A typical atom has a diameter of about 1.32 x 10-10 m. (a) what is this in inches (b) approximately how many atoms arethere along a 1.32 cm line.

Answers

Answer 1

Answer:

A ) [tex]5.197 \times 10 ^{-9} inches[/tex]

b) There are approximately 100,000,000 atoms on the 1.32 cm lines

Explanation:

To ensure the accuracy of our answer, we will have to make sure we work in the appropriate units.

A

To convert    [tex]1.32 \times 10^{-10} m[/tex] to inches, we can use this factor.

1 m = 39.3701 inches

Therefore [tex]1.32 \times 10^{-10} m[/tex] =  

B

To find the number of atoms on a 1.32 cm line, we will first of all need to convert 1.32 cm to meters.

1.32 cm = 0.0132 metres (divided 1.32 by 100 to convert to metres)

Assuming all the atoms are arranged side by side, with their edges touching on the 0.0132m line, the number of atoms present will be

[tex]0.0132/1.32\times 10^{-10}[/tex] = 100,000,000 atoms


Related Questions

Consider two balls in motion at the same time. Joe drops the first ball from rest at height h. Directly below Joe, on the ground, Hayley simultaneously tosses a second ball upward with speed v0.a. If the two balls collide at the moment the second ball is instantaneously at rest, what is the height of the collision?b. What is the relative speed of the balls when they collide?

Answers

Answer:

A. To find height of collision let's find

Speed of collision first

Using

Vf= √2g(h-g)

But for Harvey's ball we have

V²-2gy= 0

y=( v²/2g) so this is height of collision

B.

To find relative speed of the ball

Using V at y to find Vsr

Vs= √2g(h-v²/2g)

= √2gh- v²

hows a map of Olivia's trip to a coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8 mi to where Broadway turns, and then continues another 1.4mi to the shop.

Answers

The question is incomplete. Here is the complete question.

The map (in the attachment) shows Olivia's trip to the coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8mi to where Broadway turns, and then continues another 1.4mi to the shop.

What is the magnitude of the total displacement of her trip?

Whta is the direction of the total displacement of her trip?

Answer: Magnitude = 2.6mi

              Direction: 54.65° east

Explanation: Displacement is the change in postition of a moving object.

There are a few ways to determine total displacement. For this case, the Perpendicular Components of a Vector method will be used.

For this method, total displacement is given by:

[tex]\Delta d_{t}=\sqrt{(\Delta d_{x})^{2}+(\Delta d_{y})^{2}}[/tex]

[tex]\theta=tan^{-1}(\frac{\Delta d_{y}}{\Delta d_{x}})[/tex]

[tex]\Delta d_{x}[/tex] is the x-component of total displacement and it is the sum of each individual x-components;

[tex]\Delta d_{y}[/tex] is the y-component of total displacement and it is the sum of each individual y-components;

θ is the angle the resulting displacement;

For Olivia's trip, there are no x-component of the first part and for the third part, the path she bikes is a hypotenuse of a right triangle. So, that right triangle's x-component is:

[tex]sin30=\frac{x}{1.4}[/tex]

[tex]\frac{1}{2} =\frac{x}{1.4}[/tex]

x = 0.7

Then,

[tex]\Delta d_{x}[/tex] = 0 + 0.8 + 0.7

[tex]\Delta d_{x}[/tex] = 1.5

Related to y, there are no y-component in the second part of Olivia's trip and for the third part:

[tex]cos30=\frac{y}{1.4}[/tex]

[tex]\frac{\sqrt{3} }{2} =\frac{y}{1.4}[/tex]

y = 1.21

Then,

[tex]\Delta d_{y}[/tex] = 0.9 + 0 + 1.21

[tex]\Delta d_{y}[/tex] = 2.11

Total displacement is

[tex]\Delta d_{t}=\sqrt{(1.5)^{2}+(2.11)^{2}}[/tex]

[tex]\Delta d_{t}=\sqrt{6.7021}[/tex]

[tex]\Delta d_{t}=[/tex] 2.6

Magnitude of Olivia's total displacement is 2.6mi

On the map, joining the initial and final points gives a vector pointing towards east at angle:

[tex]\theta=tan^{-1}(\frac{2.11}{1.5})[/tex]

[tex]\theta=tan^{-1}(1.41)[/tex]

θ = 54.65°

Direction of total displacement is 54.65° East.

Answer:

1.9 mi, 350.7°

Explanation:

If continued another 1.2 mi*

solving for x

0 + 0.8 + 1.2(cos30) = 1.83923048 --> 1.84

solving for y

-0.9 + 0 + 1.2(sin30) = -0.3

^negative because it is going downward

a) solving for magnitude

[tex]\sqrt{(1.84)^2+(-0.3)^2} = \sqrt{3.4756} = 1.86429611... = 1.9 mi[/tex]

b) solving for direction of total displacement

[tex]tan^-1 = (\frac{x}{y} ) \\tan^-1 = (\frac{-0.3}{1.84}) = -0.16 \\[/tex]

∘, measured counterclockwise from the eastward direction

360 - 0.16 = 359.84°

*replace any of the needed values in the equation, such as 1.2 mi to 1.4 mi

A ball with a mass of 3.7 kg is thrown downward with an initial velocity of 8 m/s from a high building. How fast will it be moving after 3 seconds?

Answers

Answer:

v=37.4 m/s

Explanation:

It is given that,

Mass of a ball, m = 3.7 kg

Initial velocity of the ball is u = 8 m/s

We need to find its velocity after 3 seconds. It is moving downwards. The equation of motion is this case is

v=u+gt

[tex]v=8+9.8\times 3\\\\v=37.4\ m/s[/tex]

So, the velocity of the ball after 3 seconds is 37.4 m/s.

How long did it take me to make this?

Answers

1hour because that takes a lot of time

Answer:

Its according from what it is made with if wax it takes longer but if with a craft paper it takes lesser time if paper approximately 20 to 25 minutes

Check Concepts
4.
35. Which of the following do you calculate
when
you
divide the total distance trav-
eled by the total travel time?
A) average speed
B) constant speed
C) variable speed
D) instantaneous speed

Answers

Answer:

I think its A.........

A stone is thrown vertically upward with a speed of 28.0 m/s how much time is required to reach this height

Answers

A stone is thrown vertically upward with a speed of 17.0 m/s. How fast is it moving when it reaches a height of 11.0 m? How long is required to reach this height?

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½ at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):

v^2 = u^2 + 2as

v^2 = 17.0^2 -2(9.81)(11.0)

v = √73.18 = 8.55m/s

now from equation (3):

v = u + at

8.55 = 17.0 – 9.81t

t = (8.55 – 17.0)/(-9.81) = 0.86s

Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is your friend correct? Why or why not?

Answers

Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

Force equals mass times acceleration:

F = m*a

There must be a force.

So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.

Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.

So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

Both chemical and physical changes involve energy transfer, so your friend is not correct.

A chemical change occurs when a new substance is formed or created through a process of chemical reaction which is reversible.

The addition or removal of heat energy can affect the rate of chemical reaction. This addition or removal is known as energy transfer process.

Examples of chemical changes include;

Acid-base reaction.Rusting of iron in presence of moisture and oxygen.   Cooking any food.

A physical change on the other hand doesn't involve formation of new substance and it is can be reversible or irreversible.

The addition or removal of heat energy can affect the rate of physical changes.

Examples of such physical changes include;

vaporization of liquid (liquid to gas),freezing of liquid (liquid to solid), and condensation of gas (gas to liquid).

Thus, both process (chemical and physical changes) involve energy transfer, so your friend is not correct.

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A car originally traveling at 30.0 m/s manages to brake for 5.0 seconds while traveling 125 m along a road. After those first 5.0 seconds, the brakes fail. After an additional 5.0 seconds it travels an additional 150 m further down the road. What was the magnitude of the acceleration of the car after the brakes failed

Answers

Answer:

The magnitude of the acceleration of the car after the brakes failed is 4 m/s²

Explanation:

The car was originally traveling at 30.0 m/s, that is

The initial velocity, [tex]u[/tex] = 30.0 m/s

The time spent while the car manages to brake is 5.0 seconds, that is

time, [tex]t[/tex] = 5.0 secs

and the distance traveled during this time is

distance, [tex]s[/tex] = 125 m

From one of the equations of kinematics for linear motion,

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

Where [tex]a[/tex] is the acceleration

We can determine the deceleration of the car during the first 5.0 seconds

Hence,

From,

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

[tex]125 = 30.0(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]

[tex]125 =150.0 + 12.5a[/tex]

[tex]12.5a = 125 - 150.0[/tex]

[tex]12.5a = -25\\a = \frac{-25}{12.5}\[/tex]

[tex]a = - 2.0 m/s^{2}[/tex] (Negative sign indicates deceleration)

Now we will calculate the final velocity reached at this time

From,

[tex]v^{2} = u^{2} + 2as[/tex]

Where [tex]v[/tex] is the final velocity

[tex]v^{2} = 30.0^{2} + 2(-2.0)(125)\\v^{2} = 400\\v = \sqrt{400} \\v = 20 m/s \\[/tex]

This is the final velocity reached by the car during the first 5.0 seconds

Now, for the magnitude of the acceleration of the car after the brakes failed,

After the brakes failed,

it travels an additional 150 m further down the road, that is

s = 150m

an additional 5.0 seconds, that is

t = 5.0 seconds

Also, from

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

The initial velocity here will be the final velocity for the first 5.0 seconds, that is,

u = 20 m/s

Hence,

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

[tex]150 = 20(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]

[tex]150 = 100 + 12.5a\\12.5a = 150 - 100\\12.5a = 50\\a = \frac{50}{12.5} \\a = 4m/s^{2}[/tex]

Hence, the magnitude of the acceleration of the car after the brakes failed is 4 m/s²

Put Newton’s 1st Law in your own words

Answers

Answer:Newton’s law also states that larger bodies with heavier masses exert more gravitational pull, which is why those who walked on the much smaller moon experienced a sense of weightlessness, as it had a smaller gravitational pull. To help explain his theories of gravity and motion, Newton helped create a new, specialized form of mathematics.

Explanation:Here I dont know

Difference between calorimeter and thermometer ?

Answers

A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction. Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction.

Answer:

A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction. Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction.

Explanation:

Two equal charges are 2m2m apart. If the charges and the distance are divided by two, how is the force between the charges affected? g

Answers

Answer:

The force between the charges are not affected.

Explanation:

Given;

distance between two equal charges, R = 2m

The force between the charges is given by;

[tex]F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1[/tex]

Therefore, the force between the charges are not affected.



What quantity do units represent in a value?

A. size

Answers

Question:

What quantity do units represent in a value? A. Size B. Direction C. Magnitude D. Dimension

Answer:

D. Dimension

Explanation:

A unit is simply the measurement given to a dimension. In other words, units represent dimensions. For example, consider a ruler that is 20m long.

i. The value is 20m.

ii. The magnitude of the measurement is 20

iii. The unit of the measurement is meters(m) and that essentially represents the dimension (length) of the ruler.

A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 38.0 s. After many oscillations, he finally comes to rest 25.0 m below the level of the bridge. Calculate the spring stiffness constant and the unstretched length of the bungee cord.

Answers

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

[tex]T=\dfrac{38}{8}\\\\T=4.75\ s[/tex]

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

[tex]T=2\pi \sqrt{\dfrac{m}{k}}[/tex]

k = spring stiffness constant

So,

[tex]k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m[/tex]

When the cord is in air,

mg=kx

x = the extension in the cord

[tex]x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m[/tex]

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

The spring stiffness constant is 116.7 N/m and the the unstretched length of the bungee cord is 19.54 m.

The given parameters;

mass of the bungee jumper, m = 65 kgtime of motion, t = 38 sdistance to come to rest, d = 25 m

The period of oscillation of the bungee jumper is calculated as follows;

[tex]T = \frac{t}{n} \\\\T = \frac{38}{8} \\\\T = 4.75 \ s[/tex]

The spring stiffness constant is calculated as follows;

[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\\sqrt{\frac{m}{k} } = \frac{T}{2\pi} \\\\k = m \times \frac{T^2}{4\pi^2} \\\\k = 65 \times \frac{(4.75)^2}{4\pi ^2} \\\\k = 116.7 \ N/m[/tex]

The extension of the cord is calculated as follows;

[tex]F = kx\\\\mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{65 \times 9.8}{116.7} \\\\x = 5.46 \ m[/tex]

The unstretched length of the bungee cord is calculated as;

[tex]\Delta x = l_2-l_1\\\\l_1 = l_2 - \Delta x\\\\l_1 = 25 - 5.46\\\\l_1 = 19.54 \ m[/tex]

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20 POINTS 20 POINTS On a bright sunny day you decide to take a walk. You begin at your home and walk 1000 meters to an ice cream shop in 10 minutes. You spend 15 minutes ordering your ice cream and then return home. Since you have an ice cream cone in your hand, it takes 20 minutes to walk home. 1. Find the total displacement and total distance traveled. 2. Find your average speed and average velocity in meters/min. 3. Find your average speed and average velocity in meters/sec.

Answers

Answer:

A) Total Distance = 2000 m and Total displacement = 0 m

B) Average Speed = 44.44 m/min and Average Velocity = 0 m/min

C) Average Speed = 0.7407 m/s and Average velocity = 0 m/s

Explanation:

A) Distance to reach ice cream shop from home = 1000 meters

Therefore distance to get back home would also be 1000 meters.

Total distance traveled = 1000 + 1000 = 2000 metres

Since journey started at home and ended at home, then total displacement = 0 metres.

B) Average speed = Total distance/total time.

Total time = 10 + 15 + 20 = 45 minutes

Since total distance = 2000 m

Then;

Average speed = 2000/45

Average speed = 44.44 m/min

Average velocity = Total displacement/total time

Average velocity = 0/45 = 0 m/min

C) We now want answers in B to be in m/s.

Total time = 45 minutes.

From conversion, 60 seconds make 1 minute. Thus, 45 minutes = 45 × 60 = 2700 seconds

Thus;

Average Speed = 2000/2700

Average Speed = 0.7407 m/s

Average displacement = 0/2700 = 0 m/s

You are driving your car at a speed of 19.0 m/s and you hit the brakes. The car accelerates at -3.50 m/s2. (a) How long does it take the car to cover 10.0 m? (b) What is the final velocity of the car?

Answers

Recall the formulas,

[tex]x_f=x_i+v_it+\dfrac12at^2[/tex]

[tex]v_f=v_i+at[/tex]

(a)

[tex]10.0\,\mathrm m=\left(19.0\dfrac{\rm m}{\rm s}\right)+\dfrac12\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

[tex]\implies \boxed{t\approx0.555\,\mathrm s}[/tex]

(b)

[tex]v_f=19.0\dfrac{\rm m}{\rm s}+\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)(0.555\,\mathrm s)[/tex]

[tex]\implies \boxed{v_f\approx17.1\dfrac{\rm m}{\rm s}}[/tex]

A ship sets out to sail to a point 116 km due north. An unexpected storm blows the ship to a point 121 km due east of its starting point. (a) How far (in km) and (b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination?

Answers

Explanation:

In this problem, we are meant to slove for the resultant and the direction of the the vectors given

Given data

let the sail to a point due north be y= 116km

and the point due east be x= 121 km

(a) How far (in km)

The resultant between the two points is the distance between them

[tex]r=\sqrt{x^2+y^2} \\\\r=\sqrt{121^2+116^2} \\\\r=\sqrt{14641+13456} \\\\r=\sqrt{28097} \\\\r=167.62[/tex]

The distance between the points is 167.62

(b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination

the direction can be gotten using

tan∅= y/x

∅= tan-1 (y/x)

∅= tan-1(116/121)

∅= tan-1(0.958)

∅= 43.77°

The direction is 43.77°

Is the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth?

Answers

Answer:

Yes

Explanation:

This is because the interaction between piece of paper and earth.is gravitational while that of piece of paper and rod is electrostatic

Yes, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.

Gravitational force is the force by which earth attracts other objects by mass.The electrostatic force is the force of an object due to charge.Electrostatic forces are much stronger than gravitational forces. because gravity depends on mass, atoms have less masses so that the gravitational forces between them is close to zero. Whereas, the electrostatic force related to charges is bigger.

Therefore, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.

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2 differences between calorimeter and thermometer ?

Answers

Answer:

Calorimeter is used to measure heat in and represents that in units of joules per kelvin units J/˚C or kJ/K

A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction.

Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction. Unit measurement used by thermometer is Celsius (°C).

Explanation:

A rod attracts a positively charged hanging ball. the rod is?a) negativeb) positivec) neutrald) either negative or neutrale) either positive or neutral

Answers

Answer:

The correct option is a

Explanation:

This question seeks to test a general rule in physics (on charges) which states that like charges repel but unlike charges attract. This means that, a negatively charged substance will repel or not attract another negatively charged material and the same applies to a positively charged substance also. However, a negatively charged substance will attract a positively charged material and vice versa, hence only a negatively charged rod will attract a positively charged hanging ball.

What is the binary number system? ​

Answers

Explanation:

binary number is a number expressed in the base-2 numeral system or binary numeral system, which uses only two symbols: typically "0" and "1".

Which statement best defines inertia? An object's motion is unaffected by any external forces acting upon it. An object responds to a force by tending to move in the direction of that force. An object opposes any motion, naturally returning to a state of rest on its own. An object opposes any change in its velocity, either to its direction or to its speed.

Answers

Answer:

An object opposes any change in its velocity, either to its direction or to its speed.

Explanation:

Edmentum Answer

You are standing on top of the Empire State Building (height 330 [m]) when Superman flies past. He is headed straight down with a steady speed of 35 [m/s]. (Superman can do things like this.) At the instant he goes past, you drop a 10 [kg] lead ball over the edge. At what height above the sidewalk does the ball pass Superman?

Answers

Answer:

The ball will be overtake superman after travelled distance 248.5 m.

Explanation:

Given that,

Speed of superman = 35 m/s

Height = 330 m

Mass of ball = 10 kg

Let the height at which they are at same position be S.

For superman,

We need to calculate the time

Using equation of motion

[tex]S=ut[/tex]...(I)

For ball,

[tex]S=u't+\dfrac{1}{2}gt^2[/tex]

From equation (I) and (II)

[tex]ut=u't+\dfrac{1}{2}gt^2[/tex]

[tex]ut=0+\dfrac{1}{2}\times g\times t^2[/tex]

[tex]u=\dfrac{1}{2}\times g\times t[/tex]

[tex]t=\dfrac{2u}{g}[/tex]

Put the value into the formula

[tex]t=\dfrac{2\times35}{9.8}[/tex]

[tex]t=7.1\ sec[/tex]

We need to calculate the distance travelled

Using formula of distance

[tex]d=v\times t[/tex]

Put the value into the formula

[tex]d=35\times 7.1[/tex]

[tex]d=248.5\ m[/tex]

Hence, The ball will be overtake superman after travelled distance 248.5 m.

Assuming it is a van der Waals gas, calculate the critical temperature, pressure and volume for CO2. (a = 3.610 atm L2 mol-2, b = 0.0429 L mol-1)
pc = ___ atm
Tc = ___ K
Vc = ___ L/mol

Answers

Answer

To get critical pressure

We use

Pc = a/(27b²)

So

= 3.610/(27 X 0.0429²)

We have

= 72.7 atm

Critical temperaturewe

We use

Tc = 8a/27Rb

= 8 x 3.610/(27 x 0.0812 x 0.0429)

= 307 K

Critical volume

We use

Vc =3b =

3 x 0.0429

= 0.129L/mol

A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s 2 for 6.2 s. It then accelerates at a rate of -1.2 m/s2 until it stops. a) Find the car’s maximum speed b) Find the total time from the start of the first acceleration until the car is stopped c) What’s the total distance the car travels?

Answers

(a) For the first 6.2 s, the car has velocity at time t given by

[tex]v(t)=13.5\dfrac{\rm m}{\rm s}+\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

so that after 6.2 s, it attains a velocity of

[tex]v(6.2\,\mathrm s)=25.28\dfrac{\rm m}{\rm s}[/tex]

For any time t after 6.2 s, its velocity is given by

[tex]v(t)=25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

which tells us the velocity only falls from this point onward. This means the maximum speed is 25.28 m/s, or about 25.3 m/s.

(b) Solve for t (after 6.2 s) that makes v(t) = 0 :

[tex]25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t=0[/tex]

[tex]\implies t\approx21.067\,\mathrm s[/tex]

It takes the car about 21.2 s to come to a rest, so the car travels a total of about 6.2 s + 21.2 s = 27.4 s.

(c) For the first 6.2 s, the car undergoes a displacement at time t of

[tex]x(t)=\left(13.5\dfrac{\rm m}{\rm s}\right)(6.2\,\mathrm s)+\dfrac12\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)(6.2\,\mathrm s)^2[/tex]

[tex]\implies x\approx120.218\,\mathrm m[/tex]

For time t beyond 6.2 s, its displacement is

[tex]x(t)=120.218\,\mathrm m+\left(25.28\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

The car comes to a rest after 21.2 s (accelerating at a rate of -1.2 m/s^2), so that its total displacement is

[tex]x(21.2\,\mathrm s)\approx386.49\,\mathrm m[/tex]

so the car travels a total distance of about 387 m.

A 10 kg object is dropped from rest. a. How far will it drop in 2s? b. How long will it take a 5 kg object to drop the same distance?

Answers

Answer:

Explanation:

Using the equation of motion S = ut + 1/2at² to get the height of drop where

u is the initial velocity of the object = 0m/s

a is the acceleration due to gravity = +9.81m/s² (downwards motion of object)

t is the time it takes the object to drop = 2secs

Substituting the given parameters into the formula to get the height of drop S, we will have;

S = 0(2)+1/2(9.81)(2)²

S = 0+9.81*2

S = 19.62m

Hence the object will drop at a distance of 19.62m

b.) To determine the time it will take a 5 kg object to drop the same distance, we will use the same formula S = ut+1/2at²

Since the 5kg object also drops at the same distance, then S = 19.62m

Substituting this values into the equation we will have;

19.62 = 0(t) + 1/2(9.81)t²

19.62 = 4.905t²

t² = 19.62/4.905

t² = 4

t =2secs

This shows that it will take 5kg object 2secs to fall from the  same distance. This means that no matter the mass of the object, it will take them the same time to fall at the same distance because they are all falling under the same influence of gravity.

A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 232 km and a direction 30.0o north of east. The displacement vector B for the second segment has a magnitude of 168 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle θ with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle θ.

Answers

Answer:

a) 121 km

b) 74°

Explanation:

To start with, we assume that there exist two components, the East and the North. We'd be representing the East, by "e" and the North, by "n"

Now we start with the

First vector:

east1 = 232 cos 30 = 201

north1 = 232 sin 30 = 116

Now, that of the second vector will be

east 2 = - 168

north2 = 0

Next, we add the two together and get

East components

201 - 168 = 33 east

North components

116 + 0 = 116

Therefore, the magnitude has to be

magnitude = √(33² + 116²)

Magnitude = √14545

Magnitude = 121

tanθ = 116/33

Tanθ = 3.51

θ = tan^-1 3.51

θ = 74° North East

Common benefits of lower body endurance include improved

Answers

Answer:

Hearing, Vision and Metabolism.

A technical machinist is asked to build a cubical steel tank that will hold 310L of water. Calculate in meters the smallest possible inside length of the tank.

Answers

Answer:

0.68 m

Explanation:

Since the volume of water is 310 L, and we know that 1 litre = 1 dm³. So the volume of water is V = 310 dm³. Since this volume of water is the volume of water the cubical steel tank can contain, it equals the volume of the cubical steel tank.

We know that the volume of the cubical steel tank V = L³ where L is the length of side of the cube on the inside.

So the length of side of the cube L = ∛V = ∛310 dm³ = 6.77 dm = 6.77 dm × 1m/10 dm = 0.677 m ≅ 0.68 m

So, the smallest possible inside length of the tank is 0.68 m

What are the benefits of testing your heart rate during exercise

Answers

Answer: Heart rate provides an objective measurement of how hard your body is working. The higher the exercise intensity, the higher your heart rate will be.

Explanation:

Answer:

You're getting exercise and training your heart!

Explanation:

chemical reaction to make a foam?​

Answers

Answer: Hydrogen Peroxide

Explanation: Hydrogen peroxide breaks down into oxygen and water. As a small amount of hydrogen peroxide generates a large volume of oxygen, the oxygen quickly pushes out of the container. The soapy water traps the oxygen, creating bubbles, and turns into foam.

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