Physics, 27.02.2020 05:53 bellamore

Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the x direction.

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Answer from: hibbo579

v_f = -25.9 m/s

Explanation:

- The complete question is as follows:

" Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the +x direction.

Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's velocity just after leaving the bat if the bat applies an impulse of −8.4N⋅s to the baseball?"

Given:

- mass of baseball m = 0.145 kg

- Speed before impact v_i = 32 m/s

- Speed after impact v_f

- Impulse applied by the bat I = - 8.4Ns

Find:

What is the ball's velocity just after leaving the bat

Solution:

- Impulse is the change in linear momentum of the ball according to Newton's second law of motion:

I = m* ( v_f - v_i )

- Taking the + from pitcher to batsman and - from batsman to pitcher.

- Plug in the values:

-8.4 = 0.145* ( v_f - (32) )

v_f = -57.93103 + 32

v_f = -25.9 m/s

Answer from: Quest

explanation: in any substance/object, the particles inside it (atoms/molecules) constantly move in random directions and with random speeds (this motion is called brownian motion). as a result, the particles have some kinetic energy (which is proportional to the temperature of the substance).

explanation:

have a great day!

Answer from: Quest

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample, $v = 1.1 cm^{3}$

so, density of rock sample will be

$density = \frac{mass}{volume}$

$\rho =\frac{2.7}{1.1}g/cm^{3}\\\\\rho =2.45 g/cm^{3}$

since, the density of rock sample is $2.45 g/cm^{3}$ . hence, the rock sample is a sedimentary rock, whose densities varies between $2g/cm^{3} - 2.6 g/cm^{3}$ .

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