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3 years ago

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A luge run is straight for the first 10 m. If the velocity of the luger when he passes the start line is +25 m/s and his acceler

ation is constant at +9.5 m/s2, about how fast is he going when he hits the first turn?

1 answer:

dangina [55]3 years ago

3 0

V^2 =U^2 +2AS
V^2 = 25 ^2 + 2x9.5x10
V^2 = 625 + 1900 = 2525
V = 50.25

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a girl skateboards with a kinetic energy of 2543.2 J. if the girl and the skateboard have a total mass of 110 kg, what is her sp

zhuklara [117]

Using the definition of kinetic energy, we have:

$E_{c}= \frac{mv^2}{2} \\ 2543.2*2=110v^2 \\ \boxed {v=6.8m/s}$

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3 years ago

If you plot the distance d as an ordinate (y-axis) versus the time of fall t as an abscissa (x-axis), do you expect to get a str

const2013 [10]

Since a straight line on a distance time graph implies uniform speed, and falling is an acceleration - a non uniform speed - I woul not expect a straight line.I would expect to get a parabola.This is based on the kinematic equation s=ut+(1/2)at^2. If something is dropped from rest, u = 0; so s = (1/2)at^2Which is, I think, the equation of a parabola.

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3 years ago

What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of 1.3 cm and a uniformly di

Vlad [161]

Answer:

2.37 * 10^4 m/s

Explanation:

Constants :

Mass of electron = 9.11 * 10^(-31) kg

Electric charge of an electron = 1.602 * 10^(-19) C

Parameters given:

Radius of sphere = 1.3cm = 0.013m

Charge of sphere = 2.3 * 10^(−15) C

Using the law of conservation of energy, we have that:

K. E.(initial) + P. E.(initial) = K. E.(final) + P. E.(final)

K. E.(final) = 0, since final velocity is zero and P. E.(final) = 0 since the electron reaches a final distance of infinity.

Hence,

K. E.(initial) = P. E.(initial)

0.5mv^2 = (kqQ)/r

Where k = Coulumbs constant

Q = charge of the sphere.

r = radius of the sphere.

=> 0.5*m*v^2 = (kqQ)/r

0.5 * 9.11 * 10^(-31) * v^2 = (9 * 10^9 * 1.602 * 10^(-19) * 2.3 * 10^(-15))/0.013

4.555 * 10^(-31) * v^2 = 2550.88 * 10^(-25)

=> v^2 = 2550.88 * 10^(-25) / 4.555 * 10^(-31)

v^2 = 560 * 10^6 = 5.60 * 10^8

=> v = 2.37 * 10^4 m/s

4 0

3 years ago

A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.883 s. How much time

Salsk061 [2.6K]

Answer:

1.25 s

Explanation:

We are given that

Initial velocity of boy=u=0

Time=0.883 s

Let x be the total distance covered by boy in entire trip.

s= $\frac{x}{2}$

We know that

$s=ut+\frac{1}{2}gt^2$

Where $g=10m/s^2$

Substitute the values

$\frac{x}{2}=0\times 0.883+\frac{1}{2}(10)(0.883)^2$

$\frac{x}{2}=3.898$

$x=3.898\times 2=7.796$ m

Again using the above formula

$7.796=0(t)+\frac{1}{2}(10)t^2$

$7.796=5t^2$

$t^2=\frac{7.796}{5}=1.5592$

$t=\sqrt{1.5592}=1.25 s$

Hence, 1.25 s passes during his entire tripe from the top down to the water.

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3 years ago