A car’s velocity as a function of time is given by Vx (t) = α.t + β.t 2 , where α= 3m/s and β= 0.1m/s 3 . Calculate the average acceleration for the time interval

b) t= 5 to t = 10 s

Answers

Answer 1

The definition of average acceleration allows to find the result for the average acceleration in the given time interval is:

          [tex]a_{average}= 1.5 \ \frac{m}{s^2}[/tex]

Instantaneous acceleration is defined as the derivative of velocity with respect to time.

           a =   [tex]\frac{dv}{dt}[/tex]

Where a is the acceleration, v the velocity and t the time.

They indicate that the speed of the car is given by the relation.

          v = α t + β t²

With α = 3 m / s and β = 0.1 m / s³

Let's make  the derivative.

           a = α + 2β t

Let's substitute

            a = 3 + 2 0.1 t

Average acceleration is the change in velocity in the time interval.  

          [tex]a_{average} = \frac{\Delta v}{\Delta t }[/tex]

Let's find the velocity at the indicated time.

For t = 5 s

         v₅ = 3 + 0.1 5²

         v₅ = 5.5 m / s

For t = 10 s

          v₁₀ = 3 + 0.1 10²

          v₁₀ = 13 m / s

Let's calculate the average acceleration.

           [tex]a_{average} = \frac{13 - 5.5 }{ 10 - 5 }\\[/tex]

           [tex]a_{average}= 1.5 \ m/s^2[/tex]

In conclusion using the definition of mean acceleration we can find the result for the mean acceleration in the given time interval is:

           [tex]a_{average} =[/tex]  1.5 m / s²

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Answer:

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Answer:

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Explanation:

im doing the same one lol

For every action there is an equal and opposite reaction explains the rocket's motion.

What is rocket?

A rocket is indeed a vehicle that accelerates by employing jet propulsion rather than the surrounding air. A rocket engine generates thrust by reacting to high-speed exhaust. Since rocket engines run solely on fuel carried within the vehicle, a rocket may travel in space.

A water rocket uses an amount of water and pressurized air to send a plastic rocket several feet into the air. As the water and air rush out the tail end of the rocket, the rocket shoots into the air. For every action there is an equal and opposite reaction explains the rocket's motion.

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Answer:

Explanation:

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Answer:

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puntual
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Revisa el ejemplo de la clase del tema 6. Además, recuerda que las fuerzas no siempre actúan
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Answer:

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Answers

Answer:

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Answers

Answer:

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Recall that work is the amount of energy transferred to an object when it experiences a displacement and is acted upon by an external force. It is given a symbol of W and is measured in joules (J).

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Part A:
Why did the author most likely choose the title “Breaking Barriers” for this text?

A.
to show that Elizabeth worked hard to make people like her

B.
to show that Elizabeth accomplished things never done before

C.
to show that Elizabeth proved to be smarter than those around her

D.
to show that Elizabeth tried to show she was the better than other doctors

Part B:
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“When people in town ignored the woman with the strange ambition, Elizabeth treated them in a polite, quiet manner.”

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“She became the first woman in the country to obtain a medical degree.”

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Answer:

for part A the answer is D and for part B the answer is F

i just need help plzzzz

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Answer:

I think D it's the correct answer!! Let me know if i'm wrong

Explanation:

Thinking Mathematically: Explore the quantitative dependencies of the acceleration upon the speed and the radius of curvature. Then answer the following questions. a. For the same speed, the acceleration of the object varies _____________ (directly, inversely) with the radius of curvature. b. For the same radius of curvature, the acceleration of the object varies _____________ (directly, inversely) with the speed of the object. c. As the speed of an object is doubled, the acceleration is __________________ (one-fourth, one-half, two times, four times) the original value. d. As the speed of an object is tripled, the acceleration is __________________ (one-third, one-ninth, three times, nine times) the original value. e. As the radius of the circle is doubled, the acceleration is __________________ (one-fourth, one-half, two times, four times) the original value. f. As the radius of the circle is tripled, the acceleration is __________________ (one-third, one-ninth, three times, nine times) the original value.

Answers

The expression for the centripetal acceleration allows to find the results for the questions are:

A) The acceleration varies INVERSELY with the radius of curvature.

B) The acceleration varies DIRECTLY with the speed.

C) The acceleration becomes four times greater.

D) The acceleration becomes nine times greater.

E) The acceleration is reduced to half.

F) The acceleration is reduced to a third.

In circular motion there must be an acceleration towards the center of the circle, it is called cenripetal acceleration, in this case all the energy supplied to the system is used to change the direction of the speed even when its magnitude remains constant.

        [tex]a_c = \frac{v^2}{r}[/tex]  

Where [tex]a_c[/tex] the centripetal acceleration, v is the speed and r the radius of curvature of the circle.

Now we can answer the questions about centripetal acceleration.

A) For the same speed, the acceleration varies INVERSELY with the radius of curvature.

B) For the same radius of curvature the acceleration varies DIRECTLY with the speed.

C) The speed is doubled

         v = 2 v₀

         [tex]a_c = \frac{(2v_o)^2 }{r}[/tex]  

         [tex]a_c = 4 \ \frac{v_o^2 }{r}[/tex]  

The acceleration becomes four times greater than the original value

D) The speed is tripled

         v = 3 v₀

         [tex]a_c = 9 \frac{ v_o^2}{r}[/tex]  

Acceleration becomes nine times greater than the original

E) the radius of curvature is doubled

        r = 2 r₀  

        [tex]a_c = \frac{v_o^2}{2 r_o }[/tex]  

        [tex]a_c = \frac{1}{2} a_o[/tex]  

Acceleration is reduced to half the original value

F) The radius of curvature is tripled

       r = 3 r₀

       [tex]a_c = \frac{v_o^2 }{3 r_o} \\ \\a_c = \frac{1}{3} a_o[/tex]  

       

The acceleration is reduced to a third of the initial one.

In conclusion using the expression for the centripetal acceleration we can find the answers for the questions are:

A) The acceleration varies INVERSELY with the radius of curvature.

B) The acceleration varies DIRECTLY with the speed.

C) The acceleration becomes four times greater.

D) The acceleration becomes nine times greater.

E) The acceleration is reduced to half.

F) The acceleration is reduced to a third.

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A particle of mass m is fired upwards from the surface of a planet of mass M
and radius R with velocity
Show that the maximum height the particle attains is R/3.

Answers

At the maximum height the particles position has been shown as [tex]\frac{R}{3}[/tex].

The given parameters;

mass of the particle, = Mradius of the particle, = Rmaximum height the particle attains = R/3

The velocity of the particle is given as;

[tex]v^2 = \frac{GM}{2R}[/tex]

The total mechanical energy of the particle is given as;

[tex]-\frac{GMm}{R} + \frac{1}{2} mv^2 = - \frac{GMm}{R} + \frac{1}{2} mv^2[/tex]

At maximum height the final velocity of the particle is zero.

[tex]-\frac{GMm}{R} + \frac{1}{2} mv^2 = - \frac{GMm}{R +h}[/tex]

[tex]-\frac{GMm}{R} + \frac{1}{2} m\times \frac{GM}{2R} = - \frac{GMm}{R +h}\\\\-\frac{GMm}{R} + \frac{GMm}{4R} = - \frac{GMm}{R +h}\\\\- \frac{3GMm}{4R} = - \frac{GMm}{R +h}\\\\- \frac{3}{4R} = - \frac{1}{R+ h} \\\\3(R+ h) = 4R\\\\3R + 3h = 4R\\\\3h = 4R - 3R\\\\3h = R\\\\h = \frac{R}{3}[/tex]

Thus, at the maximum height the particles position has been shown as [tex]\frac{R}{3}[/tex].

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A boy of mass 43.2 kg runs and jumps onto a stationary skateboard.The boy lands on the skateboard with a horizontal velocity of 4.10 m/s.The skateboard has a mass of 2.50 kg.
Using ideas about conservation of momentum, calculate the combined
velocity of the boy and skateboard just after the boy lands on it.

Answers

Answer:

= 3.87 m/s

Explanation:

Since momentum is conserved, momentum right before and after the boy jumps onto the skateboard should be the same.

Initial momentum = momentum of boy = 43.2 x 4.10 = 177 kg·m/s

Final momentum = momentum of boy and skateboard = (43.2 + 2.50) x v = 45.7v

177 = 45.7v

v = 177 / 45.7 = 3.87 m/s

(Hope this helps can I pls have brainlist (crown)☺️)

The combined velocity of the boy and the skateboard just after the boy lands on it will be equal to 3.87 m/s.

What is momentum?

Momentum is the product of a particle's mass and velocity. Being a vector quantity, momentum possesses both direction and magnitude. According to Isaac Newton's second equation of motion, the force applied on a particle is equivalent to the temporal rate at which momentum changes. 

According to Newton's second law, the product of the energy and the time interval is equal to the shift in momentum if a constant force acts on a molecule for a specified amount of time.

Momentum should be the same both before and after the boy gets onto the skateboard because it is preserved.

Initial momentum = momentum of boy

= 43.2 × 4.10 = 177 kg m/s

Final momentum = momentum of boy as well as skateboard

= (43.2 + 2.50) × v = 45.7 v

177 = 45.7 v

v = 177 / 45.7 = 3.87 m/s

Therefore, the combined velocity of boy and skateboard is 3.87 m/s.

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