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Bad White [126]
3 years ago
6

A bowling ball collides with a pin, knocking it over. The ball continues to move

Physics
1 answer:
kicyunya [14]3 years ago
7 0

Answer:

idk

Explanation:

idk

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Which one is the greatest?
kipiarov [429]

Answer:

18kg dog walking

Explanation:

the dog walking has more inertia than the boy and girl because its gravity is pulling it down to the earth

6 0
3 years ago
you throw a ball vertically so it leaves the ground withe velocity of 3.71m/s. what is its acceleration at this point
tino4ka555 [31]

No matter what direction you throw it, or with what speed, its acceleration is immediately 9.8 m/s^2 downward as soon as you release it from your hand, and it doesn't change until the ball hits something.

4 0
3 years ago
A 1.5-kg mass attached to an ideal massless spring with a spring constant of 20.0 N/m oscillates on a horizontal, frictionless t
Molodets [167]

Answer:

Explanation:

a ) angular frequency ω = \sqrt{\frac{k}{m} }

k is spring constant and m is mass attached

ω = \sqrt{\frac{20}{1.5} }

= 3.6515 rad / s

frequency of oscillation n = 3.6515 / (2 x 3.14)

= .5814 s⁻¹

x = .1 mcos(ωt)

= .1 mcos(3.6515t)

b ) maximum speed = ωA , A is amplitude

= 3.6515 x .1

= .36515 m /s

36.515 cm /s

maximum acceleration = ω²A

= 3.6515² x .1

= 1.333 m / s²

c ) Kinetic energy at displacement x

= 1/2 m ω²( A²-x²)

potential energy =1/2 m ω²x²

so 1/2 m ω²( A²-x²) = 1/2 m ω²x²

A²-x² = x²

2x² = A²

x = A / √2

6 0
3 years ago
A wildebeest calf is cruising at its top speed of v= 10 m/s when it passes over a sleeping cheetah. By the time the cheetah stan
Gre4nikov [31]

The time when the cheetah catches up with the widebeest is 1.53 s

If the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.

The given parameters;

  • speed of the wildebeest calf, Vw = 10 m/s
  • distance traveled by the calf before the cheetah stands up = 7 m
  • constant acceleration of the cheetah, a = 9.5 m/s²

Let the speed of the cheetah = Vc

let the time the cheetah catches up with the wildebeest = t

a = \frac{V_c}{t}

V_c = at

Apply relative velocity formula to determine the time when the cheetah catches up with the widebeest;

Assuming the wildebeest and the cheetah are running in the same direction;

(V_c - V_w) t = 7 \\\\(at - V_w) t = 7\\\\(9.5t - 10)t = 7\\\\9.5t^2 - 10t = 7\\\\9.5t^2 -10t-7 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 9.5 \ b= -10 \ , c =-7\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-(-10)\ \ +/- \ \ \sqrt{(-10)^2-4(9.5\times -7)} }{2(9.5)}\\\\t = \frac{10 \ \ +/- \ \ 19.13}{19} \\\\t = 1.53 \ s

The time when the cheetah catches up with the widebeest is 1.53 s

If the initial separation distance approaches zero;

(V_c - V_w) t = 0\\\\(at - V_w) t = 0\\\\(9.5t - 10)t = 0\\\\9.5t^2 - 10t = 0\\\\9.5t^2 = 10t\\\\9.5t = 10\\\\t = \frac{10}{9.5} = 1.05 \ s

Thus, if the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.

Lear more here: brainly.com/question/24430414

6 0
3 years ago
Ann successfully jumped over a 25.5 m-wideriver. Assuming that she started and landed at the same level and was airborne for 2.5
dolphi86 [110]

Answer:

Explanation:

t=Time airborne=2.54 s

R= horizontal range= 25.5 m

From the projectile motion equations we know that:

H=\frac{g*t^{2}}{8}

H=\frac{9.8 \frac{m}{s^{2}} *(2.54s)^{2}}{8}

H=7.9 m

7 0
3 years ago
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