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A water skier leaves the end of an 8 foot tall ski ramp with a speed of 20 mi/hr and at an angle of 250. He lets go of the tow r

klemol [59]

Answer:

At highest point:

y1 = 10.4 ft

v1 = (26.5*i + 0*j) ft/s

When he lands:

x2 = 31.5 ft (distance he travels)

t2 = 1.19 s

V2 = (26.5*i - 25.9*j) ft/s

a2 = -44.3°

Explanation:

Since he let go of the tow rope upon leaving the ramp he is in free fall from that moment on. In free fall he is affected only by the acceleration of gravity. Gravity has a vertical component only, so the movement will be at constant acceleration in the vertical component and at constant speed in the horizontal component.

20 mi / h = 29.3 ft/s

If the ramp has an angle of 25 degrees, the speed is

v0 = (29.3 * cos(25) * i + 29.3 * sin(25) * j) ft/s

v0 = (26.5*i + 12.4*j) ft/s

I set up the coordinate system with the origin at the base of the ramp under its end, so:

R0 = (0*i + 8*j) ft

The equation for the horizontal position is:

X(t) = X0 + Vx0 * t

The equation for horizontal speed is:

Vx(t) = Vx0

The equation for vertical position is:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

The equation for vertical speed is:

Vy(t) = Vy0 + a * t

In this frame of reference a is the acceleration of gravity and its values is -32.2 ft/s^2.

In the heighest point of the trajectory the vertical speed will be zero because that is the point where it transitions form going upwards (positive vertical speed) to going down (negative vertical speed), and it crosses zero.

0 = Vy0 + a * t1

a * t1 = -Vy0

t1 = -Vy0 / a

t1 = -12.4 / -32.2 = 0.38 s

y1 = y(0.38) = 8 + 12.4 * 0.38 + 1/2 * (-32.2) * (0.38)^2 = 10.4 ft

The velocity at that moment will be:

v1 = (26.5*i + 0*j) ft/s

When he lands in the water his height is zero.

0 = 8 + 12.4 * t2 + 1/2 * (-32.2) * t2^2

-16.1 * t2^2 + 12.4 * t2 + 8 = 0

Solving this equation electronically:

t2 = 1.19 s

Replacing this time on the position equation:

X(1.19) = 26.5 * 1.19 = 31.5 ft

The speed is:

Vx2 = 26.5 ft/s

Vy2 = 12.4 - 32.2 * 1.19 = -25.9 ft/s

V2 = (26.5*i - 25.9*j) ft/s

a2 = arctg(-25.9 / 26.5) = -44.3

3 0

3 years ago

Holmes owns two suits: one black and one tweed. He always wears either a tweed suit or sandals. Whenever he wears his tweed suit

vazorg [7]

Answer:

He wore his black suit, another color of shirt (not purple) and shoes

Explanation:

Holmes owns two suits: one black and one tweed.

Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.

So, if he wore a bow tie yesterday, it means he wore his black suit, another color of shirt (not purple) and shoes because the shirt color is not purple

4 0

3 years ago

The diffusion coefficients for species A in metal B are given at two temperatures:

Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

$D = D_0e^{\frac{-Q_d}{RT} }$

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

$R = 8.314\ J/mol*K$

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

$ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}$