Physics, 31.07.2019 22:30 khadythiam6957

What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal

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Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"

Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of v_f = 16.5 m/s , so $\Delta v=16.5 m/s$ , in a time interval of $\Delta t = 181 ms=0.181 s$ . The acceleration of the ball in the horizontal direction (x-axis) is therefore

$a_x = \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2$

And the distance covered by the ball during this time interval, before it is released, is:

$S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m$

(b) For this part we need to consider also the weight of the ball, which is W=mg=2.28 N

From this, we find its mass: $m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg$

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

F_y=W=mg=2.28 N (upward)

So, the magnitude of the force is

$F= \sqrt{F_x^2+F_y^2}= \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N$

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

$\tan \alpha = \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11$

From which we find $\alpha=6.2^{\circ}$

Answer from: Quest

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Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

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