An object is projected horizontally with the velocity 4m/s from a height of 20 m. the horizontal displacement is (angle x) :

An Object Is Projected Horizontally With The Velocity 4m/s From A Height Of 20 M. The Horizontal Displacement

Answers

Answer 1

Answer:

Δx = 8 m

Explanation:

The time needed to fall from vertical rest

h = ½gt²

t = √(2h/g)

t = √(2(20)/9.8)

t = 2.0 s

the horizontal distance traveled in that time is

d = vt

d = 4(2)

d = 8 m


Related Questions

The use of non-renewable energy resources in the UK has changed in the last 30 years.
Explain how the use of energy resources has changed in the last 30 years.

Answers

Answer: Varies

Explanation: It became more relied upon, technology is responsible for this.

How can gravity's role in tectonic plate motion be described?

Answers

Answer: In ridge push, the mantle wells upward because of the convection and elevates the edges of spreading oceanic plates. Because these plates are higher at the spreading center, they are forced downhill due to gravity and eventually flatten out to the ocean floor.

the answer I need points for my math test

i need help guys please how to do it

Answers

Answering your question but a bit confusing lovesur

As a mouse sits on a desk, it has a gravitational force of 1x10^-9N pulling it towards the keyboard. If a different mouse with triple the mass was used, instead what would be the new gravitational force?

Answers

Answer:

The new gravitational pull would be three times as strong as the first which would be 0.000000003 or 3 × 10^-9N.

Explanation:

Hope this helps. . . <3

A hot air balloon rising vertically is tracked by an observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment

Answers

We have that for the Question, it can be said that

the balloon rising at [tex]0.266miles/min[/tex]

From the question we are told

An observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is , and it is changing at a rate of 0.1 rad/min.

From,

[tex]tan\theta = \frac{h}{2}[/tex]

differentiate with respect to h

[tex]sec^2\theta * \frac{do}{dz} = \frac{1}{2} * \frac{dh}{dz}\\\\\frac{dh}{dz} = 2 sec^\theta * \frac{d\theta}{dz}\\\\\theta = \frac{\pi}{6} and \frac{d\theta}{dz} = 0.1rad/min\\\\\frac{dh}{dz} = 2sec^2 (\frac{\pi}{6}) * (0.1)\\\\= 0.266miles/min[/tex]

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An astronaut travels to a star system 6.5 ly away at a speed of 0.80c . Assume that the time needed to accelerate and decelerate is negligible.How much time elapses in earth between the launch and the arrival of the first radio message from the astronaut saying that she has arrived? please 911

Answers

Answer:

14.625 years

Explanation:

Because acceleration/deceleration are negligent we can simplify this problem into 2 basic parts:

First: how long before the astronaut arrives? Since the speed of the craft is 0.8 times the speed of light and the distance to travel is 6.5 light-years, the equation to answer this part is:

0.8t=6.5

solve for t by dividing by 0.8

t=6.5/0.8 or 8.125

so, she will arrive in 8.125 years.

Then, once she sends her message, it will travel towards earth at 1 times the speed of light or:

1.0t=6.5 which is just t=6.5

so, her message will arrive back to Earth (8.125+6.5) 14.625 years after takeoff.

The time elapsed in earth will be 14.625 years.

To understand this we need to find how much time she took to travel 6.5 light years and then how much time the signal took to reach the earth.

How much time she took to travel 6.5 light year?

 0.8t = 6.5 ( because 1 light year = distance travelled by light in a year)

      t = 6.5/0.8

      t = 8.125 years

So, she travelled for 8.125 year.

How much time it took for the signal?

1.0 t = 6.5 ( because radio signal travells with the same speed of light)

     t  = 6.5/1.0

     t  = 6.5 years

Sum of these two gives us how much time elapsed on earth between two mentioned events.

So, 8.125 + 6.5 = 14.625 years

So, the time elapsed on earth between the launch and the arrivel of the first radio signal is 14.625 years.

#SPJ2

A car drives at a constant speed of 45km in 20mn. The speed of the car is..?​

Answers

Answer:

83.8851 mi/h              

The refractive index of water with respect to air is 5/3 and that of glass is 3/2.What is the refractive index glass with respect to water?

Answers

Answer:

9/8

Explanation:

50 points help

Column I Column II

______ 1. acceleration a. change in distance over time

______ 2. speed b. time interval

______ 3. velocity c. scalar

______ 4. Δt. d. change in position

______ 5. Magnitude only e. change in velocity over time

______ 6. Δx f. change in displacement over time

Answers

[tex]\\ \sf\longmapsto Acceleration\longrightarrow Change\:in\: Velocity\:over\:time[/tex]

[tex]\\ \sf\longmapsto Speed\longrightarrow Change\:in\: Distance\:over\:Time[/tex]

[tex]\\ \sf\longmapsto Velocity\longrightarrow Change\:in\: Displacement\: over\:time[/tex]

[tex]\\ \sf\longmapsto ∆t\longrightarrow Time\: interval [/tex]

[tex]\\ \sf\longmapsto Magnitude\:only\longrightarrow Scaler[/tex]

[tex]\\ \sf\longmapsto ∆x=Change\:in\: position [/tex]

The figure shows the light intensity on a screen behind a double slit. The slit spacing is 0.22 mm and the screen is 2.0 m behind the slits (Figure 1). What is the wavelength of the light?

Answers

The wavelength of the light is 550 nm

For a double slit interference pattern with slit spacing, d we have

dsinθ = mλ where d = slit spacing = 0.22 mm = 0.22 × 10⁻³ m, m = number of maximum fringe = 2(from the picture) and λ = wavelength of light.

Thus sinθ = mλ/d

Also, tanθ = L/D where L = distance between central maximum and fringe = 2.0 cm/2 = 1.0 cm = 1 × 10⁻² m and D = distance between slit and screen = 2.0 m

Since θ is small, sinθ ≅ tanθ

So, mλ/d = L/D

Making λ subject of the formula, we have

λ = dL/mD

Substituting the values of the variables into the equation, we have

λ = dL/mD

λ = 0.22 × 10⁻³ m × 1 × 10⁻² m/(2 × 2.0 m)

λ = 0.22 × 10⁻⁵ m²/4.0 m

λ = 0.055 × 10⁻⁵ m

λ = 0.55 × 10⁻⁶ m

λ = 550 × 10⁻⁹ m

λ = 550 nm

So, the wavelength of the light is 550 nm

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If a car travels 10 min 2 seconds, what is its speed?
A. 5 m/s
B. 0.2 m/s
C. 20 m/s
D. 2 m/s

Answers

If you meant the car travels 10 meters or miles then the answer would be 5m/s because v=d/t (v=10/2)

What advantage is there in measuring your distance from I1 to I1' and dividing by 2 rather than measuring the distance OI1 directly

Answers

Measuring the distance from  [tex]I_1 \ to \ I_1'[/tex] and dividing by 2 has advantage of achieving a more accurate result compared to measurement taking directly.

Measuring the distance from [tex]I_1 \ to \ I_1'[/tex] is equivalent to the double of the distance.

This method of measurement gives a more accurate result because it minimizes error that may result from taking readings form the intervals. It ensures more accurate result compared to the measurement obtained when [tex]OI_1[/tex] is measured directly.

[tex]OI_1 = \frac{I_1 + I_1'}{2}[/tex]

Thus, we can conclude that measuring the distance from  [tex]I_1 \ to \ I_1'[/tex] and dividing by 2 has advantage of achieving a more accurate result compared to measurement taking directly.

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PLEASE HELP!!!


If a 40cm rope with a 220g bob can hold a maximum tension of 3N
a) what are the maximum angular velocity and inclination angle it can reach before the rope break?
b) Angle of inclination​

Answers

Answer:

ω = 3.1 rad/s

θ = 36° from vertical

Explanation:

I will ASSUME that the bob and string is acting as a pendulum.

Please understand that the string will break when the bob is at the lowest point of the swing where the vectors of gravity and centripetal acceleration align. It will NOT break at the angle of maximum inclination measured from vertical. This angle is only a component of the maximum potential energy that gets converted to maximum kinetic energy at the lowest point of the swing.

At the bottom of the swing, the string must support the weight of the bob plus supply the required centripetal acceleration.

F = mg + mω²R

F/m = g + ω²R

F/m - g = ω²R

ω = √((F/m - g)/R)

ω = √((3/0.220 - 9.8)/0.40)

ω = 3.09691...

ω = 3.1 rad/s

Potential energy will convert to kinetic energy

       mgh = ½mv²

             h = v²/2g

R - Rcosθ = v²/2g

R(1 - cosθ) = v²/2g

   1 - cosθ = v²/2gR

        cosθ = 1 - v²/2gR

        cosθ = 1 - (Rω)²/2gR

        cosθ = 1 - Rω²/2g

        cosθ = 1 - 0.40(3.1²)/(2(9.8))

        cosθ = 0.804267

              θ = 36.46045...

              θ = 36°

What does it mean that " memory is organized in semantic
networks

Answers

Answer:

Semantic memory is a category of long-term memory that involves the recollection of ideas, concepts and facts commonly regarded as general knowledge. Examples of semantic memory include factual information such as grammar and algebra.

HELP PLS! :/

A ball thrown straight up takes 1.89s to reach a height of 41.6m.
Taking g =9.81 m/s2 the balls initial speed was:

Help plsss

Answers

Explanation:

formula: Vi = Vf - (at)

Vi: intial velocity

Vf: final velocity

a: acceleration

t: time

fill in formula with the numbers you are given

Vi= 41.6m - ((9.81 m/s^2)(1.89s))

parenthesis first according to pemdas

Vi= 41.6m - 18.54m/s

Answer: 23.06m/s

disclaimer: I havent done physics in awhile so I have no idea if this is right. just an attempt to help steer you in the right direction hopefully. good luck

which of the following is used to transport sound waves
A.medium
B.vacuum
C.mass
D.light​

Answers

I think the answer is d.right?

Change 1m2 in to cm2, mm2 and km2​

Answers

Answer:

1m² = 10000cm²

1m² = 1000000mm²

1m² = 1 × 10^-6

Explanation:

When coverting from square meter to square centimetre, multiply the area value by 10.

When coverting from square meter to millimetre, multiply the area value by 10⁶ ( 1 million )

When converting from meter square to kilometre square, divide the area value by 10^-6 ( 0.000001)

PLEASE PLEASE HELP!!

An arrow is shot straight up in the air from the ground with an initial velocity of 54.0 m/s. If on striking the ground it
embeds itself 15.0 cm into the ground, what is the acceleration required to stop the arrow when it hits the ground?

Answers

Answer:

you have patience the distance.

Explanation:

the train leaves at 6.30.

How long does Sun remain in one zodiac?

(A) 1 year (B) 1 month (C) 1 week (D) 1 day​

Answers

A. The sun takes a year to complete a zodiac cycle

Answer:  (B) 1 month

Brainliest if correct pls!

Please help if u can thanksss

Answers

Answer:

yes its okay i think so it is the correct answer

What is metamarket ?

Answers

Answer:

metamarket (plural metamarkets) A group of businesses that offer products that are related from a consumer's perspective but which have no institutional connections. quotations . A market that trades in the medium of exchange of a lower-level market, such as money, derivatives, or credit. quotations.

Explanation:

The best examples of Meta Marketing can be selling family planning ideas or the idea of prohibition. ... Let's say a car selling in a Meta market would be a website, that sells cars but you will also find car parts there, add-ons for cars, colours for cars, mechanic's reviews, etc.

A group of businesses that offer products that are related from a consumer's perspective but which have no institutional connections.

5. A quarterback throws the football to a stationary receiver who is 31.5 m
down the field. If the football is thrown at an initial angle of 40.0° to the
ground, at what initial speed must the quarterback throw the ball for it
to reach the receiver? What is the ball's highest point during its flight?

Answers

The projectile launch equations allow to find the results for the questions about the movement of the ball are:

The initial velocity is:   v₀ = 17.7 m / s. The maximum height is:   y = 16 m.

Given parameters

Horizontal distance x = 31.5 m Launch angle tea = 40º

To find

The initial speed. Maximum height.

Projectile launching is an application of kinematics, where on the x-axis there is no acceleration and on the y-axis is the gravity acceleration.

The range is the distance traveled for the same departure height, see attached.

.

          R =[tex]\frac{v_o^2 \ sin 2\theta}{g}[/tex]  

         [tex]v_o^2 = \frac{ g R}{sin 2 \theta }[/tex]  

Let's calculate.

          v₀² = [tex]\frac{9.8 \ 31.5}{sin \ (2 \ 40)}[/tex]9.8 31.5 / sin (2 40.0)

          [tex]v_o = \sqrt{313.46}[/tex]o = ra 313.46

          v₀ = 17.7 m / s

At the point of maximum height the vertical speed is zero.

          v² = v₀² - 2 g y

          0 = v₀² - 2g y

          y = [tex]\frac{v_o^2}{2g}[/tex]  

Let's calculate.

         y = [tex]\frac{17.7^2}{2 \ 9.8}[/tex]  

         y = 16 m

In conclusion, using the projectile launch equations we can find the results for the questions about the movement of the ball are:

The initial velocity is: v₀ = 17.7 m / s The maximum height is:  y = 16 m.

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Lakepointe Tower in Chicago is the tallest apartment building in the United States (although not the tallest building in which there are apartments). Suppose you take the elevator from street level to the roof of the building. The elevator moves almost the entire distance at constant speed, so that it does 1.15x10^5 J of work on you as it lifts the entire distance. If your mass is 60.0 kg, how tall is the building? Ignore the effects of friction.

Answers

Given the information supplied in the question, the height of the building is  196 m.

The work done on you is a work done in a gravitational field. Recall that the work done in a gravitational field is given by mgh where;

m = mass of the body

g = acceleration due to gravity

h = height

Given that;

Work done = 1.15x10^5 J

Acceleration due to gravity = 9.8 ms-2

Mass =  60.0 kg

Substituting values;

h = W/mg

h =  1.15x10^5 J/ 60.0 kg ×  9.8 ms-2

h = 196 m

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Does the stone remain at a constant speed? Or does it speed up?

Answers

Answer:

it would remain the same speed

Explanation: the rock isnt going down a hill or anything so therefore if gravity isnt pulling it down a slope then it would remain the same pace

Please write a paragraph explaining the bible verse below in your own words.


Exodus 16:1-3

Answers

Answer: Moses

Explanation: The Israelites are angry at Aaron and Moses, because they lead them into this desert where there is no food, or water. It was told that after the Israelites left Egypt they would wander the desert for forty years. The Israelites complain that they would have had plenty of food if they had just stayed enslaved in Egypt. They are not grateful that God has liberated them from slavery.

A bear slides down an incline which is oriented at 17 degrees above horizontal. What is the coefficient of friction between the bear and the incline if the bear moves with a constant velocity?

Answers

[tex]\mu = 0.306[/tex]

Explanation:

Assume that the direction down the incline is the +x-direction. We can apply Newton's 2nd law along the x-axis as

[tex]x:\:\:\:\:\:mg\sin17° - f_s = ma[/tex]

[tex]\Rightarrow mg\sin17° - \mu N = 0[/tex]

where m is the mass of the bear and g is the acceleration due to gravity and acceleration a is zero since the bear is moving with a constant velocity. Along the y-axis, we can write Newton's 2nd law as

[tex]y:\:\:\:\:\:N - mg\cos17° = 0 \Rightarrow N = mg\cos17°[/tex]

Combining these two equations together, we get

[tex]mg\sin17° = \mu(mg\cos17°)[/tex]

Solving for [tex]\mu, [/tex]

[tex]\mu = \dfrac{\sin17°}{\cos17°} = \tan17° = 0.306[/tex]

7 A 0.25 kg block oscillates on the end of a spring with a spring constant of 100 N/m. If the oscillation is started by elongating the spring 0.1m and giving the block a speed of 3 mis then the amplitude of the oscillation is:​

Answers

(K+U)
i

=(K+U)
f



0+
2
1

kA
2
=
2
1

mv
2
+
2
1

kx
2



2
1

(6.50 N/m)(0.100 m)
2
=
2
1

m(0.300 m/s)
2
+
2
1

(6.50 N/m)(5.00×10
−2
m)
2


3.25×10
−2
J=
2
1

m(0.300 m/s)
2
+8.12×10
−3
J

giving m=
9.0×10
−2
m
2
/s
2

2(2.44×10
−2
J)

=
0.542 kg



(b) ω=
m
k



=
0.542 kg
6.50 N/m



=3.46 rad/s

Then, T=
ω


=
3.46 rad/s
2π rad

=
1.81 s



(c) a
max

=Aω
2
=(0.100 m)(3.46 rad/s)
2
=
1.20 m/s
2

A 770-kg two-stage rocket is traveling at a speed of 6.90×103 m/s away from Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a speed of 2.60×103 m/s relative to each other along the original line of motion.What is the speed of each section (relative to Earth) after the explosion?How much energy was supplied by the explosion?

Answers

Answer:

Explanation:

Let's just have our reference frame travel along with the original un broken mass. This way the original velocity is not relevant.

Each half will have a mass of 770/2 = 385 kg

Each half will have the same magnitude of velocity (conservation of momentum) which will be 2.6 x 10³/2 = 1.30 x 10³ m/s

Now add back the reference frame velocity to get velocity relative to earth.

Section one will have velocity 6.90 x 10³ + 1.30 x 10³ = 8.2 x 10³ m/s

Section two will have velocity 6.90 x 10³ - 1.30 x 10³ = 5.6 x 10³ m/s

In the moving reference frame, each half will have kinetic energy which could only come from the explosion

KE = ½(385)(1.3 x 10³)² = 325,325,000 J

2(325,325,000) = 650,650,000 J released in the explosion.

Rounding to the three significant figures of the problem numerals

E = 6.50 x 10⁸ J or 650 MJ released

True or false. The buildings in cities are getting larger

Answers

Answer:

true

Explanation:

cuase the city is getting more people than before

An airplane has a mass of 33,000 kg and takes off under the influence of a constant net force of 44,000 N. What is the net force that acts on the plane's 70 kg pilot

Answers

Answer:

Explanation:

acceleration a = F/m = 44000/33000 = 1.3m/s²

F = ma = 70(1.3) = 93 N

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