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valkas [14]
3 years ago
7

A snowboarder glides down a 48-m-long, 15° hill. She then glides horizontally for 10 m before reaching a 30° upward slope. Assum

e the snow is frictionless. What is her velocity at the bottom of the hill?
How far can she travel up the 30° slope?

Physics
1 answer:
Masja [62]3 years ago
8 0

Answer:

Her velocity at the bottom of the hill is 15.61m/s and she travel up the 30° slope 24.85m

Explanation:

For simplicity purpose, we can analyze the section of the snowboarder's travel in the hill, in the horizontal surface and in the slope separately.

In the hill, we will say that the x-axis is parallel to the hill, and y-axis is perpendicular. Using geometry, we can see that the angle of the snowboarder's weight force from the y-axis is 15°. The velocity of the snowboarder will increase in the direction parallel to the hill, in a constant acceleration motion:

F_x: W_x = ma\\W*sin(15) = ma\\mg*sin(15) = ma\\a = g*sin(15) = 9.81m/s^2 * sin(15) = 2.54m/s^2

With the acceleration, we can use the equations for constant acceleration motion:

v_f^2 - v_o^2 = 2a*d\\v_f^2 - (0m/s)^2=2*2.54m/s^2*48m\\vf = \sqrt{2*2.54m/s^2*48m}=15.61 m/s

This would be her velocity at the bottom of the hill.

As there is no friction, she would reach the bottom of the slope with this velocity.

In the slope, the line of reasoning is similar as in the hill, with the difference that the acceleration will oppose velocity.

F_x: -W_x = ma\\-W*sin(30) = ma\\-mg*sin(30) = ma\\a = -g*sin(30) = -9.81m/s^2 * sin(30) = -4.905m/s^2

v_f^2-v_o^2=2a*d\\d=\frac{(v_f^2-v_o^2)}{2a}=\frac{((0m/s)^2-(15.61m/s)^2)}{2(-4.905m/s^2)} = 24.85m

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The question is incomplete. The complete question is :

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Solution :

Given :

$y = -x^2+4x-5$

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$x=2 $

and $\rho = 2 \ kg/m^2$ , $P_x=1 \ $ , $P_y=-2 \ $.

So,

$dI = dmr^2$

$dI = \rho \ dA  \ r^2$  ,           $r=\sqrt{(x-1)^2+(y+2)^2}$

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$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \  dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

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EX 6-1 A ball is twirled on a 0.870 - m-long string with a constant speed of 3.36 m / s . Calculate the acceleration of the ball
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Answer:

a=12.97\ m/s^2

Explanation:

Given that,

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Speed of the ball, v = 3.36 m/s

We need to find the acceleration of the ball. The acceleration acting on the ball is centripetal acceleration. It is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(3.36)^2}{0.87}\\\\=12.97\ m/s^2

So, the acceleration of the ball is 12.97\ m/s^2.

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Answer:

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Explanation:

Given,

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E =400 \times 10^{-6}\ J

we know,

E = \dfrac{1}{2}CV^2

V =\sqrt{\dfrac{2E}{C}}

V =\sqrt{\dfrac{2\times 400 \times 10^{-6}}{2\times 10^{-6}}}

V =\sqrt{400}

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so , the capacitor voltage is V = 20 V

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<u>Explanation:</u>

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