Alejandro Kirk is the catcher for the Blue Jays’ baseball team. He exerts a forward force on the 0.145-kg baseball to bring it to rest from a speed of 38.2 m/s. During the process, his hand recoils a distance of 0.135 m.
Determine is the acceleration of the ball.
Determine the force applied by Alejandro.

Answers

Answer 1

Answer:

a= -5404.6 [m/s²]; F=785.75 [N].

Explanation:

1. Determine is the acceleration of the ball:

equation of the distance (0.135m) is:

[tex]S=-\frac{at^2}{2} +v_0t,[/tex]  where S=0.135[m]; a - required acceleration; t - elapsed time; v₀ - initial velocity (38.2 m/s);

also the required acceleration is:

[tex]a=\frac{v-v_0}{t}, where[/tex] V - the end velocity (0 m/s), t - elapsed time, ₀ - initial velocity (38.2 m/s).

Using the equations of reqruired acceleration and the distance it is possible to make up and solve the next system:

[tex]\left \{ {{a=-\frac{38.2}{t} } \atop {S=-\frac{at^2}{2}+38.2t }} \right. \ => \ \left \{ {{a=-\frac{38.2}{t} } \atop {0.5at^2-38.2t=0.135}} \right. \ => \ \left \{ {{a=-\frac{38.2}{t} } \atop {19.1t-38.2t=0.135}} \right. \ => \ \left \{ {{a=-5404.6} \atop {t=0.007}} \right.[/tex]

finally, a≈-5404.6 [m/s²].

2. Determine the force applied by Alejandro.

the energy is:

[tex]E=\frac{mv^2}{2}; \ or E=FS, where \ m-the \ mass; \ v-velocity; \ F-required \ force; S-distance;[/tex]

According to these two equations, the required force is:

[tex]F=\frac{E}{S}=\frac{mv^2}{2S};[/tex]

F=0.145*38.8²/2/0.135≈785.75 [N].

note, the suggested way is not the shortest one and not the only one.

P.S. if it is possible, check the arithmetic operations and the provided answers in other sources.


Related Questions

How is "speed" defined in terms of physics?

Answers

Explanation:

In everyday use and in kinematics, the speed of an object is the magnitude of the rate of change of its position with time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.

A teacher has two radioactive sources, A and B.
Source A has a longer half-life than source B.
What can be deduced about the nuclei in source A compared with the nuclei
in source B?
Do not refer to isotopes in your answer.

Answers

Answer:

The nuclei of source have greater stability than those of source B.

Explain why we draw straight lines to show rays of light?

Answers

Answer:

Because light always travels in  straight line.

Explanation:

The fact that light travels can be demonstrated by putting an object in its path. If the object is opaque the result is a degree of blackness on the other side of it which is due to the absence of the light. ... As light does appear to travel in straight lines then light is usually modelled as straight lines in drawings.

Hope this helps :)

#Followformore

What is the correct definition of rarefaction

Answers

Answer:

Explanation:

A decrease in the density of something is rarefaction. ... Most of the time, rarefaction refers to air or other gases becoming less dense. When rarefaction occurs, the particles in a gas become more spread out. You may come across this word in the context of sound waves.

If a train travels 25 miles in 2 hours 15 minutes, what is its average speed in miles per hour? 11. 1 mph 12. 5 mph 10 mph 9. 1 mph.

Answers

Answer:

11.1

Explanation:

2 hours 15 mins = 2 1/4 hours

2 1/4 hours = 25 miles

1 hour = 25 ÷ 2 1/4 = 25 ÷ 9/4 = 25 x 4/9 = 11.1 miles

what is the chemical and physical changes if making a starchs

Answers

Answer:

Chemical modification of starch is based on the chemical reactivity of the constituent glucose monomers which are polyhydroxyl and can undergo several reactions. Starch can undergo reactions such as hydrolysis, esterification, etherification and oxidation.

Explanation:

Hope this helps!

3. If John leaves his house to walk his dog and he walks around the block, what is
his distance?
180 Meters
180 Meters
180 Meters
180 Meters

Answers

Answer:

180 meters

Explanation:

help me solve please ​

Answers

Answer:

please check the image

Explanation:

I hope this image helps you please follow the steps. Thank you

Answer:

1kg/```````2

Explanation:

solution

a force that is at rest or moves in at a constant speed and in a constant direction is called a what force

Answers

The question is fishing for "balanced force".

But the description in the question is terrible.

hannah drove 360 mi in 5.2 hours. What was her average speed?

Answers

Her average speed would be 69.2 miles per hour

Explanation:
When you take the number of miles and divide it by the number of hours you get 69.2 (when rounded)
That is how fast she went per hour

a
(2) A 800 g block is pushed up an inclined plane (angled at 18°) with a velocity of 11.8 m/s. The first block slides up the
incline a distance of 2.2 m and strikes a second block with a mass of 300 g also moving at 3.4 m/s up the incline.
The two blocks hit and stick together. Determine the following:
(i) The maximum vertical height of the two blocks when they stop.
(ii) The time needed for the blocks to reach the bottom of the incline after the moment of impact.
(u = 0.19)

Answers

this is my attachment answer hope it's helpful to you

a box takes 350 N to start moving the coefficient of static friction is 0.35. what is the weight of the box?

Answers

Answer:

101.937 kg

Explanation:

The force needed to get the box moving must just cancel the static friction force:

F = µsmg  =  0.35•m•9.81 = 350 --->   m = 350 / (0.35•9.81) = 101.937 kg

Again, with units shown, and using 1 N = 1 kg•m/s2:

0.35•m•9.81(m/s2) = 350 N

solving for m:

m = 350 N / (0.35•9.81 m/s2)  =  (350 kg•m/s2) / (3.434 m/s2 ) = 101.937 kg

______________________

(Hope this helps can I pls have brainlist (crown)☺️

Find the velocity of a body of mass 100 g having kinetic energy of 20 J.​

Answers

Answer:

Explanation:

Let v be the velocity of object ,

Given that mass of object , m=100 g=  

1000

100

kg=0.1 kg

& kinetic energy of object =20 J

K.E  =  

2

1

 mv  

2

 

20=  

2

1

×0.1v  

2

 

⟹v  

2

=400

⟹v=20 m/s

why the max Static frictional force is a Little bit bigger than the sliding frictional force​

Answers

Static Friction is more than sliding friction because, in case of STATIC FRICTION, the body needs to overcome the inertia of rest. And also in case of SLIDING FRICTION, the body posses inertia of motion since the body is already in motion.

Hope it helps!

Calculate the force applied if 100 N/m² pressure is exerted over the area of 0.2m²​

Answers

Answer: 100 pascals times 0.2 m^2= 20 N force is applied.

Explanation:

Which object has the most thermal energy?
A. A 6 kg rock at 10°C
B. A 10 kg rock at 10°C
C. A6 kg rock at 15°C
D. A 10 kg rock at 15°C

Answers

Answer:

D is the answer

Explanation:

D is the most highest one so

the answer is D

Answer:

10 kg rock at 15 degrees celcius

how many tries did it take to invent the lightbulb?

Answers

It took 1,00 attempts

I need help making a chart (preferably a line graph) with this data. I'm not sure how to show that it will be a linear line in a graph. If you don't have excel can you explain to me how to do it.


note: these datasets are separate and arent related

Spring Constant (N/m) Displacement (m)
100 0.5m
300 0.167m
500 0.1m
700 0.071m

Applied force (N) Displacement (m)
20 0.2m
30 0.3m
50 0.5m
70 0.7m

Answers

Answer:

the graph plotted based on your data, you did not mention what are the variables on the x-axis and y- axis . and the graphs can be three small graphs of different material of one quality i.e. thickness or length

hope it helps

Explanation:

solve for the BMI weight 58kg Height 1.61​

Answers

Answer:

Explanation:

BMI= weight/(height × height)          ; weight in kilogram and height in metter

     = 58kg / (1.61m  × 1.61m )

     = (58/ 2.5921) kg/[tex]m^{2}[/tex]

     = 22.375  kg/[tex]m^{2}[/tex]

     ≈ 22.4 kg/[tex]m^{2}[/tex]

Part IV Objects on an incline w/ Tension + Friction
1. A small 63 kg sleigh is pulled by a rein attached to horse up a 15'angle hill to the
horizontal. The tension of the rein is 510 N. The coefficient of kinetic friction is
0.25
a. What is the normal force that the sleigh exerts on the hill?
b. What are the magnitude and direction of the sleigh's acceleration?

Answers

(a) The normal force on the sleigh is 596.36 N.

(b) The magnitude and direction of acceleration of the sleigh is 3.2 m/s² upwards.

The given parameters;

mass of the sleigh, m = 63 kginclination of the hill, θ = 15⁰tension on the rein, F = 510 Ncoefficient of friction, μ = 0.25

The normal force on the sleigh is calculated as follows;

[tex]F_n = mg \times cos(\theta)\\\\F_n = 63 \times 9.8 \times cos(15)\\\\F_n = 596.36 \ N[/tex]

The magnitude and direction of acceleration of the sleigh is calculated as follows;

[tex]\Sigma F= ma\\\\F - mgsin(\theta) - F_f = ma\\\\F - mgsin(\theta) - \mu F_n = ma\\\\510\ - \ 63 \times 9.8 \times sin15 \ -\ 0.25\times 596.36 = 63a\\\\201 .11 = 63a\\\\a = \frac{201.11}{63} \\\\a = 3.2 \ m/s^2 \ upwards[/tex]

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Anna used a rock to drive a peg into the ground to put up her tent. If the rock applied a 9. 5 N force in 0. 50 s, what is the impulse on the peg? 4. 8 N • s 9. 5 N • s 10 N • s 19 N • s.

Answers

Answer:

4.8 N*s

Explanation:

P = F*t

P = Impulse/ momentum

F = force (Newtons)

t = time (seconds)

P = 9.5N*0.5s

P = 4.75 N*s

Hi there!

[tex]\large\boxed{I = 4.8 N}[/tex]

Impulse = Force (N) × Time (s)

We can also express this as:

I = Δp = Ft

Plug in the given values:

I = 9.5(.50) = 4.75 ≈ 4.8 Ns


What mass M will cause this system accelerate e 0.75 m/s²?

Answers

Answer:

5.54 [kg].

Explanation:

1) if m₁=6; g=9.81; a=0.75 and m₂ - reqired mass, then

2) it is possible to write: m₁*a=m₁*g-m₂*g, and

3) [tex]m_2=\frac{m_1(g-a)}{g}; \ => \ m_2=\frac{6*(9.81-0.75)}{9.81}=5.54[kg].[/tex]

SOMEONE PLEASE HELP ME IM GOING INSANE 4 Select the correct answer. A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled 2.0 x 107 meters and is only 10 meters away from its starting point. What is the numerical value of the satellite's average velocity after that one hour? O A. -3.77 x 10-2 meters/second OB. -2.77 x 10-3 meters/second OC. -2.62 x 10-2 meters/second D. -5.55 x 103 meters/second​

Answers

Hi there!

Since we are dealing with such large distances, we can approximate.

Recall that the equation for speed is:

displacement / time = velocity

OR:

d/t = v

Begin by converting one hour to seconds:

1 hr  = 3600 sec

Now, we can solve for velocity:

(-2.0 × 10⁷) / 3600 = -5555.6 m/s ⇒ D: -5.55 × 10³ m/s

how do astronomers determine the origin of a meteorite that reaches earth?

Answers

Researchers begin their study by identifying the rock type and then dating the meteorite

Explanation:

.

1 point

What is the acceleration of a bicyclist moving at a constant speed of 10

m/s for 5 seconds? *

PLEASE HELP HURRY

Answers

Answer:

0

Explanation:

The bicyclist is not speeding up or slowing down so there is no acceleration.

A force acts on an object. Which option describes an action that could prevent the object from moving

Answers

Answer:

Friction or tension

Explanation:

Friction stops an object from moving in the presence of force

A long-distance runner runs at a constant speed of 4.8 m/s. How long does it take them to run 1.5 km?

Answers

Convert km to meters:

1km = 1000 m

1.5 km x 1000 = 1500 m

1500m / 4.8 m/s = 312.5 seconds

312.5 seconds / 60 seconds per minute =5.2 minutes = 5 minutes 12.5 seconds

Consider the schematic of the molecule shown, with two hydrogen atoms, H, bonded to an oxygen atom, O. The angle between the two bonds is 106°. If the bond length r = 0.103 nm long, locate the center of mass of the molecule. The mass mH of the hydrogen atom is 1.008 u, and the mass mO of the oxygen atom is 15.9999 u. (Use a coordinate system centered in the oxygen atom, with the x-axis to the right and the y-axis upward. Give the coordinates of the center of mass in nm.)

Answers

The definition of the center of mass allows to find the result for the position of the mass center of more than the H₂O molecule is;

         [tex]x_{cm} = 0 \ and \ y_{cm} = 6.9 10^{-3 } nm[/tex]  

the concept of center of mass of a system is the point where external forces are applied, it is given by the expression

             [tex]\frac{x}{y} =\frac{1}{M_{total}} \sum m_i r_i[/tex]  

Where M is the total mass of the systemr_i and m_i sums the position and masses of the element i of the system

In the attachment we have a diagram of the system where the axis and coordinates of the molecules are shown, in this case it is indicated that the origin is in the oxygen atom, so its distance is zero.

           [tex]r_{cm} = \frac{1}{2 m + M} \ (2 m r )[/tex] )

They indicate the mass of the hydrogen atom m = 1.008 amu, the bond length r = 0.103 nm and there is an angle 106º between the two hydrogens, therefore the angle from the vertical is:

           θ = 106/2 = 53º

Let's find the position of the center of mass for each axis.

x-axis

           [tex]x_{cm} = \frac{1}{2m+ M} \ ( m x_1 - m x_2)[/tex]  

y-axis

          [tex]y_{cm} = \frac{1}{2m + M} \ ( m y_1 + m_2)[/tex]

Let's use trigonometry to find the components of the bond length.

         cos θ = [tex]\frac{y}{L}[/tex]  

         sin θ = [tex]\frac{x}{L}[/tex]  

         y = L cos θ

         x = L sin θ  

We substitute.

          [tex]x_{cm} = \frac{1}{2m+M} \ (mL (sin 53 + sin (-53)) \\y_{cm} = \frac{1}{2m + M} \ ( mL cos 53 + mL sin 53)[/tex]

we use.  

          sin θ = - sin -θ

          cos θ = cos -θ

Let's calculate.

        [tex]x_{cm} = 0 \\y_{cm} = \frac{1}{2 \ 1.008 + 15.9999} \ ( 2 \ 1.008 \ 0.103 cos 53)[/tex]

       

We see that the center of mass is on the x axis and at a distance from the y-axis of 6.9 10-3 nm

In conclusion using the definition of the center of mass we can find the result for the position of the center of mass of the H₂O molecule is;

          [tex]x_{cm}=0[/tex]  and [tex]y_{cm}[/tex]cm = 6.9 10⁻³ nm

Learn more here: brainly.com/question/8662931

Which statement indicates that motion has occurred?

A. The reference point has changed.

B. The position of the object has changed.

C.The object has not changed.

D. The object being described has changed​

Answers

The answer is b i think

1 point
A roller coaster is 20 m high and traveling 11 m/s. If the roller coaster has a
mass of 150 kg, what is its total energy? (only put the number, no units or
commas)

Answers

Answer:

ETotal=38505J

Explanation:

Known values:

m=150kg

h=20m

v=11m/s

KE=mv^2/2

PE=mgh

ETotal=PE+KE

KE=(150(11^2))/2

KE=9075

PE=150(9.81)(20)

PE=29430

ETotal=9075+29430

ETotal=38505J

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