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2 years ago

Find the average force exerted by the bat on the ball if the two are in contact for 0.00129 s. Answer in units of N.

Physics

1 answer:

Semmy [17]2 years ago

5 0

Answer:

Explanation:

Given

time of contact between bat and ball is $t=0.00129 s$

suppose u is the incoming velocity and v is the final velocity after collision

$m=mass\ of\ ball$

Impulse exerted is given by change in momentum of the particle.

Initial momentum $P_i=m\times u$

Final momentum $P_f=m\times v$

Change in momentum $\Delta P=P_i-P_f$

Impulse $J=F_{avg}\cdot t=\Delta P$

$J=F_{avg}\cdot t=m(u-v)$

$F_{avg}=\frac{m(v-u)}{t}$

$F_{avg}=775.2\times m(v-u)\ N$

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A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

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2 years ago

An object moving at a velocity of 32m/s slows to a stop in 4 seconds. What was its acceleration?

Romashka [77]

Answer:

8m/s

Explanation:

a=d/t

a=32/4

a=8 m/s

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2 years ago

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) What is the electric potential due to the nucleus of hydrogen at a distance of 7.50× 10-11 m? Assume the potential is equal to

ohaa [14]

For this, we need the formula:
V = k q / r
where k is the Coulombs law constant = 9 x 10^9 N
q is the charge of the hydrogen nucleus (proton) = <span>1.6 x 10^-19 C</span>
r is the distance
Simply plug in the values and solve for V

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A hummingbird can a flutter its wings 4800 times per minute what is the frequency of wing flutters per second

Scilla [17]

the answer i got is 80 i hope this helpss!!!!

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A diver in the pike position (legs straight, hands on ankles) usually makes only one or one-and-a-half rotations. To make two or

Korolek [52]

Answer:

from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

Explanation:

This can be explained on the basis of conservation of angular momentum.

This means the initial and the final angular velocity is conserved. Consider initial position (1)in the pike and final position in the be truck position. So there inertia's will also be different.

⇒ $I_1\omega_1 = I_2\omega_2$

$\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}$

also,

$I_1= mr_1^2$

$I_2= mr_2^2$

since, $r_2^2$

$I_2^2$

therefore,

$\omega_1^2$

So, from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

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2 years ago