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Crank
4 years ago
5

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12-m-hi

gh hill, then descends 18 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.3 m and that a loaded car will have a maximum mass of 430 kg . For safety reasons, the spring constant should be 13 % larger than the minimum needed for the car to just make it over the top.
a) what spring constant (k) should you specify ?b) What is the maximum speed of a 350 kg car if the spring iscompressed the full amount.

Physics
1 answer:
Vlada [557]4 years ago
5 0

Answer:

Vmax=11.53 m/s

Explanation:

from conservation of energy

      E_A} =E_{B}

     Spring potential energy =potential energy due to elevation

   0.5*k*x²= mg(h_{B}-h_{A} )=mgh

   0.5*k*2.3²= 430*9.81*6

         k=9568.92 N/m

For safety reason

                                 k"=1.13 *k= 1.13*9568.92

                                    k"=10812.88 N/m

agsin from conservation of energy

      E_A} =E_{C}

    spring potential energy=change in kinetic energy

   0.5*k"*x²=0.5*m*V_{max}^{2}

      10812.88 *2.3²=430*V_{max}^{2}

           V_{max}=11.53 m/s

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Answer:

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4)

5)

6)

An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of θ1 = 23° above the –x axis with a speed of v1 = 25.4 m/s. A second piece with mass m2 = 5.2 kg moves down and to the right an angle of θ2 = 28° to the right of the -y axis at a speed of v2 = 23.8 m/s.

2) What is the mass of the third piece?

3) What is the x-component of the velocity of the third piece?

4) What is the y-component of the velocity of the third piece?

5) What is the magnitude of the velocity of the center of mass of the pieces after the collision?

6) Calculate the increase in kinetic energy of the pieces during the explosion

Explanation:

Since explosions and collisions follow the law of conservation of Momentum.

1) Magnitude of the final momentum of the system = Magnitude of the initial momentum of the system

Since the body was initially at rest,

Magnitude of the initial momentum of the system = 0 kgm/s

Hence, Magnitude of the final momentum of the system is also equal to 0 kgm/s.

2) mass of the third piece

Sum of all the masses = 16.2 kg

4.7 + 5.2 + c = 16.2

c = 6.3 kg

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(4.7)×(-25.4 cos 23) + (5.2)×(23.8 cos 28) + (6.3)(v) = 0

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Hope this Helps!!!

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A 140 g air-track glider is attached to a spring. The glider is pushed in 8.2 cm against the spring, then released. A student wi
max2010maxim [7]

Answer:

Explanation:

The period of oscillation is given as

T=2π√m/k

Making k subject of the formula

Square both sides of the equation

T²=4π²(m/k)

Cross multiply

T²k=4π²m

Then, divide through by T²

k=4π²m/T²

Where

k is spring constant

m is the mass of the bob

And T is the period of the oscillation

m=140g=0.14kg

14 oscillations takes 14 seconds

Then the period is

T=time/oscillation

T=14/14

T=1sec

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4 0
4 years ago
A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. (a) Determi
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Answer:

(a) t = 3.74 s

(b) H = 136.86 m

(c) Vₓ = 41.83 m/s,  Vy = 0 m/s

(d) ax = 0 m/s²,  ay = 9.8 m/s²

Explanation:

(a)

Time to reach maximum height by the projectile is given as:

t = V₀ Sinθ/g

where,

V₀ = Launching Speed = 55.6 m/s

Angle with Horizontal = θ = 41.2°

g = 9.8 m/s²

Therefore,

t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)

<u>t = 3.74 s</u>

<u></u>

(b)

Maximum height reached by projectile is:

H = V₀² Sin²θ/g

H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)

<u>H = 136.86 m</u>

<u></u>

(c)

Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:

Vₓ = V₀ₓ = V₀ Cos θ

Vₓ = (55.6 m/s)(Cos 41.2°)

<u>Vₓ = 41.83 m/s</u>

Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.

<u>Vy = 0 m/s</u>

<u></u>

(d)

Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.

<u>ax = 0 m/s²</u>

The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:

<u>ay = 9.8 m/s²</u>

3 0
3 years ago
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