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Finger [1]
3 years ago
13

A rock is thrown vertically upward from the ground, with an initial velocity of 64 ft/sec. Its position function is s(t) = –16t2

+ 64t. What is its velocity in ft/sec when it hits the ground?
Physics
1 answer:
cestrela7 [59]3 years ago
3 0
<h2>Velocity of rock when it hits the ground is -64 ft/s</h2>

Explanation:

Its position function is s(t) = –16t² + 64t

When it reaches back ground s(t) = 0 ft

Substituting

                s(t) = –16t² + 64t = 0

                          t² - 4t = 0

                           t²  = 4t

                           t =4 seconds

Now we need to find velocity when it reaches ground, that is velocity after 4 seconds.

Differentiating s(t) equation

                   v(t) = -32t+64

            Substituting t = 4 seconds

                  v(4) = -32 x 4+64  = -64 ft/s

Velocity of rock when it hits the ground is -64 ft/s

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Answer:

a) 2nd case rate of rotation gives the greater speed for the ball

b) 1534.98 m/s^2

c) 1515.04 m/s^2

Explanation:

(a) v = ωR

when R = 0.60, ω = 8.05×2π

v = 0.60×8.05×2π = 30.34 m/s

Now in 2nd case

when R = 0.90, ω = 6.53×2π

v = 0.90×6.53×2π = 36.92 m/s

6.35 rev/s gives greater speed for the ball.

(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2

(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2

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How does gravity affect the velocity of falling objects?
sladkih [1.3K]

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The part of the flower responsible for producing pollen (sperm) is the _______. A. stigma B. anther C. style D. filament
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A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be
MrRissso [65]

The kinetic energy halfway the hill is 2.86\cdot 10^5 J

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

U_A +K_A = U_B + K_B

where

U_A=mgh_A is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

g=9.8 m/s^2 (acceleration of gravity)

h_A = 30 m (height at point A)

K_A=\frac{1}{2}mv_A^2 is the initial kinetic energy, at point A , with

v_A=17 m/s (velocity at point A)

U_B=mgh_B is the final potential energy, at point B, where

h_B = 15 m (height at point B)

K_B=\frac{1}{2}mv_B^2 is the final kinetic energy, at point B, where

v_B is the velocity at point B

Here we are interested in finding K_B, so by re-arranging the equation and substituting we find:

K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0
3 years ago
The gasoline in a car does 40,000 J of work on a car and generates a constant force of 20 N. How far did the car go?
AnnyKZ [126]

L=F•d=>d=L/F=40,000/20=2,000 m

7 0
3 years ago
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