A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 1.1 m from the axis of rotation, and he rotates with angular speed of 0.64 rad/sec. The moment of inertia of the student plus the stool is 4 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis.

Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.

Answers

Answer 1

Answer:

a) ω₁ = 0.97 rad/sec

b) K₀ = 1.31 J

   Kf  = 1.99 J

Explanation:

a)

Assuming no external torques acting on the student, total angular momentum must be conserved, as follows:

       [tex]L_{o} = L_{f} (1)[/tex]

The angular momentum of a rigid body rotating respect an axis of rotation can be written as follows:

       [tex]L = I*\omega (2)[/tex]

In order to get ωf, since ω₀ = 0.64 rad/sec, we need to find the values of the initial moment of inertia, I₀, and the final one, If:I₀ = 4 kg*m² + 1kg*(1.1 m)² + 1kg*(1.1m)² = 6.42 kg*m²  (3)If = 4 kg*m² + 1kg*(0.33m)² + 1kg*(0.33m)² = 4.22 kg*m² (4)Replacing (3), (4) in (1) we can solve for ωf:

       [tex]\omega_{f} = \frac{I_{o} *\omega_{o} }{I_{f} } = \frac{6.42kgm2*0.64rad/sec}{4.22kgm2} = 0.97 rad/sec (5)[/tex]

b)

Since the student is not translating but he is only rotating, all his kinetic energy is rotational kinetic energy.The expression for the kinetic energy of a rotating rigid body, around an axis of rotation is as follows:

       [tex]K_{rot} = \frac{1}{2} * I * \omega^{2} (6)[/tex]

The initial kinetic energy of the student, before the objects are pulled in, is as follows:

      [tex]K_{roto} = \frac{1}{2} *I_{o} * \omega_{o} ^{2} = \frac{1}{2}* 6.42kgm2*(0.64rad/sec)^{2} = 1.32 J (7)[/tex]

The final kinetic energy is given by the following expression:[tex]K_{rotf} = \frac{1}{2} *I_{f} * \omega_{f} ^{2} = \frac{1}{2}* 4.22kg*m2*(0.97rad/sec)^{2} = 1.99 J (8)[/tex]

Related Questions

The primary reason for the path of motion of an object being a smooth curve is: Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a the third derivative of parabolas is always zero. b inertia. c tangent direction unit vectors change continuously. d calculus must have continuous derivatives to apply correctly.

Answers

Answer:

the correct answer is d

Explanation:

The laws of mechanics are related

          F = m a

the acceleration of the body is given by the kinematics

         a = [tex]\frac{dv}{dt}[/tex]

         v = [tex]\frac{dx}{dt}[/tex]

         

substituting

         a = \frac{d2x}{dt^2}

         F = m [tex]\frac{d^2x}{dt^2}[/tex]

Therefore, in order to obtain the force (interaction of a body), continuous curves are needed and derivable from the position and the speed, for which all change in the trajectory of a body must be smooth where smooth is understood to have until the second derived.

Consequently the correct answer is d

The discharge of a pump is 3 m above the inlet. Water enters at a pressure of 138 kPa and leaves at a pressure of 1380 kPa. The specific volume of the water is 0.001 m3/kg. If there is no heat transfer and no change in kinetic or internal energy, what is the work per unit mass

Answers

Answer:

The answer is "[tex]1.271 \ \frac{KJ}{kg}\\[/tex]"

Explanation:

[tex]\Delta e_{mech} =\frac{P_2-P_1}{P} + \frac{v_{2}^2-v_{1}^2}{2}+g(z_2-z_1)\\\\\Delta e_{mech} =\frac{ 1380 -138 \times 1000 }{1000} + 0+g(3-0)\\\\P = \frac{1}{v}= \frac{1}{0.001} = 1000 \frac{kg}{m} \\\\ \Delta e_{mech} =1242 +9.81(3)= 1271.43 \frac{J}{kg} \\\\\text{work per unit pass}= 1.271 \ \frac{KJ}{kg}\\[/tex]

The bond order for a single covalent bond is.
A. two
B. four
C. one
D. three

Answers

Answer:

I think it should be C, which is one

From what does oil form?

A. marine organisms
B. terrestrial plants
C. dinosaurs
D. lava or magma

Answers

C dinosaurs. Once they break down and get pressurized they turn to oil

Answer: marine organisms

Explanation:

i just took the test

A virtual image larger than the object can be produced by
1) concave mitor 2) convex mutor 3)plane mirror
concave lens
O1​

Answers

Answer:

concave mirror

Explanation:

Why This is Correct Because, concave mirrors only form a virtual image  when a object is larger than the other. when the object is produced between the focus, object and the image object, it becomes a  virtual image.

Answer:

The answer is concave mirror

Jack D. Ripper flipped out after missing a Must-Do-It question for the third time on his Minds On Physics assignment. Outraged by the futility of his efforts, he flings a 4.0-gram pencil across the room. The pencil lodges into a 221.0-gram Sponge Bob doll which is at rest on a countertop. Once in motion, the pencil/doll combination slide a distance of 11.9 cm across the countertop before stopping. The coefficient of friction between the doll and the countertop is 0.325. Determine the speed at which the pencil is moving prior to striking Sponge Bob.

Answers

Answer:

Explanation:

Let the velocity after the 4 gram pencil strikes is v .

kinetic energy of the combination = 1/2 m v²

= .5 x ( 4 + 221 ) x 10⁻³ x v² = work done by friction

friction force acting on the combination = 225 x10⁻³x  .325 x 9.8 = .7166 N

work done by friction

= .7166 x .119 = .085 J

.5 x 225 x 10⁻³ v² = .085

v² = .085 / .1125 = .7555

v = .8692 m = 86.92 cm /s

Velocity of combination after collision = 86.92 cm /s

Let velocity of pencil before collision be V

Applying law of conservation of momentum at the time of collision ,

4 x V = 225 x 86.92

V = 4889.25 cm / s

= 48.9 m /s .

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.8 mm above the river, whereas the opposite side is a mere 1.3 mm above the river. The river itself is a raging torrent 53.0 mm wide.
A) How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?
B) What is the speed of the car just before it lands safely on the other side?

Answers

Answer:

A) 26.5 m/s

B) 33.0 m/s

Explanation:

A)

Once the car leaves the cliff, as no other influence than gravity acts on it, and since it causes the car an acceleration in the vertical direction only, in the horizontal direction, it keeps moving at the same speed until it reaches to the other side.So, we can apply the definition of average velocity to find this speed as follows:

       [tex]v_{x} = \frac{\Delta x}{\Delta t} (1)[/tex]

We know the value of Δx, which is just the wide of the river (53.0m), but we need to find also the value of Δt.This time is given by the vertical movement, whic.h is independent from the horizontal one, because both movements are perpendicular each other.Since the only influence in the vertical direction is due to gravity, the car is accelerated by gravity, with constant acceleration downward equal to g = -9.8m/s² (taking the upward direction as positive).Since the acceleration is constant, we can use the following kinematic equation, as follows:

       [tex]\Delta y = y_{f} - y_{o} = v_{o} * t + \frac{1}{2} * g *t^{2} (2)[/tex]

if we take the river level as our x-axis, this means that yf = 1.3 m and

       y₀ = 20.8 m.

At the same time, due to in the vertical direction the car has no initial velocity, this means that  v₀ = 0.Replacing by the values in (2) , and solving for t:

       [tex]t = \sqrt{\frac{2* \Delta y}{g} } = \sqrt{\frac{2*19.5m}{9.8m/s2} } = 2 s (3)[/tex]

If we choose t₀ =0 ⇒ Δt = t = 2 sReplacing Δx and Δt in (1):

       [tex]v_{x} = \frac{\Delta x}{\Delta t} = \frac{53.0m}{2s} = 26.5 m/s (4)[/tex]

B)

When the car is just landing in the other side, the velocity of the car has two components, the horizontal one that we just found in A) and a vertical one.Due to the initial velocity in the vertical direction was just zero, we can find the final velocity just applying the definition of acceleration, with a =g, as follows:

      [tex]v_{fy} = g*t = -9.8m/s2*2 s = -19.6 m/s (5)[/tex]

Since both components are perpendicular each other, we can find the magnitude of the velocity vector (the speed) using the Pythagorean Theorem, as follows:

       [tex]v = \sqrt{v_{x}^{2} + v_{fy}^{2} } } = \sqrt{(26.5m/s)^{2} + (-19.6m/s)^{2}} = 33.0 m/s (6)[/tex]

acceleration greater than 1,000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.63 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.3 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. hardwood floor magnitude m/s2 hardwood floor duration ms carpeted floor magnitude m/s2 carpeted floor duration ms g

Answers

Answer:
This needs to be more than 5 points man

While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 7.07 m/s. The stone subsequently falls to the ground, which is 19.3 m below the point were the stone leaves his hand. At what speed does the stone impact the ground? Ignore air resistance and use g = 9.81 m/s^2 for the acceleration due to gravity.

Answers

Answer:

the stone hits the gound with a speed of  20.7 m/s

Explanation:

Becuase gravity is constant we know that the initial upward velocity will be equal to the downward velocity when the stone has returned to its intal location.

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the
power consumed will be-
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W​

Answers

Answer:

50 W

Explanation:

Case 1

Power = V * I

100 = 220 * I

I = [tex]\frac{100}{220}[/tex] A

Case 2

P = V * I

P = 110 * [tex]\frac{100}{220}[/tex]

P = 50 W

I think the answer is 50 W

Hope it helps

A mass of 10. kg is placed on the end of a 0.50-meter pendulum. What is the period of the pendulum?

Answers

Answer:

T = 1.41 seconds

Explanation:

Given that,

The mass placed in the pendulum, m = 10 kg

The length of the pendulum, l = 0.5 m

We need to find the period of the pendulum. The relation for the period of the pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}} \\\\T=2\pi \sqrt{\dfrac{0.5}{9.8}} \\\\T=1.41\ s[/tex]

So, the time period of the pendulum is 1.41 seconds.

For anyone that needs the correct answer without POS trolls:

The answer is 1.4 s

Thank me later :)

rocket fuel contains 50j of energy in its chemical store. the rocket has a mass of 1kg. what is the maximum speed the rocket could reach​

Answers

KE = 1/2mv(squared)

rearrange the formula to:
2KE/m (square rooted) = speed

2*50=100
100/1 =100
100 (square root) = 10

ANSWER: 10m/s

At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the pilot whose weight is 676 N, maintains a constant speed of 2.25 x 102 m/s. At what speed, in m/s, will the pilot experience weightlessness

Answers

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where  v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

____made up of glucose and fructose and found in plants.
1.Xylose
2.Maltose
3.Lactose
4.Sucrose

Answers

Answer:

1.Lactose

I Hope its help for you

Have a good day

Carousel conveyors are used for storage and order picking for small parts. The conveyorsrotate clockwise or counterclockwise, as necessary, to position storage bins at the storageand retrieval point. The conveyors are closely spaced, such that the operators travel timebetween conveyors is negligible. The conveyor rotation time for each item equals 1 minute;the time required for the operator to retrieve an item after the conveyor stops rotatingequals 0.25 minute. How many carousel conveyors can one operator tend without creatingidle time on the part of the conveyors

Answers

Answer:

the number of carousel conveyors that an operator can operate without any idle time is 5

Explanation:

Given the data in the question;

first we express the equation for number of carousel conveyors that can be operated by an operator;

n' = [tex]\frac{(a + t)}{( a + b)}[/tex]

where a is the concurrent activity time ( 0.25 minute )

b is the independent operator activity time

t is the independent machine activity time( 1 )

Now independent activity time is zero as the operator is not performing any inspection or packaging tasks.

So time taken for the operator to retrieve the finished item at the end of the process is the concurrent activity and independent machine activity time, the conveyor rotation time of each item

so

we substitute

0.25min for a, 1 for t and 0min for b

n' = [tex]\frac{(0.25min + 1min)}{( 0.25min+ 0 min)}[/tex]

n' = 1.25 min / 0.25

n' - 5

Therefore, the number of carousel conveyors that an operator can operate without any idle time is 5

A block is pushed so that it moves up a ramp at constant speed. Identify from choices (a)-(e) below the appropriate description for the work done by the specified force while the block moves from point A to point B. (a) is zero. (b) is less than zero. (c) is greater than zero. (d) could be positive or negative depending on the choice of coordinate systems. (e) cannot be determined.

Answers

Answer:

*The work of the Normal (N) y Wy are zero  answer a

*The work of the applied force (F1)  is positive answer c

*The work of the friction force (fr) is negative, answer b

*The work of the Wy isnegative, answer d

Explanation:

In this exercise it is asked to identify the type of work, unfortunately the diagram cannot be seen, but in the attached we can see the diagram of a body moving upward on an inclined plane, the existing forces are shown.

As the body moves at constant speed the accelerations are zero. Let's look for the job that is defined

           W = F. d

            W = F d cos θ

where the dot represents the dot product and the bold letters are vectors.

* The work of the Normal (N) and the y component of the weight (Wy) are zero because they are perpendicular to the motion

    answer a

* The work of the applied force (F1) is positive because it is in the same direction of motion

        W = F1 Δx

answer c

* The work of the friction force (fr) is negative because the force in the displacement have opposite directions

        W = -fr Δx

answer b

* the work the x component of the weight (Wx) in this case is negative

answer d

3. What is the acceleration of a cart with a F = 80 N and m = 32 kg? *

2.0 m/s^2

O 2.5 m/s^2

2.8 m/s^2

2.7 m/s^2

Answers

Answer:

2.5 m/s^2

Explanation:

Given the following data;

Force = 80N

Mass = 32kg

To find acceleration;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

[tex] F = ma[/tex]

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

Making acceleration (a) the subject, we have;

[tex]Acceleration (a) = \frac{F}{m}[/tex]

Substituting into the equation;

[tex]Acceleration (a) = \frac{80}{32}[/tex]

Acceleration = 2.5m/s²

How long would it take for a car to travel a distance of 220 kilometers if it is traveling at a velocity of 55 km/hr South?

Your answer:

220 hours


12,100 hours


4 hours


0.25 hours

Answers

220/55=4 hours in total

What genetic test would you get if there was a specific genetic disease in your family

Answers

I believe it is called single gene testing

A student weighing 5.4 × 102 newtons takes 15 seconds to run up a hill. The top of the hill is 10 meters vertically above her starting point. What power does the student develop during her run?

Answers

Answer:

P = 360 Watts

Explanation:

Given that,

The weight of a student, [tex]F=5.4\times 10^2\ N[/tex]

It takes 15 seconds to run up a hill.

The top of the hill is 10 meters vertically above her starting point.

We need to find the power develop during her run. We know that te power developed is given by :

[tex]P=\dfrac{W}{t}\\\\P=\dfrac{mgh}{t}\\\\P=\dfrac{5.4\times 10^2\times 10}{15}\\\\P=360\ W[/tex]

So, the power develop during her run is 360 W.

diffraction of light is the __________ of light as it passes through the Edges of a barrier or a slit.

1, reflections
2, refraction
3,bending
4,absorbing

Answers

Answer:

3 bending

Explanation:

One of the fastest pitches ever thrown in Major League Baseball was by Aroldis Chapman and had a velocity of 105.1 miles/hour. How many seconds did it take this pitch to travel the 60 feet and 6 inches from the pitcher's mound to home plate

Answers

Answer:

t = 0.39 s

Explanation:

Assuming that the ball is launched horizontally, once in the air, if we neglect the resistance of the air, the ball moves at a constant speed, equal to the initial velocity, in this case, 105.1 mi/hr.In order to find time in seconds, it is advisable to convert the speed in mi/hr to m/s, as follows:

       [tex]105.1 mi/hr * (\frac{1hr}{3600s})*\frac{1609m}{1mi} = 47.0 m/s (1)[/tex]

In the same way, it's advisable to convert 60' 6'' (60.5') to m, as follows:

       [tex]60.5 ft * \frac{0.3058m}{1ft} = 18.5 m (2)[/tex]

Applying the definition of average velocity, we can find the time traveled by the ball from pitcher's mound to home plate, as follows:

       [tex]t = \frac{18.5m}{47m/s} = 0.39 s (3)[/tex]

to see if the original results are Which career field is an applied science?
geology
biotechnology
physics
chemistry

Answers

Answer:

it is chemistry

Explanation:

The (kinetic, radiant) theory explains the motion of particles in matter.
A)Kinetic
B)Radiant

Answers

Answer:

radiant

Explanation:

Answer:

kinetic

Explanation:

this is because the kinetic theory is when energy is used in motion.

Police driving with a velocity of 50 m/s decide to chase a speeder who is 3 km ahead and moving at 55 m/s. The police car accelerates at 2 m/s2. Instantly the speeder becomes aware that he is being chased and starts to accelerate at 1 m/s2. How much time (in s) passes until the police catch the speeder

Answers

Answer:

The time that passes until the police catch the speeder is 82.6204 seconds.

Explanation:

A body performs a uniformly accelerated rectilinear motion or uniformly varied rectilinear motion when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases its modulus in a uniform way.

The position is calculated by the expression:

x = x0 + v0*t + 1/2*a*t²

where:

x0 is the initial position. v0 is the initial velocity. a is the acceleration. t is the time interval in which the motion is studied.

First, let’s look at the police car’s equations of motion. In this case:

x0= 0 v0= 50 m/s a= 2 m/s²

So: x = 50 m/s*t + 1/2*2 m/s²*t²

Now for the speeder’s car’s equations of motion you know:

x0= 3 km= 3,000 m v0= 55 m/s a= 1 m/s²

So: x = 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²

When the police catch the speeder they are both in the same position. So:

50 m/s*t + 1/2*2 m/s²*t²= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²

Solving:

0= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t² - 50 m/s*t - 1/2*2 m/s²*t²

0= 3,000  + 55 *t + 1/2*t² - 50*t - 1*t²

0= 3,000  + 55 *t - 50*t - 1*t² + 1/2*t²

0= 3,000  + 5*t - 1/2*t²

Applying the quadratic formula:

[tex]x1,x2=\frac{-5+-\sqrt{5^{2}-4*(-\frac{1}{2})*3000 } }{2*(-\frac{1}{2} )}[/tex]

x1= -72.6209

and x2= 82.6209

Since you are calculating the value of a time and it cannot be negative, then the time that passes until the police catch the speeder is 82.6204 seconds.

Conductivities are often measured by comparing the resistance of a cell filled with the sample to its resistance when filled with some standard solution,such as aqueous potassium chloride. The conductivity of water is 76 mS m^(-1) at 25 C and the conductivity of 0.100 mol dm^(-3) KCl (aq) is 1.1639 S m^(-1) A cell has a resistance of 33.21ohm when filled with 0.100 mol dm^(-3) KCl (aq) and 300.0 ohm when filled with 0.100 mol dm CHaCOOH (aq). What is the molar conductivity of acetic acid at that concentration and temperature?

Answers

Answer:

1200 Sm^2mol^-1

Explanation:

Given data :

conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1

conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1

Kkcl = 1.1639 - 0.076 = 1.0879  Sm^-1

Resistance = 33.21 Ω

where conductivity can be expressed as = [tex]\frac{Cell constant}{Resistance }[/tex]

hence cell constant = conductivity * Resistance

                                 = 1.0879 * 33.21 = 36.13m^-1

conductivity of  CH3COOH ( kCH3COOH ) =  36.13 / 300

                                                                       = 0.120 Sm^-1

Determine the molar conductivity of acetic acid

= ( kCH3COOH * 1000 ) / C

C = 0.1 mol dm

=  (0.120 * 1000) / 0.1  =  1200 Sm^2mol^-1

Can I get help on this question I’m scared to get it wrong .

Answers

Pretty sure it’s KE. Hope this helps ✌️

help would be greatly appreciated

How does an unbalanced force cause a change in direction of an object?

Answers

Answer:

Explanation:

The net force is in the same direction of the acceleration. Acceleration changes the speed of an object.

Choose the words that make each statement correct.
(i) After being released from rest in a uniform electric field, a pro- ton will move [(a) in the same direction as; (b) opposite the direction of] the electric field to regions of [(c) higher; (d) lower] electric potential.
(ii) After being released from rest in a uniform electric field, an electron will move [(e) in the same direction as; (f) opposite the direction of] the electric field to regions of [(g) higher; (h) lower] electric potential.

Answers

Answer:

i). (a) in the same direction as , (d) lower

ii). (f) opposite the direction of, (g) higher

Explanation:

An proton may be defined as a sub atomic particle and it has a positive electrical charge. Its mass is slightly less than that of a neutron. When a proton is placed in an electrical field that is uniformly charged, it is at rest. When the proton first moves out from rest from the uniform electric field, it will move in a direction which is same as that of the electric field and it will move to a region of higher potential.

An electron is defined as the subatomic particle having negative electric charge. When an electron is released form rest from an uniform electric field, it will move in the opposite direction of the uniform electric field and will move to the region of lower electric potential.

A truck travels on a straight road at a velocity of 17 meters per second. Over 20

seconds, it accelerates uniformly to 27 meters per second. What distance did the truck

travel during this acceleration?

Answers

Answer:

Distance, S = 440 meters.

Explanation:

Given the following data;

Initial velocity, u = 17m/s

Time, t = 20 seconds

Final velocity, v = 27m/s

To find the distance;

First of all, we would determine the acceleration of the truck.

Acceleration = (v-u)/t

Substituting the given values into the equation, we have;

Acceleration = (27 - 17)/20

Acceleration = 10/20

Acceleration = 0.5m/s²

Now, we would use the second equation of motion to find the distance traveled.

S = ut + ½at²

S = 17*20 + ½*0.5*20²

S = 340 + 0.25*400

S = 340 + 100

S = 440m

The equations of motion can be used to obtain the distance covered as 440 m.

We have to use of the equations that are used for uniformly accelerated motion in solving the problem. The chosen equation must be;

v^2 = u^2 + 2as and v = u + at

v = final velocity

u = initial velocity

a = acceleration

s = distance

To obtain the acceleration;

27 = 17 + 20(a)

27 - 17 = 20a

a = 0.5 ms-2

Now, to obtain the distance;

v^2 = u^2 + 2as

v^2 - u^2/as = s

s = (27)^2 - (17)^2/2(0.5)

s = 440 m

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