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azamat
3 years ago
8

Your school science club has devised a special event for homecoming. You've attached a rocket to the rear of a small car that ha

s been decorated in the blue-and-gold school colors. The rocket provides a constant acceleration for 9.0s. As the rocket shuts off, a parachute opens and slows the car at a rate of 5.0m/s2. The car passes the judges' box in the center of the grandstand, 990m from the starting line, exactly 12s after you fire the rocket. Part A What is the car's speed as it passes the judges
Physics
1 answer:
Strike441 [17]3 years ago
4 0

Answer:

Explanation:

Acceleration acts for 9 s and deceleration acts for 12 - 9 = 3 s.

Total distance covered = 990 m

initial velocity u = o

Distance covered while accelerating

s₁ = 1/2 a 9² where a is the acceleration

= 40.5 a

final velocity after 9 s

v = at = 9a

while decelerating

v² = u² - 5 x s₂

0 = (9a)² - 5 s₂

s₂ = 16.2 a²

Distance covered while decelerating = 16.2 a²

s₁ + s₂ = 990

40.5 a + 16.2 a² = 990

16.2 a² + 40.5 a - 990 = 0

a = 6.5

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Answer:

Explanation:

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Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th
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The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

               the mass of planet B to be, Mₓ = M

               the radius of planet A to be, R₁

               the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁²      - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂²                           [where, R₂ = 2R₁]

   = GM/4R₁²    -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

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Answer:

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A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
BigorU [14]

Complete Question

The  complete question is  shown on the first uploaded image  

Answer:

a

  f=  -75 \ cm =  - 0.75 \ m

b

  P  =  -1.33 \ diopters

Explanation:

From the question we are told that

    The  image distance is  d_i =  -75 cm

The value of the image is negative because it is on the same side with the corrective glasses

    The  object distance is  d_o =  \infty

The  reason object distance  is because the object father than it being picture by the eye

General focal length is mathematically represented as

              \frac{1}{f}  =  \frac{1}{d_i}  -   \frac{1}{d_o}

substituting values

             \frac{1}{f}  =  \frac{1}{-75}  -   \frac{1}{\infty}

=>         f=  -75 \ cm =  - 0.75 \ m

Generally the power of the corrective lens is  mathematically represented as

        P  =  \frac{1}{f}

substituting values

       P  =  \frac{1}{-0.75}

        P  =  -1.33 \ diopters

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