Hovercraft B/C

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Re: Hovercraft B/C

UTF-8 U+6211 U+662F wrote:
Adi1008 wrote:Suppose I have a bowling ball with a diameter of 25 centimeters. What is the largest mass it can have such that it floats in corn syrup (specific gravity = 1.4)?
$\frac{m}{\frac43\pi r^3} = \frac{m}{\frac43\pi (12.5 cm)^3} = 1.4*1 \frac{g}{cm^3}$
$m = 1.4 \frac{g}{cm^3} * \frac43\pi (12.5 cm)^3 = 11453.7 g = 11.45 kg$
Looks good to me. Your turn!
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Re: Hovercraft B/C

All right! Given a graph of v vs t, how do you find the displacement traveled? What about the distance?

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Re: Hovercraft B/C

UTF-8 U+6211 U+662F wrote:All right! Given a graph of v vs t, how do you find the displacement traveled? What about the distance?
a. $\int^{b}_{a} v(t) dt$
b. $\int^{b}_{a} |v(t)| dt$
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Re: Hovercraft B/C

UTF-8 U+6211 U+662F wrote:All right! Given a graph of v vs t, how do you find the displacement traveled? What about the distance?
a. $\int^{b}_{a} v(t) dt$
b. $\int^{b}_{a} |v(t)| dt$
Area under the curve for the unitiated in calculus (signed area for a and unsigned for b)

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Re: Hovercraft B/C

UTF-8 U+6211 U+662F wrote:
UTF-8 U+6211 U+662F wrote:All right! Given a graph of v vs t, how do you find the displacement traveled? What about the distance?
a. $\int^{b}_{a} v(t) dt$
b. $\int^{b}_{a} |v(t)| dt$
Area under the curve for the unitiated in calculus (signed area for a and unsigned for b)
Suppose you have a contracting star whose new radius is 1/x as big as the old radius.

a. How much faster does the star spin?
b. By what factor does its rotational kinetic energy change?
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Re: Hovercraft B/C

UTF-8 U+6211 U+662F wrote:
a. $\int^{b}_{a} v(t) dt$
b. $\int^{b}_{a} |v(t)| dt$
Area under the curve for the unitiated in calculus (signed area for a and unsigned for b)
Suppose you have a contracting star whose new radius is 1/x as big as the old radius.

a. How much faster does the star spin?
b. By what factor does its rotational kinetic energy change?
a. x^2 faster
b. x^2
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Re: Hovercraft B/C

It's been a while, so I guess I'll ask a question.

Consider a basketball player throwing a ball into the hoop. The ball is 625 g, and the basketball player throws it at 10 m/s at 65 degrees to the horizontal. Neglect air resistance.

1) Find the force that acts on the ball once it leaves the player's hand.

2) A regulation height hoop is 10 feet tall. Find the distance from the hoop he needs to be if he shoots the ball from just above his head and he is 1.8 m tall.

Edit: Wait hovercraft is being replaced next year

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Re: Hovercraft B/C

UTF-8 U+6211 U+662F wrote:It's been a while, so I guess I'll ask a question.

Consider a basketball player throwing a ball into the hoop. The ball is 625 g, and the basketball player throws it at 10 m/s at 65 degrees to the horizontal. Neglect air resistance.

1) Find the force that acts on the ball once it leaves the player's hand.

2) A regulation height hoop is 10 feet tall. Find the distance from the hoop he needs to be if he shoots the ball from just above his head and he is 1.8 m tall.

Edit: Wait hovercraft is being replaced next year
1) The forces that are acting on the ball once the ball leaves the player’s hand is just the force of gravity.
$F_g = mg$
$F_g = 0.625 * 9.8$
$F_g = 6.125 N$

2)
$\Delta y = v_yt + \frac{gt^2}{2}$
$\frac{10}{3.28} - 1.8 = 10sin65\degree t- 4.9t^2$
$t = 0.15s, 1.7s$

$gt = v_f - v_i$
$-9.8 * 0.15 = v_f - 10sin65\degree$
$v_f = 7.593 m/s$

$-9.8 * 1.7 = v_f - 10sin65\degree$
$v_f = -7.593 m/s$

Since the velocity is negative at 1.7 seconds, $t = 1.7$ will be used to find the distance in the x direction.

$\Delta x = v_xt$
$\Delta x = 10cos65\degree * 1.7$
$\Delta x = 7.184 m$
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Re: Hovercraft B/C

Nydauron wrote:
UTF-8 U+6211 U+662F wrote:It's been a while, so I guess I'll ask a question.

Consider a basketball player throwing a ball into the hoop. The ball is 625 g, and the basketball player throws it at 10 m/s at 65 degrees to the horizontal. Neglect air resistance.

1) Find the force that acts on the ball once it leaves the player's hand.

2) A regulation height hoop is 10 feet tall. Find the distance from the hoop he needs to be if he shoots the ball from just above his head and he is 1.8 m tall.

Edit: Wait hovercraft is being replaced next year
1) The forces that are acting on the ball once the ball leaves the player’s hand is just the force of gravity.
$F_g = mg$
$F_g = 0.625 * 9.8$
$F_g = 6.125 N$

2)
$\Delta y = v_yt + \frac{gt^2}{2}$
$\frac{10}{3.28} - 1.8 = 10sin65\degree t- 4.9t^2$
$t = 0.15s, 1.7s$

$gt = v_f - v_i$
$-9.8 * 0.15 = v_f - 10sin65\degree$
$v_f = 7.593 m/s$

$-9.8 * 1.7 = v_f - 10sin65\degree$
$v_f = -7.593 m/s$

Since the velocity is negative at 1.7 seconds, $t = 1.7$ will be used to find the distance in the x direction.

$\Delta x = v_xt$
$\Delta x = 10cos65\degree * 1.7$
$\Delta x = 7.184 m$
Okay, hovercraft is gone next year, so I guess there won't be any more questions, but yep, that's right (interestingly, if you used 0.15 s, it would've given you the distance required if he wanted to shoot the basketball through the bottom of the hoop and out the top).

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