the gap between the orbits of Neptune and Uranus is large — half the distance from Uranus to the Sun.]

The ratio of gravitational attraction of the Sun on Uranus versus Neptune on Uranus is directly proportional to the ratio of the Sun’s mass to Neptune’s and inversely proportional to the ratio of the square of the distances (let’s use the closest approach of the two planets to one another to calculate a maximum attraction).

Numbers:

Sun’s mass: 2 x 10^30 kg

Neptune’s mass: 1 x 10^26 kg

Distance of Sun to Uranus: 3 x 10^9 km

Closest approach of Uranus and Neptune: 1.5 x 10^9 km

Without doing any arithmetic, we see that even at their closest approach, Uranus and Neptune are separated by about one-half of the Uranus to Sun distance. Squaring that ratio, we see that if the Sun and Neptune had the same mass, the attraction between the Sun and Uranus would only be about 1/4 of that between the Sun and Neptune; however, the Sun has 20000 times the mass of Neptune, so the attraction between Uranus and the Sun is about 5000 times stronger than the maximum attraction between Uranus and Neptune.

The explanation of the possibility of why sun exerts a greater force on Uranus than Neptune exerts on Uranus is; because the force was calculated to be greater.

The formula for calculating the Force of Gravity between two masses is:

F = G*m₁*m₂/r²

Where;

F = force of gravity

G = gravitational constant = 6.674 × 10⁻¹¹ N•m²/kg²

m₁ = mass of the larger object

m₂ = mass of the smaller object

r = the distance between the centers of the two masses

Now, from online values, we have the following;

mass of Neptune; m₁ =  102.413 × 10²⁴ kg

mass of Uranus; m₂ = 86.813 × 10²⁴ kg

average distance between the centers of Neptune and Uranus; r = 1.62745 × 10¹² m

Thus, force exerted by Neptune on Uranus is;

F = (6.674 × 10⁻¹¹ × 102.413 × 10²⁴ × 86.813 × 10²⁴)/(1.62745 × 10¹²)²

F = 2.240 × 10¹⁷ N

We are told that the force the Sun exerts on Uranus is 1.41 the force the Sun exerts on Uranus is 1.41 × 10²¹ N.

That is greater than the force Neptune exerts on Uranus.

Read more about Force of Gravity at; https://brainly.com/question/7281908

un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración necesaria para detenerlo?

Answers

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en  M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

Vf: Velocidad final Vo: Velocidad inicial a: Aceleración d: Distancia recorrida

En este  caso:

Vf: 0 m/s, porque el avión se detieneVo: 50 m/sa: ?d: 120 m

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

[tex]a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}[/tex]

a= - 10.42 m/s²

La aceleración necesaria para detener el avión es - 10.42 m/s².

If two identical objects are dropped one after 1 s delay with respect to another then in the absence of the air resistance *

A. distance between two falling objects will keep increasing
B. distance between two falling objects will keep decreasing
C. will stay the same
D. All of the above

Answers

Answer:

A.No. Assuming no other factors (such as air resistance) the first object will have a velocity of 32 feet/second when the other is dropped. Since they will both have the same acceleration, the first distance between them will increase by 32 feet per second.

Explanation:

Is it true or false that the displacement always equals the product of the average velocity and the time interval?

Answers

Answer:

True.

Explanation:

Applying the definition of average velocity, we know that we can always write the following expression:

        [tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)

By definition, Δx is just the displacement, and Δt is the time interval.So, just rearranging terms in (1), we get:

        [tex]\Delta x} = v_{avg}* {\Delta t}[/tex]

I need help with this answer

Answers

The answer is Synthesis

An ideal gas increases in temperature from 22°C to 42°C by two different processes. In one process, the temperature increases at constant volume, and in the other process the temperature increases at constant pressure. Which process requires more heat or are the required amount of heat same in both?

Answers

Answer:

a- More heat is required for the constant-pressure process than for the constant-volume

Explanation:

we have to solve using the thermodynamic first law. this is the heat applied to the system

dQ = dU + dW

definition of terms:

dU = change in internal energy

dW = work done

we have it that

change in internal energy dU is directly proportional to work done dW

but when we are in constant volume process, work done of the gas is zero

therefore

dQ of constant pressure is > than that of constant volume

so constant pressure process requires more heat

The process that requires more heat is the constant-pressure process than the constant-volume process.

According to the first law of thermodynamics, the heat that's applied to the system will be the addition of the change in internal energy and the work done.

In a constant-volume process, the work done on the gas is equal to zero. More heat will be required for the constant-pressure process than for the constant-volume process.

Also, it should be noted that the change in the thermal energy of the gas will be the same for the constant-pressure process and the constant-volume process.

Read related link on:

https://brainly.com/question/16951562

A motorcyclist goes around an un-banked (i.e., flat) circular turn of radius 31m, at a constant speed of 110km/hr (convert this to m/s). What is the minimum coefficient of static friction needed to keep the tires from slipping? Explain why the answer is (or is not) plausible.

Answers

Answer:

[tex]\mu_{s} = 3.071[/tex]

This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.

Explanation:

From Second Newton's Law we understand that centripetal acceleration experimented by motocyclist is due to force derived from static friction. And normal force of the ground on motocyclist equals weight of motocyclist due to the flatness of circular turn. The equations of equilibrium of the motocyclist is:

[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (Eq. 1)

[tex]\Sigma F_{y} = N-m\cdot g = 0[/tex] (Eq. 2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

[tex]m[/tex] - Mass of the motocyclist, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v[/tex] - Speed of the motorcyclist, measured in meters per second.

[tex]R[/tex] - Radius of the circular turn, measured in meters.

The static coefficient of friction is cleared in (Eq. 1):

[tex]\mu_{s} = \frac{m\cdot v^{2}}{N\cdot R}[/tex]

From (Eq. 2) we get that normai force is:

[tex]N = m\cdot g[/tex]

And we expand the resulting expression in (Eq. 1):

[tex]\mu_{s} = \frac{m\cdot v^{2}}{m\cdot g\cdot R}[/tex]

[tex]\mu_{s} = \frac{v^{2}}{g\cdot R}[/tex] (Eq. 3)

If we know that [tex]v = 30.556\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 31\,m[/tex], the expected static coefficient of friction is:

[tex]\mu_{s} = \frac{\left(30.556\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (31\,m)}[/tex]

[tex]\mu_{s} = 3.071[/tex]

This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.

what is the meaning of relative as a noun?

Answers

Answer:

noun. a person who is connected with another or others by blood or marriage. something having, or standing in, some relation or connection to something else. something dependent upon external conditions for its specific nature, size, etc. (opposed to absolute).

An airline employee tosses two suitcases in rapid succession with a horizontal velocity of 7.2 ft/s onto a 50-lb baggage carrier which is initially at rest. Problem 14.003.a Conservation of momentum: two colliding suitcases Knowing that the final velocity of the baggage carrier is 4.8 ft/s and that the first suitcase the employee tosses onto the carrier has a weight of 30 lb, determine the weight of the other suitcase. (You must provide an answer before moving on to the next part.) The weight of the other suitcase is lb.

Answers

Answer:

m₁ = 70 lb

Explanation:

Here we will use the law of conservation of momentum:

m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃

where,

m₁ = mass of first suitcase = ?

m₂ = mass of second suitcase = 30 lb

m₃ = mass of baggage carrier = 50 lb

u₁ = initial speed of first suitcase = 7.2 ft/s

u₂ = initial speed of second suitcase = 7.2 ft/s

u₃ = initial speed of baggage carrier = 0 ft/s

v₁ = Final speed of first suitcase = 4.8 ft/s

v₂ = Final speed of second suitcase = 4.8 ft/s

v₃ = Final speed of baggage carrier = 4.8 ft/s

because after collision all three will have same speed

Therefore,

(m₁)(7.2 ft/s) + (30 lb)(7.2 ft/s) + (50 lb)(0 ft/s) = (m₁)(4.8 ft/s) + (30 lb)(4.8 ft/s) + (50 lb)(4.8 ft/s)

(m₁)(7.2 ft/s) + (216 lb ft/s) + (0 lb ft/s) = (m₁)(4.8 ft/s) + (144 lb ft/s) + (240 lb ft/s)

(m₁)(7.2 ft/s) - (m₁)(4.8 ft/s) = 168 lb ft/s

m₁ = (168 lb ft/s)/(2.4 ft/s)

m₁ = 70 lb

Find the work done by a 75.0 kg person in climbing a 2.50 m flight of stairs at a constant speed.

Answers

Answer:

1,839.375 Joules

Explanation:

Work is said to be done is the force applied to an object cause the object to move through a distance.

Workdone = Force * Distance

Workdone = mass * acceleration due to gravity * distance

Given

Mass = 75.0kg

acceleration due to gravity = 9.81m/s²

distance = 2.50m

Substitute the given parameters into the formula:

Workdone = 75.0*9.81*2.50

Workdone = 1,839.375Joules

Hence the workdone is 1,839.375 Joules

An 80-kg bungee jumper jumps off a bridge. Rubber bungee cords act as a large spring attaching the jumper to the bridge. A bear standing in the river below catches the jumper. If the spring constant of the bungees is 20 N/m and they stretch 50 m. How much force must the bear apply to keep the jumper from moving?

Answers

Answer:

The force is [tex]F_b =  216 \  N [/tex]

Explanation:

From the question we are told that

   The  mass of the bungee jumper is  m  = 80 kg

    The spring constant is  [tex]k = 20 \ N/ m[/tex]

    The extension of the rubber bungee cords is  x = 50 m

Generally the weight of the jumper is

      [tex]W =  m *  g[/tex]

=>   [tex]W =  80 *  9.8 [/tex]

=>   tex]W =  784  \  N  [/tex]

Generally  the returning force of the rubber bungee cords is mathematically represented as

      [tex]F  =  k  *  x[/tex]

=>     [tex]F  =  20   * 50 [/tex]

=>     [tex]F  =  1000 \  N [/tex]

The force to be applied by the bear is

    [tex]F_b =  F -  W[/tex]

=>    [tex]F_b =  100 -  784[/tex]

=>  [tex]F_b =  216 \  N [/tex]

Alejandro made 6.4 liters of punch using half apple juice and half orange juice. How many milliliters of apple juice are in the punch?

Answers

Answer:

3.2

Explanation:

I hope that this helps! Have a good day!!

How many strings of length 10 over the alphabet (a, b, c, d] have at least one b somewhere in the string?
a) 310
b) 410 - 310
c) 10.4
d) 10.39

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  B

Explanation:

   The number  of alphabet is  n= 4  (a , b , c , d )

Generally the total  number of  string of length 10 over the 4 alphabets is  

     [tex]N  =  4^{10}[/tex]

Gnerally the number of string of length 10 that does not include b is  

     [tex]T =  3^{10}[/tex]    

Generally the number of strings of length 10 over the 4  alphabets that have at least one alphabet b  somewhere in the string is  

        [tex]G  =  N - T[/tex]

=>    [tex]G  =  4^{10} -  3^{10}[/tex]

A 12 kg bowling ball would require what force to accelerate it down an alley at a rate of 2.5 m/s ^ 2

Answers

Answer:

hi

Explanation:

hiijjjjjjjjjjjjjjj

A region around the nucleus of an atom where electrons are likely to be found​

Answers

Answer:

The region where an electron is most likely to be is called an orbital. Each orbital can have at most two electrons. Some orbitals, called S orbitals, are shaped like spheres, with the nucleus in the center.

Explanation:Hope this helps :)

By definition, a region around the nucleus of an atom where electrons are likely to be found​ is called an orbital.

First of all, an atom is the smallest constituent unit of ordinary matter that has the properties of a chemical element.

All atoms are made up of subatomic particles: protons and neutrons, which are part of their nucleus, and electrons, which revolve around them. Protons are positively charged, neutrons are neutrally charged, and electrons are negatively charged.

In other words, every atom consists of a nucleus in which neutrons and protons meet and energy levels where electrons are located.

This is, the atomic nucleus is the central part of the atom that is made up of protons and neutrons, while the orbitals or peripheral region is an area where electrons are found.

In summary, a region around the nucleus of an atom where electrons are likely to be found​ is called an orbital.

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as a result, the net electric force experienced by each negatively charged particle is reduced to F/2. The value of q is

Answers

Answer:

The value of q is [tex]\dfrac{Q}{8}[/tex]

Explanation:

Given that,

Each charge = -Q

Distance between charges = L

Reduced force = [tex]\dfrac{F}{2}[/tex]

Suppose, Two particles each with a charge -Q are fixed a distance L apart as shown above. Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the original two particles.

We know that,

The force on each end is

[tex]F=\dfrac{kQ^2}{L^2}[/tex]...(I)

If the charge q is placed at mid point then

The  force on each end charge is

[tex]\dfrac{F}{2}=F+F'[/tex]....(II)

We need to calculate the value of q

Using equation (II)

[tex]\dfrac{F}{2}=F+F'[/tex]

Put the value of F into the formula

[tex]\dfrac{\dfrac{kQ^2}{L^2}}{2}=k\dfrac{Q^2}{L^2}+k\dfrac{q\times(-Q)}{(\dfrac{L}{2})^2}[/tex]

[tex]\dfrac{kq(-Q)}{(\dfrac{L}{2})^2}=-\dfrac{kQ^2}{2L^2}[/tex]

[tex]\dfrac{q}{\dfrac{1}{4}}=\dfrac{Q}{2}[/tex]

[tex]q=\dfrac{Q}{8}[/tex]

Hence, The value of q is [tex]\dfrac{Q}{8}[/tex]

Given that water at standard pressure freezes at 0∘C, which corresponds to 32∘F, and that it boils at 100∘C, which corresponds to 212∘F, calculate the temperature difference ΔT in degrees Fahrenheit that corresponds to a temperature difference of 1 K on the Kelvin scale. Give your answer to two significant figures.

Answers

Answer:

In two significant figure 360K

Explanation:

The temperature difference (ΔT) can be calculated as the boiling temperature minus the freezing temperature in Fahrenheit.

Hence,

ΔT = 212 - 32

ΔT = 180°F

To convert to °F to kelvin, we use the formula below

= (°F - 32) × 5/9 + 273.15

= (180°F - 32) × 5/9 + 273.15

= 355.37K ⇔ 360K

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