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Genrish500 [490]
3 years ago
13

baseball player hits a line drive estimated to have traveled 145 meters the ball leaves the bat with a horizontal velocity of 40

.0 m/s . how much time was the ball in the air
Physics
1 answer:
Taya2010 [7]3 years ago
8 0

Answer:

3.63 s

Explanation:

Since distance d = vt where v = velocity and t = time taken to cover the distance.

Now, the baseball traveled a distance of d = 145 m with a horizontal velocity of 40.0 m/s.

So, making t subject of the formula,

t = d/v

Substituting the values of d and v we have

t = 145 m/40.0 m/s

= 3.625 s

≅ 3.63 s

So the time the ball was in the air is 3.63 s

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Explanation:

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An applied force varies with position according to F = k1 x n − k2, where n = 3, k1 = 3.6 N/m3 , and k2 = 76 N. How much work is
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Answer:

The work done is 205 kJ.

Explanation:

Hi there!

Work can be calculated using the following equation:

W = F · Δx

Where:

W = work

F = applied force

Δx = displacement

In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:

W = ∫ F · dx  

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W = ∫ (3.6 x³ - 76)dx

W = 0.9 x⁴ - 76x

Evaluating from xi to xf:

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3 years ago
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Consider a block on frictionless ice. Starting from rest, the block travels a distance din
sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

\displaystyle a=\frac{F}{m}

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

\displaystyle d = vo.t+\frac{at^2}{2}

If the block starts from rest, vo=0:

\displaystyle d = \frac{at^2}{2}

Substituting the value of the acceleration:

\displaystyle d = \frac{\frac{F}{m}t^2}{2}

Simplifying:

\displaystyle d = \frac{Ft^2}{2m}

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

\displaystyle a'=\frac{4F}{m}

And the distance is now:

\displaystyle d' = \frac{4Ft^2}{2m}

Dividing d'/d:

\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}

Simplifying:

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2 years ago
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