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

murzikaleks [220]

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 $\alpha=6.2^{\circ}$

7 0

2 years ago

Read 2 more answers

A plane has an average air speed (this is the speed the plane moves through air) of 750 mph. The plane flies a route of 5000 mil

Digiron [165]

Answer:

6 hours 15 minutes

Explanation:

On the trip from L.A. to London, the plane travels at 750 mph against a headwind of 50 mph, and that makes the net 700 mph (in aviation speak, 750 is the airspeed, while 700 is the groundspeed). 5000 miles divided by 700 mph results in about 7.14 hours, or about 7 hours and 9 minutes. On the return trip, ASSUMING THE SAME WIND, the plane travels at 750 mph, but this time the wind of 50 mph is a tail wind. So the net (groundspeed) is 800 mph. Traveling 5000 miles at 800 mph only takes 6.25 hours, or 6 hours and 15 minutes.

Outbound flight 7 hours 9 minutes

Return flight 6 hours 15 minutes

6 0

1 year ago

How much heat must be absorbed by 375 grams of water to raise its temperature by 25°c

Nesterboy [21]

<span>the formula q = 375 g * 25 C * 4.186 J / (g*C) = 39,243.75 J q represents the heat in Joules , m the mass in grams, difference of temperature in Celsius degree, and 4.186 J/(g*C) is the specific heat of water( I assume the water is in liquid from and will remain liquid). Approximately 39.24 kJ once you round and transform to kJ..1 kJ=1000J</span>

7 0

2 years ago

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You wad up a piece of paper and throw it into the wastebasket. How far will

vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, $g=9.8 m/s^2$ )

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

$d=\frac{u^2 sin 2\theta}{g}$