A ball is thrown straight upward with a velocity of 14 m/s. What is the maximum height
reached by the ball?
(1 Point)
10 m
20 m
14 m
32 m

Answers

Answer 1

Hello!

[tex]\large\boxed{d = 10 m}[/tex]

When the ball reaches its maximum height, the velocity equals 0. We can use the kinematic equation:

vf = vi + at, where:

vf = final velocity (0)

vi = initial velocity (14 m/s)

a = acceleration due to gravity, or 9.8 m/s (downwards)

t = time, which will be solved for.

Plug in the given values:

0 = 14 - 9.8(t)

-14 = -9.8(t)

t ≈ 1.429 sec

Solve for the distance traveled using the kinematic equation:

d = 1/2(vf + vi)t

Plug in the values:

d = 1/2(14 + 0)(1.429)

d = 1/2(14 · 1.429)

d ≈ 10 m


Related Questions

A pinball bangs against a bumper of a pinball machine with a speed of 0.9 m/s. If the ball has a mass of 0.056 kg, what is the ball's kinetic energy?

Answers

It would be 200 newtons of force
It would be 2.356

A ball is thrown up into the air. Ignore air resistance. When it is rising and reaches half of its maximum height,the net force acting on it is

Answers

Answer:

The net and only force acting on the ball thrown up into the air at half of its maximum height is the weight  of the ball

Explanation:

When the ball is rising in the air, the force, F, acting on it is given by the product of the mass, m × acceleration, a

The acceleration of a body thrown in the air = Gravitational acceleration = g = Constant

Therefore;

The force acting on the body thrown in the air F = Constant = m × g (downwards)

The force acting on the ball thrown up into the air at half of its maximum height = The mass of the ball × The acceleration due to gravity = The weight  of the ball.

13. Austin rode his bike 10 m/s for two minutes. How far did he travel? A. 200 meters B. 1200 meters C. 1000 meters D. 20 meters​

Answers

Answer:

B. 1200

Explanation:

60 sec in one min in 2 min there will be 120 sec. 10x120=1200

HELP ASAP!! WILL TRY TO GIVE BRAINLIESTT!!

Please define the prefixes hyper-, hypo-, & iso- AND describe a solution that is hypertonic, hypotonic, & isotonic as well as compare how a living cell reacts when placed in a solution of each of the 3 types.

Answers

If a cell is placed in a hypertonic solution, water will leave the cell, and the cell will shrink. In an isotonic environment, there is no net water movement, so there is no change in the size of the cell. When a cell is placed in a hypotonic environment, water will enter the cell, and the cell will swell.

When an object moves, where does the energy come from?

Answers

Ans: Kinetic and potential energies are found in all objects. If an object is moving, it is said to have kinetic energy (KE). Potential energy (PE) is energy that is "stored" because of the position and/or arrangement of the object. The classic example of potential energy is to pick up a brick.

Answer:

Kinetic energy.

Explanation:

This is the energy of motion, observable as the movement of an object or subatomic particle. Every moving object and particle have kinetic energy.

It takes serina 0.25 hours to drive to school. Her route is 16km long . What is serina’s average speed on her drive to school

Answers

Explanation:

distance=16km=16000m

time=0.25hours=1/4hr=15min=900sec

speed=dist./time

=16000/900=160/9=17.777m/s

speed=17.78m/s (rounded)

Answer:

64

Explanation:

Avg. speed = total distance/total time

Avg. speed= 16km/0.25h

Avg. speed= 64km/h

HELP ASAP!! WILL TRY TO GIVE BRAINLIESTT!!

Please define the prefixes hyper-, hypo-, & iso- AND describe a solution that is hypertonic, hypotonic, & isotonic as well as compare how a living cell reacts when placed in a solution of each of the 3 types.

Answers

Answer:

Hypertonic, isotonic, and hypotonic solutions and their effect on cells. ... There are some different explanations out there. ... Tonicity is a concern for all living things, particularly those that lack rigid cell walls and live in ...

What the distance between 2 particles is doubled and the force between them is quartered, what is that called

Answers

Answer:

Coulomb's law.

Explanation:

If the distance between two particles is doubled and the force between them is quartered, this is known as Coulomb's law. Coulomb presented a law in which he states that when there is increase occur in distance between two objects, the force between them is decreases so if the distance between two particles or objects is doubled so the force between them is also quartered.

A 0.15 kg ball is moving with a velocity of
35 m/s. Find the momentum of the ball.

Answers

Answer:

5.25 kg.m/s

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 0.15 × 35

We have the final answer as

5.25 kg.m/s

Hope this helps you

What is the height of a horizontal projectile if it takes 3 seconds to hit the ground?
a
29.4 m
b
3.0 m
C
44.1 m
d
88.3 m

Answers

Answer:

B

Explanation:

A 81.9 kg person stands on her toes. The surface area of her toes in contact with the ground is only .002^ - m ^2. How much pressure is exerted between her toes and the ground, IN ATMOSPHERES ?

Answers

Answer:

3.77202

Explanation:81.9 kg×9.8N/kg÷0.0021 Pa ×(1atm/101325)=3.77202

The pressure exerted between her toes and the ground is 3.77 atm.

What is pressure?

The stress at a location within a confined fluid or the perpendicular force per unit area are the two definitions of pressure used in the physical sciences. A 42-pound box with an area of 84 square inches at the bottom will apply pressure on a surface equal to the force divided by the surface area.

Given:

An 81.9 kg person stands on her toes. The surface area of her toes in contact with the ground is only 0.00210 m²,

Calculate the pressure on the toe as shown below,

The pressure on the toe = Force of man / area of toe

The pressure on the toe = mass of man × 9.8 / 0.00210

The pressure on the toe = 81.9 × 9.8 / 0.00210

The pressure on the toe = 802.62 / 0.00210

The pressure on the toe = 382200 N/m²

The pressure on the toe = 382200 / 101325 (1 atm = 101325 N / m²)

The pressure on the toe = 3.77 atm,

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A softball pitcher rotates a 0.250 kg ball around a vertical circular path of radius 0.5 m before releasing it. The pitcher exerts a 33.0 N force directed parallel to the motion of the ball around the complete circular path. The speed of the ball at the top of the circle is 13.0 m/s. If the ball is released at the bottom of the circle, what is its speed upon release

Answers

Answer:

The value is  [tex]u = 27.93 \ m/s[/tex]

Explanation:

From the question we are told that

   The  mass of the ball is [tex]m = 0.250 \ kg[/tex]

    The radius of the circle is  [tex]r = 0.5 \ m[/tex]

    The force exerted is  [tex]F = 33 \ N[/tex]

    The  speed of the ball on top of the circle is  [tex]v = 13.0 \ m/s[/tex]

Gnerally from the law of energy conservation we have that

     [tex]PE + W = \Delta KE[/tex]

Here PE  is the potential energy of the pitcher which is mathematically represented as

      [tex]PE = m * g * (2 r )[/tex]

=>   [tex]PE = 0.250 * 9.8 * (2 * 0.5 )[/tex]  

=>   [tex]PE = 24.5\ J[/tex]

And  W  is the workdone by the force applied which is mathematically represented as

       [tex]W = F * \pi * r[/tex]

=>    [tex]W = 33 * 3.142 * 0.5[/tex]

=>    [tex]W = 51.84 \ J[/tex]

And   [tex]\Delta KE[/tex] is the change in kinetic energy which is mathematically represented as

       [tex]\Delta KE = \frac{1}{2} * m * (v^2 - u^2 )[/tex]

=>    [tex]\Delta KE = \frac{1}{2} * 0.250 * ( 13^2 - u^2 )[/tex]

=>    [tex]\Delta KE = 0.125 * ( 13^2 - u^2 )[/tex]

=>    [tex]\Delta KE = 21.125 - 0.125u^2[/tex]

So

      [tex]24.5 + 51.84 = 21.125 - 0.125u^2[/tex]

=>    [tex]u = 27.93 \ m/s[/tex]

The velocity of the ball released at the bottom is required.

The velocity of the ball released at the bottom is 24.56 m/s

F = Force = 33 N

r = Radius of loop = 0.5 m

m = Mass of ball = 0.25 kg

u = Initial velocity of ball = 13 m/s

v = Final velocity

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

The energy balance of the system is given by

[tex]F\pi r+mg2r=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow v=\sqrt{\dfrac{2}{m}(F\pi r+mg2r)+u^2}\\\Rightarrow v=\sqrt{\dfrac{2}{0.25}(33\times \pi\times 0.5+0.25\times 9.81\times 2\times 0.5)+13^2}\\\Rightarrow v=24.56\ \text{m/s}[/tex]

The velocity of the ball released at the bottom is 24.56 m/s.

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2. Heavier football players tend to play on the front line. Why?
What law is it?

Answers

Answer: They are put in front for defense so so they can block the opponents from getting the ball

Explanation:

Answer:

Heavier football players are usually on the front line because they have a low point in gravity therefore they are able to block the back running players in a better way. And its Newton's second law.

Explanation:

mass does not vary from place to place why? ​

Answers

Answer:

It is always constant unlike weight

Weight is the mass of a system multiplied by the gravitational force exerted by the planet

Why do women typically tend to have slightly greater stability than men?
A.
Women are shorter than men, and shorter people are more stable.
B.
Men have more muscle mass in their lower bodies that makes them stiff and less stable.
C.
Women have lower centers of gravity, and lower centers of gravity provide more stability.
D.
The increased muscle mass in their upper bodies makes their centers of gravity difficult to find.



Please select the best answer from the choices provided.


A
B
C
D

Answers

Answer:

C

Explanation:

I figured it sounded more accurate

What would the mechanical advantage of a ramp be if the length is 25 feet and the height is 5 feet?
10
125
5
30

Answers

Answer:

5.

Explanation:

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

Length (L) of ramp = 25 feet

Height (H) of ramp = 5 feet

Mechanical advantage (MA) of ramp =?

Mechanical advantage of an inclined plane is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is expressed as:

Mechanical advantage = length /height

MA = L/H

With the above formula, we can simply calculate the mechanical advantage of the ramp as follow:

Length (L) of ramp = 25 feet

Height (H) of ramp = 5 feet

Mechanical advantage (MA) of ramp =.?

MA = 25 / 5

MA = 5

The, the mechanical advantage of the ramp is 5.

Answer:

5.

Explanation:

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

Length (L) of ramp = 25 feet

Height (H) of ramp = 5 feet

Mechanical advantage (MA) of ramp =?

Mechanical advantage of an inclined plane is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is expressed as:

Mechanical advantage = length /height

MA = L/H

With the above formula, we can simply calculate the mechanical advantage of the ramp as follow:

Length (L) of ramp = 25 feet

Height (H) of ramp = 5 feet

Mechanical advantage (MA) of ramp =.?

MA = 25 / 5

MA = 5

The, the mechanical advantage of the ramp is 5.

An atom of lithium (Li) forms an ionic bond with an atom of chlorine (Cl) to form lithium chloride. How are the valence electrons of these atoms rearranged to form this bond?

A few valence electrons are shared between the atoms.
Many valence electrons are shared between the atoms.
Electrons are transferred from the chlorine atom to the lithium atom.
Electrons are transferred from the lithium atom to the chlorine atom.

Answers

Answer:

electrons are transferred from the lithium atom to the chlorine atom

Answer:

D

Explanation:

I just took the test :)

How are seeds different from spores?

Seeds have endosperm which provides nourishment for a new plant, but spores do not have any stored food supplies.

Spores have an endosperm and involve sperm and egg cells, while seeds do not.

Spores involve egg and sperm cells, while seeds do not.

Spores will germinate and must find good soil to survive. Seeds can survive in extremely harsh conditions.

Answers

Explanation:

1.Seeds have endosperm which provides nourishment for a new plant, but spores do not have any stored food supplies.

hope it helpful

A projectile is fired into the air. Three characteristics of its subsequent motion are
1) the horizontal component of velocity
II) the vertical component of velocity
III) the acceleration
Neglecting air resistance, which characteristic(s) change(s) whilst the projectile is in the air.
(1 Point)
I only
Il only
II and III only
I and II only

Answers

Answer:

The answer is C) II and III only

Explanation:

How long will it take a sample of I-131 to decay to 12.5% assuming it's half-life is 8.07 days?

Answers

Answer:

If you take A0 to be the initial sample of iodine-131, you will be left with. 12⋅A0=A02→ ... m24 hours=12.5%mo=12.5⋅10−2⋅mo

Explanation:

is that what u look for

Sample I-131 decays to 12.5% in 24.21 days.

What is half-life?

Half-life, in radioactivity, is the amount of time needed for half of a radioactive sample's atomic nuclei to decay (change spontaneously into other nuclear species by emitting particles and energy), or, alternatively, the amount of time needed for a radioactive material's rate of disintegrations per second to decrease by half.

Give parameters:

Half-life of  the radioactive element I-131:T = 8.07 day

It decays to 12.5% means it decays to 1/8.

According to definition of half-life,

The radioactive element I-131 decays to half in 8.07 days.

So, the radioactive element I-131 decays to 1/8 in 3×8.07 days = 24.21 days.

Hence,  the radioactive element I-131 decays to12.5% in 24.21 days.

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n Olympic diver is on a diving platform 3.60 m above the water. To start her dive, she runs off of the platform with a speed of 1.3 m/s in the horizontal direction. What is the diver's speed, in m/s, just before she enters the water

Answers

Answer:

8.5m/s

Explanation:

Using the equation of motion

v² = u² + 2gH

v is the final speed of the diver

u  is the initial speed of the diver

g is the acceleration due to gravity

H is the height of the object

Given the following

V = 1.3m/s

H = 3.60m

g = 9.8m/s²

Required

Initial speed of the diver u

Substitute the given values into the formula:

v² = u² -2gH

1.3² = u²  - 2(9.8)(3.60)

1.69 = u²-70.56

u² = 1.69+70.56

u² = 72.25

u = √72.25

u = 8.5m/s

Hence the diver's speed, in m/s, just before she enters the water is 8.5m/s

a car travelling at 20 ms-1 accelerates uniformly for 8 s until its velocity reaches 36 ms-1.what are the acceleration and the average velocity of the car during this period​

Answers

Answer:

Acceleration of the car = 2 m/s²

Average velocity of the car =28 m/s

Explanation:

Initial velocity = 20 m/s

Final velocity = 36 m/s

Time taken = 8 s

Acceleration = Δ velocity / time taken

Acceleration = 36-20 / 8 = 16/8 = 2 m/s²

Average velocity of the car =[ Initial velocity + Final velocity ] /2

Average velocity of the car = [ 20+36] / 2 = 28 m/s

You will get the most accurate resting heart rate if you take your pulse for ___ consecutive mornings and average the number
A: 3
B: 4
C: 5

Answers

The answer will be 3

While delivering 125 kg blocks of ice to a local village, Kristoff and his family (and Sven too) come upon a cliff that is 5.7 m above them. To solve their problem, they build a catapult that will launch their blocks of ice with an initial velocity of 15 m/s and at an angle of 45 degrees above the ground.


Prove that the catapult will successfully launch the ice blocks up to the top of the cliff. Quantities you will need to solve for will be initial vertical velocity, horizontal velocity, the distance the catapult is from the cliff, and the amount of time it takes the ice to reach the cliff. Assume that they are experts and have arranged everything so that the ice blocks just barely reach the top of their parabolic path when they land at the top of the cliff and make a smooth landing. Draw a picture of the scene to help you visualize what is happening.


After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.

What is the rate of acceleration while the blocks slow to a stop?


How long do the blocks take to slow to a stop?


What is the amount of friction between the ice and the snowy ground?



How long do the blocks take to slow to a stop?


What is the amount of friction between the ice and the snowy ground?

Answers

Answer:

a. Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.

b. -2.81 m/s²

c. 3.78 s

d. -351.25 N

Explanation:

a. After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.

For the block of ice to reach the top of the cliff, its maximum height, h should be greater than or equal to 5.7 m. That is, h ≥ 5.7 m.

The maximum height of a projection h, projected with an initial velocity v at an angle Ф is h = v²sin²Ф/2g where g = acceleration due to gravity = 9.8 m/s².

For the block of ice, v = 15 m/s and Ф = 45°. So,

h = v²sin²Ф/2g

= (15 m/s)²sin²45/(2 × 9.8 m/s²)

= 225 (m/s)²(1/√2)²/19.6 m/s²

= 225 (m/s)²(1/2)/19.6 m/s²

= 112.5 (m/s)²/19.6 m/s²

= 5.74 m

Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.

The graph is in the attachment.

b. What is the rate of acceleration while the blocks slow to a stop?

Using v² = u² + 2as where u = initial horizontal velocity of block = 15m/scos45° = 10.61 m/s, v = final velocity of block = 0 m/s since it stops, a = acceleration and s = distance block moves = 20 m

So, a =  (v² - u²)/2s

substituting the variables into the equation, we have

a =  ((0 m/s)² - (10.61 m/s)²)/2(20 m)

= - 112.57 (m/s)²)/40 m

= -2.81 m/s²

c. How long do the blocks take to slow to a stop?

Using v = u + at where u = initial horizontal velocity of block = 10.61 m/s v = final velocity of block = 0 m/s since it stops, a = acceleration = -2.81 m/s² and t = time it takes block of ice to stop

So, making t subject of the formula,

t = (v - u)/a

substituting the values of the variables, we have

t = ( 0 m/s - 10.61 m/s)/-2.81 m/s²

= -10.61 m/s/-2.81 m/s²

= 3.78 s

d.  What is the amount of friction between the ice and the snowy ground?

The frictional force, f = net force on block of ice

f = ma where  m = mass of bock = 125 kg and a = acceleration of block = -2.81 m/s²

f = ma

= 125 kg(2.81 m/s²)

= -351.25 N

A go-cart and rider have a mass of 14 kg. If the cart accelerates at 6m/s^2 during a 40 m sprint in 100 seconds, how much power did the cart do? MUST SHOW ALL WORK!!!!!!!!!

Answers

Answer:

P = 33.6 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = forces [N]

m = mass = 14 [kg]

a = acceleration = 6 [m/s²]

[tex]F = 14*6\\F = 84 [N][/tex]

In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.

[tex]W = F*d[/tex]

where:

W = work [J]

F = force = 84 [N]

d = displaciment = 40 [m]

[tex]W = 84*40\\W = 3360 [J][/tex]

Finally, the power can be calculated by the relationship between the work performed in a given time interval.

[tex]P=W/t\\[/tex]

where:

P = power [W]

W = work = 3360 [J]

t = time = 100 [s]

Now replacing:

[tex]P=3360/100\\P=33.6[W][/tex]

The power is given in watts

If you run North for 5 meters and then East for 15 meters and then
South for 2 meters all in 22 seconds. What was your speed?
0 points

Answers

Answer:

1m/s

Explanation:

D = 22m

T = 22s

S = 22÷22

= 1 m/s

what is the total magnification of a microscope with an eyepiece magnification of 10x and an objective lens magnification of 50x quizziz

Answers

Answer:

500x

Explanation:

A microscope had an eyepiece magnification of 10x

The objective lens magnification is 50x

Therefore the total magnification can be calculated as follows

= 10 × 50

= 500x

Hence the total Magnificattion is 500x

A child kicks a ball horizontally with a speed of 2.8 m/s from the end of a deck that is 8.5 m high.
a. How long will the ball take to hit the ground?

Answers

Answer:

Explanation:

time =d/s

therefore, t =8.5/2.8

t=3.02s

At a sports event, the car starts from iosi, in 5.0 siis acceleration is 5.
Calculate the distance travelled by ban
100 m
25 m
63 m
O bag 6.
The Score Tower hingga laglomery 4
werissa 13​

Answers

Question:

At a sports event, the car starts from rest. in 5.0 s its acceleration is 5.0 m/s2.  Calculate the distance travelled by car.

Answer:

62.5 metres

Explanation:

Given

[tex]u = 0[/tex] -- Initial Velocity

[tex]a = 5.0m/s^2[/tex] --- acceleration

[tex]t =5.0s[/tex] -- time

Required

Calculate the distance

This question will be solved using the following equation of motion

[tex]S = ut + \frac{1}{2}at^2[/tex]

Where S represents the distance

Substitute values for u, t and a

[tex]S = 0 * 5.0 + \frac{1}{2} * 5.0 * 5.0^2[/tex]

[tex]S = 0 + \frac{1}{2} * 5.0 * 25.0[/tex]

[tex]S = \frac{1}{2} * 5.0 * 25.0[/tex]

[tex]S = \frac{1}{2} * 125.0[/tex]

[tex]S = 62.5m[/tex]

Hence, the distance travelled is 62.5m

A car slows from 22 m/s to 3.0 m/s at a constant rate of 2.1 m/s2. How many seconds are required before the car is traveling at a forward velocity of 3.0 m/s?

Answers

Answer: 9.05 s

     

Explanation: v = u +at

v = 3 m/s

u = 22 m/s

Deceleration = 2.1 m/s^2

∴ 3 = 22 - 2.1 ( t )

∴ t = 9 .05 s

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