En una práctica un beisbolista lanza verticalmente la bola con una velocidad de 12m/s en dirección ascendente. ¿Cuál será la altura máxima que alcanza la pelota?

Answers

Answer 1

Answer:

Thus, the maximum height is 7.35 m.

Explanation:

initial velocity, u = 12 m/s

acceleration due to gravity, g = 9.8 m/s^2

Velocity at maximum height, v = 0 m/s

Let the maximum height is h.

Use third equation of motion

[tex]v^{2}=u^{2}-2 g h\\0 = 12\times 12 - 2 \times 9.8\times h\\h =7.35 m[/tex]


Related Questions

PLEASE HELP WILL MARK BRAINLIEST *

Answers

THE ANSWER IS FALSE AND GOOD LUCK ON YOUR TEST


A skipper on a boat notices wave crests passing the anchor chain every 6.0 seconds. The skipper estimates the distance between crests at 30.0
m. What is the speed of the water waves?

Answers

Explanation:

given Time Period = 5 Sec

Amplitude = 1m, Wave length = 15 m

using time period = 1/ frequency , Frequency = 1/5 cycles per sec = 0.2 sec = 12 per min

wave speed= Frequency * Wave length

speed = 0.2*15 = 3m/s

Who's Buzz Aldrin?
______________​

Answers

Answer:

Buzz Aldrin was American astronaut, engineer and fighter pilot. he did three spacewalks as pilot of the 1966 Gemini 12 mission, and as the lunar modulbe pilot on the 1969 Apollo 11 mission, he and mission commander Neil Armstrong were the first two people to land on the Moon.

I hope it helps

have have a great day

convert 11 milliseconds into seconds

Answers

Answer:

0.011

Explanation:

0.011

divide the time value by 1000

Damian is texting while driving and ends up slamming into the back end of the car in front of him, which then strikes the car in front of it. By texting while driving, Damian has created a

Answers

The answer for this question is negative externality

what is force? and give its formula​

Answers

Force is a push or pull that changes or attempts to change the shape or size of an object's position.

Formula Of Force

F = m × a

Here

F= Force

m= mass of object

a= Acceleration

Hope This Helps You ❤️


[tex]what \: is \: refraction \: of \: light \: {?} [/tex]

Answers

When light passes through one transparent medium to another transperent medium it bends and that beding of light is know as refraction of light !

Refraction of light is the bending of light as it passes from one medium to another.

After being assaulted by flying cannonballs, the knights on the castle walls (12.0 m above the ground) respond by propelling flaming pitch balls at their assailants. One ball lands on the ground at a distance of 81.1 m from the castle walls. If it was launched at an angle of 53.0° above the horizontal, what was its initial speed?

Answers

Answer:

28.6 m/s

Explanation:

Using the equation for the range of a projectile,

R = U²sin2θ/g where U = initial speed of flaming pitch balls, θ = launch angle = 53° and g = acceleration due to gravity = 9.8 m/s²

Making U subject of the formula, we have

U = √(gR/sin2θ)

substituting the values of the variables into the equation given that R = 81.1 m, we have

U = √(9.8 m/s² × 81.1 m/sin2(53°))

U = √(794.78 m²/s²/sin106°)

U = √(794.78 m²/s²/0.9613)

U = √(826.78 m²/s²)

U = 28.75 m/s

U ≅ 28.6 m/s

If person A weights less then Person B, and they both push away from each other with 10N of force *

Answers

Answer:

The system will tend to person A

Explanation:

A force is defined as either a pull or a push, in this scenario person A weighs lass that person B, so the resultant effect of the 10N interactive force will tend towards person A.

This is solely because person A is less than person B in weigh

A ball of 10kg falls from rest from a height of 150m, Neglating air resistance, calculate its kinetic energy after falling a distance of 40m. Take g=10m/s

Answers

Answer: [tex]3920\ J[/tex]

Explanation:

Given

mass of ball m=10 kg

It is placed at a height of 150 m

It is dropped from the height and allowed to free fall for 40 m

Velocity acquired by the ball during this fall is given by [tex]v^2-u^2=2as[/tex]

Insert u=0, a=g

[tex]\Rightarrow v^2-0=2\times 9.8\times 40\\\Rightarrow v=\sqrt{784}\\\Rightarrow v=28\ m/s[/tex]

Kinetic energy at this instant

[tex]K.E.=\dfrac{1}{2}\times 10\times 28^2\\\\\Rightarrow K.E.=3920\ J[/tex]

A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time did the skateboarder spend in the air and how far will be his final position from the bridge?

Answers

Answer:

  t = 0.714 s and  x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

         vₓ = 7.0 m / s

Let's find the time it takes to get to the river

         y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

        0 = y₀ + 0 - ½ g t²

        t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]

        t = ra 2 2.5 / 9.8

        t = 0.714 s

the distance traveled is

       x = vₓ t

       x = 7.0 0.714

       x = 5.0 m

An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object then slides at a constant velocity for 6.0 s until it reaches a rough section that causes the object to stop in 2.5 s.

Answers

Answer:

[tex]D_T=18.567m[/tex]

Explanation:

From the question we are told that:

Acceleration [tex]a=8.0 m/s^2[/tex]

Displacement [tex]d=1.05 m[/tex]

Initial time [tex]t_1=6.0s[/tex]

Final Time [tex]t_2=2.5s[/tex]

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

 [tex]V^2=2as[/tex]

 [tex]V=\sqrt{2*6*1.05}[/tex]

 [tex]V=4.1m/s[/tex]

Generally the equation for Distance traveled before stop is mathematically given by

 [tex]d_2=v*t_1[/tex]

 [tex]d_2=3.098*4[/tex]

 [tex]d_2=12.392[/tex]

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity [tex]v_3=0 m/s[/tex]

Initial velocity [tex]u_3=4.1 m/s[/tex]

Therefore

Using Newton's Law of Motion

 [tex]-a_3=(4.1)/(2.5)[/tex]

 [tex]-a_3=1.64m/s^2[/tex]

Giving

 [tex]v_3^2=u^2-2ad_3[/tex]

Therefore

 [tex]d_3=\frac{u_3^2}{2ad_3}[/tex]

 [tex]d_3=\frac{4.1^2}{2*1.64}[/tex]

 [tex]d_3=5.125m[/tex]

Generally the Total Distance Traveled is mathematically given by

 [tex]D_T=d_1+d_2+d_3[/tex]

 [tex]D_T=5.125m+12.392+1.05 m[/tex]

 [tex]D_T=18.567m[/tex]

?????? ?????help please

Answers

the answer is d. cassani

why does a child look different to both parents?​

Answers

Answer:

Each parent has a set of genes, and a combination of the genes from the mom and the dad could come together to make the child look totally different.

A child looks different to both parents because children usually look like a "combination" of both their parents. Each parent gives their child their genes.  So a child wouldn't look exactly like one parent.

Please Help!!!!
A scientist defects an earthquake wave traveling at a speed of 6.5 km/s in rock with a density of 2.8 g/cubic cm. Based on the data, a scientist can BEST conclude that the wave is traveling through which type of rock?

Answers choices:
A. limestone
B. shale
C. anhydrite
D. dolomite​

Answers

I think is limestone

77. The first law of motion applies to
a. Only objects that are moving
b. Only objects that are not moving
C. All objects whether moving or not
No obiect, whether moving or not

Answers

In the first law, an object will not change its motion unless a force acts on it. So I think the answer is A. Only objects that are moving.

A ball of 5.2 kg is swings around on a rope with radius 0.86
meters. If the ball is traveling at 1.8 m/s, what is its angular
momentum? (Round to the nearest hundredth).

Answers

Answer:

Explanation:

Angular momentum is L = mvr:

L = (5.2)(1.8)(.86) so

L = 8.05 kg*m/s

Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by dE dt = q2a2 6pP0c3 where c is the speed of light. (a) Verify that this equation is dimensionally correct. (b) If a proton with a kinetic energy of 6.0 MeV is traveling in a particle accelerator in a circular orbit of radius 0.750 m, what fraction of its energy does it radiate per second? (c) Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second?

Answers

Answer:

a)  [tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]

b)  [tex]x=\frac{6.0}{4.3*10^{-32}}[/tex]

c) [tex]x'=\frac{6.0}{2.22*10^{-3}}[/tex]

Explanation:

From the question we are told that:

Kinetic energy of Proton[tex]K.E_p= 6.0 MeV[/tex]

Radius [tex]r=0.750[/tex]

Energy [tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]

a)

[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]

b)

Generally the equation for acceleration of proton is mathematically given by

[tex]a_p=\frac{v^2}{r}[/tex]

Where

Speed of Proton particle is

[tex]V_p=2.12 *10^3 m/s[/tex]

[tex]a_p=\frac{(2.12 *10^3)^2}{0.750}\\a_p=1.27*10^{10}m/s^2[/tex]

Therefore

[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]

[tex]\frac{dE}{dt}=\frac{9*10^9*(1.6*10^{-19}^2*(1.27*10^{10})^2)}{(3*10^8)^3}[/tex]

[tex]\frac{dE}{dt}=6.881810^{-51}J/s[/tex]

Energy radiated per sec

[tex]mev=\frac{6.881810^{-51}}{1.6*10^{-19}}\\mev=4.3*10^{-32}ev[/tex]

Therefore the Fraction of its energy  it radiates per second is given as

[tex]x=\frac{ K.E_p}{mev}\\[/tex]

[tex]x=\frac{6.0}{4.3*10^{-32}}[/tex]

c)

Generally the equation for acceleration of proton is mathematically given by

[tex]a_p=\frac{v^2}{r}[/tex]

Where

Speed of Proton particle is

[tex]V_p=2.5 *10^7 m/s[/tex]

[tex]a_p=\frac{(2.5 *10^7)^2}{0.750}\\a_p=0.33*10^{14}m/s^2[/tex]

Therefore

[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]

[tex]\frac{dE}{dt}=\frac{9*10^9*(1.6*10^{-19}^2*(0.33*10^{14})^2)}{(3*10^8)^3}[/tex]

[tex]\frac{dE}{dt}=59.26^{-25}J/s[/tex]

Energy radiated per sec

[tex]mev=\frac{59.26^{-25}}{1.6*10^{-19}}\\mev=2.22*10^{-3}ev[/tex]

Therefore the Fraction of its energy  it radiates per second is given as

[tex]x'=\frac{ K.E_p}{mev}\\[/tex]

[tex]x'=\frac{6.0}{2.22*10^{-3}}[/tex]

13. When you say that something is a factor, you mean that it is an) (factor)
a. problem.
b. obstacle.
c. ingredient.
d. fact

Answers

Answer:

C.

A factor of something is an ingredient, or it makes up that thing. For example, in math, we always try to find the factors of a number. We are basically trying to find the "ingredients" of a number.

What are the four components to a workout program?

Answers

Answer:

Aerobic Fitness. Aerobic fitness improves overall health and well-being. ...

Muscular Fitness. Strength training improves your muscle and bone health and helps with weight loss. ...

Flexibility. Flexibility allows you to move your body freely. ...

Stability and Balance

Explanation:

UP

ayuda porfa es de actividades fisicas

Answers

Translate to English please

A forensics investigator discharged an assault rifle-replica such that the bullet fired at an angle of 30 (degrees) off the horizontal with an initial velocity
of 28
m/s northwest. What is the maximum height the bullet will reach?
O 14 m/s
10 m
O 30 km
O 0.4351 seconds

Answers

Answer:

Initial y-component of speed

Vy = 28 * sin 30 = 14 m/sec vertically

1/2 m Vy^2 = 2 m g h    conservation of energy of y-component

h = Vy^2 / (2 * g) = 14^2 / (2 * 9.8) = 10 m

If 150 Joules of work is needed to move a box 1000 cm, calculate the force that was used?

Answers

Answer:

[tex] \large{ \tt{☄ EXPLANATION}} : [/tex]

We're provided : Work ( W ) = 150 J & Displacement ( D ) = 1000 cm & We're asked to find out the force that was used. Firstly, Notice that we're provided the unit of Displacement as centimetre so we have to convert 1000 cm into m. Displacement ( D ) = [tex] \frac{1000}{ 100} = 10[/tex] m

[tex] \large{ \tt{♡ \: LET'S \: GET \: STARTED}} : [/tex]

Work is defined as the product of force and Displacement. By definition ,

[tex] \large{ \boxed{ \tt{❃ \: WORK ( \: W \: ) = FORCE\: ( \: F) \times DISPLACEMENT \: ( \: D \: )}}}[/tex]

- Plug the values and then simplify !

[tex] \large{ \bf{↦ \: 150 = F \times 10}}[/tex]

[tex] \large{ \bf{↦F\times 10 = 150}}[/tex]

[tex] \large{ \bf{↦ \: F = \frac{150}{10}}} [/tex]

[tex] \large {\bf{↦F = \boxed{ \bf{15} \: N}}}[/tex]

[tex] \boxed{ \boxed{ \large{ \tt{✤ \: OUR \: FINAL \: ANSWER : \boxed{ \bf{15 \: n}}}}}}[/tex]

KEEP READING , KEEP STUDYING , KEEP WORKING , KEEP PUSHING , KEEP TAKING CARE OF YOURSELF. YOUR HARD WORK WILL PAY OFF ! ♪

ツ Hope I helped ! ۵

☪ Have a wonderful day / evening ! ☼

# StayInAndExplore ! ☂

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a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after 6 seconds​

Answers

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

What is the net force acting on the airplane?
740 N right -->
700 N right -->
100 N left <--
760 N right -->

Answers

[tex]\huge{ \mathrm{  \underline{ Answer }\:  \:  ✓ }}[/tex]

Total force acting on right side = 800 N

Total force acting on left side :

60 N + 40 N

100 N

Now, equivalent force acting on the plane is :

greater force - minor force

800 N - 100 N 700 Newtons

And the direction of equivalent force will be the direction of greater force, that is right direction.

Hence, Correct option is :

700 N right -->

_____________________________

[tex]\mathrm{ \:TeeNForeveR\:}[/tex]

Give your answer to 2 dp
When taking off a plane accelerates at 2.7m/s2 down the runway. It accelerates from a stationary position for 25 seconds before leaving the ground. What
is the planes speed when it leaves the ground?​

Answers

Answer:

67.5

Explanation:

The plane accelerates at 2.7m/s,^2

Time is 25 seconds

The velocity can be calculated as follows

= 25×2.7

= 67.5

Hence the speed f the plane is 67.5


As the wavelength of light decreases,
What happens

Answers

Answer:

Waves. Refraction is an effect that occurs when a light wave, incident at an angle away from the normal, passes a boundary from one medium into another in which there is a change in velocity of the light. ... The wavelength decreases as the light enters the medium and the light wave changes direction.

Explanation:

As a wavelength increases in size, its frequency and energy (E) decrease. From these equations you may realize that as the frequency increases, the wavelength gets shorter. ... Mechanical and electromagnetic waves with long wavelengths contain less energy than waves with short wavelengths.

Waves. Refraction is an effect that occurs when a light wave, incident at an angle away from the normal, passes a boundary from one medium into another in which there is a change in velocity of the light. ... The wavelength decreases as the light enters the medium and the light wave changes direction.

11. In a new science fiction movie an enemy spaceship explodes. Will an observer on a nearby
moon hear the explosion, see the explosion, or both? Explain

Answers

Answer:

Sound is transmitted by longitudinal waves thru air - the observer cannot hear the explosion because no air is present between the moon and the ship

Light is transmitted by electromagnetic waves so the light from the explosion can travel between the  moon and the observer.

(only answer if you know the answer or I'll report) Solve it w the steps tysm​

Answers

Answer:

 v_f = 20 m / s

Explanation:

For this exercise let's use the relationship between momentum and moment

        I = Δp

        F t = m v_f - m v₀

as the body starts from rest v₀ = 0

        F t = m v_f

        v_f = [tex]\frac{F t}{m}[/tex]

let's calculate

        v_f = 4 2 / 0.4

         v_f = 20 m / s

If a 70 kg man is standing on the surface of the earth, then what is his weight on the moon

Answers

Answer:

P = 113.4 N

Explanation:

We know that if it is on the Moon it has another gravity. Therefore, it is known through the International System that the gravity of the Moon is 1.62 m/s², so..

Data:

m = 70 kgg = 1.62 m/s²P = ?

Use formula:

[tex]\boxed{\bold{P=m*g}}[/tex]

Replace and solve:

[tex]\boxed{\bold{P=70\ kg*1.62\frac{m}{s^{2}}}}[/tex][tex]\boxed{\boxed{\bold{P=113.4\ N}}}[/tex]

The man's weight is 113.4 Newtons.

Greetings.

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