A 70 kg bicyclist rides his 9.8 kg bicycle with a speed
of 16 m/s.
How much work must be done by the brakes to bring the bike and rider to a stop?

Answers

Answer 1
Answer:

Hello,

QUESTION)

✔ We have Ek = 1/2m x v²

Ek = 1/2 x 79.8 x 16²Ek = 10 214.4 J

In order to come to a complete stop, the cyclist must convert all his kinetic energy into thermal energy. Given that the braking force opposes movement, the work is therefore resistant, i.e.  W = -10 214.4 J.

A 70 Kg Bicyclist Rides His 9.8 Kg Bicycle With A Speedof 16 M/s. How Much Work Must Be Done By The Brakes

Related Questions

Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader

Answers

Answer:

7.55 g

Explanation:

Using the relation :

Δt = temperature change = (6° - 0°) = 6°

Q = quantity of heat

C = specific heat capacity = 4190 j/kg/k

1000 J = 1kJ

333 KJ = 333000 j

The quantity of ice that will melt ;

= 0.419 * 6 * 100 / 333000

= 2514000 / 333000

= 7.549 g

The mass of ice that will melt :

2.514 / 0.333

= 7.549 g

plz help me with my career!!!
part one...

Answers

Answer:

#1 Yes

Explanation: #1: The rest of them are used mainley by farmers, and crops are used by common citizens in the world.

Question 1: Crops.

Question 2: Diagnostic Services.

Question 3: A cable company needs to lay new fiber optic cable to reach its customers across a large lake.

Question 4: A bachelor's degree in energy research.

Question 5: Environmental Resources.

If any of these answers are incorrect, please tell me, so I can fix my mistake. Thank you.

The Sun is divided into three regions.
True оr False?

Answers

Answer:

false I think

Explanation:

hope that help

so it's not divided in 3 regions

Part D Here is one last question as a final check on your understanding of your work for this problem, looking at this problem as an example of the Conservation of Energy. The action in this problem begins at location A , with the block resting against the uncompressed spring. The action ends at location B, with the block moving up the ramp at a measured speed of 7.35 m/s . From A to B, what has been the work done by non-conservative forces, and what has been the change in the mechanical energy of the block-Earth system (the ramp is a part of the Earth)

Answers

Answer:

The answer is "39.95 J".

Explanation:

Please find the complete question in the attached file.

[tex]\to W_{AC}=(\mu \ m \ g \ \cos \theta ) d[/tex]

            [tex]=(0.45 \times 1.60 \times 9.8 \times \cos 26^{\circ}) 6.30 \\\\=(7.056 \times \cos 26^{\circ}) 6.30 \\\\=6.34189079\times 6.30\\\\=39.95 \ J\\\\[/tex]

[tex]\therefore \\\\\bold{\Delta E =39.95 \ J}[/tex]

A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.

Answers

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2               ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1                  ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Which graph represents the relationship between the magnitude of the gravitational force exerted by earth on a spacecraft the distance between the center of the spacecraft the center of earth

Answers

Answer:

B as distance increase force decrease, but it is not a linear relationship.

Car X is travelling at 30m/s north. Its driver looks at car Y approaching on another road and he estimates it is moving at 15m/s south-west relative to his car. Calculate the velocity of car Y relative to the ground.

Answers

Answer: 22.1 m/s

Explanation:

The velocity of Car traveling 30 m/s towards the north

In vector form it is

[tex]v_x=30\hat{j}[/tex]

The velocity of car Y w.r.t X is

[tex]\Rightarrow v_{yx}=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}][/tex]

Solving this

[tex]\Rightarrow v_{yx}=v_y-v_x\\\Rightarrow v_y=v_{yx}+v_x[/tex]

putting values

[tex]\Rightarrow v_y=15[-\cos 45^{\circ}\hat{i}-\sin 45^{\circ}\hat{j}]+30\hat{j}[/tex]

[tex]\Rightarrow v_y=-10.606\hat{i}+19.39\hat{j}[/tex]

absolute velocity relative to ground is

[tex]\left | v_y\right |=\sqrt{(-10.606)^2+(19.39)^2}\\\left | v_y\right |=22.101\ m/s[/tex]

A trough is 10 meters long, 1 meters wide, and 2 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 2 meters, and base, on top, of length 1 meters). The trough is full of water (density 1000kg/m3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use g

Answers

Answer:

The amount of work required to empty the trough by pumping the water over the top is approximately 98,000 J

Explanation:

The length of the trough = 10 meters

The width of the through = 1 meter

The depth of the trough = 2 meters

The vertical cross section of the through = An isosceles triangle

The density of water in the through = 1000 kg/m³

Let 'x' represent the width of the water at a depth

x/y = 1/2

∴x = y/2

The volume of a layer of water, dV, is given as follows;

dV = 10 × y/2 × dy = 5·y·dy

The mass of the layer of water, m = ρ × dV

∴ m = 1000 kg/m³ × 5·y·dy m³ = 5,000·y·dy kg

The work done, W = m·g·h

Where;

h = The the depth of the trough from which water is pumped

g = The acceleration due to gravity ≈ 9.8 m/s²

[tex]\therefore \, W \approx \int\limits^2_0 {5,000 \times y \times 9.8 \, dy} = \left[24,500\cdot y^2 \right]^2_0 = 98,000[/tex]

The work done by the pump to pump all the water in the trough, over the top W ≈ 98,000 J

Let's assume raspberries are 10 wt% protein solids and the remainder water. When making jam, raspberries are crushed and mixed with sugar, in a 45:55 berry to sugar ratio, by mass. Afterward, the mixture is heated, boiling off water until the remaining mixture is 0.4 weight fraction water, resulting in the final product, jam. How much water, in kilograms, is boiled off per kilogram of raspberries processed

Answers

Answer:

The mass of water boiled off is [tex]0.0 \overline{185}[/tex] kg

Explanation:

The given percentage by weight of protein solids in raspberries = 10 weight%  

The ratio of sugar to raspberries in ja-m = 45:55

The mass of the mixture after boiling = 0.4 weight fraction water

Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry

The mass of raspberry, r = 1 kg

The percentage by weight of water in raspberry = 90 weight %

The mass of water in 1 kg of raspberry =  90/100 × 1 kg = 0.9 kg

The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55

∴ s = 1 kg × 55/45 = 11/9 kg

The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg

The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction

Let 'w' represent the mass of water boiled off, we have;

(0.9 - w)/(20/9 - w) = 0.4

(0.9 - w) = 0.4 × (20/9 - w)

0.9 - w = 8/9 - 0.4·w

9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w

(81 - 80)/(90) = (6/10)·w

1/90 = (6/10)·w

w = ((10/6) × 1/90) = 1/54

w = 1/54  

The mass of water boiled off, w = (1/54) kg = [tex]0.0 \overline{185}[/tex] kg

Fat Albert (the TV show character) runs up the stairs on Monday. On Tuesday, he walks up the same set of stairs. Which day did he do more work?

Answers

Answer:

Tuesday bc instead of running he/she was walking bc he/she might not have as much energy

Explanation:

PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE



Which best explains why species living in Australia are found nowhere else on Earth? This is an example of Geologic Evolution.

A.
Australia has an ecosystem different from any other area on Earth.

B.
Humans have genetically altered many Australian species in laboratories.

C.
Australian species were genetically altered after a comet hit the landmass.

D.
Australia separated from other continents and species there evolved independently.

Answers

Australia separated from other continents and species there evolved independently

A ball is dropped off the side of a bridge,
After 1.55 S, how far has it fallen?
(Unit=m)

Answers

Answer:

Distance S = 11.77 m (Approx.)

Explanation:

Given:

Time t = 1.55 Second

Gravity acceleration = 9.8 m/s²

Find:

Distance S

Computation:

S = ut + (1/2)(g)(t)²

S = (0)(1.55) + (1/2)(9.8)(1.55)²

S = (0)(1.55) + (1/2)(9.8)(1.55)²

Distance S = 11.77 m (Approx.)

Which one is it? Help ASAP

Answers

Answer:

extreme heat, because no physical damage can demagnetize a magnet

Explanation:

Answer:

the 3rd one

Explanation:

A spring in a toy gun has a spring constant of 10 N/m and can be compressed 4 cm.
It is then used to shoot a 1 g ball out of the gun. Find the velocity of the ball as it
leaves the gun

Answers

F=ma
We know from this that a=10,000m/s2

V=at
X=vt


You end up with v^2=ax

Plug in 10,000 and 0.04 and solve for v =20m/s

What is one disadvantage of sending information over long distances
wirelessly using digital signals?
O A. The signals become weaker the farther the receiver is from the
source.
B. The farther the signals travel, the more slowly they move.
C. The signals become stronger the farther the receiver is from the
source.
O D. The farther the signals travel, the easier they are to detect.

Answers

The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."

Since most of the signal which we use for communication are radio signal Radio signal are basically electromagnetic waves.As the wave moves forward it looses its amplitude.So basically radio signal becomes weaker for long distance.

What are radio signals?Radio signals or radio waves are a form of electromagnetic wave. Although this may sound complicated, it is possibly sufficient to say that these waves have both electric and magnetic components. They are the same as light rays, ultra-violet and infra-red. The only difference is in the wavelength of the waves.

Thus , The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."

Learn more about Radio waves here -

https://brainly.com/question/69373

#SPJ2

Fossil clues are one of the _____________ clues that support the theory of continental drift.
A. crust B. resource C. climate D. rock

Answers

Answer:

a

Explanation:

I think don't get mad if I'm wrong

A 0.6 kg basketball is 3.0 high calculate its potential Energy PE=mgh​

Answers

.6(9.81)(3)

=17.65
18 J

2 significant figures

Jim and Sally both do identical jobs. Jim works quickly while Sally works slowly. Which of the following is true?

A) Sally uses more energy.
B) Jim uses more energy.
C) Jim uses more power.
D) Sally uses more power.

Answers

The answer should be C

10. A person is about to kick a soccer ball. Consider the leg and foot to be a single rigid body and that it rotates about a fixed axis through the knee joint center. Immediately prior to impact the leg and foot are in the vertical direction and the distal end of the foot has an acceleration of 10 g in the horizontal direction. The muscle force vector makes an angle of 15 degrees with the vertical and has a moment arm of 5 cm from the knee joint center. Assume the person is 1.7 m tall and has a mass of 75 kg. Find the force in the muscle

Answers

Answer:

i don't know but i hope you get it right

Explanation

A toy projectile is fired vertically from the ground upward with a velocity of +29 meters per second. It arrives at its maximum altitude in 3.0 seconds. How high does the projectile go?

Answers

Answer:

[tex]\boxed{\text{\sf \Large 42 m}}[/tex]

Explanation:

Use height formula

[tex]\displaystyle \sf H=\frac{u^2 sin(\theta)^2}{2g}[/tex]

u is initial velocity

θ = 90° (fired vertically upward)

g is acceleration of gravity

[tex]\displaystyle \sf H=\frac{29^2 \times sin(90 )^2}{2 \times 10}=42.05[/tex]

Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 lb and was traveling eastward. Car B weighs 11251125 lb and was traveling westward at 42.042.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.517.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.7500.750 . How fast (in miles per hour) was car A traveling just before the collision

Answers

Answer:

v = 28.98 ft / s

Explanation:

For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision

In the exercise they indicate the weight of each car

          Wₐ = 1500 lb

          W_b = 1125 lb

Car B's velocity from v_b = 42.0 mph westward, car A travels east

let's find the mass of the vehicles

             W = mg

             m = W / g

             mₐ = Wₐ / g

             m_b = W_b / g

             mₐ = 1500/32 = 46.875 slug

             m_b = 125/32 = 35,156 slug

Let's reduce to the english system

             v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s

We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved

we assume the direction to the east (right) positive

initial instant. Before the crash

           p₀ = mₐ v₀ₐ - m_b v_{ob}

final instant. Right after the crash

           p_f = (mₐ + m_b) v

the moment is preserved

           p₀ = p_f

           mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v

           v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]

we substitute the values

           v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]

           v = 0.559 v₀ₐ - 26.40                  (1)

Now as the two vehicles united we can use the relationship between work and kinetic energy

the total mass is

              M = mₐ + m_b

              M = 46,875 + 35,156 = 82,031 slug

starting point. Jsto after the crash

              K₀ = ½ M v²

final point. When they stop

             K_f = 0

The work is

             W = - fr x

the negative sign is because the friction forces are always opposite to the displacement

Let's write Newton's second law

Axis y

           N-W = 0

           N = W

the friction force has the expression

            fr = μ N

we substitute

            -μ W x = Kf - Ko

             

            -μ W x = 0 - ½ (W / g) v²

            v² = 2 μ g x  

            v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5  

            v = 28.98 ft / s

1) A man leans against the wall and...
a) The man's shoulder pushes against the wall
b) The wall pushes against the man
c) Bricks in the wall push against each other
d) The ground pushes against the man, supporting him.

Answers

Answer:

I thinks its a, but its really about gravity im not sure

Explanation:

:)

A cat dozes on a stationary merry-go-round, at a radius of 7.0 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.9 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding

Answers

Answer:

0.6

Explanation:

Given that :

Radius, R = 7m

Period, T = 6.9s

The Coefficient of static friction, μs can be obtained using the relation :

μs = v² / 2gR

Recall, v = 2πR/T

μs becomes ;

μs = (2πR/T)² / 2gR

μs = (4π²R² / T²) ÷ 2gR

μs = (4π²R² / T²) * 1/ 2gR

μs = 4π²R / T²g

μs = 4π²*7 / 6.9^2 * 9.8

μs = 28π² / 466.578

μs = 276.34892 / 466.578

μs = 0.5922887

μs = 0.6

Activities:
1. Name the instrument that is used to measure Air Pressure.
2.Explain what is Cyclone and Anticyclone

Answers

Answer: barometer.

A cyclone is a storm or system of winds that rotates around a center of low atmospheric pressure. An anticyclone is a system of winds that rotates around a center of high atmospheric pressure.

A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe very large accelerations.
Part A) What is the acceleration at the end of a test tubethat is 10 cm from the axis of rotation?
Part B) For comparison, what is the magnitude of theacceleration a test tube would experience if dropped from a heightof 1.0 m and stopped in a 1.0-ms-long encounter with a hardfloor?

Answers

Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²

A kite 40 ft above the ground moves horizontally at a constant speed of 10 ft/s, with a child, holding the ball of kite string, standing motionless on the ground. Assume the kite is flying away from the child. At what rate is the child releasing the string when (a) 50 ft of the string is out

Answers

Answer:

 v = 27.28 m /s, θ = 63.9º

Explanation:

For this exercise we can approximate the movement to a projectile launch, let's analyze the situation.

* We must find the horizontal speed, for this we will find the descent time and the horizontal distance

* We look for the vertical speed

At the highest point the speed is horizontal

Let's find the time it takes for the kite to reach the ground

             y = y₀ + v_{oy} t - ½ g t²

             0 =y₀ + 0 -1/2 gt²

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

             t = √(2 40/32)

             t = 2.5 s

to find the horizontal velocity we must know the horizontal distance, let's use trigonometry

          sin θ = y / l

          θ = sin⁻¹1 y / l

          θ = sin⁻¹ 40/50

          θ = 53.1º

therefore the horizontal distance is

          x = l cos 53.1

          x = 50cos 53.1

          x = 30 m

let's use the equation

          x = v₀ₓ t

          v₀ₓ = x / t

          v₀ₓ = 30 / 2.5

          v₀ₓ = 12 m / s

we look for the vertical component of the velocity

          v_y = v_{oy} - g t

          v_y = 0 - g t

          v_y = - 9.8 2.5

          v_y = -24.5 m / s

the negative sign indicates that the speed is directed downwards, because it is the arrival point, as they indicate that there is no friction, the exit speed is the same, worse with the opposite sign

We already have the two components of the velocity, let's use the Pythagorean theorem to find the modulus

          v = [tex]\sqrt{v_x^2 + v_y^2}[/tex]

          v = [tex]\sqrt{12^2 + 24.5^2}[/tex]

          v = 27.28 m /s

we use trigonometry for the angle

          tan θ = v_y / vₓ

          θ = tan⁻¹ v_y / vₓ

          θ = tan⁻¹ 24.5 / 12

          θ = 63.9º

HELP PLEASE DUE IN 3 MINUTES

Answers

Answer:

Tectonic Plate Movement

Explanation:

Each continent and ocean sits on its own tectonic plate which floats on the Earths upper mantle. They move very little over time.

Answer:

tectonic plates movement

A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 140 m above the launch point. (a) Calculate the horizontal component of its velocity. (b) Calculate the vertical component of its velocity just after launch. (c) At one instant during its flight the vertical component of its velocity is found to be 65 m/s. At that time, how far is it above or below the launch point

Answers

Answer:

a). 53.78 m/s

b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

In the c part, The negative distance is telling us that the project went below the lunch point.

In an effort to be the star of the half-time show, the majorette twirls a highly unusual baton made up of four mases fastened to the ends of light rods. Each rod is 1.0 m lone. Find the moment of inertia of the system about an axis perpendicular to the page and passing through the point where the rods cross.

Answers

Answer:

"0.25 kg-m²" is the appropriate answer.

Explanation:

The diagram of the question is missing. Find the attachment of the diagram below.

According to the diagram, the values are:

m₁ = 0.2

m₂ = 0.3

m₃ = 0.3

m₄ = 0.2

d₁ = d₂ = d₃ = d₄ = 0.5 m

As we know,

The moment of inertia is:

⇒  [tex]I=\Sigma M_id_i^2[/tex]

then,

⇒  [tex]I=m_1d_1^2+m_2d_2^2+m_3d_3^2+m_4d_4^2[/tex]

⇒     [tex]=d^2(m_1+m_2+m_3+m_4)[/tex]

On substituting the values, we get

⇒     [tex]=0.5^2\times (0.2+0.3+0.3+0.2)[/tex]

⇒     [tex]=0.25\times 1[/tex]

⇒     [tex]=0.25 \ Kg-m^2[/tex]

Suppose you are standing in front of a flat mirror which is mounted to a vertical wall. For this problem you may suppose that your height is 1.70 m and your eyes are 12 cm below the top of your head. What is the smallest mirror that will still allow you to see the full length of your body

Answers

Answer:

The right approach is "0.85 m".

Explanation:

According to the question, the diagram will is provided below.

So that as per the diagram,

The values will be:

My height,

AO = 1.70 m

My eyes at,

AB = 12 cm

i.e.,

     = 0.12 m

As we can see, the point of incidence lies between the feet as well as the eyes, then

BO = 1.58 m

Now,

⇒ [tex]O'D = \frac{1.58}{2}[/tex]

           [tex]=0.79 \ m[/tex]

The point of incidence of the ray will be:

⇒ [tex]CO'=1.70-\frac{0.12}{2}[/tex]

           [tex]=1.70-0.06[/tex]

           [tex]=1.64 \ m[/tex]

Hence,

The smallest length of the mirror will be:

= [tex]CO'-O'D[/tex]

On substituting the values, we get

= [tex]1.64-0.79[/tex]

= [tex]0.85 \ m[/tex]

Other Questions
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