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3 years ago

7

A ball is shot straight up from the surface of the earth with an initial speed of 19.6 m/s. Neglect any effects due to air resis

tance. How much time elapses between the throwing of the ball and its return to the original launch point?

1 answer:

Pavel [41]3 years ago

5 0

Answer:

4 s

Explanation:

u = 19.6 m/s, g = 9.8 m /s^2

Let the time taken to reach the maximum height is t.

Use first equation of motion.

v = u + at

At maximum height, final velocity v is zero.

0 = 19.6 - 9.8 x t

t = 19.6 / 9.8 = 2 s

As the air resistance be negligible, is time taken to reach the ground is also 2 sec.

So, total time taken be the ball to reach at original point = 2 + 2 = 4 s

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Answer:

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2 years ago

Estimate the average force that a baseball pitcher's hand exerts on a 0.145-kg baseball as he throws a 40-m/s pitch. Assume the

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Answer:

38.67 N

Explanation:

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∴ Average force that a baseball pitcher's hand exerts on the ball is 38.67 N

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3 years ago

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3 years ago

Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m/s^2, where t is in seconds. What

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To solve this problem we will use the given expression and derive it in order to find the algebraic expressions of velocity and position. These equations will be similar to those already known in the cinematic movement but will be subject to the previously given function. We start deriving the equation for velocity

$a = 2t-6$

$\frac{dv}{dt} = 2t-6$

Integrate acceleration equation

$\int dv = (2t-6)dt$

$v = 2(\frac{t^2}{2})-6t+C_1$

$v=t^2-6t+C_1$

At $t = 0, v = 0$

Replacing,

$0 = 0^2-6*0+C_1$

Therefore the value of the first Constant is

$C_1 = 0$

The expression can be escribed as,

$v = t^2-6t$

Calculate the velocity after 6s,

$v=t^2-6t$

$v = 6^2-6*6$

$v = 0m/s$

Now using the same expression we can derive the equation for distance

$v = t^2-6t$

$\frac{dx}{dt} =t^2-6t$

$\int dx = \int (t^2-6t)dt$

$x = \frac{t^3}{3}-6\frac{t^2}{2}+C_2$

At t=0, x=0

$0 = \frac{0^3}{3}-6(\frac{0^2}{2})+C_2$

Therefore the value of the second constant is

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3 years ago

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Mars -No
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c. Neptune - Neptune has several faint rings around it.
d. Saturn - Saturn has bright ice rings,
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3 years ago

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