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inessss [21]
3 years ago
11

A baseball leaves a bat and flies upward and toward center field. Do not ignore air resistance. What forces are exerted on the b

all just after it leaves the bat?
Physics
1 answer:
Elina [12.6K]3 years ago
4 0

After the ball leaves the bat, the only forces on it are air resistance and gravity.
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What new characteristic did John Dalton add to the model of the atom? An atom is a round, solid mass. An atom has a massive, pos
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You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi
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Answer:

a)  x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3  (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

                              v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

                              dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

                             x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

                              2 = 10t + (2/3)*t^3

                              0 = 10t + (2/3)*t^3 + 2

- solve for t:

                              t = 0.1875 s

- Evaluate x* at t = 0.1875 s

                              x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

                              x*(0.1875) = 10.18 m

3 0
3 years ago
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
bogdanovich [222]

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.

Learn more about free-fall here:

<u>brainly.com/question/13299152</u>

#SPJ4

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D because friction will slow it down going up hills (for safety reasons) and you go really fast going down.
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