1) An object located 33.9 cm in front of a lens
forms an image on a screen 8.57 cm behind
the lens.
Find the focal length of the lens.
Answer in units of cm
2) What is the magnification of the object?

Answers

Answer 1

Answer:

Explanation:

Given:

d = 33.9 cm

f = 8.57

___________

F - ?

G - ?

The focal length of the lens:

F = d*f / (d + f) = 33.9*8.57 / (33.9 + 8.57) ≈ 6.84 cm

The magnification of the object:

M = f / d = 8.57 / 33.9 ≈ 0.25

Answer 2

d= 33.9 cm
f= 857
F-?
G-?
The focal length of the lens:
F= d* / (d + f) = 33.9*8.57 / (33.9 + 8.57) = 6.84 cm
The magnification of the object:
M=f/d =8.57 / 33.9 ÷ 0.25

Related Questions

A 131 kg electric motorcycle moving at 30 mph accelerates by 0.03 m/s^2 for 60 s in the same direction that it’s moving. What is the average velocity during this period? Answer in km/hr.

Answers

We would calculate the distance travelled by the motor by applying one of Newton's formula for calculating motion which is expressed as

s = ut + 1/2at^2

where

s is the distance travelled

u is the initial velocity

t is the time

a is the acceleration

From the information given,

u = 30 mph

We would convert it to m/s

Recall,

1 mile = 1609.344 m

1 hour = 3600 s

Thus,

30mph = (30 x 1609.344)/3600

u = 13.4112 m/s

t = 60 s

a = 0.03m/s^2

Thus,

s = 13.4112 x 60 + 1/2 x 0.03 x 60^2

s = 804.672 + 54

s = 858.672 m

We would convert 858.672 m to km and 60s to hours. It becomes

858.672 m = 858.672/1000 = 0.858672 km

60s = 60/3600 = 1/60 hr

Average velocity = 0.858672 km/(1/60 hr)

Average velocity = 51.5 km/hr

An ocean fishing boat is drifting just above a school oftuna on a foggy day. Without warning, an engine backfireoccurs on another boat at a distance of d = 1.55 km(Figure 1). The speed of sound in air is 343 m/s, and insea water is 1560 m/s.How much time elapsed before the backfire is heard by the fishermen? Assume that the fishermen hears the sound that travels only through air.

Answers

For the fisher man use speed of sound in air as v = 343

d= vt

t= d/v

where:

t= time

d= distance = 1.55 km = 1550 m

v= speed= 343 m/s

t= 1550m / 343 m/s = 4.52 s

Biggs and Smalls play tug-of-war with a 12-meter rope in P.E. class. Both are in their socks on a waxed gym floor, which is nearly frictionless. After a brief time, they meet. If Biggs has twice the mass of Smalls, how far has Biggs moved?

Answers

After a brief time, Biggs and smalls would meet at 4m from Biggs' side if the rope is 12 meter long and the floor is frictionless.

[tex]X_{cm}[/tex] = ( m1 x1 + m2 x2 ) / ( m1 + m2 )

[tex]X_{cm}[/tex] = Center of mass

After a brief time, Biggs and smalls would meet at the point of centre of mass. Let Biggs be 1 and smalls be 2 and the distances be measured from Biggs' side assuming that Biggs is standing on the left side.

m1 = 2 * m2

x1 = 0 m

x2 = 12 m

[tex]X_{cm}[/tex] = [ ( 2 * m2 * 0 ) + ( m2 * 12 ) ] / [ ( 2 * m2 ) + m2 ]

[tex]X_{cm}[/tex] = 12 m2 / 3 m2

[tex]X_{cm}[/tex] = 4 m

Center of mass is the average position of all the objects on a system, weighed according to the mass of individual objects. It is a position relative to an object.

Therefore, after a brief time, Biggs and smalls would meet at 4 m from Biggs' side.

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Identify the time interval during which the velocity of shuttle bus is zero.​

Answers

Explanation:

during D and E, so between 13 and 16 (whatever time unit this is - hours, minutes, ... ?).

Answer:

Explanation:

The time interval 13 to 16 seconds

A ball is thrown up in the air, it reaches its maximum height and falls back. It’s acceleration is:A - same up and down but zero at the topB- less on the way up than on the way downC- the same ally all points of motionD- less on the way down than on the way up

Answers

When the ball is thrown up in the air, it is called a Freefall motion.

In freefall motion, the acceleration remains constant at all times because the acceleration due to gravity is always constant.

The acceleration due to gravity is 9.81 m/s^2

The ball has zero velocity (not acceleration) at the maximum height.

Therefore, the correct answer is option C

"Its acceleration is the same at all points of motion".

What space telescope is used to look deep into the universe?
O A.
OB.
O C.
Hubble
Galileo
Penzias
OD. Edwin

Answers

Hubble is the correct answer

What is the first step in scientific method?
A) Gather information
B) state the problem
C) form a hypothesis
D) perform the experiment

Answers

The first step in scientific method is to state the problem. That is option B.

What are scientific methods?

The scientific methods are those methods that are used by scientists and researchers to discover answers to specific questions that are being asked and it involves the use of scientific steps and guidelines.

These steps include the following:

State the problem: This is the first step where the problem to be investigated is being defined.Form a hypothesis: This is the step that involves the making of predictions based on the research.Gather information: The data that is related to the problem are being collected.Perform the experiment: Analysis are being carried out on the collected data.

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An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.0 m/s/s. Find the final speed and the displacement after 5.0s

Answers

The final speed of the automobile with an initial speed of 4.30 m/s that rates uniformly at the rate of 3.0 m/s/s for 5 s is 19.3 m / s and its displacement will be 59 m

v = u + at

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

u = 4.3 m / s

a = 3 m / s / s

t = 5 s

v = 4.3 + ( 3 * 5 )

v = 19.3 m / s

s = ut + 1 / 2 at²

s = Displacement

s = ( 4.3 * 5 ) + ( 1 / 2  *3 * 5 * 5 )

s = 21.5 + 37.5

s = 59 m

Therefore,

Final speed = 19.3 m / sDisplacement = 59 m

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The inside of the Earth is hot due to heat left over from when the Earth formed and due to heat generated by the radioactive decay of elements in the Earth's mantle and crust. Heat from these sources flows to the surface at a total rate of of 47 terrawatts. If the Earth were instantly teleported into deep space so that absorption of any outside radiation stopped, what would the equilibrium surface temperature of the Earth be, using e=0.6 for the emissivity, and R = 6400km for the radius? Write the temperature in Kelvin, just the number without units.

Answers

The temperature at the earth's surface would be

40.48K

This is further explained below.

What is Stefan Boltzmann's law:?

Generally, In accordance with the Stefan-Boltzmann equation, the total radiant heat output that is radiated from a surface is proportional to the fourth power of that surface's absolute temperature.

According to the Stefan-Boltzmann law, the total energy radiated per unit surface area of a black body across all wavelengths per unit of time is directly proportional to the fourth power of the black body's thermodynamic temperature T: j=.

To be more specific, the Stefan-Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit of time is directly proportional to the black

[tex]P=\sigma e AT^4[/tex]

Where P stands for power, e stands for emissivity, A stands for the surface area of the earth, and T stands for the temperature at the surface.

[tex]&P=47 * 10^{12} W, \sigma \\\\=5.67 * 10^{-8} W / m^2 . K^4, e ,[/tex]

=0.6

[tex]\\\\&A=4 \pi R^2 \\\\=4 \pi *\left(6400 * 10^3\right)^2 \\\\[/tex]

=5.147 * 10^{14}m^2

[tex]T=\left(\frac{P}{\sigma e A}\right)^{1 / 4} \\\\=\left(\frac{47 * 10^{12}}{5.67 * 10^{-8} * 0.6 * 5.147 * 10^{14}}\right)^{1 / 4}[/tex]

=40.48K

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The gravitational force between two masses is 36 N. What is the gravitational force if the distance between them is tripled? (G = 6.673 x10 (Power of-11)

Answers

If the distance between them is tripled, the gravitational force is F2= 36/9=4N. The force of attraction between any two bodies is proportional to the product of their masses and inversely proportional to the square of their distance.

WHAT IS GRAVITATIONAL FORCE?

It is the force that connects all masses in the universe, particularly the pull of the earth's mass on objects near its surface. It follows the inverse square law. The force of attraction exerted on a body by the earth is known as gravitational force. For example, the leaves and fruits of a tree fall to the ground, water in a river flows down streams, and a ball thrown up travels to a height before returning to the ground are all examples of motion caused by gravitational force.

The formula for Gravitational force:

F=G m1*m2/r² ,

G: gravitational constant = 6.67x10⁻¹¹ N*m²/kg²

m1: is the mass of the first object

m2: is the mass of the second object

r: is the distance between the center of the masses of the objects

F∝1/r²

so F1/F2=( r2/r1)².......(1)

Given F1= 36N F2=?

let r1=r, r2= 3r

putting all this value in equation (1)

36/F2= (3r/r)²

⇒F2= 36/9=4N

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A 3.5-inch floppy disk in an old computer rotates with a period of 2.00 x 10^-1 s. Calculate the angular speed of the disk, the linear speed (in inches/sec) of a point on the rim of the disk, and the linear speed (in inches/sec) of a point 0.750 inches from the center of the disk. (Hint: a 3.5 inch floppy disk has a 3.50 inch diameter)

Answers

We have the next information

T=2.00 x 10^-1 s=0.2s

r=3.5 inch

For the angular speed

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

where omega is the angular speed, T is the period

We substitute

[tex]\omega=\frac{2\pi}{0.2}=31.41\text{ rad/s}[/tex]

For the linear speed on the rim of the disc, we will use the next formula

[tex]v=\omega\cdot r[/tex]

in this case r= 3.5/2=1.75 inch

[tex]v=31.41(1.75)=54.97\text{ }\frac{inches}{\text{sec}}[/tex]

Then for the linear speed on the point at 0.750 inches from the center of the disk.

[tex]v=31.41(0.750)=23.56\frac{inches}{\text{sec}}[/tex]

ANSWER

ω=31.41 rad/sec

v on the rim= 54.97 inches/sec

v on the point=23.56 inches/sec

Look at the diagram below.

Encoding technique - Decoding technique
"Written language
Pictures
Billboards
Advertisements
E-mails
Reading
Viewing
Interpreting

Answers

The communication process, which involves understanding parts and meanings, consists of eight key elements: Source Message Channel Receiver Feedback Environment Context, and Interference

Common examples of character encoding systems are Morse Code, Bodo Code, American Standard Code ASCII for Information Interchange, and Unicode. The purpose of encoding is to transform data so that it can be properly and safely processed by another type of binary system data transmitted via e-mail or special characters that appear on web pages.

The goal is not to keep information secret, but to make it available for proper consumption. Coding is the process of transforming data into the formats required for various information processing needs, such as compiling and running programs. Coding uses different patterns of voltage or current levels to represent the 1s and 0s of a digital signal on a transmission line. Common types of line coding are unipolar, polar, bipolar, and Manchester.

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Change 72km/h into m/s

Answers

Answer:

20 m/s

Explanation:

72 km/hr *  1  hr/ 3600sec *  1000m/km     =  20 m/s

List three ways that you can use water in each of its states.

Answers

You can use water as a solid to keeps things cool, like a drink. You can use water as a liquid to drink, or to clean yourself . You can use water as a gas to help humidify a room.

Solving For Temperature (T)
7. How many degrees would the temperature of a 450 g piece of iron increase if 7600 J of
energy are applied to it? (The specific heat of iron is0.4494 J/g x °C)

Answers

Temperature is a unit used to represent how hot or cold something is. It can be stated using the Celsius or Fahrenheit scales, among others. Temperature shows which way heat energy will naturally flow, i.e., from a hotter (body with a higher temperature) to a colder (body with a lower temperature) (one at a lower temperature)

Additionally, we are provided the iron's particular heat as well as the iron skillet's mask.

We must determine the required quantity of heat based on this information. It will be Q = c x mass x delta P using this method.

That is the temperature change. Therefore, we can replace the value here. The c represents the iron's particular temperature. That comes to 0.4499.

The supplied mass is 5.07, and the resulting temperature change will be 250 - 25 °C. If we can solve this, we can determine what Q is worth.

Alternately, we might state that the required amount of heat is 512.19 Joules.

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A soccer ball kicked with a force of 13.5 N accelerates at 6.5 m/s^2. What is the mass of the ball?

Answers

ANSWER:

2.08 kg

STEP-BY-STEP EXPLANATION:

The force is given by the multiplication of the mass and the acceleration, like this:

[tex]\begin{gathered} f=m\cdot a \\ f=13.5\text{ N} \\ a=6.5\frac{m}{s^2} \end{gathered}[/tex]

We replace and calculate for m:

[tex]\begin{gathered} 13.5=m\cdot6.5 \\ m=\frac{13.5}{6.5} \\ m=2.08\text{ kg} \end{gathered}[/tex]

The mass is 2.08 kg

A car accelerates from rest at -1.60 m/s2. (Include the sign in your answers.)(a) What is the velocity at the end of 5.6 s? m/s(b) What is the displacement after 5.6 s?

Answers

Given data

*The given initial velocity of the car is u = 0 m/s

*The given acceleration of the car is

[tex]undefined[/tex]

Can you please explain to me how a steam engine is a example of the 2nd law of thermodynamics

Answers

To find:

How a steam engine is an example of the second law of thermodynamics?

Explanation:

The second law of thermodynamics has more than one statement.

Kelvin-Plank statement:

No process is possible whose sole result is to convert the heat absorbed completely into useful work.

Clausius statement:

No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.

Both of the statements are equivalent.

A steam engine takes the heat from a hot reservoir and converts some parts of it into useful work and the rest of the heat absorbed will be sent to a cold reservoir. Thus the steam engine is an example of the 2nd law of thermodynamics.

The position of a particle is given by the vector < 3.3t2 , 3.2 , - 2.4t >. What is the magnitude of the acceleration at t=0?

Answers

The position of a particle is given by the vector < 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0 is 3.3 m/sec².

What is acceleration?

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.

Given in the question The position of a particle is given by the vector

< 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0.

Differentiating position vector for two times we get acceleration vector,

so acceleration is 3.3 m/sec²  

.The position of a particle is given by the vector < 3.3t² , 3.2 , - 2.4t >. The magnitude of the acceleration at t=0 is 3.3 m/sec²

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Kelley is playing pool and hits the cue into the nine ball, which is directly in the center of the pool table. The nine ball travels a total of 7 meters before stopping, 1 meter to the left of the initial starting point of the nine ball. Which sentence reflects the correct statement regarding distance and displacement when considering the motion of the nine ball?(1 point)

I NEED THE ANSWER NOW. . .I really don't understand displacement

Answers

The displacement of the nine balls is 1 m while the distance travelled by the nine balls is 7 m.

What is the displacement of he ball?

The displacement of the ball is the change in the position of the ball. Displacement describes the shortest distance between the initial and final position of an object.

Mathematically, the displacement of the nine balls is given as;

Δx = xf - xi

where;

xf is the final position of the objectxi is the initial position of the object

If the balls travels 1 meter to the left of the initial starting point, then the displacement of the balls is 1 m. The 1 meter is the shift from the initial position.

The distance of the balls is the total path covered by the balls.

distance = 7 m

Thus, the displacement of the nine balls is the change in the position of the balls while the distance of the nine balls is the total path covered.

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Which is the best statement of the Law of Conservation of Energy? (1 point)

The amount of energy put out by a simple machine is always less than the amount of energy put into the machine

The amount of energy put into a simple machine should be conserved Item

The amount of energy put out by a simple machine depends on its efficiency

The amount of energy put out by a simple machine must equal the amount of energy put into the machine ​

Answers

The best statement of the Law of Conservation of Energy is option D.  The amount of energy put out by a simple machine must equal the amount of energy put into the machine.​

The regulation of conservation of energy states that the entire electricity of an isolated system remains steady; it's far stated to be conserved over time.

The law of conservation of power is a physical law that states strength cannot be created or destroyed but can be modified from one form to some other. any other manner of pointing out this regulation of chemistry is to mention the full electricity of a remoted device stays regular or is conserved within a given body of reference.

The law of conservation of energy states that power can neither be created nor destroyed - best converted from one form of energy to some other. this means that a device constantly has an equal quantity of power unless it is added from the out of doors.

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Exactly 3.3 s after a projectile is fired into the air from the ground, it is observed to have a velocity v⃗ = (8.6 i^ + 4.7 j^)m/s, where the x axis is horizontal and the y axis is positive upward. aDetermine the horizontal range of the projectile.bDetermine its maximum height above the ground.cDetermine the speed of motion just before the projectile strikes the ground.dDetermine the angle of motion just before the projectile strikes the ground.= ∘ below the horizontal

Answers

a. Horizontal range is given by,

[tex]Range=\frac{u^2sin2\theta}{g}[/tex]

where u is the initial speed, and theta is the angle of projection,

Now velocity vector as a function of time is given by,

[tex]\vec{v}(t)=(ucos\theta)\hat{i}+(usin\theta-gt)\hat{j}[/tex]

Given,

[tex]\begin{gathered} \vec{v}(3.3s)=8.6\text{ }\hat{i}+4.7\text{ }\hat{j} \\ \Rightarrow ucos\theta=8.6,usin\theta-(10)(3.3)=4.7 \\ \Rightarrow ucos\theta=8.6,usin\theta=37.7 \\ Dividing, \\ \Rightarrow tan\theta=\frac{37.7}{8.6}=4.3837 \\ \Rightarrow\theta=77.15\degree \\ Hence \\ ucos77.15\degree=8.6 \\ \Rightarrow u=\frac{8.6}{0.2224}=38.67\text{ m/s} \end{gathered}[/tex]

And so, the horizontal range will be,

[tex]Range=\frac{(38.67)^2sin(154.3\degree)}{(10)}=64.85\text{ m}[/tex]

b. Maximum height reached by a projectile is given by,

[tex]\begin{gathered} H_{max}=\frac{u^2sin^2\theta}{2g} \\ \Rightarrow H_{max}=\frac{(38.67)^2sin^2(77.15\degree)}{(2)(10)}=71.07\text{ m} \end{gathered}[/tex]

c. Now time of flight of the projectile is given by,

[tex]t_{flight}=\frac{2usin\theta}{g}=\frac{(2)(38.67)sin(77.15\degree)}{(10)}=7.54\text{ s}[/tex]

this is the time at which the projectile strikes the ground, and so its velocity then would be,

[tex]\begin{gathered} \vec{v}(7.54s)=8.6\text{ }\hat{i}+(37.7-10\times7.54)\hat{j} \\ \Rightarrow\vec{v}(7.54s)=8.6\text{ }\hat{i}-37.7\text{ }\hat{j} \end{gathered}[/tex]

As expected the vertical component of velocity has just been flipped, and therefore its speed will be same as its initial speed i.e.

[tex]v_{strikes\text{ }ground}=u=38.67\text{ m/s}[/tex]

d. Below the horizontal, let the angle be φ, then

[tex]tan\varphi=-\frac{v_y}{v_x}[/tex]

extra -ve sign since its below the horizontal, when the projectile strikes the ground,

[tex]\begin{gathered} v_y=-37.7\text{ m/s} \\ v_x=8.6\text{ m/s} \\ \Rightarrow tan\varphi=-\frac{(-37.7)}{8.6}=\frac{37.7}{8.6}=tan\theta \\ \Rightarrow\varphi=\theta \\ \Rightarrow\varphi=77.15\degree \end{gathered}[/tex]Result: a. 64.85 m, b. 71.05 m, c. 38.67 m/s, d. 77.15°.

A 20.0 N block slides over a horizontal table. If it takes 5.0 N to move the block at constant velocity, what is the coefficient of friction?

A. 0.50
B. 4.0
C. 0.10
D. 0.25

Answers

A 20.0 N block slides over a horizontal table. If it takes 5.0 N to move the block at a constant velocity, the coefficient of friction is D. 0.25.

Given,

Frictional force (F) is 5.0 N

Normal force (N) is 20.0 N

The coefficient of friction (µ), a numerical value, is obtained by dividing the resistive force of friction (F) by the normal or perpendicular force (N) pushing the objects together.

i.e., µ = F/N

Calculating the coefficient of friction :

The frictional force is equal to and opposing the applied force since the box is traveling at a constant speed.

µ = F / N

µ = 5.0 N / 20.0 N

µ = 1/4 N

µ = 0.25 N

Thus, the coefficient of friction is D. 0.25.

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Find the answer to the problem

Answers

The box is moving at a 2.06 m/s rate.

What causes acceleration?

Acceleration is the rate at which velocity changes over time in both the direction and the speed. Anything is said to have been accelerated when it goes faster or slower in a straight line. Even travel on a circle accelerates steadily since the direction is continuously changing.

What metrics are used to quantify acceleration?

Acceleration is measured in meters per second per second (m/s2).

Formulas for

F NET are F NET = m and F NET = F1 + F2.

3 + 3 + 7 Equals 13 when F NET is used.

F NET = m * a then,

where a = F NET/m and

a = 13N / 6.3 kg and 2.06m/s respectively.

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A football is kicked with a velocity of 32.0m/s and at an angle of 42 degrees it takes 4.40s. How high does it travel?

Answers

23.4 m

Explanation

to solve this we need to use the expression:

[tex]h=v\sin (\theta)t-\frac{gt^2}{2}[/tex]

where

[tex]\begin{gathered} h\text{ is the heigth} \\ v\text{ is initial velocity} \\ \theta\text{ is the angle} \\ t\text{ is the time} \end{gathered}[/tex]

then, let

[tex]\begin{gathered} v=32\text{ m/s} \\ \theta=42\text{ \degree} \\ t=4.4\text{ s} \\ g=9.8\text{ }\frac{m}{s^2} \end{gathered}[/tex]

now, replace

[tex]\begin{gathered} h=v\sin (\theta)t-\frac{gt^2}{2} \\ h=32\sin (42)(4.40)-\frac{(9.8)(4.4^2)}{2} \\ h=94.21-94.864 \\ h=-0.654 \end{gathered}[/tex]

let's check the time of flight

[tex]\begin{gathered} t=\frac{2v\sin\theta}{g} \\ t=\frac{2\cdot32\cdot\sin(42)}{9.8} \\ t=4.36 \end{gathered}[/tex]

it means after 4.36 the ball is on the ground.

Step 2

so, to find the maximum heigth we need to use the expression

[tex]\begin{gathered} y_{\max }=\frac{v^2\sin ^2\theta}{2g} \\ \text{replace} \\ y_{\max }=\frac{(32)^2\sin ^2(42)}{2(9.8)} \\ y_{\max }=\frac{1024\cdot0.4477}{2(9.8)} \\ y_{\max }=\frac{458.48}{19.6} \\ y_{\max }=23.39 \end{gathered}[/tex]

therefore, the answer is

23.4 m

I hope this helps you

Energy stored in the kinetic energy container is fundamentally different from energy stored in the elastic energy container.True or false

Answers

The kinetic energy is given by:

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

Where:

m = mass of the object

v = Speed of the object

The elastic potential energy is given by:

[tex]U=\frac{1}{2}k\Delta x[/tex]

Where:

k = Constant of the spring

Δx = Displacement

As we can see, both energy depends on different factors, therefore we can conclude the statement is true

Answer:

True

a lot of this content is conceptual but I'm having a hard time understanding this

Answers

If momentum is denoted by p and wavelength by

[tex]\lambda[/tex]

Then, the wavelength and momentum are related as

[tex]p\propto\frac{1}{\lambda}[/tex]

Thus if momentum is doubled, then the wavelength is halved.

A 1,116 kg space vehicle is traveling along a straight-line at a constant speed of 800 m/s. What is the magnitude of the net force on the space vehicle?

Answers

The magnitude of the net force on this space vehicle moving at constant speed is zero (0) Newton.

What is a net force?

A net force can be defined as the vector sum of all the forces that are acting on a physical object or body. This ultimately implies that, a net force is a single force that substitutes the effect of all the forces acting on a physical object or body.

Generally speaking, the amount of net force that keeps a physical object or body moving at constant speed and in the same direction is always equal to zero (0) Newton in accordance with Newton's First Law of Motion.

In this context, we can reasonably infer and logically deduce that the net force on a physical object or body would be equal to zero (0) Newton when it moves at constant speed.

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The velocity-time table represents the motion of a leftward moving car.

What is the magnitude and direction of the acceleration?

Answers

Given velocity-time table represents the motion of a leftward moving car.

Magnitude of acceleration = -4 m/s^2

Direction is Left.

When any object is slowing down then the acceleration is in opposite direction. Then the object will have a negative acceleration.

Deceleration is the opposite of acceleration. The deceleration is calculated by dividing the difference  final and initial velocity by the amount of time taken for the drop in velocity. The formula for acceleration with a negative sign is used to identify the deceleration value.

Deceleration = (Final Velocity–Initial Velocity)/Time taken

It is written' –a,' where a is acceleration.

When initial velocity, final velocity and time are given, then Deceleration Formula is given by,

Given in the graph, u= 20m/s and v= 4m/s

t= 4sec

a = (v−u)/t

= (4 -20)/ 4

=- 16/4

a= -4 m/s^2

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If Mercury's average orbital distance from the sun is 0.312 AU and k for our sun is 1.00 AU3/yr2, then what is Mercury's orbital period?

Answers

Take into account that by the Kepler's Third Law, you can use the following formula:

[tex]k^{}=\frac{a^3}{T^2}[/tex]

where,

T: period of Mercury

a: average distance from Mercuty to Sun = 0.312AU

k: Kepler's constant = 1.00 AU^3/yr^2

Then, by replacing the previous values of the parameters, you obtain for T:

[tex]T=\sqrt[]{\frac{a^3}{k}}=\sqrt[]{\frac{(0.312AU)^3}{1.00\frac{AU^3}{yr^2}}}\approx0.56yr[/tex]

Hence, the Mercury's orbital period is approximately 0.56 yr

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