"A soccer ball is kicked horizontally off a 22 m high hill and lands a distance of 35 m from the edge of the hill." Which variable is 35 m? *

Answers

Answer 1

Answer:Let’s assume that, after the soccer ball is kicked and moves through its trajectory, it first makes contact with level ground a horizontal distance of 35 meters from where it was kicked. Let’s also assume that we can neglect air resistance. The time, t, that the soccer ball is in the air until it first contacts the ground can be found from the equation h = (1/2)gt^2 which can be rewritten as t = sqrt(2h/g) where h is the vertical distance the ball falls which is the height of the hill since the ball was kicked horizontally, and g is the acceleration of gravity which is 9.8 m/s^2. So t = sqrt(2(22)/9.8) = 2.12 seconds. In that time, the ball travelled 35 meters so its horizontal velocity was 35 meters/2.12 seconds = 16.5 meters/second.

Explanation:


Related Questions

A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency of the first harmonic mode (m

Answers

Answer:

Explanation:

f = [tex]\sqrt{T/(m/L)} / 2L[/tex]

T = 120 N

L = 3.00 m

(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m

                                                  (wow that's massive for a "rope")

f = [tex]\sqrt{120/12} /(2(3))[/tex])

f = [tex]\sqrt{10\\}[/tex]/6 = 0.527 Hz

This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.

A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N

What causes the difference in the angle of the sun on the Earth's surface throughout the year?

Answers

Answer:

The axis is tilted and points to the North Star no matter where Earth is in its orbit. Because of this, the distribution of the Sun's rays changes. ... It also means that the angle at which sunlight strikes different parts of Earth's surface changes through the year.

Explanation:

Pls sub to bdoggaming if this helped

Define the term dimension

Answers

Answer:

Q1. A measurable extent of a particular kind, such as length, breadth, depth, or height.

Q2. A dimensional constant is a physical quantity that has dimensions and has a fixed value. Some of the examples of the dimensional constant are Planck's constant, gravitational constant, and so on.

Q3. Physical quantities which posses dimensions and have variable are called dimensional variables. Examples are length, velocity, and acceleration etc.

Q4. Dimensionless variables are the quantities which doesn't have any dimensions the the value is a variable. Eg: angle = arc/ radius. Dimensions = L/L. = 1. So angle does not have any dimensions and the value can vary.

Q5. Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another.

Q6. Dimensional analysis has been around a long time, Newton called it the "Great principle of Similitude", but the modern form can be traced back to James Clerk Maxwell. It was Maxwell who distinguished mass [A/], length [£], and time [7"] as the independent dimensions from which others could be derived.

Q7. Mass, length, time, temperature, electric current, amount of light, and amount of matter.

Q8. Dimensional analysis is used to convert the value of a physical quantity from one system of units to another system of units. Dimensional analysis is used to represent the nature of physical quantity. The expressions of dimensions can be manipulated as algebraic quantities.

Hope that helps. x

have a difinite shape and do not easily take the shape of their containers

Answers


Ok well that’s something to think about

yayy here you are f, r, e, e, p, o, i, n, t, s

Answers

Answer:

Albert Einstein Albert Einstein was a German-born theoretical physicist, widely acknowledged to be one of the greatest physicists of all time. Einstein is best known for developing the theory of relativity, but he also made important contributions to the development of the theory of quantum mechanics.

Explanation:

Thank you so much buddy !!

A mars surface exploration vehicle drops a rock off a 1.00 I'm high vertical Cliff. The sound of the rock landing at the base of the cliff is recorded by instruments on the vehicle 27.1 seconds later. Calculate the acceleration due to gravity on Mars given that the speed of sound on Mars is 320 m/s

Answers

The acceleration due to gravity on Mars is 11.81 m/s².

The given parameters:

Height of the cliff, h = 1 mTime of motion of the sound wave, t = 27.1 sSpeed of sound in mass, v = 320 s

The equation of motion to determine the acceleration due to gravity on the moon is calculated as follows;

[tex]s = vt + \frac{1}{2} gt^2[/tex]

where;

s is the distance traveledt is the time of motion

Since the time measured is two way time, the new equation for the total distance traveled is calculated as;

[tex]v = \frac{2d}{t} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{320 \times 27.1}{2} \\\\d = 4,336 \ m[/tex]

The acceleration due to gravity is calculated as follows;

[tex]s = vt + \frac{1}{2} gt^2\\\\4,336 = 0 \ + \ \frac{1}{2} \times g \times (27.1)^2\\\\4,336 = 367.21g\\\\g = \frac{4,336}{367.21} \\\\g = 11.8 1 \ m/s^2[/tex]

Thus, the acceleration due to gravity on Mars is 11.81 m/s².

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An object accelerates from rest to 93 m/s over a distance of 49 m. What acceleration did it experience?

Answers

Answer:

Explanation:

acceleration=  change in velocity/time taken

acceleration=  93/49

=2.02

A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m when it begins to roll down the incline. If rolling and sliding friction are neglected, what is the linear velocity, in m/s, of the center-of-mass of the wheel when it reached the bottom of the incline?

Answers

Answer:

Explanation:

If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.

mgh = ½mv²

v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s

However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½(½mR²)(v/R)²

2gh = v² + ½v²

2gh = 3v²/2

v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s

If a 35 kg box collides with a stationary 120 kg box with a force of 90 N, what must be true of the magnitude of the reaction force?

Answers

Newton's third law allows to find the result for the value of the reaction force during the collision is:

The reaction force is F = 90 N and is applied to the lighter body.

Newton's third law stable that the forces appear in pairs or ea that when two bodies interact, the interaction forces appear in the two bodies simultaneously, in general they are called action and reaction forces.

These furas are of the same magnitude, but in the opposite direction, each one applied to one of the bodies.

They indicate that the most lighter body collides with the one with the greatest mass with a force of F = 90 N. If we call this the action, the larger body must react with a force of equal magnitude on the lighter body.

Consequently, the reaction force is F = 90 N directed towards the lighter body.

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The diagram below shows a 5.00-kilogram block
at rest on a horizontal, frictionless table.
5.00-kg
block
Table
Which of the following is the correct name and strength of the force holding the block up?

Answers

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

Mass of the block, m = 5 kg

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

g is acceleration due to gravity = 10 m/s²

W = 5 x 10

W = 50 N (downwards)

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N (upwards)

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

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The block will remain on the table because the normal force balances with the weight of the block. The correct answer is  50 N upward normal force

From the diagram shown a 5.00-kilogram block at rest on a horizontal, frictionless table. The weight of the block will act downward which will be

Weight W = mg

let g = 10 m/[tex]s^{2}[/tex]

W = 5 x 10

W = 50 N

The block will also produce an equal but in opposite direction of a normal force which is equal to the weight of the block. That is,

Normal force N = 50 N

The block will remain on the table because the normal force balances with the weight of the block.    

Therefore, the correct name and strength of the force holding the block up is 50 N upward normal force.

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A 21 newton force keeps a 3 kg object in uniform circular motion. The speed of the object is 9 m/s. The magnitude of the centripetal acceleration is


3 m/s^2
7 m/s^2
27 m/s^2
30 m/s^2

Answers

Answer:

Explanation:

F = ma

a = F/m = 21/3 = 7 m/s²

Explain how the linear rate spring operates?

Answers

A linear spring has the same diameter along its entire length, and this uniform diameter gives it a constant spring rate.

A wagon of dog treats (combined mass 55 kg) is rolling at 2.1 m/s. A dog with mass 21 kg dives into the wagon, colliding with just enough momentum to make both stop. If the collision between the dog and the wagon lasts 0.1 s, what is the magnitude of the average force that will be exerted on the dog by the collision with the wagon

Answers

Answer:

Explanation:

An impulse results in a change of momentum

If the wagon and dog both stop, they must have had equal and opposite momentums

FΔt = mΔv

F = mΔv/Δt = m(v₁ - v₀)/(t₁ - t₀)

v₁ = t₀ = 0

F = m(v₀)/t₁

F = 55(2.1)/0.1 = 1155 N

We could have also figured the dog's initial velocity and used the dog's mass in the equation as well. Result would be identical.

During take-off a 8kg model rocket is burning fuel causing its speed to increase
at a rate of 4m/s2 despite experiencing a 90N drag.

What’s is the strength of the thrust?
(Answer unit is in N)( and the answer isn’t 212)

Answers

The strength of the thrust is 122 newtons.

The motion of the rocket is described by the second Newton's law, whose model is shown below:

[tex]\Sigma F = F - D = m\cdot a[/tex] (1)

Where:

[tex]F[/tex] - Thrust, in newtons[tex]D[/tex] - Drag, in newtons[tex]m[/tex] - Mass of the rocket, in kilograms[tex]a[/tex] - Net acceleration of the rocket, in meters per square second

If we know that [tex]D = 90\,N[/tex], [tex]m = 8\,kg[/tex] and [tex]a = 4\,\frac{m}{s^{2}}[/tex], then the strength of the thrust is:

[tex]F = D + m\cdot a[/tex]

[tex]F = 90\,N + (8\,kg)\cdot \left(4\,\frac{m}{s^{2}} \right)[/tex]

[tex]F = 122\,N[/tex]

The strength of the thrust is 122 newtons.

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explain the process of convergence and divergence ! HELPPP

Answers

Answer:

Divergence generally means two things are moving apart while convergence implies that two forces are moving together. ... Divergence indicates that two trends move further away from each other while convergence indicates how they move closer together.

Explanation:

i just want an answer please

Answers

Answer: An answer on what? I’ll never ignore you!

Explanation:

Answer:

an answer on what?

Explanation: Im here to help!!

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!pls

Answers

I think this is the solution:

1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
4: In-9, N-1
5: N-2, H-4, S-1, C-1
6: Ge- 15, N-4
7: N-1, H-4, C-1, I-1, O-3
8: H-7, F-1
9: N-1, O-5, H-1, S-1
10: H-8
11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4

The planar simple harmonic wave travels in the positive direction of x axis with wave velocity u=2m/s, and the vibration curve of the particle at the origin in cosinusoidal form is shown in the figure.
Try to find (1) the vibration function of the particle at the origin, (2) the wave function of the planar simple harmonic wave according to the origin.

Answers

The planar simple harmonic wave travels in the positive direction of x axis with wave velocity u=2m/s, and the vibration curve of the particle at the origin in cosinusoidal form is shown in the figure.

Try to find (1) the vibration function of the particle at the origin, (2) the wave function of the planar simple harmonic wave according to the origin.

Answer:

Figure 16.8 The pulse at time

t

=

0

is centered on

x

=

0

with amplitude A. The pulse moves as a pattern with a constant shape, with a constant maximum value A. The velocity is constant and the pulse moves a distance

Δ

x

=

v

Δ

t

in a time

Δ

t

.

The distance traveled is measured with any convenient point on the pulse. In this figure, the crest is used.

I need a short answer ?

Answers

Answer:

Explanation:

7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s

7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m

  or

  h = (45sin14.5)² / (2(9.81)) = 6.47 m

which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.

A 200-kg, 2.0-m-radius, merry-go-round in the shape of a flat, uniform, circular disk parallel to level ground is rotating at 1.2 cycles/second about an axis through its center of mass and perpendicular to the ground. A 50-kg boy jumps onto the edge of the merry-go round and lands at a fixed point. What is the angular velocity of the merry-go-round after the boy lands on it

Answers

Answer:

Explanation:

Conservation of angular momentum.

Disk            I = ½MR²

Point mass I = mR²  (boy)

Initial angular momentum

L₀ = Iω = ½MR²ω₀

Final angular momentum

L₁ = Iω = (½MR² + mR²)ω₁

as momentum is conserved, these are equal

(½MR² + mR²)ω₁ = ½MR²ω₀

                       ω₁ = ω₀(½MR²/ (½MR² + mR²))

                       ω₁ = ω₀(½M/ (½M + m))

                       ω₁ = 1.2(½(200)/ (½(200) + 50))

                       ω₁ = 1.2(⅔)

                      ω₁ = 0.8 cycles/second or 0.8(2π) = 1.6π rad/s

A 2-kg object moving at 10 m/s has a 4-N force applied to it. Can you predict how the force will affect the object? A) yes, it will slow it down at a rate of 2 m/s2. B) yes, it will speed it up at a rate of 2 m/s2. yes, it will speed it up at a rate of 4 m/s2. D) It cannot be determined without more information.​

Answers

Answer:

D) It cannot be determined without more information.​

Explanation:

Velocity is a vector meaning it has both magnitude and direction.

Force and acceleration are also vectors.

Without knowing the directions of each, we cannot know if the mass has a speed (scalar value) change.

We can know that the mass will have a velocity change at the rate of 2 m/s² in the direction of the applied force.

If that force is applied in the direction of initial velocity, the velocity (and speed) will increase in magnitude in the same direction.

If that force is applied opposite of the initial velocity, the  the velocity (and speed) will decrease in magnitude in the same direction. If the acceleration lasts long enough, velocity will eventually become zero and then become negative. At the same time, speed will become zero, and then increase again as speed is the absolute value of the magnitude of velocity.

If the force is applied at any other angle, both the velocity and the speed will change in both magnitude and direction.

two factors affecting the magnitude of force of gravity betwwn 2 objects are A. mass and matter B. mass and distance C. weight and mass D. distance and weight

Answers

Answer:

B. MASS & MATTER

Explanation:

Newton's law also states that the strength of gravity between any two objects depends on two factors: the masses of the objects and the distance between them. Objects with greater mass have a stronger force of gravity between them.

13) A mass attached to the free end of a spring executes simple harmonic motion according to the equation y = (0.50 m) sin (18π t) where y is in meters and t is seconds. What is the period of vibration?

Answers

Hi there!

The period is given by:

[tex]T = \frac{2\pi}{w}[/tex]

T = Period (sec)

w = angular frequency (rad/sec)

According to the equation for SHM in terms of position:

y(t) = Asin(ωt + φ)

A = Amplitude (m)

ω = angular frequency (rad/sec)

t = time (sec)

φ = phase angle

In this instance, the angular frequency is given as 18π.

Plug this value into the equation for T:

[tex]T= \frac{2\pi}{18\pi} = \frac{1}{9} = \boxed{0.111 s}[/tex]

12) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart? [hint: use work – energy principle] [3 marks]

Answers

Hi there!

We can use the following:

W = ΔKE = F · d

Find the work done on the cart:

W = 200 · 10 = 2000 J

Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:

2000J = KEf - KEi

KE is given as:

[tex]KE = \frac{1}{2}mv^2[/tex]

2000J = 1/2(55)v²

4000 = 55v²

√(4000/55) = 8.53 m/s

An object, initially traveling at a velocity of 73 m/s, experiences an acceleration of -9.8 m/s^2. How much time will it take it to come to rest?

Answers

7.4 s

Explanation:

Given:

[tex]v_0 = 73\:\text{m/s}[/tex]

[tex]v = 0[/tex]

[tex]a = -9.8\:\text{m/s}^2[/tex]

[tex]t = ?[/tex]

To solve the time it takes for the object to come to a stop, we are going to use the equation below:

[tex]v = v_0 + at \Rightarrow t = \dfrac{v - v_0}{a}[/tex]

Using the given values above, we get

[tex]t = \dfrac{0 - 73\:\text{m/s}}{-9.8\:\text{m/s}^2}[/tex]

[tex]\;\;\;\;= 7.4\:\text{s}[/tex]

HELPPP!! Thanks!

If you only wanted to increase the particle motion of a gas without increasing any of its other properties, which would the most correct situation?


a. Keep the gas at a constant pressure and keep the temperature constant, but increase the volume of the gas

b. Keep the gas in a fixed container at constant pressure and increase the temperature

c. Keep the gas in a fixed container at constant pressure and decrease the temperature

d. Keep the gas at a constant volume and keep the temperature constant, but decrease the pressure of the gas

Answers

Answer:c i think

Explanation: not sure

How much potential energy does an 8 kg flower pot hanging 5 m above the ground have

Answers

Answer:

Explanation:

Relative to ground level it has

PE = mgh = 8(10)(5) = 400 J

PE = mgh => 8x5x9.81 => PE = 392.4 N

What are sound detectors?

Answers

Answer:

A sound detector comes in the shape of a rectangular board which comprises a microphone as well as a processing circuitry.

Answer:

Sound detection sensor works similarly to our Ears, having diaphragm which converts vibration into signals.

A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 25 J. The maximum speed of the block is:

Answers

Answer:

Explanation:

easy way

when system is all kinetic energy, velocity is at a maximum

E = ½mv²

v = √(2E/m) = √(2(25)/0.5) = √100 = 10 m/s

harder way

ω = √(k/m) = √(80/0.5) = √160 rad/s

When the system is entirely spring potential, the amplitude A is

E = ½kA²

A = √(2E/k) = √(2(25)/80) = 0.790569... = 0.79 m

maximum velocity is ωΑ = 0.79√160 = 10 m/s

5.000 km =
3.125
mi
8.000 fl oz =
mL

Answers

Answer:

236.588 mL

Explanation:

The formula for an approximate result is to multiply the volume value by 29.574

[tex]8.000 \times 29.574 = 236.588[/tex]

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