What is the horizontal acceleration of a ball that is launched horizontally with a velocity of 5. 6 m/s? â€""9. 8 m/s2 0 m/s2 5. 6 m/s2 9. 8 m/s2.

Answers

Answer 1

The horizontal acceleration of a ball that is launched horizontally is equal to: B. 0 [tex]m/s^2[/tex]

Given the following data:

Horizontal velocity = 5.6 m/s

To determine the horizontal acceleration of a ball that is launched horizontally:

Acceleration can be defined as the rate of change of velocity with respect to time. In Physics, the acceleration of an object or body is calculated by subtracting its initial velocity from the final velocity and dividing by the time.

Mathematically, acceleration is given by the formula;

[tex]Acceleration = \frac{V\; -\;U}{t}[/tex]

The projectile motion of an object launched horizontally has a constant velocity and as such its horizontal acceleration would be equal to zero meter per seconds square ([tex]0\;m/s^2[/tex]).

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Related Questions

A transformer is being built to triple the voltage produced. Which data are collected to investigate whether the transformer can achieve the goal? (1 point)

A) the ratio of the number of turns in the primary and secondary coils

B) only the number of turns in the secondary coil and the magnetic field of the core

C) only the number of turns in the primary coil and the magnetic field of the core

D) the ratio of the number of turns in the primary and secondary coils and the magnetic field of the core

Answers

Answer:

the ratio of the number of turns in the primary and secondary coils.

Explanation:

hope i'm not too late!!

how do i do this question in science?

Answers

simple subtract and ummmmmm.....

Can u plz help me with my hw

Answers

Answer:

1) 4 x 25m = 100m

2) 0 because after 4 lengths, he's back at the starting block.

3) speed is distance over time so speed here is 100m/125s = 0.8m/s

4) ...

5) 100m / 1.25m/s = 80s

6.a) 100m / 0.5m/s = 200s

7.b) ... can't draw this now..

how have astronomers interpreted the unexpectedly fast rotation of galaxies

Answers

Answer:

There must be a lot of dark matter that can be felt but not seen

The astronomers interpreted the unexpectedly fast rotation of galaxies that there must be a large quantity of dark energy whose gravity is detectable yet invisible.

What is a galaxy?

Any system of stars plus interstellar material that makes up the cosmos is referred to as a galaxy. Such assemblages are common, and many of them are so massive that they hold tens of trillions of stars.

A vast variety of galaxies, from dim, hazy dwarf objects to spectacular spiral-shaped giants, have been created by nature. Almost all galaxies seem to have formed shortly after the universe started, and they are everywhere in space, even at the farthest limits of the universe that can be seen by the most advanced telescopes.

The majority of galaxies are found in clusters, many of which are further organized into clusters that span hundreds of billions of light-years.

Since there are almost empty spaces between these so-called super clusters, the universe's overall structure resembles a network of sheets or chains of galaxies.

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4) An 873kg dragster, starting from rest, attains a speed of 26.3m/s in 0.590s.
A) What was the acceleration of the dragster?
B) What was the net force needed to accelerate the dragster?
C) Assume that the driver has a mass of 68.0kg. What horizontal force does the seat exert on
him?

Answers

Answer:

A.)a=44.57m/s

B.)f=38,909.61

C.)f=666.4

Explanation:

A.) A=v/t 26.3/0.590=44.57

B.) 873kg x 44.57= 38,909.61

C.) 68kg x 9.8=666.4

A particular roller coaster has a mass of 3500 kg, a height of 4.0, and a velocity of 12m/s. What is the potential energy? If needed, use g=10.m/s^2

Answers

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Gravitational Potential Energy of an object is calculated by formula ~

[tex] \large\boxed{\sf P = mgh}[/tex]

where,

m = mass of the object = 3500 kg

g = Acceleration due to gravity = 12 m/s²

h = height attained by the object = 4 m

Now, let's calculate its potential energy ~

[tex]3500 \times 10 \times 4[/tex]

[tex]140000 \: \: joules[/tex]

[tex]140 \: \: kj[/tex]

Answer:

Potential Energy of an object is calculated by formula:

Potential Energy (P.E)=m×g×h

Where,

m=mass of bodyg=acceleration due to gravityh=height from the earth surface

Now, let's solve the question.

Given,

mass(m)=3500 kgheight (h)=4mvelocity (v)=10m/s²

Now,

We know that,

Potential Energy (P.E)=mgh

[tex] = 3500 \times 10 \times 4[/tex]

[tex] = 3500 0\times 4[/tex]

[tex] =140000 joules [/tex]

[tex]\mathfrak{\blue{DisneyPrincess29}}[/tex]

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

The kinematics and Newton's second law allow to find the results for the questions about the motion of the ball are:

The acceleration is: a = 5.4 10³ m / s² The force is: F = 7.84 10² N

Given parameters

Mass of the ball m = 0.145 kg Ball starts speed vo = 38.2 m / s Setback distance x = 0.135 m

To find

Acceleration The force

Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.

         v² = v₀² - 2 ax

Wher vo es la initial veloicity, a the acceleration y x is the distance,

When the ball stops the velocity is zero.

         0 = v₀² - 2ax

         a = [tex]\frac{v_o^2}{2x}[/tex]  

Let's calculate

         a = [tex]\frac{38.2^2}{2 \ 0.135}[/tex]  

         a = 5.4 10³ m / s²

Newton's second law establishes a relationship between force, mass, and acceleration of the body.

             F = ma

             F = 0.145 5.4 10³

             F = 7.84 10² N

In conclusion using kinematics and Newton's second law we can find the results for the questions about the motion of the ball are:

The acceleration is: a = 5.4 10³ m / s² The force is: F = 7.84 10² N

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easy one - giving brainly if correct.​

Answers

Gas.

HOPE YOU GET 100!

when light enters water, it bends. what does the amount of bending depend on?

Answers

Answer:

the change of speed!

Explanation:

if light speeds up or slow down more it bend more

Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, this gun accelerates a 5.6-kg projectile from rest to a speed of 3.5 × 103 m/s. The net force accelerating the projectile is 2.5 × 105 N. How much time is required for the projectile to come up to speed?

Answers

Answer:

7.84 s

Explanation:

Ok, one step at a time. We know the mass of the projectile, we know the force. Acceleration is easily obtained from Newton second law:

[tex]\vec F = m\vec a[/tex][tex]2.5\times 10^5 = 5.6a \rightarrow a = (2.5/5.6) \times 10^5 = 4.46 \times 10^4 ms^-^2[/tex].

At this point, we know the acceleration, we know initial and final velocity, we can time the ammount of time it took to get there.

[tex]v= v_0 +at \rightarrow 3.5\times10^3 = 0 + 4.46\times10^4 t\\t= \frac{3.5\times10^3}{4.46\times10^4} = 7.84 s[/tex]

Help, please. I am not sure what to do.

Answers

Answer:

option D) -3m

Explanation:

if 6m is diplaced by -3m then it would be -3+6=3m

feel free to ask if you are confused

How fast is an object going if it travels from San Diego to Anaheim in 1.25 hours (hr)? The distance from San Diego to Anaheim is 93 miles (mi).

a) 74.4 mi/hr

b) 116.25 mi/hr

c) 0.013 mi/hr

d) 84.25 mi/hr

e) None of the answers

Answers

Answer:

74.4 mph

Explanation:

Since we have the distance in miles and the time it took in hours, we can divide the two to get miles per hour.

speed = mi/hr

speed = 93/1.25

speed = 74.4 mph

which wave has a higher frequency and why?

Answers

Explanation:

the figure in the left side has higher frequency.

because it has more nos. of wave in 1sec.

A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One knot = 0.51 m/s.)


show all steps

Answers

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

[tex]\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\[/tex]

To get the required work done, we will divide the mass by the speed of one knot to have:

[tex]w=\frac{7230}{0.51}\\w= 14,176.47Joules[/tex]

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

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Calculate torque using angular momentum

Answers

Answer:

The equation net τ=ΔLΔt net τ = Δ L Δ t gives the relationship between torque and the angular momentum produced.

if you have 3 moles of iron, how many grams of iron do you have?

Answers

Answer:

55.84 look at periodic table

so 55.84*3= 167.52gFe

Acceration is a change in motion over time:
True False

Answers

The answers is definitely true for a fact

A bob attached to a string of length L = 1.25 m, initially found at the equilibrium
position, is given an initial velocity v = 0.8 m/s. The maximum displacement angle,
Omax, is: (Take g=10 m/s)
17
3.2
13
10.2
6.4
Clear selection
A hoh attached to a string of length 2 m is displaced by an angle of 8º and then

Answers

The maximum displacement angle of the bob is 13⁰.

The given parameters;

Length of the pendulum, L = 1.25 mInitial velocity of the bob, v = 0.8 m/s

The maximum displacement of the bob is calculated by applying the principle of conservation of energy;

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m[/tex]

The maximum displacement angle is calculated as follows;

[tex]cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0[/tex]

Thus, the maximum displacement angle of the bob is 13⁰.

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I need help in question 7, a and b.

Answers

Answer:

The graph for 7a is shown in the attachment. For question 7b she walks a distance of 16 meters. (m)

Explanation:

. Quỹ đạo chuyển động của một vật là

Answers

Answer:

you are very kind Po lang

7. A bus covers a certain distance in 60 minutes if it runs at a speed of 60 km/hr.
What must be the speed of the bus in order to reduce the time of journey by 40
minutes?

Answers

Answer:

90 Km/h

Explanation:

60 mins is 1 hr

so in an hour bus covers 60 Km

so new speed:

d/t

(60km/40mins)*60mins/h

90Km/h

Which of the following is NOT true of a hotspot?

Answers

Given what we know about hotspots and their characteristics, we can confirm that the option that is false regarding hotspots is "A. Hotspots are constantly moving forming chains of volcanoes".

In regards to the study of geology, we refer to hotspots as spots in an oceanic mantle where magma located below causes them to become intensely hot. This heat forces magma and therefore the mantle itself upwards as a tectonic plate moves across the spot, and therefore results in the formation of volcanic chains. Some very popular examples of volcanoes and volcanic islands formed by this method include:

Yellowstone National ParkHawaiiThe country of Iceland

Despite the other options being true, option A is false given that hotspots are stationary elements of the oceanic mantle. Although some scientists believe that there is evidence of the movement of hotspots, this movement is slow enough to not be taken into account.

This question was answered in regards to the complete question found online which states:

Which of the following is NOT TRUE about hotspots? a. Hotspots are constantly moving forming chains of volcanoes.

b. Hotspots are stationary.

c. Hotspots form volcanic island arcs

d. Hotspots lie along an oceanic plate

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can someone sent some gravity questions

Answers

What do you mean ???

Answer:

can gravity from waves

Explanation:

yes gravity can froms way

a bus increase its velocity from 30m/s to 45m/s and covers 300m distance. Find the time​

Answers

Answer:

3.5 m/s²

Hope this helps, have a nice day/night! :D

Se aplican dos fuerzas concurrentes a un objeto de 4N a la derecha y 5N a la izquierda. ¿Hacia donde se movió y con cuanta fuerza?

Answers

The forces move strongly towards the left by 1N

Given the following

Force towards the right = 4N

Force towards the left = 5N

Note that the force acting towards the left is negative, hence the force acting towards the left is -5N

Take the sum of force

Resultant force = -5N + 4N

Resultant force = -1N

This shows that the forces move strongly towards the left by 1N

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Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an
an unbalanced force of 42 Newtons has been applied?
A 75.60 kg
B 00.04 kg
C 23.33 kg
D 43.80 kg

Answers

Answer:

23.33 kg

Explanation:

The mass of the object can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{42}{1.8} = 23.3333... \\ [/tex]

We have the final answer as

23.33 kg

Hope this helps you

Answer:

[tex]\boxed {\boxed {\sf C. \ 23.33 \ kg}}[/tex]

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

[tex]F=ma[/tex]

The mass of the ball is unknown. The ball is accelerating at 1.8 meters per second squared. An unbalanced force of 42 Newtons is applied to the ball.

Convert the units of force. 1 Newton is equal to 1 kilogram meter per second squared, so our answer of 42 Newtons is equal to 42 kg*m/s².

F= 42 kg*m/s² a= 1.8 m/s²

Substitute the values into the formula.

[tex]42 \ kg*m/s^1 = m * 1.8 \ m/s^2[/tex]

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 1.8 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 1.8 m/s².

[tex]\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} = \frac{a*1.8 \ m/s^2}{1.8 \ m/s^2}[/tex]

[tex]\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} =m[/tex]

The units of meters per second squared cancel.

[tex]\frac {42 \ kg}{1.8 } =m[/tex]

[tex]23.3333333 \ kg=m[/tex]

Round to the hundredth place. The 3 in the thousandth place tells us to leave the 3 in the hundredth place.

[tex]23.33 \ kg \approx m[/tex]

The mass of the ball is approximately 23.33 kilograms.

what is Secular Music's instrument's?​

Answers

Answer:

Drums, harps, recorders, and bagpipes.

Explanation:

A car traveling at 35.6m/s crashes into a concert barrier and comes to a stop in 0.35 seconds. Calculate the average force applied to the 75kg driver.A 3.2kg steel ball traveling at 4.1m/s strikes a second ball of a mass 2.3kg Initially at rest. Calculate the velocity of the second ball when the first one continues traveling in the same direction with a speed of 1.5m/s2 balls of putty are shot towards one another. Ball 1 has a mass of 4.3kg and is moving at 18.6m/s . Ball 2 has a mass of 5.8kg and is moving at 9.5m/s. They collide and stick together. Calculate their final combine velocity.I really appreciate those attempting the problems. I do know the answers but I’m unaware of the steps to get there. Please include all formulas in your response and steps so I can learn and understand.Check your answer:7629N3.6m/s2.46m/sThank you all!

Answers

The force on the driver is 7629 N. The velocity of the second ball is 3.6 m/s. The combined velocity of the balls is 13.37 m/s.

We have to find the acceleration using;

v = u - at

v = final velocity = 0 m/s

u = initial velocity = 35.6m/s

a = acceleration = ?

t = time = 0.35 s

u = at

a = u/t = 35.6m/s / 0.35 s

a = 101.7 ms-2

The force on the driver =  75kg ×  101.7 ms-2 = 7629 N

Using the principle of conservation of momentum;

Momentum before collision = momentum after collision

m1u1 +m2u2 = m1v1 + m2v2

Hence

(3.2 × 4.1) + 0 = (3.2 × 1.5) + 2.3v2

13.12 = 4.8 + 2.3v2

13.12 - 4.8 = 2.3v2

v2 = 13.12 - 4.8/2.3

v2 = 3.6 m/s

Using the principle of conservation of linear momentum;

m1u1 + m2u2 = m1v1 + m2v2

(4.3 × 18.6) + (5.8 × 9.5) = (4.3 + 5.8) v

v = 79.98 + 55.1/10.1

v = 13.37 m/s

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The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at 3. The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at atmospheric pressure, would have to be admitted into the space to cause the column of the mercury
to drop to 59 cm?

Answers

The ideal gas equation and the pressure in barometer allows us to find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:

The  variation  of the volume is: ΔV = 7.67 cm³

Pressure is defined by the relationship between force and area.

       P = F / A

The ideal gas equation establishes a relationship between pressure, volume, and temperature of an ideal gas.

          PV = nR T

Where P is pressure, V is volume, and T is temperature.

Let's write this equation for two points assuming that the temperature has not changed.

          P₀ V₀ = P₁ V₁

          V₁ = [tex]\frac{P_o}{P_1} \ V_o[/tex]                 (1)

The subscript "o" is used for the start point and the subscript "1" for the end point.

The pressure in a barometer is:

         P = ρ g y

They indicate the initial height of the barometer y₀=75 cm, the distance from empty space y'₀ = 9 cm and the final height of the barometer y₁ = 59 cm.

 

The volume of the cylinder is

         V = π r² y

Let's calculate the initial volume.

         V₀ = π 1 9

         V₀ = 28.27 cm³

We substitute in equation 1.

         V₁ = [tex]\frac{\rho \ g \ y_o}{\rho \ g \ y_1} \ V_o[/tex]  

         V₁ = [tex]\frac{y_o}{y_1} \ V_o[/tex]  

Let's calculate.

        V₁ = [tex]\frac{75}{59} \ 27.27[/tex]  

        V₁ = 35.94 cm³

The volume to be incremented is

         ΔV = V₁ - V₀

         ΔV = 35.94 - 28.27

         ΔV = 7.67 cm³

Using the ideal gas equation and the pressure in barometer we can find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:

The change of the volume is: ΔV = 7.67 cm³

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What symbols are these?

Answers

Answer:

the bottom one is wollsiegel

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